-
PEARSON
l l: ll: CtL~Q Ll tCS 1 Q \ 1\_1l: L'-
v
I
Contents viii
Abnnt the Author
Chapter 1
I n tro d uction
1
1.1
I Applications of Power Electronics 1.1.l History of Power E lectronics
]2
Po,ver Semiconductor Devices 12 1 1.2.2
Power D iodes Thyristors 6
2
s
5
1 2 3 Pn,vcr Transistors 9 1 3 Con trol Characteristics of Power Devices 1.4
Characteristics and Specifications of Switches I 4 1 Ideal Cbaractetistics 16 1 4 2 Cbaracleristics nf Practical Devices 1.4.3 Switch Specifications 18
]44 Device Choices 1.5 1.6 1.7 1.8 19
1.10 1 11
10 16
]7
12
Types of Power E lectronic Circuits 20 Design o r Power E lectronics Equipment 23 Determining the Root-Mean-Square Values of Waveforms 24 Peripheral E ffects Power Modules 26 Intelligent Modules 26 Power Electronics Iaurnals and Conferences ?8 S ummary 29
References 29 Review Questions
24
30
Ch a pter 2 Power Semiconductor Diodes and Ci rcuits ? 1 lo1codur tian 31 2 2 Sen1iconductar Basics 31 ? 3 Diode Cbaractccistirs 11
31
vii
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viii
Contents
2.4 2.5
Reve rse Recovery Characteristics 35 Power D iode Types 38 2.5.l General-Purpose Diodes 38 2.5.2 Fast-Recovery Diodes 38 2.5.3 Schollky Diodes 39
? 6 Si licon Carbide Diodes 2.7
Spice Diode Model
39 40
2 8 Secies-Cooncc1ed Diodes 2 9 Parallel-Connected Diodes
42 4S
2 IQ Diodes ,Yi tb RC and RI I oads 46 2 11 Diodes with I C and RI CI oads 49 2.12 Freewheeling D iodes 56 2.13 Recovery of Trapped Energv with a Diode Summarv
References 61 Revje,v Questions Proh)em, 64 Chapter 3
61
Diod e Rectifiers
68
1]
ln1rod11rtioo
3.2
Single-Phase Half-Wave Rectifiers
6.8
11 Performance Parameters 3.4
3.5 3.6 3.7 · 3.8 3.9 3.10 3.11 3.12
119
Power lransistors
Intcadurtino
82
119
Review Questions Prablcros 12Q
4) 4.2
68
69
Single-Phase Full-Wave Rectifiers 77 Single-Phase Full-Wave Rectifier with RL Load Multiphase Star Rectifiers 87 Three-Phase Bridge Rectifiers 92 Three-Phase Bridge Rectifier with RL Load 95 Comparison of Diode Rectifiers 101 Rectifier Circuit Design 101 Output Voltage with LC Filter 112 Effects of Source and Load Inductances 1 6
ummarv References
Chapter 4
58
62
122
122
Bipolar Junction Transistors 123 4.2.1 Steady-State Characteristics 124 4.2.2 Switching Characteristics 128 4.2.3 Switching Limits 135
43 Power MQSEEis 4.3.1 4.3.2
J'.37
Steadv-State Characteristics 141 142 Switching Characteristics
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Contents 44
ma, MOS
45
STis
48 49
4.10
)44
J45
46 IOBis 4.7
142
Series and Parallel Operation 150 dVdt and du/dt I imitations 151 SPICE Models 155 4 9 I BU SPICE Madel I 'i5 4 2 2 MQSEEI SPICE Madel t •;, 4 9 3 JOBI SPICE Model 158 160 Comparisons of Transistors Summarv 160
References
16'
Review Questions
Pcnbleros Chapter 5 51
ix
163
164
DC-DC Converters Introd uctinn
166
166
5.2
Principle of Step-Down Operation 166 5.2.1 Generation of Dutv Cycle 170 171 5.3 Step-Down Converter with RL Load 176 5.4 Principle of Step-Up Operation 5.5 Step-Up Converter with a Resistive Load 179 "l 6 Performance Parameters 181 5 7 Converter Classifica tion 182 5.8 Switching-Mode Regulators 186 5.8.1 Buck Regulators 186 5.8.2 Boost Regulators 190 5.8.3 Buck- Boost Regulators 194 I98 5.8.4 Cuk Regulators 5.8.5 Limitations of Single-Stage Conversion 205 5.9 Comparison of Regulators 5.10 Multioutput Boost Converter 206 5 11 Diode Rec1 ificr·Eed Boast Converter 208 5.1 2 Chopper Circuit Design 211 5.13 State-Space Analysis of Regulators 217 Summary 221 Bcfc rc orcs 221 Review Questions 223 Problems 224 · Chapter 6 6]
6.2
Pulse-Wid th-Modulated Inverters Int roduction 226 Principle of Operatio n
63 Pcrforn1anc:c Pora1nercrs 6.4
204
226
227
230
Sinelc-Phasc Bridec Inverters
232
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Contents
6 5 Three-Phase inverters
237
6.5.l 6.5.2
6.6
6.7 6.8
180-Degree Conduction 237 120-Degree Conduction 246 Voltage Control of Single-Phase Inverters 248 6.6.1 Single-Pulse-Width Modulation 248 6.6.2 Multiple-Pulse-Width Modulation 250 6 6 3 Sinusoidal Pulse-Width Modulation 253 6.6.4 Modified Sinusoidal Pulse·Width Modulation 6.6.5 Phase-Displacement Control 258 Advanced Modulation Techniques 260 Voltage Control of Three -Phase Inverters 264 6 8 ] Sinusoidal PWM 265 6.8.2 60-D egrec PWM 268
6 8 3 Thjrd-Harmooic PWM 6.8.4 6.8.5
257
268
Space Vector Modulation 271 Comparison of PWM Techniques
279
6 9 Harmonic Reductions 280 610 Cuaeot-Source Inverters 285 6 ] ] Variable DC-I ink In verter 288 6.12 Boost Inverter 289 6.13 In verter Circuit Design Summary
294
299
References
299 Review Questions
PcobJems Chapter 7
Thyristors
7 1 Introduction 7.2 7.3
7.4 7.5 7.6
300
301 304 304
304 Thyristor Characteristics Two-Transistor Model ofThvristor 307 Thyristor Tum-O n 309 Thyristor Turn-Off 311 Thyristor Tvpes 3 13 7.6.1 Phase-Controlled Thyristors 314
Z6,2 BCTs 7.6.3 764 7.6.5 766
767
ll 4
Fast-SwitchingThyristors 315 I A$CBs 316 Bidirectional Triode Thyristors 31 6 BCTs 111
oms
3]8
MTQs
123
7.6 8 FET-CTHs 7.6 9
122
7.6.10 ETOs 2 6 11 ,acrs
321
Z 612
325
MCfs
324
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Contents
xi
7 613 SIIlis 328 7.6.14 Comparisons of1l1yristors 330 330 7.7 Series Operation of'Thvristors 7.8 Parallel Operation of Thyristors 337 7 9 dildt Protection 338 7 1Q du/dt Protection 339 7.11 SPICE Thyristor Model 341 7.11.1 Thyristor SPICE Model 341 2 I I 2 GTQ SPICE Madel 343 2 ]] 3 MCI SPICE Madel 345 7 ]] 4 S ITH SPICE Model 345 Summary 346 References 347 Review Questions 350 350 Problems Chapter 8
BI
Reso nant Pulse Inverters lntrodnction
352
352
352 8 2 Series Resonant Inverters 8 7 1 Series Resonant lnverrers with 1Jnidirectianal Switches 8.3
353 8 2 2 Series Resonant lnverrers with Bidirectional Switches 361 Frequency Response of Series-Re'sonant Inverters 368 8.3.1 Frequency Response for Series Loaded 368 8.3.2 Frequency Response for Parallel Loaded 370 8.33 Frequency Response for Series-Parallel Loaded 372
84
Parallel Resonant Inverters
8.5 86
Voltage Control of Resonant Inverters Oass E Resonant lnvecrec 380
8 7 Class E Resonant Rectifi er 8.8 8.9 8.10
374 383
Zero-Current-Switching Resonant Converters 388 8.8.1 L-Type ZCS Resonant Converter 389 8.8.2 M-Type ZCS Resonant Converter 391 393 Zero·Voltage-Switching Resonant Converters Comparisons Between ZCS and ZVS Resonant Converters
8 J1 Two-Ouadrant zvs Rewoam Converters 812 Resonant DC·l ink Inverters 399 Summary
References Problems
91 9.2
396
396
402
401
Review Questions
Chapter 9
377
403
404
406
Multilevel Inverters Introd11ctiao 406 Multilevel Concept
407
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Contents
9.3 9.4
9.5
JYpes of Multilevel Inverters 408 Diode-Clamped Multilevel Inverter 409 9.4.1 Principle of Operation 410 9.4.2 Features of Diode-Clamped Inverter 411 9.4.3 Improved Diode-Clamped Inverter 412 Flying-Capacitors Multilevel Inverter 414 9.5.1 Principle of Operation 415 9.5.2 Features of Flying-Capacitors Inverter 417
96 Cascaded Multilevel Inverter 9.6.1 9.7
9. 11
Chapter 10 JO I 10.2 10.3
10.4
10.5 10.6
10.7 10.8
10.9
418
9,6.2 features of Cascaded lovecler
412
Applications 421 9.7.1 Reactive Power Compensation
422
9 72 9.8 9.9 910
417
Principle of O peration
Rack·tn·Back lnlcrtic
423
9.7.3 Adjustable Speed Drives 424 Switching Device Currents 424 DC· Link Capacitor Voltage Balancing 425
Feau1res of Multilevel Inverters
427
Comparisons of Multilevel Converters Summary 428 References 429 Review Questions 430 Problems 430
Controlled Rectifiers lntrod11c1ion
428
431
431
Principle of Phase-Controlled Converter Operation 432 Single-Phase Full Converters 434 10.3.1 Single-Phase Full Converter with RL Load 438 Single-Phase Dual Converters 440 Principle of Three-Phase Half· Wave Converters 443 Three-Phase Full Converters 447 45] JO 6 I Three-Phase f un Convener with RI J oad Three-Phase Dual Converters 453 Power Factor Improvements 456 10.8.1 Extinction Angle Control 456 457 10.82 Symmetric Angle Control 10.8.3 PWM Control 461 10.8,4 Single-Phase Sinusoidal PWM 463
JO.a 5
Three-Phase PWM Rectifier
465
Single-Phase Semiconvcrtcrs 467 10.9.J Single-Phase Scmiconvcrter with RL Load 474 IO.IO Three-Phase Semiconverters 10.10.1 Three-Phase Semiconverters with RL Load 480 JO.I 1 Single-Phase Series Converters' 10.12 Twelve-Pulse Converters 485
472 479
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Contents
10.13
D esign of Converter Circuits
10 14
Effects oft oad and Source lnd11crances Summary 494 Rcfen:nccs 495 R eview Questions
Problems
487
49?
496
497
11 I
AC Voltage Controllers Introduction ~00
11.2 11.3 11.4 11.5
Principle of On-Off Control 501 Principle o f Phase Control 503 Single-Phase Bidirectional Controllers with Resisti ve Loads Single-Phase Controllers with Inductive Loads 509
Chapter 11
500
) ).6 Three-Phase Fun. Wave Conrronecs 11.7 11.8 11.9
11.10 JI JI 11.12 11.13
Chapter 12 )2
Matrix Converter
12 3 Three-Phase AC Switches
554
12.9
545
551 551
12.5 12.6 12.7 12.8
537
551
Single-Phase AC Switches
12.4
518
i:;1n
Design of AC Voltage-Contro ller Circuits Effects of Source an d Load I nductances Summary 546 547 References Review Questions 54 7 Problems 548
Static Switches
506
SJ4
Three-Phase Bidirectional Delta-Connected Controllers Single-Phase Transformer Connection C hangers 522 Cycloconverters 526 11.9.l Single-Phase Cycloconverters 527 11.9.2 Three-Phase Cycloconvcrtcrs 530 ll .9.3 Reduction of Output H armonics 530 AC Voltage Controllers with PWM Control 534
1 lnrroductioo
12.2
xiii
Three-Phase Rcversine Switches 554 AC Switches fo r Bus Transfer 556 D C Switches 557 Solid-State Re lays 561 Microelectronic Relays 563 563 12.8.1 Photovoltaic Re lay I 2.8.2 Photovoltaic Isolators 565 Design of Static Switches 566 Summary 567 References 567 568 Review Questions Problems 568
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xiv
Contents Chapter 13
13.1 13.2 13.3 13.4
13.5 13.6
13.7 13.8 13.9 13.10
Ch apter 14
Flexible AC Transmission Systems
570
Introduction 570 Principle of Power Transmission 571 Principle of Shunt Compensation 573 Shunt Compensators 575 13.4.1 Thyristor-Controlled Reactor 575 13.4.2 Thyristor-Switched Capacitor 577 13.4.3 Static VAR Compensator 580 13.4.4 Advanced Static VAR Compensator 581 Principle of Series Compensation 582 Series Compensators 585 13.6.1 Thyristor-Switched Series Capacitor 585 13.6.2 Thyristor-Controlled Series Capacitor 586 13.6.3 Forced-Commutation-Controlled Series Capacitor 13.6.4 Series Static VAR Compensator 589 589 13.6.5 Advanced SSVC Principle of Phase-Angle Compensation 592 Phase-Angle Compensator 594 Unified Power Flow Controller 596 Comparisons of Compensators 597 Summary 598 References 598 Review Questions 599 Problems 599 Power Supplies
601
14 I
Introduction
14.2
14.4
DC Power Supplies 602 14.2.1 Switched-Mode DC Power Supplies 602 14.2.2 Flyback Converter 602 14.2.3 Forward Converter 606 14.2.4 Push-PulJ Converter 611 14.2.5 Half-Bridge Converter 613 14.2.6 Full-Bridge Converter 616 14.2.7 Resonant DC Power Supplies 619 14.2.8 Bidirc'ctional Power Supplies 619 AC Power Supplies 621 14.3.1 Switched-Mode AC Power Supplies 623 14.3.2 Resonant AC Power Supplies 623 14.3.3 Bidirectional AC Power Supplies 624 Multistage Conversions 625
)4 5
Control Circuits
14.6
Magnetic Design Considerations 630 14.6.1 Transformer Design 630 14.6.2 • DC Inductor 634
14.3
587
©J
626
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Conti,nts
14.6.3 Magnetic Saturation Summary 636 Rcfcrcnrr:s
1S 1
635
636
Review Questions Problems 637
Chapter 15
xv
DC Drives Introduction
637
640 640
15 2 15.3 15.4
Basic Cba ractccislics of DC Motors 641 Operating Modes 645 Single-Phase Drives 648 }5.4.1 Single-Phase Half-Wave-Converter Drives 649 15.4.2 Single-Phase Semiconverter Drives 650 15.4.3 Single-Phase Full-Converter Drives 651 15.4.4 Single-Phase Dual-Converter Drives 652 15.5 Three-Phase Drives 656 15.5.1 Three-Phase Half-Wave-Converter Drives 657 15.5.2 Three-Phase Semiconverter Drives 657 15.5.3 Three-Phase Full-Converter Drives 657 15 5 4
15 6
15.7
Three·Phase Dnal·C-Onverter Drives
ReviewQuestions Problems
Chapter 16
658
DC-DC Converter Drives 662 15.6.1 Principle of Power Control 662 15.6.2 Principle of Regenerative Brake Control 664 15.6,3 Principle of Rheostatic Brake Control 667 15.6.4. Principle of Combined Regenerative and Rheostatic Brake Control 668 15.6.5 1\vo- and Four-Quadrant DC-DC Converter Drives 15.6.6 Multiphase DC-DC Converters 670 Closed-Loop Control of DC Drives 673 15.7.1 Open-Loop lransfer Function 673 15.7.2 a osed-Loop Transfer Function 678 15.7.3 Phase-Locked-Loop Control 684 15.7.4 Microcomputer Control of DC Drives 685 Summary 6'i!/7 References 682
669
688
688
AC Drives
692
16 1 Introduction 692 16 2 Induction Motor Drives
693
16.2.1 Performance Characteristics 694 16.2.2 Stator Voltage Control 701 16.2.3 Rotor Voltage Control 703 16.2.4 Frequency Control 711
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Contents
16.3 16.4
16.2.5 Voltage and Frequency Control 713 16.2.6 Current Control 716 16.2.7 Voltage. Current, and Frequency Control 720 Closed-Loop Control of Induction Motors 721 Vector Controls 726 16.4.I Basic Principle of Vector Control 727 16.4.2 Direct and Quadrature-Axis Transformation 728 16 4 3
16.5
16.6
I ndi rect Yertor Control
References Problems
745
756
Review Questions
Chapt er 17
734
16.4.4 Director Vector Control 736 Synchronous Motor Drives 738 16.5.l Cylindrical Rotor Motors 738 16.5.2 Salient-Pole Motors 741 16.5.3 Reluctance Motors 743 16.5.4 Permanent-Magnet Motors 743 16.5.5 Switched Reluctance Motors 744 16.5.6 Closed-Loop Control of Synchronous Motors 16.5.7 Brushless DC and AC Motor Drives 747 Stepper Motor Control 749 16.6. I Variable-Reluctance Stepper Motors 750 16.6.2 Permanent-Magnet Stepper Motors 753 Summarv 756 757
758
Gate Drive Circuits
761
17.1 Introduction 761 17.2 MOSFET Gate Drive 761 17.3 BJT Base Drive 763 17.4 Isolation of Gate and Base Drives 767 17.4. 1 Pulse Transformers 769 17.4.2 Op1ocouplcrs 769 17.5 Thyristor Firing Circuits 770 17.6 Unijunction Transistor 772 17.7 Programmable Unijunction Transistor 775 17.8 Thyristor Converter Gating Circuits 777 17.9 Gate Drive ICs 777 17.9.1 Drive IC for Converters 781 17.9.2 High-Voltage IC for Motor Drives 784 Summary 788 References 789 Review Questions 789 Problems 790
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Contents Chapter 18
Protection of Devices and Circuits
791
18.l lntroduction 791 18.2 C-Ooling and Heat Sinks 791 18.3 Thermal Modeling of Power Switching Devices 797 18.3.1 Electrical Equivalent Thermal Model 798 18.3.2 Mathematical Thermal Equivalent Circuit 800 18.3.3 Coupling of Electrical and Thermal Components 803 18.4 Snubber Circuits 804 18.5 Reverse Recovery Transients 18.6 Supply- and Load-Side Transients 810 18.7 Voltage Proteclion by Selenium Diodes and Metal Oxide Varistors 813 18.8 Current Protections 815 18.8.1 Fusing 815 18.8.2 Fault Curreot with AC Source 822 18.8.3 i'ault Current with DC Source 824 18.9 Electromagnetic Interference 827 18.9.1 Sources of EMI 828 18.9.2 Minimizing EMI Generation 828 18.9.3 EMI Shielding 829 18.9.4 EMI Standards 829 Summary 830 References 831 Review Questions 831 Problems 832 Appendix A
Three-Phase Circuits
Appendix B
Magnetic Circuits
AppendixC
Switching Functions of Converters
AppendixD
DC Transient Analysis
AppendixE
Fourier Analysis
Bibliography
801
835 839
847
853
857
860
Answers to Selected Problems Index
xvii
863
871
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Preface The third edition of Power Electronics is intended as a textbook for a course on power electronics/static power converters for junior or senior undergraduate students in elec· trical and electronic engineering. It can also be used as a textbook for graduate stu· dents and as a reference book for practicing engineers involved in the design and applications of power electronics. The prerequisites are courses on basic electronics and basic electrical circuits. The content of Power Electronics is beyond the scope of a one-semester course. The time allocated to a course on power electronics in a typical undergraduate curriculum is normally only one semester. Power electronics has aJ. ready advanced to the point where it is difficult to cover the entire subject in a onesemester course. For an undergraduate course, Chapters 1 to 11 should be adequate to provide a good background on power electronics. Chapters 12 to 16 could be left for other courses or included in a graduate course. Table P.1 shows suggested topics for a one-semester course on "Power Electronics" and Table P.2 for one semester course on "Power Electronics and Motor Drives."' TABI f PJ Suggested Topics for One Scme.stcr Course on Pnwcr Elcarnoics
t mutes
ln1mduccinn Power srmimndnctnr diodes and circuits
J 110 1 12 2.1102.4. 2.7.2.10 102.13
3
Diode rtttifica
3.11039
4
Power rransistnrs PC..-DC con·•erter:s J>WM iovcr1ca Thyristors
l 2
s
6 1 8
lll
u
12
Reson;1nt pulse inverters
Controlled u:ctifica
AC ,•oltage conuoUen Static switcbcs Mid·tcrm exams and quizzes
final exam Tntal tcctnrcs in a IS·s·ctk serotstt r
4.2 4.10 4.11 5110 S7 6.l 10 6.6.6.8 10 6.ll
2 2
s
2.
5 1
7 Jlo 75. 7.9.7 JO
2
81 to 85 JQJ 10 106 11 I to JI S 12 I to 128
3
6 l
2 3 l
45 xix
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xx
Preface TARI E P2 Suggested Topics ror One Semester Course on fnwer EJecrrnnirs and Morar Prlxes I ectures
2 3
Introduction Power stmironduc1nr dindes and circuits Diode rcclifica
1 1 10 1 12 2.1 to 2.4, 2.7,2.10 10 2.13
2 2
31to38
Q
Power transisrnrs
4.2.4.10, 4.11 5.1 to 5.1 6.t 10 6.6, 6.8 10 6.11 7.1 to 1.5. 1.9. 7.10
4 I 4
I
PC-PC ron,.·crtGU
~
Appendix
PWM inverters Thvristors Controlled rcctificn AC voltage controllers Magnetic circuits
DC dci ..·cs
1SJtolS7
Appendix
Three-phase circuits
u
AC drives
A 1.610 16.6 16.1 to 16.6
6 7 Ul
u
u
s I
101 ro 107
5
11.110 11.5
2 I
8 1610 166
~tid·tcrm exams and qu.i.zzcs
fi.nal exam Total lectures in a 15-week semester
s I
6 3 3
45
The fundamentals of power electronics are well established and they do not change rapidly. However, the device characteristics are continuously being improved and new devices arc added. Power Eltctronics, which employs the bottom-up approach, covers device characteristics conversion techniques first and then applications. It emphasizes the fundamental principles of power conversions. This third edition of
Power Electronics is a complete revision of the second edition, and (i) features bottom· up approach rather than top-down approach; (ii) introduces the state-of-the-art advanced Modulation Techniques: (iii) presents three new chapters on "Multilevel Inverters" (Chapter 9), "Flexible AC Transmission Systems" (Chapter 13), and "Gate Drive Circuits" (Chapter 17) and covers state-of-the-art techniques; (iv) integrates the industry standard software, SPICE, and design examples that are verified by SPICE simulation; (v) examines converters with RL-loads under both continuous and discon· tinuous current conduction: and (vi) has expanded sections and/or paragraphs to add explanations. The book is divided i.n to five parts: 1. 2. 3. 4. S.
Introduction-Chapter 1 Devices and gate-drive circuits-Chapters 2, 4, 7, and 17 Power conversion techniques-Chapters 3, 5, 6, 8, 9, 10, and 11 Applications-Chapters 12.13, 14, 15, and 16 Protection and thermal modeling-Chapter 18
Topics like three-phase circuits, magnetic circuits, switching functions of converters, DC transient analysis, and Fourier analysis arc reviewed in the Appendices. Power electronics deals with the applications of solid-state electronics for the control and conversion of electric power. Conversion techniques require the switching oo and off of power semiconductor devices. Low-level electronics circuits, which nor· malJy consist of integrated circuits and discrete componenu, generate the required
Copyrighlcd malcria
Preface
xxi
gating signals for the powe r devices. Integrated circuits and discrete components are being replaced by microprocessors and signal processing !Cs. An ideal power device should have no switching-on and -off limitations in terms of tum-on time, turn-off time, current, and voltage handling capabilities. Power semiconductor technology is rapidly developing fast switching power devices with increasing voltage and current limits. Power switching devices such as power BJTs, power MOSFETs. S!Ts. IGBTs. MCTs. SITHs. SCRs, T RIACs. GTOs. MTOs, ETOs, IGCTs, and other semiconductor devices are finding increasing applications in a wide range of products. With the availability of faster switching devices, the applications of modem microprocessors and digital signal processing in synthesizing the control strategy for gating power devices to meet the conversion specifications are widening the scope of power electronics. The power electronics revolution has gained momentum, since the early 1990s. Within the next 20 years, power electronics will shape and condition the electricity somewhere between its generation and all its users. The potential applications of power electronics are yet to be fully explored but we've made every effort to cover as many applications as possible in this book. Any comments and suggestions regarding this book are welcomed and should be sent to the author. Dr. Muhammad H. Rashid Professor and Director Electrical and Computer Engineering University of West Florida 11000 University Parkway Pensacola, FL 32514-5754 E-mail:
[email protected]
PSPICE SOFTWARE AND PROGRAM FILES The student version PSpice schematics and/or Orcad capture software can be obtained or downloaded from Cadence Design Systems, Inc. 2655 Seely Avenue San Jose, CA 95134 Websites: http://www.cadence.com http://www.orcad.com http://www.pspice.com The website http://uwf.edu/mrashid contains all PSpice circuits, PSpice schematics, Orcad capture, and Mathcad files for use with this book. Important Note: The PSpice circuit files (with an extension .CIR) are selfcontained and each file contains any necessary device or component.models. However, the PSpice schematic files (with an extension .SCH) need the user-defined model library file Rashld_PEJ...MODELLIB, which is included ,vith the schematic files, and must be Included from the Analysis menu of PSpice Schematics. Similarly, the Orcad
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xxll
Preface schematic files (with extensions .OPJ and .DSN) new the user-defined model library file Rashld..)'EJ...J,IODELUB, which is included with the Orcad schematic files, must
be Included from the PSpice Simulation settings menu of Orcad Capture. Without these files being included while running the simulation, it will not run and will.give errors.
ACKNOWLEDGMENTS Many people have contributed to this edition and made suggestions based on their classroom experience as a professor or a student. I would like to thank the following persons for their comments and suggestions: Mazen Abdel-Salam, King Fahd University of Petroleum and Minerals, Saudi Arabia Johnson Asumadu, Western Michigan University Ashoka K. S. Bhat, University ofVictoria, Canada Fred Brockhurst, Rose-Hu/man Institution of Technology Jan C Cochrane, Tlze University of Melbourne.Australia Ovidiu Crisan, University of Houston Joseph M. Crowley, University of lllinois, Urbana-Champaign Mehrad Ehsani, Texas A&M University Alexander E. Emanuel, Worcester Polytechnic Institute George Gela, Ohio State University Herman W. Hill, Ohio University Constantine J. Hatziadoniu, Sowhern lllinois University, Carbondale Wahid Hubbi, New Jersey lnstitllle of Technology Murrija Ilic-Spong, University of Illinois, Urbana-Champaign Shahidul I. Khan, Concordia University, Canada Hussein M. ~ojabadi, Sahand University of Technology, Iran Peter Lauritzen, University ofWashington Jack Lawler, University of Tennessee Arthur R. Miles, North Dakota State University Medhat M. Morcos, Kansas State University Hassan Moghbelli, Purdue University Calumet H. Ramezani-Ferdowsi, University of Mashhad, Iran Prasad Enjeti, Taas A &M University Saburo Mastsusaki, TDK Corporation, Japan Vedula V. Sastry, Iowa State University Elias G. Straogas, Michigan State University Selwyn Wright, The University of Huddersfield, Queensgate, UK S. Yu.varajan, North Dakota State University It has been a great pleasure working with the editor, Alice Dworkin and the production edi· tor, Donna Kmg. Finally, I would thank my family for their love, patience, and understanding.
MUHAMMAD H. RAsHID
P_ensacola, Florida
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About the Author Muhammad ff. Rashid received the B.S<:. degree in electrical engineering from the Bangladesh University of Engineering and Technology and the M.Sc. and Ph.D. degrees from the University of Birmingham, UK. Currently, he is a Professor of electrical engineering with the University of Florida and the Director of the UF/UWF Joint Program in Electrical and Computer Engineering. Previously, he was a Professor of electrical engineering and the Chair of the Engineering Department at Indiana University-Purdue University at Fort Wayne. In addition, he was a Visiting Assistant Professor of electrical engineering at the University of Connecticut, Associate Professor of electrical engineering at Concordia University (Montreal, Canada), Professor of electrical engineering at Purdue University, Calumet, and Visiting Professor of electrical engineering at King Fahd University of Petroleum and Minerals, Saudi Arabia. He has also been employed as a design and development engineer with Brush Electrical Machines Ltd. (UK), as a Research Engineer with Lucas Group Research Centre (UK), and as a Lecturer and Head of Control Engineering Department al the H igher Institute of Electronics (Malta). He is actively involved in teaching, researching, and lecturing in power electronics. He bas published 14 books and more than 100 technical papers. His books have been adopted as textbooks all over the world. His book Power Electronics has been translated into Spanish, Portuguese, Indonesian, Korean and. Persian. His book Microelectronics has been translated into Spanish in Me,rlco and Spain. He has had many invitations from foreign governments and agencies to be a keynote lecturer and consultant, from foreign universities 10 serve as an external Ph.D. examiner, and from funding agencies to serve as a research proposal reviewer. His contributions in education have been recognized by foreign governments and agencies. He has previously lectured and consulted for NATO for Turkey in 1994, UNDP for Bangladesh in 1989 and 1994, Saudi Arabia in 1993, Pakistan in 1993, Malaysia in 1995 and 2002, and Bangkok in 2002, and has been invited by foreign universities in Australia, Canada, Hong Kpng, India, Malaysia, Singapore 10 serve as an external examiner for undergraduate, master's and Ph.D. degree examinations, by funding agencies in Australia, Canada, United Stales, and Hong Kong to review research proposals, and by U.S. and foreign universities to evaluate promotion cases for professorship. He has previously authored seven books published by Prentice HaU: Power Electronics-Circuits, Devices, and Applications (1988, 2/e 1993), SPICE For Power Electronics (1993), SPICE for Circuirs and Electronics Using PSpice xxill Copyrighted maleria
JOtlv
About the Author
(1990, 2/e 1995), Electromechanical and Electrical Machinery (1986), and Engineering Design for Electrical Engineers (1990). He has authored five IEEE self-study guides: Self-Study Guide on Fundamentals of Power Electronics, Power Electronics Laboratory Using PSpice, Selected Readings on SPICE Simulation of Power Electronics, and Selected Readings on Power Electronics (IEEE Press, 1996) and Microelectronics Laboratory Using Electronics Workbench (lEEE Press, 2000). He also wrote two books: Electronic Circuit Design using Electronics Workbench (January 1998), and Microelectronic Circuits- Analysis and Design (April 1999) by PWS Publishing). He is editor of Power Electronics Handbook published by Academic Press, 2001. Dr. Rashid is a registered Professional Engineer in the Province of Ontario (Canada), a registered Chartered Enginee r (UK), a Fellow of the Institution of Electrical Engineers (IEE, UK) and a Fellow of the Institute of Electrical and Electronics Engineers (IEEE, USA). He was elected as an IEEE Fellow. with the citation •'Leadership in power electronics education and contributions to the analysis and d~· sign methodologies of solid-state power converters." He was the recipient of th·e /991 Outstanding Engineer Award from The Institute of Electrical and Electronics Engineers (IEEE). He received the 2002 IEEE Educational Activity Award (EAB) Meritorious Achievement Award in Continuing Education with the citation "for contributions 10 the design and delivery of co111iJ111ing education in power electronics and computer-aided-simulation". He was also an ABET program evaluator for electrical engineering from 1995 to 2000 and be is currently an engineering evaluator for the Southern Association of Colleges and Schools (SACS, USA). He has been elected as an IEEE-Industry Applications Society (lAS) Distinguished Lecturer. He is the Editor-in-Chief of the Power Electronics anti Applications Series, published by CRC Press. ·
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CH A PTER
1
Introduction The /earning objectives of this chapter are as fo/loK's: • To get an overview of power electronics and its history of development • To get an overview of different types of power semiconductor devices and their switching characteristics • To learn about the types of power converters • To know about resources for finding manufacturers of power semiconductors • To know about resources for fmding published articles of power electronics and applications 1.1
APPLICATIONS OF POWER ELECTRONICS
The demand for control of electric power for electric motor drive systems and industrial controls existed for many years, and this led to early development of the Ward-Leonard system to obtain a variable de voltage for the control of de motor drives. Power electronics have revolutionized the concept of power control for power conversion and for control of electrical motor d rives. Power electronics combine power, electronics, and control. Control deals with the steady-state and dynamic characteristics of closed-loop systems. Power deals with the static and rotating power equipment for the generation, transmission, and distribution of electric energy. Electronics deal with the solid-state devices and circuits for signal pr<><;essing to meet the desired control objectives. Power electronics may be defined as the applications of solid-state electronics for the control and conversion of electric power. The interrelationship of power electronics with power, electronics, and control is shown in Figure 1.1. Power electronics are based primarily on the switching of the power semiconductor devic.es. With the development of power semiconductor technology, the powerhandling capabilities and the switching speed of the power devices have improved tremendously. The deve lopment of microprocessors and microcomputer technology has a great impact on the control and synthesizing the control strategy for the power semiconductor devices. Modem power electronics equipment uses (!} power semiconductors that can be regarded as the muscle, and (2) microelectronics thai have the power and intelligen·ce of a brain. 1
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2
Chapter 1
Introduction
Power
Control Analog I Digital
Elecuon.ics Devices I Circuits
Power
equipment
Static I Rotalin
Electronics
FIGURE 1.1 Relationship of power electronics to power. electronics. and control.
Power electronics have already found an important place in modern technology and are now used in a great .variety of high-power products, including beat controls, light controls, motor controls, power supplies, vehicle propulsion systems, and highvoltage direct-current (HVDC) systems. It is difficult to draw the flexible ac transmissions (FACTs) boundaries for the applications of power electronics, especia.lly with the present trends in the development of power devices and microprocessors. Table 1.1 shows some applications of power electronics (5). 1.1.1
History of Power Electronics
The history of power electronics began with the introduction of the mercury arc rectifier in 1900. Then the metal tank rectifier, grid-controlled vacuum-tube rectifier, ignitron, phanotron, and thyratron were introduced gradually. These devices were applied for power control until the 1950s. The first electronics revolution began in 1948 with the invention of the silicon transistor at BeU Telephone Laboratories by Bardeen, Brattain, and Schockley. Most of today's advanced electronic technologies are traceable 10 that invention. Modem microelectronics evolved over the years from silicon semiconductors. The next breakthrough, in 1956, was also from Bell Laboratories: the invention of the PNPN triggering transistor, which was defined as a thyristor or silicon-controlled rectifier (SCR). The second electronics revolution began in 1958 with the development of the commercial thyristor by the General Electric Company. That was the beginning of a Copyr°ghlcd malcria
1.1 Applications of Power Electronics
3
TABLE 1.1 Some Applications of Power EJectronics Advertising Air-conditioning
M3g.nets M8.$$ tr·ansits
Aircrall power supplies
Mercury-arc lamp ballasts
Alarms
Mining
Appliances
Model trains
Audio amplifiers
Motor controls Motor drives MO\'ie projectors Nuclear reactor control rod Oil w,:11 drilling Qycn con1rols
Battery charger Blenders
Blowers Boilers Burglar al.anns Ccmcot kiln
Paper mills
Chemical processing Oothes dryers
Pan.icle accelerators People movcn
Computers ConVeyors Cranes and hoists Dimmers Displays Electric bla9kets Elecuic door openers Electric dryers
Phonograp~s Photocopies
Photographic supplies
Electric \'chicles
Po"'ct supplies Printing press Pumps and compressors Radar/sonar power supplies Range surface unit Rcfrigcraton Rcg·ulators
Electromagnets
RF amplifiers
Electromechanical electroplating Electronic ignHion
Security systems Servo systems Sewing mac:hines Solar power supplies Solid,sta1e contactors Solid-stale rel3ys
Electric fans
Electrostatic precipitators
Elt\'ators
Fans Flashers
Food mixers food warmer trays Forklift trucks Furnaces Games Garage door openers
Gas turbine starting Generator exciters Grinders
Hand po"A·er tools Heat rot1trols
Space power supplies S1atic circuit breakers Static relays Steel n1ills
Synchronous machine staning Synthetic fibers Television circuits Temperature controls
nmcrs Toys
Laser P<>""'cr supplies
Traffic: signal controls Trains T\' de ncciions Uhrasonic generalors Uninterruptible power supplies
Latching relays
Vacuum cleaners
Light dimmers
Linear induction n1olo r controls
Volt-ampere reactive (VAR) compensation Vending machines Very IO\\' frequency (VLF) lransmittcn
Locomo1ivc-.s Machine tools Magnc1ic rcrordings
Washing m;1ebincs Welding
Higb·frequcncy lighting
High-voltage de (HVDC) Induction heating
Light n..hm
Voltage regulators
Sflurct : Re( 5.
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4
Copyr ghl Im erid
1.2 Power Semiconductor Devices
5
new era of power electronics. Since then, many different types-of power semiconductor devices and conversion techniques have been introduced. Toe microelectronics revolution gave us the ability to process a huge amount of information at incredible speed. Toe power electronics revolution is giving us the ability to shape and control large amounts of power with ever-increasing efficiency. Due to the marriage of power electronics, the muscle, with microelectronics, the brain, many potential applications of power electronics are now emerging, and this trend will continue. Within the next 30 years, power electronics will shape and condition the electricity somewhere in the transmission network between its generation and all its users. Toe power electronics revolution bas gained momentum since the late 1980s and early 1990s (1). A chronological history of power electronics is shown in Figure 1.2.
1.2
POWER SEMICONDUCTOR DEVICES Since the first thyristor SCR was developed in late 1957, there have been tremendous advances in the power semiconductor devices. Until 1970, the conventional thyristors had been exclusively used for power control in industrial applications. Since 1970, various types of power semiconductor devices were developed and became commercially available. Figure 1.3 shows the classification of the power semiconductors, which are made of either silicon or silicon carbide. Silicon carbide devices are, however, under development. A majority of the devices arc made of silicon. These devices can be divided broadly into three types: (1) power diodes, (2) transistors, and (3) thyristors. These can be divided broadly into five types: (1) power diodes, (2) thyristors, (3) power bipolar junction transistors (BJTs), (4) power metal oxide semiconductor field-effect transistors (MOSFETs), and (5) insulated-gate bipolar transistors (IGBTs) and static induction transistors (SITs).
1.2.1
Power Diodes
A diode bas two terminals: a cathode and an anode. Power diodes are of three types: general purpose, high speed (or fast recovery), and Schottky. General-purpose diodes are available up to 6000 V, 4500 A, and the rating of fast-recovery diodes can go up to 6000 V, 1100 A. The reverse recovery time varies between 0.1 and 5 µs. The fast-recovery diodes arc essential for high-frequency switching of power converters. Schottky diodes have low on-state voltage and very small recovery time, typically nanoseconds. The leakage current increases with the voltage rating and their ratings are limited to 100 V, 300 A. A diode conducts when its anode voltage is higher than that of the cathqde; and the forward voltage drop of a power diode is very low, typically 0.5 and.1.2 V. If the cathode voltage is higher than its anode voltage, a diode is said to be in a blocking mode. Figure 1.4 shows various configurations of general-purpose diodes, which basically fall into two types. One is called a stud, or st11d-mo1mted type, and the other is called a disk, press pak, or hockey puck type. In a stud-mounted type, either the anode or the cathode could be the stud.
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6
Chapter 1
Introduction
Si1k:on
Diodes
Schottky-Diode Epltaxi.JJ. Diode (PIN) Dooble Diffused Diode PIN'
Transistors Bipolar Junction Tnmistor
NPN
PNP
MOSFET
Silicon Q.rbidc Thyrinon
Thyristors (or Phase: Control
MOSFET
f as1 Thyristor
Symmetric: AtJ'ffimctric
PIN-Diode
Rcvenc
N.CbW>
~ Conventional
t S-FET Cool-MOS P.cbanncl· Enhancement
Conducting
OTO S)'lllmCtric: As)'fflfflClric
Reverse Conducting
JOBT IGCT As)-m.mctrk Revene
Conduccina
~tcr
r - - - - -- - - , IDJ Low lmportanoc UD Marl:ci I
r :l)P< N·Typc:
MTO FIGURE 1.3
Oassification of the power semiconductors. [Ref. 2. S. Bernet)
1.2.2
Thyristors A 1hyristor has three terminals: an anode, a cathode, and a gate. When a small current is passed through the gate terminal to cathode, the thyristor conducts, provided that the anode terminal is at a higher potential than the cathode. The thyristors can be subdivided into eleven types: (a) forced-commutated thyristor, (b) line-commutated thyristor, (c) gate-tum-off thyristor (GTO), (d) reverse-conducting thyristor (RCT), (e) static induction thyristor {SITH), ({) gate-assisted turn-off thyristor (GATT), (g) light-activated silicon-controlled rectifier (LASCR), (h) MOS turn-off (MTO)
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1.2
Power Semiconduelor Devices
7
FIGURE 1.4
Various gcncral-purpo$C diode configurations..
(Counesy or Powerex. lnc.)
thyristor, (i) emitter tum-off (ETO) thyristor, (j) integrated gate-commutated thyristor (IGCT), and (k) MOS-controlled thyristors (MCfs). Once a thyristor is in a conduction mode, ihe gate circuit has no control and the thyristor continues to conduct. When a thyristor is in a conduction mode, the forward voltage drop is very small, typically 0.5 to 2 V. A conducting thyristor can be turned off by making the potential of the anode equal to or less than the cathode potential. The line-commutated thyristors arc turned off due to the sinusoidal nature of the input voltage, and forccd-<:ommutated thyristors are turned off by an extra circuit called com11111tatio11 circuitry. Figure 1.5 shows various configurations of phase control (or line-commutated) thyristors: stud, hockey puck, flat, and pin types. Natural or line-commutated thyristors are available with rating,,; up to 6000 V, 4500 A. The tum·off time of high-speed reverse-blocking thyristors has been improved substantially and it is possible to have 10 to 20 µsin a 3000-V, 3600-A thyristor. The turnoff time is defined as the time interval between the instant when the principal current has decreased to zero after external switching of the principal voltage circuit, and the instant when the thyristor is capable of supporting a specified principal voltage without
FIGURE 1.~
Various thyristor configurations. (Courtesy of Pi:,~·c rcx, Inc.)
8
Chapter 1
Introduction
turning on. RCTs and GATis are widely used for high-speed switching, especially in traction applications. An RCT can be considered as a thyristor with an inverse-parallel diode. RCTs are available up to 4000V,2000A (and 800A in reverse conduction) with a switching time of 40 µs. GATfs are available up to 1200 V, 400 A with a switching speed of 8 µs. LASCRs, which are available up to 6000 V, 1500 A, with a switching speed of 200 to 400 µs, are suitable for high-voltage power systems, especially in HVOC Foflow-power ac applications, TRIACs are widely used in all types of simple beat controls, light controls. motor controls, and ac switches. The characteristics ofTRIACs are similar to two thyristors connected in inverse parallel and having only one gate terminal. The current flow through a TRIAC can be controlled in either direction. GTOs and SITHs are self:tumed-off thyristors. GTOs and SITHs are turned on by applyiug a short positive pulse to the gates and are turned off by the applications of short n"gative pulse to the gates. They do not require any commutation circuit. GTOs are very attractive for forced commutation of converters and arc available up to 6000 V, 6000 A. SITHs. whose ratings can go as high as 1200 V, 300 A, arc expected to be applied for medium-power converters with a frequency of several hundred kilohertz and beyond the frequency range of GTOs. Figure 1.6 shows various configurations of GTOs. An MTO (3] is a combination of a GTO and a MOSFET, which together overcome the limitations of the GTO tum-off ability. Its structure is similar to that of a GTO and retains the GTO advantages of high voltage (up to 10 kV) and high current (up to 4000 A). MTOs can be used in high power applications ranging from 1 to 20 MVA. An ETO is a MOS-GTO hybrid device that combines the advantages of both the GTO and MOSFET. ETO bas two gates: one normal gate for tum-on and one with ~ series MOSFET (or turn~off. ETOs with a current rating of up to 4 kA and a voltage rating of up to 6 kV have been demonstrated. An IGCT [4] integrates a gate-commutated thyristor (GCT) with a multilayered printed circuit board gate drive. The GCT is a hard-switched GTO with a very fast and large gate current pulse, as large as the full-rated current, that draws out aU the current from the cathode into the gate in about 1 µs to ensure a fast tum-off. Similar to a GTO,
FIGURE 1.6
Gatc·turn-ofr thyristors. (Courtesy or ln1cmutional Rrttiriers.)
Coµyr ·Ill
m \~rl
I
1.2 Power Semiconductor Devices
9
the IGCT is turned on by applying the turn-on current to its gate. The IGCT is turned off by a multilayered gate-driver circuit board that can supply a fast-rising turn-off pulse (i.e., a gate current of 4 kAfµs with gate-cathode voltage of20V only).An MCT can be turned "on" by a small negative voltage pulse on the MOS gate (with respect to its anode), and turned "off' by a small positive voltage pulse. It is like a GTO, except that the tum-off gain is very high. MCTs are available up to 4500 V, 250 A.
1.2.3
Power Transistors Power transistors are of four types:(!) BJTs, (2) power MOSFETs, (3) IGBTs, and (4) SITs. A bipolar transistor has three terminals: base, emitter, and collector. It is normally operated as a switch in the common-emitter configuration. As long as the base of an NPN-transistor is at a higher potential than the emitter and the base current is suffi. ciently large to drive the transistor in the saturation region, the transistor remains on, provided that the collector-to-emitter junction is properly biased. High-power bipolar transistors are commonly used in power converters at a frequency below 10 kHz and are effectively applied in the power ratings up to 1200 V, 400 A. The various configurations of bipolar power transistors are shown in Figure 4.2. The forward drop of a conducting transistor is in the range 0.5 to 1.5 V. If the base drive voltage is withdrawn, the transistor remains in the nonconduction (or off) mode. Power MOSFETs are used in high-speed power converters and are available at a relatively low power rating in the range of 1000 V, 100 A at a frequency range of several tens of kilohertz. The various power MOSFETs of different sizes are shown in Figure 4.24. IGBTs are voltage-controlled power transistors. They are inherently faster than BJTs, but stiU not quite as fast as MOSFETs. However, they offer far superior drive and output characteristics to those of BJTs. IGBTs are suitable for high voltage, high current, and frequencies up to 20 kHz. IGBTs are available up to 1700 V, 2400 A. COOLMOS [8) is a new technology for high-voltage power MOSFETs, and it implements a compensation structure in the vertical drift region of a MOSFET to improve the on-state resistance. It has a lower on-state resistance for the same package compared with that of other MOSFETs. The conduction losses are at least 5 times less as compared with those of the conventional MOSFET technology. COOLMOS is capable of handling two to three times more output power as compared to the conventional MOSFET in the same package. The active chip area of COOLMOS is approximately 5 times smaller than that of a standard MOSFET. The on-state resistance of a 600 V, 47 A COOLMOS is 70 mil. A SIT is a high-power, high-frequency device. It is essentially the solid-state version of the triode vacuum tube. and is similar to a junction field-effect transistor (JFET). It has a low-noise, low-distortion, high-audio-frequency power capability. The tum-on and tum-off times are very short, typically 0.25 µs. The norrnaUy on-characteristic and the high cin-state drop limit its applications for general power conversions. The current rating of S!Ts can be up to 1200 V, 300 A, and the switcliing spe'ed can be as high as 100 kHz. S!Ts are most suitable for high-power, high-frequency applications (e.g.,audio, VHF/ultrahigh frequency [UHF), and microwave amplifiers). Figure 1.7 shows the power range of commercially available power semiconductors. The ratings of commercially available power semiconductor devices arc shown in Table 1.2, where the on-voltage is the on-state voltage drop of the device at
Copyrghlcd matcria
·10
Chapter 1
Introduction 6SOO V/«IJ A 12000 V/1500 A (Eupec) (Mitsubishi) 7500 V/1650 A (Eupec)
V(V) 12000
IO' 7500 «IJO 5500
3300 2500
_/ l~ ~ ~ / 2 6 S O A I 5500V/2300A I ~ ( A B B) IGBT (market) I 1----------; __ o «IJQV/«IJOAGTO ·······-········-······- ---------·-····-·····-··· ···· ····· · · · ··----- - ---• (Mittubishi) IOCT(market) ,.':::- i «IJOV/«IJOAIGCT i (Mittubishi IGCT announced) 3300 V/1200 A 4800 V/5000 A Module (Eupec)
----- - --SCR___ ----
··-··-!\
1700
2500 V/1800 A
P,c..·P•clt (Fuji)
u>'
·-r·"·
1700 V/2400 A Module (Eupcc)
(Wcstcodc)
4500 V/4000 A (Mittubishi)
1000 V/100 A '-, (SanRc,) ·~ower MOSFET 0 '-,
200
200V/500A "-.,_/ (Scmikron)
\ \
10'
L
60V/1000 A
(Semilcron) l-41:~, .,-
I
\,
~2001--~-5~00~~,.1 --,~--1-~-,..J'-+~-I-~ .,2400 4000 «IJO 10' I (A)
FIGURE 1.7
Po\\'eJ ranges orcommercially available power semiconductors. [Re!. 2. S. BerneI]
the specified current. Table 1.3 shows the symbols and the v-i characteristics of commonly used po,yer semiconductor devices. Figure 1.8 shows the applications and frequency range of power devices. A superpower device should (1) have a zero on-state voltage, (2) withstand an infinite off-state voltage, (3) handle an infinite current, and (4) turn on and off in zero time, thereby having infinite switching speed. 1.3
CONTROL CHARACTERISTICS OF POW ER DEVICES
The power semiconductor devices can be operated as switches by applying control signals to the gate termihal of thyristors (and to the base of bipolar transistors). The required output is obtained by varying the conduction time of these switching devices. Figure 1.9 shows the output voltages and control characteristics of commonly used power switching devices. Once a thyristor is in a conduction mode, the gate signal of · either positive or negative magnitude has no effect and this is shown in Figure l.9a.
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1.3
Control Characteristics of Power Devices
11
TABLE 1.2 Ratings of Power Semiconductor Devices Upper
Frequency
Rating
(Hz)
4000 V/4500 A
6000 V/3500 A
lk Ik
600V/9570 A 2800 V/1700 A 4500 V/1950 A 6000 V/ 1100 A 600V/l7 A ISOV/80A
lk 20 k 20k 20k 30k 30k
50-1()0 50-ICIO 50-100 5-10 5-10 5-10 0.2 0.2
25 k 30k
9 6
35 k
Darlington
400V/2SOA 400V/40A 630V/SOA ! 200V/400A
20 k
2 30
MOSFETs
Single
soovn.S A
IOOk
1.6
COOi.MOS
Single
s oovn .s A 600V/40A IOOOV/6. 1 A
125 k 125 k 125 k
2 I
1.5
2500 V/2400 A 1200V/52A 1200V/25 A 1200V/80A 1800 V/2200 A
IOO k IOOk IOOk IOO k ! OOk
5-10 5-10 5-10 5-10 5- 10
2.3m 0.13 0.14 44m 1.2
Dc\'iccs
Type Power Diodes
Pov."Cr diodes
General purpose
High speed
Schonky
Po"·cr Transislors
Bipolar
Single
transistors
IGBTs
Single
SITs Thyristors (SiliconControlled
On.State Resistance (0 )
Voltage/Current
Device
Phase control thyristors
Line.
commut.atcd low speed
Rcccificn) Forcedturned-off thyristors
Reverse blocking high
speed Bidirectional R CT GAIT Llgh1 triggered
Sell-turned· off thyristors
OTO HD-GTO Pulse OTO SITH MTO
ETO IGCT
Switching lime (µs)
0.32 m 0.6m O. lm 0.4m 1.2 m 1.96 m
0.14 8.63 m
4m 31 m !S m !Om I l.2 m 0.12 m 20
l.76m
1200V/300A
IOOk
o.s
6500 V/4200 A 2800 V/ 1500 A 5000 V/4600 A 5000 V/3600 A 5000 V/5000 A
60 60 60 60 60
100-400 I 00-400 I 00-400 100-400 I 00-400
0.58 m 0.72m 0.48 m 0.50 m 0.45m
2800 V/ 1850 A 1800V/2100A 4500V/3000A 6000 V/2300 A 4500 V/3700 A 4200V/1920A 2500 V/1000 A 1200V/400A 6000V/ 1500A
20 k 20 k 20k 20 k 20 k 20k 20 k 20 k 400
20-100 20-100 20-100 20-100 20-100 20-100 20-100 10-50
0.87m 0.78 m 0.Sm 0.52 m 0.53 m
2ocµ()()
2.1 m 2.2 m 0.53 01
4500 V/4000 A 4500 V/3000 A 5000 V/4600 A 4000 V/2200 A 4500V/SOO A 4500 V/4000 A 4500 V/3000 A
!Ok IOk !Ok 20 k 5k Sk 5k
50-110 50-llO 50-UO 5-10 80-llO 80-llO 80-llO
1.07 m 0.48m 5.6m 10.2m 0.5 m 0.8m
TRIA Cs
Bldireajonill
1200V/300A
MCTs
Single
4500V/250A 1400V/65 A
400 Sk Sk
0.77 n1
l.07m
200-400
3.6m
50-110 50-110
·10.4m 28m
Copyrghlcd matcria
TABLE 1.3 Cbancteristics ond Symbols of Some Pow,,r Devices Devices
Characteristics
Symbols A
.. +
'
A
'·
+
K
~ - -·-·. .... ,,_ _ _ v..
v.,
~("a
" I
v.. 0
srrn
A:
OTO
' +'·
A
~·~ '" ~r
Gate trigtttd
0
v,..
0
K
t'.
(
v
..
MCT
C.,hodc
"",.':t,._~ C..hodc
fl=lbm<>llp1, -
ETO
Tbm-on pie
~
t
Cathode IOC'r
O.i• (111.n1-on A h•,.-olfl
~
TIUAC A
A
' NPN 8JT
Ilka
B
' K
'
''tr==='•· '•>,.. I
I
IOBT
l'i-Oamid
'•
MOSFET
0
srr
12 Copyrghlcd malcria
1.3 Control Characteristics of Power Devices 13 r-• • •
•• ••
lOOM
••• •• •
~
•!:!
~
ukm
HY.DC.
-
-
CIUTtnl
Product R,U lit
• • • • • • ru111re {)e,;d, Plan
IOM IM
-?
?.- lOOK
'O
n
0.
e.
u"
~
\\'dditr
IOK
Iron mill, 1':l•n ,uppl)' f0tttitmlt.al
LK 100
10 10
100
IK IOK Operalion frequency (Hz)
lOOK
lM
FIGURE 1.8
Applications of po"·cr dc\'iCCS. (C.Ourtesy of Powcrex, Inc.)
When a power semiconduc1or device is in a normal conduction mode, there is a small voltage drop across the device. In the o u1put vohage waveforms in Figure 1.9, these voltage drops are considered negligible, and unless specified this assumption is made throughout the following chapters. The power _semiconductor switching devices can be classified on the basis of:
1. Uncontrolled turn on and off (e.g., diode); 2. Controlled tum on and uncontrolled tum off (e.g., SCR); 3. Controlled turn-on and -off characteristics (e.g.. BJT, MOSFET, GTO, SITH, IGBT, SIT, MCT); 4. Continuous gate signal requirement (BIT, MOSFET, IGBT, SIT); 5. Pulse gate r equire ment (e.g.•SCR. GTO. M CT): 6. Bipolar voltage-withstanding capability (SCR, GTO); 7. Unipolar voltage-withstanding capability (BJ'f, MOSFET, GTO, IGBT. MCT); 8. Bidirectional curre nt capability (TRIAC. RCT); 9. Unidirectional current capability (SCR, GTO, BJT, MOSFET. MCT, IGBT, SITH, SIT, diode).
Table 1.4 shows the switching characteristics in terms of its voltage, current, and gate signals.
Copyrghicd matcria
14
Chapter 1
Introduction
Oate
signal +
Vo
+er--~ ~ -t>f:.._..::....--, Input
Thyris1or
voltage
R
v,
+ Outpu1 voltage
••
r
-~ _________o__ n
n
' t
vr
(•)Thyristor •~i•ch SITH..)) A~
K
+
J~_ _ _ _ _ n vr
,,
n
[.
(b) GTOIMTO/ETO/IGCT/MCT/SITH swilch (For MCT, the polarity olVo is reversed as shown)
+ <>-----
v,
+ R
(c) TrantiSlor switch
(d) MOSFET/JGBT swi1ch
,,
[.
FIGURE 1.9
Control characteristics or power sv.itching de\'ices.
Copyrighted m~lcria
TABLE 1.4 Switching Characteristics of Po"·cr Semiconductors Continuous
Device Type
Device
Diodes
Power diode BIT MOSFET COOLMOS IGBT SIT SCR RCT TRIAC OTO MTO ETO IGC r SITH MCT
Transistors
1hyristors
() 0
~
c5::r ~ 3
~
§.
"'-
~
u,
Gate
Pulse Gate
Controlled
'I\Jm-On
Turn·Orl
x
x x x
x x x x
x x
x x x x x
Unipolar Volt<1ge
Dipolar
Unidirectional
Bidirectional
Vol lage
Current
Current
x x x x
x
x x x
x
x
x
x
x
x
x
x
x x
x
x
x
x
x
x
x x x x x
x
x x
x
Conuolled
x x x
x x x
x x x x x x
x
x x x
x x x x x
16 1.4
Chapter 1
Introduction
CHARACTERISTICS A ND SPEOFICATIONS OF SWITCHES
There are many types of power switching devices. Each device, however, has its advantages and disadvantages and is suitable to specific applications. The motivation behind the development of any new device is to achieve the characteristics of a "super device." Therefore, the characteristics of any real device can be compared and evaluated with reference to the ideal characteristics of a super device .
.
1.4.1
Ideal O,aracteristlcs
The characteristics of an ideal switch are as follows: 1. In the on-state when the switch is on, it must have (a) the ability to carry a high forward current JF, tending to infinity; (b) a low on-state forward voltage drop .VoN, tending to zero; and (c) a low on-state resistance RoN, tending to zero. Low RoN causes low on-state power loss Pm,. These symbols are normally referred to under de steady-state conditions. 2. In the off-state when the switch is off, it must have (a) the ability to withstand a high forward or reverse voltage VoR, tending to infinity; (b) a low off-state leak· age current foFF, tending to zero; and (c) a high off-state resistance RoFF, tending to infinity. High RoFF cause low off-state power loss Poff· These symbols are nor· maUy referred to under de steady-state conditions. 3. During the tum-on and tum-off process, it must be completely turned on and off instantaneously so that the device can be operated at high frequencies. Thus, it
4.
S.
6. 7. 8. 9. 10.
must have (a) a low delay time IJ, tending to zero; (b) a low rise time,,. tending to zero; (c) a low storage time,,, tending to zero; and (d) a low fall time ,1, tend· ing to zero. For tum-on and turn-off, it must require (a) a low gate-drive power Pc, tending to zero; (b) a low gate-drive voltage Ve, tending to zero; and (c) a low gate-drive ~urrent le, tending to zero. Both turn-on and turn-off mus t be controllable. Thus, it must turn on with a gate signal (e.g., positive) and must tum off with another gate signal (e.g., zero or negative). For turning on and off. it should require a pulse signal only, that is, a small pulse with a very small width rlV, tending to zero. It must have a high d vltlr, tending to infinity. That is, the switch must be capable of handling rapid changes of the voltage across it. It must have a high dildr, tending to infinity. That is, the switch must be capable of handling a rapid rise of the current through it. It requires very low thermal impedance from the internal junction to the ambient RJA, tending to zero so that it can transmit heat to the ambient easily. The ability to sustain any !a ult current for a long time is needed; that is, it must have a high value of i 2r, tending to infinity.
Copyrghlcd malcria
1.4 Characteristics and Specifications of Switches
17
lL Negative temperature coefficient on the conducted current is required to result in an equal current sharing when the devices are operated in parallel. 12. Low price is a very imponant consideration for reduced cost or the power electronics equipment.
1.4.2
Characteristics of Practical Devices During the tum-on and -off process, a practical switching device, shown in Figure l . l Oa, requires a finite delay time (td), rise time (t,), storage time (1,), and fall time (11). As the device current i,., rises during tum-on, the voltage across the device v.., falls.As the de· vice current falls during tum-off. the voltage across the device rises. The typical wave· forms of device voltages v.... and currents i,,. are shown in Figure l.!Ob. The turn-on
VSW(1,,1.g
1--i..,---1
. I-- too --J
'""'
lsws - r-,--- -
lswo 0
I I I I
I I I I
,,
1.. 1, Vee
t,
I I I I I
·I
t,
t,
,,
·I
t.,
io
las RL 0
T = Uf,
isw + io
Ysw
V()
Voc~u)
0 switch
-
Ps,..,, : I I I I I
I
0
(a) ControUcd switch
(b) $\\·ilch wa.,.crorms
FIGURE 1.10
Typical waveforms of device vot1agcs and currcn1s.
Copyrghlcd matcria
18
Chapter 1
Introduction
time (100 ) of a device is the sum of the delay time and the rise time, whereas the turnoff time (lo11) of a device is the sum of the storage time and the fall time. In contrast to an ideal, lossless switch, a practical switching device dissipates some energy when conducting and switching. Voltage drop across a conducting power device is at least on the order of 1 V, but can often be higher, up to several volts. The goal of any new device is to improve the limitations imposed by the switching parameters. The average conduction power loss, PoN is given by
1 {'"" PoN = Tslo P dt
(1.1)
where Ts denotes the conduction period and p is the instantaneous power loss (i.e., product of the voltage drop v,,. across the switch and the conducted current i,..). Power losses increase during turn-on and turn-off of the switch because during the transition from one conduction state to another state both the voltage and current have significant values. The resultant switchi.n g power loss Psw during the turn-on and tum-off periods, is given by
Psw = J, ( lof" p dt +
f" p dt + lof'' p dt ) lo
(1.2)
fs = 1/Ts is the switching frequency; 1,, 1,, and f I are the rise time, storage time, and fall time respectively. Therefore, the power dissipation of a switching device is given by: Po = PoN
+ Psw +
(1.3)
Pa
where Po is the gate driver power.
1.4.3
Switch Specifications The characteristics of practical semiconductor devices differ from those of an ideal device. The device manufacturers supply data sheets describing the device parameters and their ratings. There are many parameters that are important to the devices. The most important among these are:
Voltage ratings: Forward and reverse repetitive peak voltages, and an on-state forward voltage drop. Current ratings: Average, root-mean-square (nns), repetitive peak, nonrepetitive peak, and off-state leakage currents. Switching speed or frequency: Transition from a fully nonconducting to a fully conducting state (tum-on) and from a fully conducting to a fully nonconducting state (tum-off) arc very important parameters. The switching period Ts and frequency fs are given by 1
ls = -
Ts
1 = - - - - - - - -- Id
+ ,, + loo + ,, + ,, +
'• rt
(1.4)
where 10 rt is the off time during which the switch remains off.
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1.4 Characteristics and Specifications of Switches
19
dildt rati11g: The device needs a minimum amount of time before its whole conducting surface comes into play in carrying the full current. If the current rises rapidly, the current flow may be concentrated to a certain area and the device may be damaged. The di/tit or the current through the device is normally limited by connecting a smal.1inductor in series with the device, known as a series s11ubbu dvldr rating: A semiconductor device has an inte rnal junct ion capacitance C1 . If the voltage across the switch changes rapidly during turn-on, turn-off and also while connecting the main supply the initial current, the current C1 dvldt flowing through C1 may be too high, thereby causing damage to the device. The dvldt or the voltage across the device is limited by connecting an RC circuit across the device, known as a shum snubber, or simply snubber. Switd1i11g losses: During turn-on the forward current rises before the forward voltage falls, and during turn-off the forward voltage rises before the current falls. Simultaneous existence or high voltage and current in the device represents power losses as shown in Figure 1.lOb. Because of their repetitiveness, they represent a significant part of the losses, and often exceed the on-state conduction losses. Gate drive requirements: The gate-drive voltage and current are important parameters to turn-on and -o ff a device. The gate-driver power and the energy requirement are very important parts of the losses a nd total eq uipment cost. With large and long current pulse requirements for turn-on and turn-off. the gate drive losses can be significant in relation to the total losses and the cost of the driver circuit can be higher than the device itself. Safe operating area (SOA): The amount of heat generated in the device is proportional to the power loss, that is, the voltage-current product. For this product to be constant P = vi and equal to the maximum allowable value, the current must be in.verse proportional to the voltage. This yields the SOA limit on the allowable steady-state operating points in the voltage-current coordinates. 1211 for fusing: This parameter is needed for fuse selection. The 121 of the device must be les.~ than that of the fuse so that the device is protected under fault cur· rent conditions. Temperalllres: Maximum allowable junction, case and storage temperatures, usually between 150°C and 200°C for junction and case, and between -so•c and 175°C for storage. Thermal resistance: Junction·to-case thermal resistance, Q1c; case.to-sink thermal resistance, Qcs: and sink-ambient thermal resista nce. QsA· Power dissipation must be rapidly removed from the internal wafer through the package and ultimately to the cooling medium. The size of semiconductor power s,vitches is small, not exceeding 150 mm, and the thermal capacity of a bare device is too low to safely remove the heat generated by internal losses. Power devices are generally mounted on heat sinks. Thus, removing heat represents a high cost of equipment. 1.4.4
Device Choices
Although, there are many power semiconductor devices, none of them have the ideal characteristics. Continuous improvements are made to 1hc existing devices and new
Copyrghlcd matcria
20
Chapter 1
Introduction
devices are u·nder development.. For high power applications from the ac 50- to 60-Hz main supply, the phase control and bidirectional thyristors are the most economical · choices. COOLMOSs and IGBTs are the potential replacements for MOSFETS and BJTs, respectively, in low and medium power applications. GTOs and IGCTs are most suited for high-power applications requiring forced commutation. With the continuous advanceme nt in technology, IGBTs arc increasingly employed in high-power applications and MCTs may find potential applications that require bidirectional blocking voltages.
1.5
TYPES OF POWER ELECTRONIC CRCUITS
For the cObtrol of electric power or power conditioning, the conversion of electric power from one form to another is necessary and the switching characteristics of the power devices permit these conversions. The static power converters perform these functions of power conversions. A converter may be considered as a switching matrix. The power electronics circuits can be classified into six types: L Diode rectifiers 2. Ac-de converters (controlled rectifiers) 3. Ac-ac converters (ac voltage controllers) 4. De-de converters (de choppers) S. Dc-ac converters (inverters) 6. Static switches The devices 'in the following converters are used to illustrate the basic principles only. The switching action of a converter can be performed by more than one device. The choice of a particular device depends on the voltage, current, and speed requirements of the converter. Diode Rectifiers. A diode rectifier circuit converts ac voltage into a fixed de voltage and is shown in Figure 1.11. The input voltage to the rectifier v1could be either single phase or three phase. Ac--
Copyrghlcd malcria
1.5 Types of PowerEle
21
v,
Diode 0 1
+
+
Vi• V 111 sinwc
vi
11
ac
$Uppiy
~---Loa __ d.,.rcsNi,.,.ta_n_cc_ -. ~+
R
+
v, Diode 0, (a} Circuil diagram
(b) Voltage waverorms
FIGURE 1.11
Slngle·pbase diode rectifier drcuil.
Dc--ac converters. A dc-ac converter is also known as an inverter. A single· phase transislor inverter is shown in Figure 1.15. If transistors M1 and M2 conduct for one half of a period and M3 and M 4 conduct for the other half, the output voltage is of the alternating form. The output voltage can be controlled by varying the conduction time of transistors.
v,
v ••
__,
Thyrisr~r T I
+ ac supply
+
I I 1
11\.
\,.v,
,-
I I I
- Vm
v~ =Vmsinwt :
I
R
Load !:S:..-~tance - vo
R
.,,
Thyris1or T2 (a) Circuit diagram
+
--r---------I
·:~ 1
~
ff
2g
..,
(b) Vol\;lge waveforms
FIGURE 1.12
Single-phase ac-dc converter.
Copyrighted maleria
22
Chapter 1
Introduction
v.,
v,
"• I
I I
---------,---I
- v.. TRIAC
V.,
170
: I I I
+ ac
supply
v1 ·
+
I O
Resistive
·v'" sin'->t
load. R
- v .. (b) Voltage WB\'elorms
(a) Cucuil diagram FIGURE 1.13
Single-phase ao-.ac con,·erter.
Static switches. Because the power devices can be operated as s1atic switches or contactors, the supply 10 lhese swi1cbes could be either ac or de and the switches arc ca.lied as ac static switches or de switches. A number of conversion stages are often cascaded to produce the desired output as shown in Figure 1.16. Mains 1 supplies the normal ac supply to the load through the static bypass. The ac-dc converter charges the standby battery from mains 2. The dc-ac
:D
+
~ +
v,
Transistor
o,
Vo e -
de
+
wpply
Dm
L
•• d
••
~
T
6= ~ ·
,.u----f---f--p·-
w.
0
(a) Circuil diagram
1
I II
11
T
t
(b) Voltage waveforms
FIGURE 1.14
De-de con,·encr.
Copyrghlcd matcria
1.6 Design of Power Electronics Equipment 23
+
J
de
supply
to
v,, +
v
-
.,.,
I
~
T
+
"·
Co
Ot-~~~-tr:--~~~T.TC--
f
- v,
--------..L..~~~--' (b) Voltage waveforms
(a) Circuil diagr.1ni FIGURE 1.15
Slngle-pba.se dc-ac con\'ener.
converter supplies the emergency power to the load through an isolating transformer. Mains l and mains 2 arc normally connected to the same ac supply. 1.6
DESIGN OF POWER ELECTRONICS EQUIPMENT
The design of a power electronics equipment can be divided into four parts: L 2. 3. 4.
Design of power circuits Protection of power devices Determination of control strategy Design of logic and gating circuits
Mains 1
"'-~~~~~~~~~~~~~~~~~~--l
Load
Mains 2
Rcctifier/cba.rgcr
ln\'cner
l.solalion transformer
S1a1ic bypass sv.itch
Battery
FIGURE 1.16
Block diagram or nn unin1crruptiblc power supply (UPS).
Copyrghtcd matcria
24
Chapter t
Introduction
In the chapters that follow, various types of power electronic circuits are described and analyzed. In the analysis, the power devices are assumed to be ideal switches unless stated otherwise; and effects of circuit stray inductance, circuit resistances, and source inductance are neglected. The practical power devices and circuits differ from these ideal conditions and the designs of the circuits are also affected. However, in the early stage of the design, the simplified analysis of a circuit is very useful to understand the operation of the circuit and to establish the characteristics and control strategy. Before a prototype is built, the designer should investigate the effects of the circuit parameters (and devices imperfections) and should modify the design if necessary. Only after the prototype is built and tested, the designer can be confident about the validity of the design and can estimate more accurately some of the circuit parameters (e.g., stray inductance).
1.7
DETERMINING THE ROOT·MEAN· SQUARE VALUES OF WAVEFORMS
To accurately determine the conduction losses in a device and the current ratings of the device and components, the rms values of the current waveforms must be known. The current waveforms arc rarely simple sinusoids or rectangles, and this can pose some problems in determining the rrns values. Therms value of a waveform i(t) can be calculated as
(15) where Tis the time period. If a waveform can be broken up into harmonics whose rrns values can be calculated individually, the rms values of the actual waveform can be approximated satisfactorily by combining the rms values of the harmonics. That is, the rms value of the waveform can be calculated from
Inns =
Y13., + /~
(1.6)
where I& = the de component. lnns(t) and /=(• ) are therms values of the fundamental and nth harmonic components, respectively. Figure 1.17 shows the rrns values of different waveforms that are commonly encountered in power electronics. 1.8
PERIPHERAL EFFECTS
The operations of the power converters are based mainly on the switching of power s.e miconductor devices; and as a result the converters introduce current and voltage harmonics into the supply system and on the output of the converters. These can cause problems of distortion of the output voltage, harmonic generation into the supply sys.tern, and interference with the communication and signaling circuits. It is normally nee,, essary to introduce filters on the input and output of a converter system to reduce the
Copyrghlcd matcria
1.8 Peripheral Effects
25
I,,
I R)l~ - ~V2
'• -~1--------'• - -,,
[_.tS •
PHASE CONTROLLED SINUSOIDAL
1, .!. + sin To(! - k) co,,,.(1 - k) 2 2,r
[
]"'
k = l -.!L T
(c)
·o-- n--n f- T-j
IRMS = l p\'k
T k =~ T
PULSE
O
(d)
1-T-j
,
:lJ"TI't1 To
IIL\fS =
k
J
(k(f +
112
l:,.l b +
•lV3)
T = .!9.
T
RECTANGULAR (e)
FIGURE 1.17
The rms values or commonly encountered "'aveforms.
Copyrighlcd maleria
26
Chapter 1
Introduction Power
Source
lnpul filter
r+--
Po\\·er
convener
~
Output filter
Output
i Switching control signal generator
FIGURE 1.18
Generalized po"'cr converter systen1.
harmonic level to an acceptable magnitude. Figure 1.18 shows the block diagram of a generalized power converter. The application of power electronics to supply the sensitive electronic loads poses a challenge on the power quality issues and raises problems and concerns to be resolved by researchers. The input and output q uantities of converters could be either ac or de. Factors such as total harmonic distortion (THO). displacement factor (OF), and input power factor (lPF) are measures of the quality of a waveform. To determine these factors, finding the harmonic content of the waveform.s is required. To evaluate the performance of a converter, the input and output voltages and currents of a converter are expressed in a Fourier series. The quality of a power converter is judged by the quality of its voltage and current waveforms. The control strategy for the power converters plays an important part on the harmonic generation and output waveform distortion, and can be aimed to minimize or reduce these problems. The power converters can cause radio-frequency interference due to electromagnetic radiation, and the gating circuits may generate erroneous signals. lbis interference can be avoided by grounded slrieldi11g. 1.9
POWER MODULES
Power devices are available as a single unit or in a module. A power converter often requires two, four, or six devices, depending on its topology. Power modules with dual (in half-bridge configuration) or quad (in full bridge) or six (in three phase) are available for almost all types of power devices. The modules offer the advantages of lower onstate losses, high voltage and current s,vitching characteristics, and high.er speed than that of conventional devices. Some modules even include transient protection and gate drive circuitry. 1.10
INTELLIGENT MODULES
Gate drive circuits are commercially available to drive individual devices or modules. !11tellige111 modules, which are the state-of-the-art power electronics, integrate the power module and the peripheral circuit. The peripheral circuit consists of input or
Copyrighlcd maleria
1.1 O Intelligent Modules
27
output isolation from, and interface with, the signal and high-voltage system, a drive circuit, a protection and diagnostic circuit (against excess currcnt,short circuit, an open load.overheat ing, and an excess voltage), microcomputer control, and a control power supply.1l1c users need only to connect external (floating) power supplies. An intelli· gent module is also known as smart power. These modules are used increasingly in power electronics [6]. Sman power technology can be viewed as a box that interfaces power source to any load. The box interface function is realized with high-density complementary metal oxide semiconductor (CMOS) logic circuits, its sensing and protection function with bipolar analog and detection circuits, and its power control function with power devices and their associated drive circuits. The functional block diagram of a smart power system [71 is shown in Figure 1.19. The analog circuits arc used for creating the sensors necessary for self-protection and for pro,>iding a rapid feedback loop, which can terminate chip operation harmlessly when the system conditions exceed the 11om1al operating conditions. For example, smart power chips must be designed 10 shut down wi1hou1 damage when a short circuit occurs across a load such as a motor winding. With sman power technology, the load current is monitored, and whenever this current exceeds a preset limit, the drive voltage to the power switches is shut off. In addition to this over-current protection features such as
Sniart po,vcru:cl1001ogy
,
I
,
,,
I
/
i
,I ,,
I
l""'IU1pol:i.r power transistors
1
l_ PoY.·er
control
Power ~ HPower MOSFETs
dc\lite~
f
-lnsulated-uatr binolnr transiuors
YMOS-controlled thvristors
I
JO-VCMOS l.fitb·VOIHll,!C level shirt
Load
'
I Analog 1 OICUll'i Sonsmg and r,otection I-
Ooerauonal nmnlifiers
~ - - ~ r-lQvcrvolt:ii.!.elundef'\Oh:.te_c
: loc1~11rinl
\
1
\
\
l-fie.h·5nrcd bin,,,(a.r 1ru1l!U1ors
''
''
C'IJCUllS J HO\"cttem"'"r'J.turc
YOvercunenl/no,lo.1d
\
'
\ \
'
Interface
Logic L.rH"oh..S ' CMOS ciicult~ 1 ensov
' 'i..::=::=::::'.........::====:_~~~~~~~_J
FIGURE 1. 19 FunC'l.ionnJblock diagram or a snu,n po"·c r. l Rcf. 7. J. Balig-.11
Co ,yr ;hi
m
ri,1
28
Olapter 1
Introduction
overvollage and overtemperature protection are commonly included to prevent destructive failures. Some manufacturers of devices and modules and their Web sites are as follows: Advanced Power Technology, Inc. ABB Semiconductors Eupec Fuji Electric Collmer Semiconductor, Inc. Dynex Semiconductor Harris Corp. Hitachi, Ltd. Power Devices Infineon Technology International Rectifier Marconi Electronic Devices, Inc. Mitsubishi Semiconductors Mite! Semiconductors Motorola, Inc. National Semiconductors, Inc. Nihon International Electronics Corp. On Semiconductor Philips Semiconductors Power Integrations. Inc. Powerex, Inc. PowerTech, Inc. RCA Corp. Rockwell Inc. Reliance Electric Siemens Silicon Power Corp. Semikron International Siliconix, Inc. Tokio, Inc. Toshiba America Electronic Components, Inc. Unitrode Integrated Circuits Corp. Westcode Semiconductors Ltd .
1.11
www.advancedpowcr.com/ www.abbsem.com/ www.eupec.com/p/index.hlm www.fujielectric.co.jp/eng/denshi/scd/index.htm www.collmer.com www.dynexsemi.com www.harris.com/ www.hitachi.co.jplpse www.infineon.com/ www.irf.com www.marconi.com/ www.mitsubishielectric.com/ www.mitelsemi.com www.motorola.com www.national.com/ wwv.:,abbscm.corn/english/salesb.htm www.onsemi.com www.semiconductors.philips.com/catalog/ www.powerint.com/ www.pwrx.com/ www.powcr-tech.com/ www.rca.com/ www.rockwell.com www.reliance.com www.siemens.com www.siliconpower.com/ www.semikron.com/ www.siliconix.com www.tokin.com/ www.toshiba.com/taec/ www.unitrode.com/ www.westcode.com/ws-prod.html
POWER ELECTRONICS JOURNALS AND CONFERENCES There are many professional journals and conferences in which the new developments are published. The Institute of Electrical and Electronics Engineers (IEEE) c·library Explore is an excellent tool in finding articles published in the IEE journals
Copyrghlcd malcria
References
29
and magazines, and in the IEEE journals. magazines, and sponsored conferences.
Some of them are: IEEE e_Library ieeexplore.ieee.org/ IEEE Transactions on Aerospace and Systems www.iece.org/ IEEE Transactions on Industrial Electronics www.ieee.org/ IEEE Tnmsactions 011 Industry Applications www.ieee.org/ IEEE Transactions on Power Delivery www.iece.org/ IEEE Transactions on Power Electronics www.iccc.org/ IEE Proceedings 011 Electric Power www.ice.org/Publish/ Journal of Electrical Machinery 111,d Power Systems Applied Power Electronics Conference (APEC) European 'Power Electronics Conference (EPEC) IEEE Industrial Electronics Conference (IECON) IEEE Industry Applications Society (!AS) Annual Meeting International Conference on Electrical Machines (lCEM) International Power Electronics Conference (IPEC) International Power Electronics Congress (CIEP) International Telecommunications Energy Conference (INTELEC) Power Conversion Intelligent Motion (PCIM) Power Electronics Specialist Conference (PESC)
SUMMARY As the technology for the power semiconductor devices and integrated circuits develops, the potential for the applications of power electronics becomes wider. There arc already many power semiconductor devices that arc commercially available; however, the development in this direction is continuing. 11te power converters fall generally into six categories:(!) rectifiers, (2) ac-dc conver1ers, (3) ac- ac converlers, (4) de-de converters, (5) dc-ac converters. and (6) static switches. The design of power electronics circuits requires designing the power and control circuits. The voltage and current harmon.ics that arc generated by the power conver1crs can be reduced (or minimized) with a proper choice of the control strategy.
REFERENCES
.
[l) E. I. Carroll. "Power electronics: where next?" Power E11gi11eeri11g /011rnal, December 1996, pp. 242- 243. (2) S. Bernet. ''Recent developments of high power converters for industry and traction appli· cations." IEEE Tra11sactiom 011 Power Electronics, Vol. 15, No. 6, November 2000, pp. 1102-1117. [3] E. l. Carroll. "PO\VCT electronics for very high power applications,.. Power Engineering l our·
1101.April 1999,pp.81-87. (4) P. K. Steimer. H. E. Gruning. J. Wcminger. E. Carroll. S. Klada. and S. Linder, "IGCT-a new emerging ror hig.h po,1,·er. lo,v cos1 inverters;· IEEE Industry Applico1ions Maga7.ine, July/August 1999, pp.12-18. [SI R. G. Horr, Se111ico11tl11c1or Po1vtr £tcc1ro11ics. Ne\\' York: Van Nostrand Reinhold. 1986.
Copyrghtcd matcria
30
Chapter 1
Introduction
(6) K. Gndi. "Power electronics in action," IEEE Sp,ctrum,July 1995, p. 33. (7) J. Baliga, " Power )Cs in the daddlc." IEEE SptcJnim, July 1995, pp. 34-49. [8) I. Zvcrev and J. Hancock, "CoolMOS Select.ion Guide; Application note: AN-CoolMOS· 02, lnfm,on Tcdmologies, June 2000. [9J "Power Electronics Books." SMPS Technology Knowledge Base, March I , 1999. www.smpstech.com/bookslbooklist.htm (10) "Power Communities; Darnell.Com Inc., March I, 2002. www.damell.com/
REVIEW QUESTIONS Lt 1.2 1.3 1A
1.6 1.7 1.8 1.9 1.10 1,11 LU 1.13 Ll4
What are power electronics? What are the various types of thyristors? What is a commutation circuit? What are the conditions for a thyristor to conduct? How can a conducting thypstor he turned off? What is a line commutation? What is a forced commutation? What is the difference. between a thyristor and a TRIAC? What is the gating characteristic of a GTO? WbAt is the gating characteristic of an MTO? What is the gating characteristic of an ETO? What is the gating characteristic of an IGCT? What is tum-off time or a thyristor? What is a converter?
1..15 1.16 Ll7 L18 1.19 1.20 l.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 1.29
What is the principle o( ac-dc conversion? What is the principle of ac-ac conversion? What i.s the principle of de-de conversion? Whal is the p rinciple of dc-ac conversion? Whal are the steps invo lved in designing power electronics equipment? What arc the peripheral effects of power electronics equipment? Whal are the differences in the gating characteristics of GTOs and thyristor..? What arc the djffcrcnccs in the gating characteristics of thyristors and transistors? What arc the differences in the gating characteristics of BJTs and MOSFETs? Whal is the gating characteristic of an IGBT? Whal is the gating characteristic of an MCT? Whal is the gating characteristic of an SIT? Whal arc the differences between BJTs and JGBTs? What are the diflcrcnccs between MCT, and GTOs? Whal are the differences between SITHs and GTOs?
LS
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CHAP T ER
2
Power Semiconductor Diodes and Circuits The /earning objectives of this chapter are as follo ws: • • • • • • 2.1
To understand the diode characteristics and its models To learn the types of diodes To learn the series and parallel operation of diodes To learn the SPICE diode model To study the efrects of a unidirectional device like a diode on RLC circuits To study the applications of diodes in freewheeling and stored-energy recovery INTRO DUCTION
Many applications have been found for diodes in electronics and electrical engineering circuits. Power diodes play a significant role in power electronics circuits for conversion of electric power. Some diode circuits that are commonly encountered in power electronics for power processing are reviewed in this chapter. A diode acts as a switch to perform various functions. such as switches in recti· Ciers, freewheeling in switching regulators. charge reversal of capacitor and energy transfer between components, voltage isolation. energy feedback from the load to the power sou,ce, and trapped energy recovery. Power diodes can be assumed as ideal switches for most applications but practical diodes differ from the ideal characteristics and have certain limitations. The power diodes are similar to p11-j unction signal diodes. However, the power diodes have larger power-, voltage-, and current-handling capabilities than those of ordinary signal d iodes. The frequency response (or switching speed) is low compared witli that of signal diodes. 2.2
SEMICONDUCTOR BA SICS
Power semiconductor devices are based on high-purity. single-crystal silicon. Single crystals of several meters long and "ith the required diameter (up to 150 mm) arc 31 Copyrighlcd maleria
32
Chapter 2
Power Semiconductor Diodes and Circuits
grown in the so-called float w11e furnaces. Each huge crystal is sliced into thin wafers, which then go through numerous process steps to turn into power devices. Silicon, is a member of Group IV of the periodic table of elements, that is. having four electrons per atom in its outer orbit. A pure silicon material is known as an intri11sic semico11ductor with resistivity that is too low to be an insulator and too high to be a conductor. It has high resistivity and very high dielectric strength (over 200 kV/cm). The resistivity of an intrinsic semiconductor and its charge carriers that arc available for conduction can be changed, shaped in layers, and graded by implantation of specific impurities. The process of adding impurities is called doping, which involves a single atom of the added impurity per over a million silicon atoms. With different impurities, levels and shapes of doping, high technology of photolithography, laser cutting, etching, insulation, and packaging, the finished power devices are produced from various structures of n-type and p-type semiconductor layers. n· Type material: If pure silicon is doped with a small amount of a Group V ele· ment, such as phosphorus. arsenic, or antimony, each atom of the dopant forms a covalent bond within the silicon lattice, leaving a loose electron. These loose electrons greatly increase the conductivity of the material. When the silicon is lightly doped with an impurity such as phosphorus, the doping is denoted as n doping and the resultant material is referred to as n·type semiconductor. When it is heavily doped, it is denoted as n + doping and the material is referred to as n +-type semiconductor. p· Type material: If pure silicon is doped with a small amount of a Group HI element. such as boron, gallium. or indium. a vacant location called a hole is introduced into the silicon lattice. Analogous to an electron, a hole can be considered a mobile charge carrier as it can be filled by an adjacent electron, which in this way leaves a hole behind. These holes greatly increase the conductivity of the material. When the silicon is lightly doped with an impurity such as boron, the doping is denoted as p-dopi11g and the resultant material is referred to as p -type semiconductor. When it is heavily doped, it is denoted as p+ doping and the material is referred to as p+-type semiconductor.
Therefore, there are free eleclrons available in an rt-type material and free holes available in ap-type material. In a p-type material the holes are called the majority carriers and electrons are called the minority carriers. In the 11-type material, the electrons are called the majority carriers, and holes are called the minority carriers. These carriers are continuously generated by thermal agitations, \hey combine and recombine in accordance to their lifetime, and they achieve an equilibrium density of carriers from about 1010 to 1013/cm3 over a range of about O •c to 1000 •c. Thus, an applied c lcclric field can cause a current flow in an n-type or p -type material. Key Points of Section 2.2
• Free e lectrons or holes are made available by adding impurities to the pure silicon or germanium lhrough a doping process. The electrons are the majority carriers in the n-type material whereas the holes are the majority carriers in a p-type
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2.3
Diode Characteristics
33
material. Thus, the application of electric field can cause a current flow in an n-type or a p-type material. 2.3
DIODE CHARACTERISTICS
A power diode is a two-terminal pn-junction device [l, 2) and a pn-junction is normally formed by alloying, diffusion, and epitaxial growth. The modern control techniques in diffusion and epitaxial processes permit the desired device characteristics. Figure 2.1 shows the sectional view of a pn-junction and diode symbol. When the anode potential is positive with respect to the cathode, the diode is said to be forward biased and the diode conducts. A conducting diode has a relatively small forward voltage drop across it; and the magnitude of this drop depends on the manufacturing process and junction temperature. When the cathode potential is positive with respect to the anode, the diode is said to be reverse biased. Under reverse-biased conditions, a small reverse current (also known as leakage currelll) in the range of micro- or milliampere, flows and this leakage current increases slowly in magnitude with the reverse voltage until the avalanche or zener voltage is reached. Figure 2.2a shows the steady-state v -i characteristics of a diode. For most practical purposes, a diode can be regarded as an ideal switch. whose characteristics are shown in Figure 2.2b. ' The v-i characteristics shown in Figure 2.2a can be expressed by an equation known as Schockley diode equation. and it is given under de steady-slate operation by lo = l,(eV,{nVr - I)
(2.1)
where 10 = current through the diode, A; V0 = diode voltage with anode positive with respect to cathode, V; I, = leakage (or reverse saturation) current, t)'Pically in the range 10-6 to 10-15 A; n = empirical constant known as emission coefficiem, or idealiry factor, whose value varies from 1 to 2. Toe emission coefficient n depends on the material and the physical construction of the diode. For germanium diodes, 11 is considered to be 1. For silicon diodes, the predicted value of II is 2, but for most practical silicon diodes, the value of II falls in the range 1.1 to 1.8.
Anode ___..,. p
Cathode n ,__ ..
+ " (a) pn·junclion
Anode
Cathode
"
+ 1, -
'----1,1 , ----' (b) Diode symbol
FIGURE Z.1 pn·lu.nction and diode symbol.
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34
Chapter 2
Power Semiconductor Diodes and Circuits
lo -VoR
O
v
0
v
Rc\'ersc
leakage currcnl
FIGURE 2.2
(b) Ideal
(•) Practical
v-i Otaracteristics of diode.
V7 in Eq. (2.1) is a constant called thermal voltage and it is given by kT Vr= IJ
(2.2)
where q = electron charge: 1.6022 x 10- 19 coulomb (C); T = absolute temperature in Kelvin ( K = 273 + °C}; k ~ Boltzmann's constant: 1.3806 x 10- 23 J/K. At a junction temperature of 25 °C, Eq. (2.2) gives
kT
Vr
= -
1.3806
x
10-23
1.6022
x (273 + 25)
X
10- 19
"' 25.7mV
At a specified temperature, the leakage current I, is a constant for a given diode. The diode characteristic of Figure 2.2a can be divided into three regions: Forward-biased region, where VO > 0 Reverse-biased region, where VO < 0 Breakdown region. where V0 < -V8 R
Forward-biased region. In the forward-biased region. V0 > 0. The diode current 10 is very small if the diode voltage V0 is lcs.5 than a specific value Vro (typically 0.7 V). The diode conducts fully if V0 is higher than this value Vro, which is referred to as the threshold voltage. c11t-i11 voltage, or 111m-011 voltage. Thus, the threshold voltage is a voltage at which the diode conducts fully. Let us consider a small diode voltage V0 = O.l V, 11 = I, and Vr = 25.7 mV. From Eq. (2.1) we can find the corresponding diode current I O as lo = l,(ev,,inv,. - 1) = l , [eO.tl( l x0.02S7 ) - lj = /,(48.96 - 1) = 47.9615
which can be approximated to 10 a. l,ev.,nv, = 48.96 /,. that is, with an error of c.1 %. As v 0 increases. the error decreases rapidly.
Copyrighlcd maleria
2.4 Reverse Recovery Characteristics
Therefore, for Vo > 0.1 V, which is usually the case, lo be approximated within 2.1 % error to
>>
35
l,. and Eq. (2.1) can
10 = l,(ev,;,v, - 1) ::: l,ev,;,v,
(2.3)
Reverse-biased region. In the reverse-biased region, V0 < 0. If Vo is negative and !Vol >> V7 , which occurs for V0 < - 0.1 V, the exponential term in Eq. (2.1) be· comes negligibly small compared with unity and the diode current / 0 becomes lo
Q
l ,(e-!VvllnV, - I):::- /,
(2.4)
which indicates that the diode curre nt 10 in the reverse direction is constant and equals l ,. Breakdown region. In the breakdown region, the reverse voltage is high, usually with a magnitude greater than IOOOV. The magnitude of the reverse voltage may exceed a specified voltage known as the breakdown voltage V88, with a small change in reverse voltage beyond VsR· The reverse current increases rapidly. The operation in the break· down rcgjon will not be destructive, provided that the power dissipation is within a "safe level " that is specified in the manufacturer's data sheet. However, it is often necessary to limit the reverse current in the breakdown regjon to limit the power dissipation within a permissible value.
Example 2.1
Finding the Saturation Current
The forward voltage drop of a power diode is v,, • 1.2 Vat / 0 and Vr = 25.7 mV, find the reverse saturation current / J.
•
300 A. Assuming that " • 2
Solution Applying Eq. (2.1), we can find the leakage (or saturation) current I, from 300
which gives I, = 2. 17746 x
-=
/ l (e1 ..?l(.?x 1S.1x10·1, _
IJ
w-• A.
Key Points of Section 2.3 • A diode exhibits a nonlinear v -i characteristic, consisting of three regions: forward biased, reverse-biased, and breakdown. In the forward condition the diode drop is small, typically 0.7 V. If the reverse voltage exceeds the breakdown voltage, the diode may be damaged.
2.4
REVERSE RECOVERY CHARACTERISTICS The current in a forward-biased junction diode is due to the net c((cc.i of majority and minority carriers. Once a diode is in a fo rward conduction mode and then its forward cur· rent is reduced to zero (due to the natural behavior of the diode circuit or application of a reverse voltage), the diode continues to conduct due to minority carriers that remain stored in the p11-junction and the bulk semiconductor material. The minority carriers require a certain time to recombine with opposite charges and to'be neutralized. This time
Copyrghtcd matcria
36
Chapter 2
Power Semiconductor Diodes and Circuits
IRR - - - - - - -
----!
'• (b) Abrupt recovery
(a) Soft recovery FIGURE 2.3
Reverse rcco,·cry charac.tc·ristics.
is called the reverse recovery time of the diode. Figure 2.3 shows two reverse recovery characteristics of junction diodes. Toe soft-recovery type Is more common. Toe reverse recovery time is denoted as t,, and is measured from the initial zero crossing of the diode current to 25% of maximum (or peak) reverse current IRR· The t,, consists of two components, t0 and tb. Variable t0 is due to charge storage in the depletion region of the junc· tion and represents the time between the zero crossing and the peak reverse current IRR· "The tb is due to charge storage in the bulk sernioonductor material. Toe ratio t,Jt0 is known as the softness factor (SF}. For practical purposes, one needs be concerned with the total recovery time t,, and the peak value of the reverse current IRR·
,,, = t,, +
lb
(2.5)
The peak reverse current can be expressed in reverse dildt as di IRR = la dt
(2.6}
Reverse recovery time r,,. may be defined as the time interval between the instant the current passes through zero during the changeover from forward conduction to reverse blocking condition and the moment the reverse current has decayed to 25% of its peak reverse value I RR· Variable r,, is dependent on the junction temperature, rate of fall of forward current, and forward current prior to commutation, IF· Reverse recovery charge QRR, is the amount of charge carriers that flows across the diode in the reverse direction due to changeover from forward conduction to reverse blocking condition. Its value is determined from the area enclosed by the path of the reverse recovery current. The storage charge, which is the area enclosed by the path of the recovery current, is approximately (2.7)
or (2.8)
Copyrighted m~lcria
2.4
Reverse Recovery Characteristics
37
Equating I RR in Eq. (2.6) to IRR in Eq. (2.8) gives 2QRR
(2.9)
t,,t. ; dildt If tb is negligible becomes
as compared to 1., which is usually the case,'" "' 1., and Eq. (2.9)
1 "
~fW} dildt
(2.10)
and IRR;
~
(2.11)
It can be noticed from Eqs. (2.10) and (2.11) that the reverse recovery time 1,, and the peak reverse recovery current IRR depend on the storage charge QRR and the reverse (or reapplied) dildt. The storage charge is dependent on the forward diode current IF. The peak reverse recovery current IRR• reverse charge QRR, and the SF are all of interest to the circuit designer, and these parameters are commonly included in the specification sheets of diodes. If a diode is in a reverse-biased condition, a leakage current flows due to the minority carriers. Then the application of forward voltage would force the diode to carry :urrent in the forward direction. However, it requires a certain time known as fonvard recovery (or tum-on) time before all the majority carriers over the whole junction can contribute to the current flow. If the rate of rise of the forward current is high and the forward current is concentrated to a small area of the junction, the diode may fail. Thus, the forward recovery time limits the rate of the rise of the forward current and the switching speed. Example 2.2
Finding the Reverse Recovery Current
The reverse recovery time or a diode is 1,, = 3 µs and the rate or fall of the diode current is dildt = 30 Nµs. Determine (a) the storage charge QHR, and (b) the peak re,•erse current /R R·
Solution ,,, • 3 µsand dildt • 30 Nµs. L
From Eq.(12.10),
Q•• = } dfi =·0.5 x 30 N µs x (3 x 10· )2 = 135 µC 1~,
6
((
b. From Eq. (2. ll),
r:::,fi
, •• = y
V2 X 135 X t0-6 X 30 x
l
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38
Chapter 2
Power Semiconductor Diodes and Circuits
Key Points or Sedion 2.4 • During the reverse recovery lime 1,,. the diode behaves effectively as a short circuit and is not capable of blocking reverse voltage, allowing reverse current flow, and then suddenly disrupting the current. Parameter 1,, is important for switching applications. · 2.5
POWER DIODE TYPES Ideally, a diode should have no reverse recovery time. However, the manufacturing cost of such a diode may increase. In many applications, the effects of reverse recovery time is not significant, and inexpensive diodes can be used. Depending on the recovery characteristics and manufacturing techniques, the power diodes can be classified into the following three categories:
1. Standard or general-purpose diodes 2. Fast-recovery diodes 3. Schottky diodes The characteristics and practical limitations of these types restrict their applications. 2.5.1
General-Purpose Diodes The general-purpose rectifier diodes have relatively high reverse recovery time, typically 25 µ s: and arc ·used in low·speed applications, where recovery time is not critical (e.g., diode rectifiers and converters for a low-input frequency up to I -kHz applications and line-commutated converters). These diodes cover current ratings Crom less than I A to several thousands of amperes, with voltage ratings from 50 V to around 5 kV. These diodes are generally manufactured by diffusion. Howe,•er, alloyed types of rectifiers that ure used in we lding power supplies are most cost-effective and rugged. and their ratings can go up to 1500 V, 400 A. Figure 2.4 shows various configurations of general-purpose diodes, which basically fall into two types. One is called a s tud, or stud-moumed type; the other one is called a disk, press pak, or hockey-puck type. In a stud-mounted type, either the anode or the cathode could be the stud.
2.5.2
Fast-Recovery Diodes Toe fast-recovery diodes have low recove ry time, normally less than 5 11s. They are . used in dc-
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2.6
Silicon Carbide Diodes
39
FIGURE 2.4 Fast.reco\·e,ry diodes. (Courtesy of
Pov.·erex. lnc.)
2.5.3
Schottky Diodes The charge storage problem or a p11-junc1ion can be eliminated (or minimized) in a Schottky diode. ll is accomplished by setting up a "barrier potential" with a contact between a metal and a semiconductor. A layer or metal is deposited on a thin epitaxial layer or 11-type silicon . The potential barrier simulates the behavior or a p11-junction. The rectifying action depends on the majority carriers only, and as a result there are no excess minority carriers to recombine. The recovery effect is due solely to the selfcapacitancc of the semiconductor junction. The recovered charge of a Schottky diode is much less than that of an equivalent p11-junction diode. Because it is due only lo the junction capacitance, it is largely independent of the reverse dildt. A Schottky diode has a relatively low forward voltage drop. Toe·Jeakage current of a Schottky diode is higher than that of a p11-junction diode. A Schottky diode with relatively low-conduction voltage has relatively high leakage current, and vice versa. As a result. the maximum allowable voltage of this diode is generally limited to 100 V. The current ratings of Schonky diodes vary Crom 1 to 400 A.11,e Schottky diodes arc ideal for high-currenl and low-voltage de power supplies. However, these diodes arc also used in low-current power supplies for increased efficiency. In Figure 2.5, 20- and 30-A dual Schottky rectifiers are shown. Key Points or Section 2.5
• Depending on the switching recovery lime and the on-state drop, the power diodes are of three types: general purpose, fast recovery, and Schottky. 2.6
SILICON CARBIDE DIODES
Silicon Carbide (SiC) is a new material for power electronics. lls physical properties outperform Si and GaAs by far. For example. the Schottky SiC diodes manufaclurcd
Co yr ,Jh!
m
'1
rir.1
40
Chapter2
Power Semiconductor Diodes and Circuits
FIGURE 2.5
Dual Schouky center r«tifi
by lnfineon Technologies (3) have ullralow power losses and high reliability. They also have the following features:
• No reverse recovery time; • Ultrafast switching behavior; • No temperature inOuence on the switching behavior. Toe typical storage charge QRR is 21 nC for a 600-V, 6-A diode and is 23 nC for a 600-V, 10-A, device. 2.7
SPICE DIODE MODEL
The SPICE model of a diode (4-6) is shown in Figure 2.6a.The diode current / 0 that depends on its voltage is represented by a current source. R, is the series resistance, and it is due to the resistance of the semiconductor. R,, also known as bulk resista11ce, is dependent on the amount of doping. The small-signal and static models that are generated by SPICE are shown in Figures 2.6b and 12.6c, respectively. C 0 is a nonlinear function of the diode voltage v 0 and is equal 10 C0 = dqidv0 , where qd is the depletion-layer charge. SPICE generates lhe small-signal parameters from the operating point. Toe SPICE model statement of a diode has the general forrn
.MODEL ONAME D (Pl=Vl P2 =V2 P3=V3 ..... PN=VN) DNA ME is the model name and it can begin with any character: however.its word size is normally limited lo 8. Dis the type symbol for diodes. Pl, P2, .. . and VI, V2, ... are the model parameters a.n d their values, respectively. Among many diode parameters, the important parameters (5) for power switching arc: IS BV JBV TT CJO
Saturation current Reverse breakdown voltage Reverse breakdown current Transit time Zero-bias pn capacitance
Coµyr ·Ill
m \~rl I
SPICE Diode Model
2. 7
41
A
A
+
K
(a) Diode
K (b) SPICE model
A
A
+ +
K
K
(c) Small-signal model
(d) Static model
FIGURE 2.6 SPICE diode model with reverse-biased diode.
Because the SiC diodes use a completely new type of technology. the use of SPICE models for silicon diodes may introduce a significant amount of errors. The manufacturers (3) are, however, providing the SPICE models of SiC diodes.
Key Points of Section 2.7 • The SPICE parameters. which can be derived from the data sheet, may significantly affect the transient behavior of a switching circuit.
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42
2.8
Chapter 2
Power Semiconductor Diodes and Circuits
SERIES-CONNECTED DIODES In many high-voltage applications (e.g., high-voltage direct current (HVDC] transmission lines), one commercially available diode cannot meet the required voltage rating, and diodes are connected in series to increase the reverse blocking capabilities. .Let us consider two series-connected diodes as shown in Figure 2.7a. Variables i 0 and v0 are the current and voltage, respectively, in the forward direction; v01 and v 02 are the sharing reverse voltages of diodes D1 and Di, respectively. In practice, the v -i characteristics for the same type of diodes differ due to tolerances in their production process. Figure 2.7b shows two v-i characteristics for such diodes. In the forward-biased condition, both diodes conduct the same amount of current, and the forward voltage drop of each diode would be almost equal. However, in the reverse blocking condition, eQcb diode bas to carry the same leakage current, and as a result the blocking voltages may differ significantly. A simple solution to this problem, as shown in Figure 2.Sa, is to force equal voltage sharing by connecting a resistor across each diode. Due to equal voltage sharing, the leakage current of each diode would be different, and this is shown in Figure 2.8b. Because the total leakage current must be shared by a diode and its resistor,
I,
= I,1 + IR1 = la + Im
(2.12)
However, /RI = V01 I R1 and Im = Vrnl R2 = V01 /R2• Equation (2.12) gives the relationship between R 1 and R2 for equal voltage sharing as (2.13)
If the resistances are equal, then R = R 1 = R2 and the two diode voltages would be slightly different depending on the dissimilarities of the two v-i characteristics. The
- Vo,
- Voi
I
I I
---.""---r''--:::::::=-:::"""r.----+ v., ,.,. 0
,. ,.
o,
+
+
I
I
_,.,. _. /
--- /~--- -- - - - -- -l, 1 I I
D, I
I,
+
I
I
(a) Circuit diagram
I
I
I
(b) 11- i Characccristics
FIGURE 2.7
Two series-connected diodes ""'ilh re\'erse bias.
Copyrghlcd malcria
2.8
Serles-Connected Diodes
in)
,., Vu1
-
R,
+
ii)
- V o,
..,
R,
+
y'
v/
- Vu.?
~ ~~'--.-~-'-~--::::=="'!"~;;-~ ~ - vn
o,
,.,.,. ,.' '
o,
0
___ ..,.... _______ _ - 1,,
+
1,,
,/
Vo
,.,
\'02
c=:
43
,, __________ ___ I ,' I ,'
I
I,
I
I
I
,.
- I ,
I
(b) &.1 - i Chanu:tcrislics
(a) Circuit diagram FIGURE 2.8
Series-conn«.1ed diodes v.•ith ste3
S1eady·
. 'R I
R,.;
·:,
c,,= Transient
! o,
stale \'Oltage sharing
.~R2 '
,·oh age 1 c~..
~ Dz
I""'
...,>
sh,ring
R
FIGURE 2.9
Series diodes with \'Oltagc-sharing netv.'Ol'ks under steady-stale and ttansie nl conditions.
values of V01 a nd V02 can be determi ned from Eqs. (2.14) a nd (2. 15):
,,, +
1101
\/02
If = ,,2 + If
(2. 14)
v,
(2. 15)
v0 1 + Vm =
The voltage sharings unde r 1rnnsie n1 conditio ns (e.g., d ue to switching loads, the initial applications of 1he inp ut voltage) arc accomplished by connecting capacitors across each diode, which is shown in Figure 2.9. R, limits the ralc of rise of the block· in g ..-oltagc.
Example 2.3
Finding t he Voltage Sharing Resistors
Two diodes are t'Onncctcd in series. sho\vn in Figure 2.Sa to share a tota l de reverse voltage or VI)= 5 kV. The re\•crsc leakage curre nts o f the t,vo d iodc.s a rc / 11 = 30 mA a nd l.~i = 35 011\. (n) Find t he diode vo hag<'s if the vohage-shnring resistances arc equa l. R1 = R2 = R = 100 ktl . (b) Find 1hc voltage-sharing resistances 111 a nd 112 if t he d iode voltages are equal. Vm = v,,, = V1f2. (c) Use PSpice 10 check your results of part (a). PSpicc model p:11amc 1crs or the diodes are l)V = 3 kV and IS = 30 mA for di ode 0 1• a nd IS = 35 mA for d iode D,.
Copy(ghlcd matcria
44
Chapter 2
Power Semiconductor Diodes and Circuits
Solution a. 1, 1 = 30 mA, 1,2 = 35 mA, and R 1 V02 = V 0 - V01 . From Eq. (2.14), V01
1, + -
'
R
= R2 = R = 100 kO.
-V0 = -V01 - Vm or
Voi
= /- + -
"
R
Substituting Vm = V0 - V01 and solving for the diode vollage Di, we gel Vo, =
Vo
R
2 + 2(/,2 -
/, 1)
5 kV + IOO kO (35 x 10-l - 30 x 10- 3) = 2750 V
G
2
(2.16)
2
and V02 = V0 - Vm = 5 kV - 2750 = 2250V. b. I,. a 30 mA, l ,z = 35 mA, and V01 = Vo, a Vof2 = 2.5 kV. Frnm Eq. (2.13),
v)
v
I + - '- ' = I +..!E. "'
R1
R2
,2
which gives the resistance R2 for a known value or R, as R
_
,-
V02R,
(2.17)
Vo , - R1( l,1 - l ,1)
Assuming that R 1 = 100 kO. we get
R
2
=
2.5 kV x 100 kO = l2S kO 2.5 kV - 100 kO x (35 x 10-1 - 30 x 10-1)
c. The diode circuit for PSpice simulation is shown in Figure
2.10. The list or the circuit
file is as follows: Exa:rple 2 .3
vs
0
R
1 1
Rl
2
R2 Dl lJ2 . MODEL
3
3 0
3 0
2
2
3
Diode Voleage- Sharing Circuit SKV 0. 01
cc
lOOK lOOK M:lDl M:)02 0
M:JDl D ( IS=30MA BV=3KV)
.!1:lDEL !1'02 D (IS=35l>!A BV-3KV) .OP
~
Diode model parameters Diode model parameters De operating' point analysis
. END
.Copyrghlcd :• malcria
2.9
Parallel-Connected Diodes
45
R
o.ot n R,
D, +
v.
5kV
iOOkfl
3.-- --. R:
iOOkfl
0--------------
FIGURE
2.10
Diode circuil for PSpicc simulation ror Example 2.3.
The results of PSpicc simulation arc NAME
ID VD
01 -3. 00E- 02 1 01 • - 30 mA - 2. 75E+03 V01=-2750 V expected -2750 V
l .OOE+l2 R01 =l
en
02 -3 . SOE- 02 I a2=-JS mA -2. 25E+03 Vc,=- 2250 V expected - 2250 V 1. 00E+12 Ro,=l en
Note: 11te SPICE gives lhe same vollages as expecled. A small resislance of R = 10 mn is insened 10 avoid SPICE error due to a zero-resistance voltage loop. Key Points of Section 2.8 • When diodes of lhe same 1ype are connec1ed in series. they do not share the same reverse vollage due to mismalches in !heir reverse v-i characleristics. Voltagesharing networks are needed to equalize the voltage sharing. 2.9
PARALLEL-CONNECTED DIODES
In high-power applications, diodes are connected in parallel to increase 1be currentcarrying capability to meet the desired current requirements. The current sharings of diodes would be in accord with their respective forward voltage drops. Uniform current sharing can be achieved by providing equal inductances (e.g., in the leads) or by connecting current-sharing resistors (which may not be practical due to power losses}; and this is depicted in Figure 2.11. It is possible to minimize this problem by selecting diodes with equal forward vohage drops or diodes of the same type. Because the diodes are connected in parallel. the reverse blocking voltages of each diode would be lhe same. The resistors of Figure 2.1 la help current sharing under slcady-statc conditions. Current sharing under dynamic conditions can be accomplished by connecting coupled induciors as show'n in Figure 2.llb. lf the current through Di rises, the L dVdt across Li increases. and a corresponding vohage of opposite polarily is induced across induclor L 2• The result is a low-impedance path through diode Di and the current is shifled to Di, The induclors may gcnerale voltage spikes and they may be expensive and bulky, especially al high currents.
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46
Chapter 2
Power Semiconductor Diodes and Circuits
o,
o,
+ Vo
R,
o,
o,
R,
R,
+ Vo
R,
FIGURE 2.11 Parallel-connt.'Ctcd diodes.
(•) S1cady-statc
(b) Dynamic sharing
Key Points of Section 2.9 • When diodes of the same type are connected in parallel, they do not share the same on-state current due to mismatches in their forward v-i characteristics. Current sharing networks are needed to equalize the current sharing. 2.10
DIODES WITH RC AND RL LOADS
Figure 2.12a shows a diode circuit with an RC load. For the sake of simplicity the diodes are considered to be ideal. By "ideal" we mean that the reverse recovery timer,, and the forward voltage drop VD arc negligible. That is,,,, = 0 and VD = 0. The source voltage Vs is a de constant voltage. When the switch S1 is closed at r = 0, the charging current i that nows through the capacitor can be found from
11''•
V, = VR + Ve = VR + C
i dt
+
(2.18)
v,(t = 0)
(2.19)
v,K__ +
-+
t • 0'
R
"•
o.~~Lh V, 0.632 v,
I
-----L---~-~-,-~~~
v,
1
Q IC...~~.I....~~~~~ ·
1-T-l
(a) O rcuit diogram
(b) W:t\•eforms
FIGURE 2.12
Diode circuit with an RC load.
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2.10
Diodes with RC and RL Loads
47
With initial condition v,(t Q 0) a 0, the solution of Eq. (2.18) (which is derived in Ap· pendix D. Eq. D.l) gives the charging current i as
v
i(t) = ..2. e-•I RC R
(2.20)
vc(r) = Cl {; dr = V,(1 - e-,,RC) = V,( l - c-''')
(2.21)
The capacitor voltage v, is
JO
where T = RC is the time constant of an RC load. The rate of change of the capacitor voltage is
dv, V, - 1/RC = e dr RC
-
(2.22)
and the initial rate of change of the capacitor voltage ( at r • 0) is obtained from Eq. (2.22)
dv, 1
tit
_ V, RC
,.o-
(2.23)
A diode circuit with an RL load is shown in 1-igure 2.13a. When switch S 1 is closed at t = 0, the current i through the inductor increases and is expressed as
di R,' V =vl+vR= L -+ '
dt
(2.24)
With initial condition i(r s 0) = 0, the solution of Eq. (2.24) (which is derived in Appendix D, Eq. D.2) yields (2.25)
s, o, +
-
1= 0 R
v,
+
v,
+
•• +
L
•c
(•) Circuit diogram FIGURE 2.13
Oiodecircui1 ~i•h an RL load.
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48
Chapter 2
Power Semiconductor Diodes and Circuits
The rate of change of this current can be obtained from Eq. (2.25) as di = V, e - tRI L
dt
(2.26)
L
and the initial rate of rise of the current (at 1 = 0) is obtained from Eq. (2.26):
dil
d1 ,ao =
v.
(2.27)
I
The voltage vL across the inductor is vL(t}
d' = L_!_ = V,e-tRIL dt
(2.28}
where LIR = ~ is the time constant of an RL load. The waveforms for voltage vl and current are shown in Figure 2.13b. If I >> UR, the voltage across the inductor tends to be zero and its current reaches a steady-state value of I, = V,JR. If an attempt is then made to open switch Si, the energy stored in the inductor ( = 0.5Li2) will be transformed into a high reverse voltage across the switch and diode. This energy dissipates in the form of sparks across the switch; and diode D1 is likely to be damaged in this process. To overcome such a situation, a diode commonly known as a fruwlreeling diode is connected across an inductive load as shown in Figure 2.21a. Note: Because the current i in Figures 2.12a and 2.13a is unidirectional and does not tend to change i ts polarity, the diodes have no effect o n circuit operation.
Key Points of Section 2.10
• TI1e current or an RC or RL circuit that rises or falls exponentially with a circuit time constant does not reverse its polarity. The initial dvldt of a charging capaci· tor in an RC circuit is V/ RC, and the initial dildt in an RL circuit is V/L. Example 2.4
Finding the Peak Current and Energy Loss in an RC Circuit
A diode circuit is shown in Figure 2.14a with R Q 44 n and C = 0.1 µF. The capacitor has an ini· tial voltage. V,o = V,(t = 0) = 220 V. I.I switch S 1 is closed at t = 0, determine (a) the peak diode current. (b) the energy dissipated in the resistor R. and (c) the capacitor voltage at I = 2 µs.
Solution The waveforms are shown in Figure 2.14b. a. Equation (2.20) can be used with V, = V,o and the peak diode current IP is I = V,o = 220 =SA
'
R
44
b. The e nergy IV dissipated is W • O.SCV;.,
a
O.S
x 0.1 x 101 x 2202 = 0.00242 J = 2.42 mJ
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Diodes with LC and RLC Loads
2.11
49
s, +
v•
R
D,
(b) Wavelonns
(a) Circuil diagram
FIGURE 2.14
Diode circuit with an RC load.
c. For RC =. 44 X 0.1 µ = 4.4 µsand t = r1 = 2 µs, the capaci1or vohage is
v, (t = 2 µs) = V,ot- llRC = 220 X ,-vu = 139.64 V
Nore: Because the current is unidirectional, the diode does not affect circuit operation. 2.11
DIODES WITH LC AND RLC LOADS
A diode circuit with an LC load is shown in Figure 2.15a. The source voltage V. is a de constant voltage. When switch S1 is closed at t = 0, the charging current i of the capac· itor is expressed as
11'
V, = L d' d: + C
i de
+ vc(c ~ 0)
(2.29)
10
+
s, D, 0;, ~*---i;+
t•O
L
Vt.
+
.F--~ VJ \ O
l2
11
v,
I I l1 '
2V, -----------:....•,- -
' \ILC :•1='r.'
v,
I
(a) Circuit diagrams
(b) Wa,·clonns
FIGURE 2.15
Diode cin:uit with an LC lo•d.
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50
Chapter '2
Power· Semiconducto.r Diodes and Circuits
With initial conditions i(t = 0) = 0 and v,(1 = 0) capacitor current i as (in Appendix 0, Eq. D.3)
where
wo =
= 0, Eq. (2.29) can be solved forthe
i(1) = V,{isinWot
(2.30)
= Ip sin Wo1
(2.31)
1/'I/LC and the peak current
1; is
v,,./f
Ip=
(2.32)
The rate of rise of the current is obtained from Eq. (2.30) as
v,
di dt
- = - cosw0t l
(2.33)
and Eq. (2.33) gives the initial rate of rise of the current (at t = 0) as
di ' d1 ,.0 =
v,
"'i
(2.34)
The voltage v, across the capacitor can be derived ~ l
['
v,(1) = C Joi dt = V,( l - cos ~ 01)
(2.35)
At a time 1 = 11 = "'v'Lc, the diode current i falls to zero and the capacitor is charged to 2V,. The waveforms for the voltage vL and current i are shown in Figure 2.15b. Note: Because there is no resistance in the circuit, there can be no energy loss. Thus, in the absence of any resistance, the current of an LC circuit oscillates and the energy is transferred from C to l and vice versa. Example 2.5
Find\ng the Voltage and Current in an LC Circuit
A diode circuit with an LC load is shown in Figure 2.16a with the capacitor having an initial volt· age: V.(1 = 0) = = V0 - 220 V, capacitance, C = 20 µF; and inductance, L = 80 µH. If
-v..
switch.S1 is.closed alt • 0, determine (a) the peak current through the diode.(b) the conduction time o! the diode, apd (c) the final steady-state capacitor voltage.
Solution a. Using Kircl,hoffs voltage low (KVL). we can write the equation for the current i as
'J'·
L -di + dt C
1
tit + v,(t = 0) = 0 .
10
Copyrghtcd matcria
Diodes with LC and RLC Loads
2.11
't7I\
. s, +
1=0
L
O
"L
I
1112
I I
Ve
o,
v,, --- --- ---:;-- --'t, = ,,v'[c
+
+
I
-
.
0
_Vto v,
C
51
1
- V,0 1+--- -
(a) Circuit diagram
(b) Waveforms
FIGURE 2.16
Diode circuil with an LC load. and the current i with initial conditions of i(r = O) = 0 and v,(r solved as
a
0) = -Veil is
i(r) = v"'.JfsinWQI where "lo=
1/VLC = !06/V20 x 80 I,
b. At 1 = ,, =
V«l
m
=
{20 • vz~ • 220 vso
"'v'ZX, the diode current
diode is I,
=
25.000 rnd/s. The peak current 1, is
'IT\/LC
= 'IT V20
IIOA
becomes zero and the conduction lime ,, of ·
x 80 = 125.66 µs
c. The capacitor voltage can easily be shown to be
v,(r) = C I
Fort =
11
J.'i dr -
V«l = - V,,, cos .,,,,
O
= 125.66µs ,v,(r = r,) = -220oos1r
~
220 V.
A d iode circuit with an RLC load is shown in Figure 2.17. If switch S1 is closed at t = 0, we can use the KVL to wrile 1he equation for the load current i as
L :; + Ri + ~Ji tit +,vc(t
=. 0)
= V,
(2.36)
with initial conditions i (r = 0) and v,(r = 0) = V,o. Differentiating Eq. (2.36) and dividing both sides by L gives the characteristic equation ti2i
-
t/12
R tli
i
+- - +- = 0 L tit LC
(2.37)
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52
Chapter 2
Power Semiconductor Diodes and Circuits
s,
R
+
+ L
+
FIGURE 2.17
Diode circuit with an RLC load.
v._ TL--v<>·-----------'cJ_v'°".'. 0
-
+
+
Under final steady-state conditions, the capacitor is charged to the source voltage V, and the steady-state current is zero. The forced component of the current in Eq. (2.37) is also zero. The current is due to the natural component. The characteristic equation in Laplace's domain of s is
s2+Rs+-l-=O L LC
(2.38)
and the roots of quadratic equation (2.38) are given by sii
~(2~Y - Zc
= -2~ ±
(2.39)
Let us define two important properties of a second-order circuit: the damping factor,
R
a =-
(2.40)
2L
and the resona/11 frequency , 1
WO=
(2.41)
VLC
Substituting these into Eq. (2.39) yields
s1i = - a ±
Va 2 -
(2.42)
(Oiji
The solution for the current, which depends on the values of a and one of the three possible cases.
WO,
would follow
Case l. If a = wo- the roots are equal, s1 = s 2, and the circuit is called critically damped. The solution takes the form
i(t} = (A 1 + A 2t )e'•'
(2.43)
Case 2. If a > wo. the roots are real and the circuit is said to be over-damped. The solution takes the form (2.44) Case 3. If a < wo. the roots arc complex and the circuit is said to be 1mderdamped. The roots arc s 1.,2
=
-o ± jw,
(2.45)
Copyrighlcd maleria
2.11
Diodes with LC and RLC Loads
53
where w, is called the ri11gi11g freq11e11cy (or damped resonant frequency) and
w, : ~ · The solution takes the form
i(t) = e""'( A 1 cos ••,I
+ A2sin w,t)
(2.46)
which is a damped or decaying si1111soidnl.
Note:The constants A 1 and A 2 can be determined from the initial conditions of
the circuit. The ratio of cx.lwc, is commonly kno,vn as the da,nping rat;o, 6 = R/2 \lcii.
Power e lectronic circuits are generally underdamped such that the circuit current becomes near sinusoidal, to cause a nearly sinusoidal ac output or to turn off a power semiconductor device. Example 2.6
Finding the Current in an RLC Circuit
The second-order RLC circuil or Figure 2. 17 has the de source \'Oltage V, = 220 V, inductance L = 2 mH. capacitance C = 0.05 µF. and resistance R = 160 n. The initial value or the capacitor "oltage is v,(1 = 0) = Vc0 = Oand conductor current /(1 = 0) = O. If switch S1 is closed at 1 = 0. determine (a) an cxprc.. ion for 1hc currcn1 i(1). and (b) 1hc conduction time or diode. (c) Draw a ske1ch or i(r). (d) Use PSpice to plot the instantaneous current i for R = 50 n. 160 n, and 320
n.
Solution ~ 160 x lo'/(2 x 2) = 40.000 rad/s, and from Eq. (2.41), w0 = 1/vLC = 10-' rad/s. 1l1e ringing frec1uency becomes
a. From Eq. (2.40), " = RJ2L w, =
\/1010
-
16 X 10" • 91.652 rad/s
Bcc11usc o < wi:r ii is an undcrdnmpcd circuit and the solution is of the Corn, i(1) =
l'- 0 '( A 1cosw,t
+ A?sinw,t}
Att = 0. i(r = 0) = 0 and 1his gives A1 = 0. The solution becomes
The dcrivati"c of i(r) becomes
\Vhcn the s,vitch is closed at t = 0. the capacitor offers a lo,v impedance and the inductor orrcrs a high impedance. The initinl ra1e or rise of the current is 1in,i1ed only by
the inductor l. Thus at 1 • 0. the circuit dVdt is ~IL. Therefore,
didi l
t• O
=
w A, ; V, r
-
L
which gives the constant as
v,
220 x 1,000
A , = w,I., = 91.652 X 2 ; 1. 2 /\
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54
Chapter 2
Power Semiconductor Diodes and Circuits
i.amp
1.2 0.8 0.4 O i<-----+--~;,,..----+---1--~
..,,
-0.4
- 0.8 FIGURE 2.18
- 1.2
Current wavcrorm for Ex:amplc 2.6.
The final expression for the current i(r) is 1(1) = '1.2 sin(91,652t),""°·
' .
b. The conduc1ion ii.me , 1 or the diode is obtained when i = 0. That is. w, 11
=1r
or
11
= 91."'652 = 34.27 µs
c. The sketch for the current waveform is shown in Figure 2.18.
d. The circuit for PSpicc simulation [4) is shown in Figure 2.19. The list of the circuit file
is as follows:
·
Exairt>l e 2 . 6
RLC Circuit with Di ode
.PARA."!
VN,;J = 160
. STEP VS
PARAM l O
R
2
3
{VALU}
L
3
c
4 1
4 0 2
2MH O. OSUF
Dl
.MODEL DMOD . TRAN O. lUS .PROBE
VALU LISI' 50 160 320 !'tit. ( 0 O INS 220V
tM)O
D(IS•2.22E- 15 BV=l800V) 60US
;Define parameter VALU ; Vary para.-neter VALU 1MS 220V) ; Piecewise linear
; Varillbl e resi stance
Diode with model 1:1'.0D Diode rrodel parameters Transient anal ysis
Graphics postproce.ssor
.END The PSpice plot of the current /(R) through resistance R is shown in Figure 2.20. The current response depends on the resistance R. With a higher value of R. the current becomes more damped; and with a lower value. it tends more toward sinusoidal. For R = o. the · peak current becomes V,(C/L) s 220 X (0.05 µ12m) m 1.56 A.
Copyrghlcd m~leria
2.11
o,
R
2
son u,oo
+
L
3
55
4
2mH
3200
••
Diodes with LC and RLC Loads
c
0.05µ1'
0 (a) Circuit
••
Ins !ms (b) Input voiUtgc
l,ms
FIGURE 2.19 R LC circuit ror PSpiOC simulation.
An RLC Circuit with a Diode
l.OA
Tcmpcrt1turc: 27.0
-1------- --1-----+-----+----1------+ --------------------- '------------------ _,_ ·-
0.8 A
- r -·
1-----·
0.6A
,
L
0.2 A
O.OA
.J-----J-__..i_--i--- - -+.:J~ --1....:_::::;=-,.---~ 10,.. 401'5 so,.. 301'5 601'5
0 ,..,
@J •
0
l(L)
n me
Ci = 14.385,..
C2 =
913.522 m
0.000.
0.000
dif • 14.385 I',
913.522m
FIGURE 2.20 Plots (or Example 2.6.
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56
Chapter 2
Power Semiconductor Diodes and Circuits
Key Points of Section 2.11
• The current of an LC circuit goes through resonant oscillation with a peak value of V. (CIL). The diode Di stops the reverse current flow and the capacitor is charged to 211,. • The current of an RLC depends on the damping ratio 8 = (R/2) (C/L). Power electronics circuits are generally underdamped such that the circuit current becomes near sinusoidal. 2.12
FREEWHEELING DIODES
If switch Si in Figure 221a is closed for time 11, a currcnf is established through the load: and then if the switch is opened, a path must be provided for the current in the inductive load. Otherwise, the inductive energy induces a very high voltage and this energy is dissipated as heat across the switch as sparks. This is normally done by connecting a diode Dm as shown in Figure 2.21a, and this diode is usuaUy called a fruwhuli11g diode. The circuit operation can be divided into two modes. Mode I begins when the switch is closed at t • 0, and mode 2 begins when the switch is then opened. The equivalent circuits for the modes are shown in Figure 2.21b. Variables i 1
s, o, +
I
+
~o
i,
R
v,
+
v,
L
v, +
Dm ;,
v,
L
R
Modc 2 Mode I (b) Equivalent circuits
~
T:------~.-------:~:sj . :
o~
I
i •.
;, R
(a) Circuit d iagram
1~;
L
I
I
o
(c) W11,·cforms FIGURE Z.21 CU'cuil " 'ith II freewheeling diode.
Copyrighlcd m~lcria
2.1 2
Freewheeling Diodes
57
and i2 a.re defined as the instantaneous currents ror mode I and mode 2, respectively, 1, and 12 arc the rorresponding durations of these modes.
Mode 1. During this mode, the diode current ii, which is similar to Eq. (2.25), is i 1(1) =
v
_..!
R
(1 - e-,Rlt.)
(2.47)
When the switch is opened at r = r1 (at the end of this mode), the current at that time becomes /1
= i 1(r = ti) = V, (1 - e•,Rlt.) R
(2.48)
lf the time 11 is sufficiently long, the current practically reaches a steady-state current of I, = VJR nows through the load. Mode 2. This mode begins when the switch is opened and the load current starts to now through the freewheeling diode D.,. Redefining the time origin at the beginning of this mode, the current through the freewheeling diode is found from di2
0 = L dr
.
+ R11
(2.49)
with initial condition i 2(t = 0) = / 1• The solution of Eq. (2.49) gives the freewheeling current i1 = i 2 as
(2.50) and at t = r2 this current decays exponentially 10 practically zero provided that >> UR. The waveforms for the currents are shown in Figure 2.21c. Note: Figure 2.2l c shows that at 11 and r2 , the currents have reached the steadystale conditions. These arc the extreme cases. A circuit normally operates under conditions such that the current remains continuous.
12
Example 2.7
Finding the Stored Energy in an Inductor w ith a Freewheeling Diode
In Figure 2.2la, the resistance is negligible (R = 0), the source voltage is V, = 220 V (constant lime),and the load inductance is L = 220 µH. (a) Draw the waveform for the load current if the switch is closed for a time ,, = 100 µsand is then opened. (b) Determine the final energy stored in the load inductor.
Solution •· The circuil diagran, is sho\\'n in Figure 2.22a with a zero initial current. \Vhcn the switch is closed ot , = 0. the load current rises linearly and is expressed as
v.
i(r) • - , l and at t
•
11• / 0 •
\~11/L
1:11
220 x 100/220
=
100 A.
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58
Chapter 2
Power Semiconductor Diodes and Circuits
s, +
v,
+
t
o,
=0 ;.,
v,
Dm
L 1,
(a) Circuit diagram
~~
o~:3
:
1
I
l
1,
(b)Wavelonm
FIGURE 2.22
Diode circuit with an L load. b. When switch S, is opened at a time t - 11, the load current starts to Oow through diode D,,.. Because there is no dissipative (resistive) element in the circuit, the load curre,n t remains constant at /0 = 100 A and the energy stored in the inductor is = 1.1 J. The current waveforms ore shown in Figure 2.22b.
o.sut
Key Points or Section 2.12
• If the load is inductive, an antiparaUel diode known as the freewheeling diode must be connected across the load to provide a path for the inductive current to .. flow. Otherwise, energy.may be trapped into an inductive load. 2.13
RECOVERY OF TRAPPED ENERGY WITH A DIODE
of Figure 2.22a, the ·energy stored in the inductor is trapped there because no resistance exists in the circuit. In a practical circuit it is desir-· able to improve the efficiency by {etuming the stored energy into the supply source. 1bis can be achieved by adding to the inductor a second winding and connecting a diode D 1 as shown in Figure 2.23a. The inductor and the secondary winding behave as a transformer. The transformer secondary is connected such that if tit is positive, Vi is negative with respect to vi, and vice versa. The secondary winding th~t facilitates re. turning the stored energy to the source via diode D1 is known as a feedback winding. Assuming a transformer with a magnetizing inductance of Lm, the equivalent cin.,uit is as shown in Figure 2.23b. If the diode and secondary voltage (source voltage) are referred to the primary side o f the transformer, the equivalent circuit is as shown in Figure 2.23c. Parameters i 1 and i2 define the primary and secondary currents of the transformer, respectively. Tri the ideal lossless circuit [7)
Copyrghlcd matcria
2.13
Recovery of Trapped Energy with a Diode
59
+ llo -
,___,___...,,- o~s_,_ _ ___ o.;.,-, N, : N,,,__ _, iz
+ v,
+
t
=0
i,
•,
v,
II •"'+ N 1 :N2
(a) Omiit diagn,m
s, +
~
I s
O
ai2 i,
i,
•,
L.,
N, N,
•= - -
-
II "'•
o,
•u +
+
v,
+
Ideal lran.sfonncr (b) Equivalent circuit
v,
+
ai2
+
i,
v,
Lm inmodc2
+
o,
·-N,
N,
+
+
•of• v,,a
(c) Equh,alent circuit, referTed to primary side FIGURE 2.23
Circuit with an energy recovery diode. (ReC. 7. S. Dewan]
The 111ms ratio of an ideal transfonner is defined as
a ;Ni Ni
(2.51)
The circuit operation can be divided into two modes. Mode 1 begins when switch S1 is closed at 1 = 0 and mode 2 begins when the switch is opened. The equivalent circuits for the modes arc shown in Figure 2.24a, with 11 and 12 the durations of mode 1 and mode 2, respectively.
Copyrighted m~lcria
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Chapter 2
Power Semiconductor Diodes and Circuits ai2 = O
v,
+
~
+ +
v,
v,
o,
!.,,,
+
i,
+ v0 /a V/a
ai2
o,
.....
- v, +•
i1
Mode I
v,,=O
+
Mode2
(a) Equivalent circuit
f
~-~ ·:;.?]---•,---', v
i,
v,
-•, L,,,
: ,,
l
I
I
I
I
I
I I I
,,
I I
I
0
v,
L,,, ,,
~ --------
0
v, v, 0 - Vila
av,
-------v,
-v, I
V,(I
I
+ a) ,__ _ __
I
+
-
av, v, ---------
----------v,
o~----'--------'--~ (b) Wavclorms
FIGURE 2.24
Equivalent circuits and waveforms.
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2.13
Recovery of Trapped Energy with a Diode
61
Mode L During this mode switch S 1 is closed at I = 0. Diode 0 1 is reverse biased and the current through the diode (secondary current) is ai2 = 0 or i2 = 0. Using the KVL in Figure 2.24a for mode I, V, a ( v0 - V, )la. and this gives the reverse diode voltage as
Vo = V,(l
+ a)
(2.52)
Assuming that there is no initial current in the circuit, the primary current is the same as the switch current i, and is expressed as (2.53) which gives
v.
i1(1) = i, (1) = - ' I Lm
for
O S I S 11
(2.54)
This mode is valid for O :S t :S 11 and ends when the switch is opened at r = 11. At the end of this mode the primary current becomes lo =
v,
Lm ,,
(2.55)
Mode 2. During this mode the switch is opened, the voltage across the inductor is reversed, and the diode D1 is forward biased. A current Dows through the transformer secondary and the energy stored in the inductor is returned to the source. Using the KVL and redefining the time origin at the beginning of this mode, the primary current is expressed as dit
v,
L -+- = 0 m dt a with initial condition i 1(1
(2.56)
= 0) = 10 , and we can solve the current as
v,
i 1(1) = - - - 1 + l o for O s t s r2
aLm
(2.57)
The conduction time of diode D1 is found from the condition i 1(1 = 12) = 0 of Eq. (2.57) and is (2.58) Mode 2 is valid for O s t s 12• At the end of this mode at t = 12, all the energy stored in the inductor Lm is returned to the source. TI1e various waveforms for the currents and voltage arc shown in Figure 2.24b for a = 10/6.
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62
Chapter 2
Example 2.8
Power Semiconductor Diodes and Circuits
Finding the Recovery Energy in an Inductor with a Feedback Diode
For the energy recovery circuit of figure 2.23a, the magnetizing inductance of the transformer is L., = 250 µH. N1 - 10, and N, • 100. The leakage inducta.nees and resistances of the tran•formcr are negligible. The source voltage is V, = 220 V and there is no initial current in the cir· cuit. If switch S1 is closed for a time 11 • SO JJS and is then opened, (a) determine the reverse voltage of diode Di, (b) calculate the peak value of primary current, (c) calculate the peak value of secondary current,(d) determine the conduction time of diode D,, and (c) determine the energy suwlicd by the source.
Solution The turns ratio is a - N,tN1 = 100/10 = 10. L
From Eq. (2.52) the reverse voltage of the diode,
Vo• V,(1 +a) • 220
X
(I+ 10) • 2420V
b. From Eq. (2.55) the peak value
so
v,
lo - - 11 L,.
l;I
220 x • 44 A 250
c. The peak value of the secondary current / 0 = /,la = 44/10 = 4.4 A. cL From Eq. (2.58) the conduction time of the diode
aL.,lo
,, = - -
v,
• = 250 x 44 x
10 = 500 JJS 220 · ·
e~ The.source energy, W =
lo'•
vi dt =
lo'' V o sl.,,.
1
V' V _s t dt = -I -
2L"'
tf
Using / 0 from Eq. (255) yields W = O.SL.,./5 = 0.5
x 250 x 10-6 x 442 = 0.242 J = 24i mJ
Key Points of Section 2.13
• The trapped energy of an inductive load can be fed back ·to the input supply through a diode known as the feedback diode.
SUMMARY The characteristics of practical diodes differ from those of idea~ diodes. The reverse recovery time plays a significant role, especially at high-speed switching applications. Diodes can be classified into three types: (1) general-purpose diodes, (2) fast-recovery diodes, and (3) Schottky diodes. Although a Schottky diode behaves as a pn.-junction
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Review Questions
63
diode, there is no physical junction; and as a result a Schottky diode is a majority carrier device. On-the other hand, a pn-junction diode is both a majority and a minority carrier diode. If diodes are connected in series to increase the blocking voltage capability, voltage-sharing networks under steady-state ·and transient conditions are required. When diodes are connected in parallel to increase the current-carrying ability, current-sharing elements are also necessary. In this chapter we have seen the applications of power diodes in voltage reversal of a capacitor, charging a capacitor more than the de input voltage, freewheeling action, and energy recovery from an inductive load. REFERENCES (1) M. H. Rashid, Microeltctronic Circuits: A.nalysis and Design. Boston: PWS Publishing. I 999, Chapter 2. (2) P. R. Gray and R. G. Meyer, Analysis and Design ofAnalog lnttgrattd Circuits. New York: John Wiley & Sons. I 993, Chapter I. (3) Infineon Technologies: Power Stmicanductors. Gennany: Siemens, 2001. www.infincon.com/ (4) M . H. Rashid, SPICE for Circ11i1s and Eltctronics Using PSpict. Englewood Cliffs, NJ: Prentice·Hall lnc.1995.
(SJ M. H. Rashid, SPICE for Po wer Electronics and Electric Po wtr. Englewood Cliffs, NJ: Prentice-Hall 1993. (6) P. W. Tuincnga, SPICE: A guide 10 Circuit Simulation and Analysis Using PSpice. Engle· wood Cliffs, NJ: Prcnlicc-Hall. 1995.
[7] S. B. Dewan and A . Straughen, Power Stmiconductor Circuits. New York: John Wiley & Sons. 1975, Chapter 2.
REVIEW QUESTIONS 2.1 What 'lire the types of power diodes? 2.2 What is a leakage current of diodes? 2.3 What is a reverse recovery time of diodes? 2.4 What is a reverse recovery current of diodes? 2.S What is a softness factor of diodes? 2.6 What arc the recovery types of diodes? 2.7 What is the cause of reverse recovery time in a pn-junction diode? 2.8 What is the effect of reverse recovery time? 2.9 Why is it necessary to use fast-recovery diodes for high·spccd switching? 2.10 What is a forward recovery time? 2.11 What arc the main differences between pn·junclion diodes and Scbollky diodes? 2.12 What are the limitations of Schollky diodes? 2.13 What is the typical reverse recovery time of general-purpose diodes? 2.14 What is the typical reverse recovery time of fast.recovery diodes? 2.15 What arc the problems of series-connected diodes, and what arc the possible solutions? 2.16 What are the problems of parallel-connected diodes, and what arc ·1be possible solutions? 2.17 If two diodes are connected in series with equal-voltage sbarings, why do the diode leak·
age currents differ?
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64
Olapter 2
Power Semiconductor Diodes and Circuits
2.18 What is the time constant of an RL circuit? 2.lll What is the time constant of an RC circuit? 1.20 What is the resonant frequency of an LC circuit? 2.21 What is the damping factor of an RLC circuit? 2.22 What is the difference between the resonant frequency and the ringing frequency or an RLC circuit? 2.23 What is a freewheeling diode, and what is its purpose? 2.24 What is the trapped energy or an inductor? 2.25 How is the trapped energy recovered by a diode?
PROBLEMS 2.1 The reverse recovery time or a diode is 1,, • S µs. and the rate of fall of the diode current is di/dt = 80 Nµs. If the softness factor is SF = 0.5, detctmine (a) the storage charge QRR• and (b) the peak reverse current /RR· 2.2 The measured values of a diode al a temperature of25 "Care
Vo • l.OV at/o • 50A = 1.5 Vat / 0 = 600 A
Determine (a) the emission coefficient n, and (b) the leakage current I,. 2.3 Two diodes are connected in series and the voltage across each diode is ma.intained the same by connecting a voltage-sharing resistor. such that V01 • Vm • 2000 V and R 1 = 100 kO. lbe v- i characteristics or the diodes arc shown in Figure P2.3. Determine the leakage currents of each diode and the resistance R2 across diode D,.
ISO
100 2200
2000
1600
1200
800
50
400 200
-+~~~1--~--~_:,_::_:_:_::!-~-:::-~-:-:::=:::::=!:!!ia..~t;;o~s~1tc.oi"i2~-;-3~v .,,
/.,.,"' I I
I
I
I
I
" ""
I
I
S mA !OmA 15mA
20mA 2SmA
30mA
I
I
I I
I
FIGURE PZ.3
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Problems
65
2.4 Two diodes are connecled in parallel and the forward voltage drop across each diode is 1.5 V. The v- i charac1cris1ics of diodes arc shown in Figure P2.3. Dclcrminc the forward currents through each diode. 23 'Iwo diodes arc connected in parallel as shown in Figure 2.1 la, with current-sharing resistances. The v- i characteristics arc shown in Figure P2.3. The total current is I r = 200 A. The voltage across a diode and ils resistance is v = 2.5 V. Determine the values of resis· tana:s R 1 and R2 if the currcnl is shared equally by 1hc diodes. 2.6 Two diodes are connected in series as shown in Figure 2.8a. The resistance across the diodes is R1 = R1 = 10 kn. The inpul de vollage is 5 kV. The leakage currcnu arc 1,1 = 25 mA and /,2 = 40 mA. Dclermine lhe voltage across the diodes. 2...7 The current waveforms of a capacitor are shown in Figure P2.7. Determine the average, rool mean square (rms). and peak currenl ralings of 1hc capacilor.
,, - 100 .., t2 = 300µs t1 •
,,
r-
2so Hz
5001,1.s
,,
- 200 -------------- .....~ ~ ~ - - ' FIGURE P2.7
2.8 The waveforms ol 1be current flowing through a diode are shown in Figure P2.8. Deter·
mine 1he average, rms. and peak currenl racings of the diode.
300
i.A 11 = t3 •
100 .. ~ 12 = 200 µ.s, 400 µs, t.. .. 800 µ.s., t 5 • 1 ms r - 200 Hz
-----~-----J------ -----..----, I
I
I
I
,,
,,
FIGURE P2.8
2.9 A diode circuit U. shown in Figure. P2.9 with R - 22 0 nnd C - 10 1,1F. If switch S 1 is
closed a11 = 0, determine the expression for the voltage across the capacilor and the en· ergy }0$1 in the circuit.
.e5-~ C
I
R
D1
----l)I-- - - - '
FIGURE P2.9
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66
Chapter 2
Power Semiconductor Diodes and Circuits
2.10 A diode circuit is shown in Figure P2.10 with R • 10 n, L • S mH. and V, • 220 V. If a load current of 10 A is flowing through freewheeling diode D., and switch S 1 is closed at 1 = 0, dctennine the expression for the current i through the switch.
s,.?'\
o - - - e f ~--1-- - - , l• 0"i +
v, L
!
IOA
FIGURE P2.10
2.U If the inductor of the circuit in Figure 2.15 has an initial current of / 0• determine the ex- · pression for the voltage across the capacitor. 2.U If switch S1 of Figure P2.12 is closed at 1 = 0, determine the expression for (a) the current flowing through the switch i(I), and (b) the rate of rise of the current dildt. (c) D raw sketches of i(1) and dildt. (d) What is the value of initial dildt? For Figure P2.12e, find the initial dildt only.
s,
s, +
v,- v,
s,
R
+ t=O
t• O
+ t• O
+
+
L
R
+
v._T ~·
cJ_v 0
v, v, +
L
0
(b)
(a)
s, +
i
t• O
+
o,
+
s,
L,
t~o
20 µH
+
L
v,_1:·
v, v,
(c)
+
cT_v
IOµF
0
H
=o.s n
c Lz •
IOµH
0
(d)
(c)
FIGURE P2.12
2.13 The second-order circuit of Figure 2.17 has the source voltage V, = 220V, inductance L • 5 mH. capacitance C • 10 µF, and resistance R • 22 n. The initial voltage of the ca· pacitor is V«> 50 V. If the switch is closed at 1 0, detennine (a) an expression for the current. and (b) the conduction time of the diode. (c) Draw a sketch of i(1).
=
=
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Problems
67
2.14 For the energy recovery circuit or Figure 2.23a, the magnetizing inductance of the transformer is L,. = 150 µH, N1 = 10, and N, = 200. The leakage inductances and resis· tances of the transformer are negligible. The source voltage is V, = 200 V and there is no initial current in the circuit. If SMlch S1 is closed for a lime ,, = 100 µs and is then opened,(•) determine the reverse voltage or diode D,. (b) calculate the peak primary current, (<) calculate the peak secondary current, (d) determine the time fo r which diode lJ., conducts. and (e) determine the energy supplied by the source. 2.15 A diode circuit is shown in Figure P2.15 where the load current is flowing through diode D,.. If switch S, is closed at a time t = 0, determine (a) expressions for v,(1), iit), and id(t); (b) lime 11 when the diode 0 1 stops conducting;(<) time'• when the voltage acroSll the capacitor becomes zero; and (d) the time required for capacitor to recharge to the supply voltage V,. D,
L
s, +
v,
t=O
+
ia
u,
-v,+
i,
i,
Dm
11, FIGURE P2. 15
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C HAPTER
3
Diode Rectifiers Tlit learning objtctivu of titis chapter an as follows: • • • • • • 3.1
To understand the operation and characteristics of diode rectifiers To learn the types of diode rectifiers To understand the performance parameters of diode rectifiers To learn the techniques for analyzing and design of diode rectifier circuits To learn the techniques for simulating diode rectifiers by using SPICE To study the effects of load inductance on the load current INTRODUCTION
Diodes are extensively used in rectifiers. A rectifier is a circuit that converts an ac signal into a unidirectional signal. A rectifier is a type of dc-ac convener. Depending on the type of input supply, the rectifiers are classified into two types: (1) single phase and (2) three phase. For the sake of simplicity the diodes arc considered to be ideal. By "ideal" we mean that the reverse recovery time'" and the forward voltage drop V0 arc negligible. Thal is, 1" = 0 and V0 = 0. 3.2
SINGLE-PHASE HALF-WAVE RECTIFIERS
A single-phase half-wave rectifier is the simplest type, but it is not normally used in industrial applications. However, it is useful in understanding the principle of rectifier operation. The circuit diagram with a resistive load is shown in Figure 3.la. During the positive half-cycle of the input voltage, diode Di conducts and the input voltage appears across the load. During the negative half-cycle of the input voltage, the diode is in a blocking condition and the output voltage is zero. The waveforms for the input voltage and output voltage are shown in Figure 3.lb. Key Points or Section 3.2 • The half-wave rectifier is the simplest power electronics circuit that is used for low-cost power supplies for electronics like radios. 68
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. P 3.3
Performance Parameters
69
\ 'n1 .. _ ..
O
i,
1·
t
I
Wl
D,
-1
I I I I
0
•
+ =
wt
vR,,b '-- ii Vo
+ v,
· , :;_;; ;'\_ / , 211 2 : "'---./' :
v .. sinwt
11
1r
,r
, 2ff
2
:
:
:
:
1
1
"o
R.' v.
.'
i
O
- v,. (a) Circuit diagram
I
wt
wl
--------~~ 2 " (b) \Va.\'erorms
FIGURE 3.1
Single-phase halJ-wa\'C rectifier.
3.3
PERFORMANCE PARAMETERS Although the output voltage as shown in Figure 3.lb is de, it is discontinuous and contains harmonics. A rectifier is a powe r processor that should give a de output voltage with a minimum amount of harmonic contents. At the same time, it should maintain the input current as sinusoidal as possible and in phase with the input voltage so that the power factor is near unity. l11e power-processing quality or a rectifier requires the determination of harmonic contents of the input current, the output voltage, and the output current. We can use Fourier series expansions to find the harmonic contents of voltages and currents. There are different types of rectifier circuits and the performances of a rectifier are normally evaluated in terms of the following parameters:
v.,,
The average value or the output (load) voltage, The average value of the output (load) current . /"" The output de power,
P.,, = V.,,,!.,,
(3.1)
The root-mean-square (rms) value of the output voltage, V,m, lbc nns value of the outpul current, lrms
The output ac power (3.2)
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70
Chapter 3
Diode Rectifiers
The efficiency (or rectification ratio) of a rectifier. which is a figure of me tit and permits us to compare the effectiveness, is defined as
Pde 11 = -
(3.3)
P..,
The output voltage can be considered as composed of two components: (1) the de value, and (2) the ac component or ripple. The effective (rms) value of the ac component of output voltage is
Vac =
VY~ - Vac
(3.4)
Toe form factor, which is a measure of the shape of output voltage, is
FF= Vmu vde
(
3.5
)
Toe ripple factor, which is a measure of the ripple content, is defined as
RF= V,.
(3.6)
vdc
Substituting Eq. (3.4) in Eq. (3.6), the ripple factor can be expressed as
RF =
J( ;::r-
2
(3.7)
1 = yFF - 1
The transformer uri/ization factor is defined as
TUF = Pde
(3.8)
v,1,
where V, and I, are the rms voltage and rms current of the transformer secondary, respectively. Let us consider the waveforms of Figure 3.2, where v, is the sinusoidal input voltage, i, is I.he instantaneous input current, and;,. is its fundamental component .
.. t
Fundamcnta.l current FIGURE 3.2
Waveforms for input.voltage and current.
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3.3
Performance Paramete~
71
If q, is the angle between the fundamental components of the input current and voltage, q, is called the displacement angle.The displaccmem factor is defined as
DF
= cos.j,
(3.9)
The harmonic factor (HF) of the input current is defined as
HF
c
(/;~./;,ya= [(2.Y - l ra
(3.10)
where /, 1 is the fundamental component of the input current I,. Both 1, 1 and I, are ex· pressed here in rms. The input power factor (PF) is defined as
V,1, 1
1, 1
PF= - - cosq, = - cosq, V,l , I,
(3.11)
Crest factor (CF), which is a measure of the peak input current /,(peak) as compared with its rrns value I,, is often of interest to specify the peak current ratings of devices and components. CF of the input current is defined by
(3.12)
Notes L HF is a measure of the distortion of a waveform and is also known as rora/ har·
monic distortion (THD). 2. If the input current i, is purely sinusoidal, /, 1 = I, and the power factor PF equals the displacement factor DF. The displacement angle cl> becomes the impedance angle 9 = tan- 1(wUR) for an Rl load. 3. Displacement factor DF is often known as displacemem power factor (DPF). 4. An ideal rectifier should have 11 = !00%. v.. = 0, RF = 0. TUF = 1, HF = THO = 0, and PF = DPF = I. Example 3.1
finding the Performance Parameters of a Half-Wave Rectifier
The rectifier in Figure 3.la has a purely resistive load of R. Determine (a) the eflicieocy,(b) the FF. (e) the RF, (d) the TUF, (e) the PJV or diode D 1, (() the CF of the input current, and (g) input PF. Solution The average oulput vohage VJc is defined as
I!.'
v.,, • T
v,.(r) dt
O
We can notic:c from Figure 3.lb that v,.(r) = 0 for T/2 s r s T. H ence. we have
I/.., , .
""'
T
-v.,( cos -wT -· wT 2
V sinw1d1 = - -
V,. _ = -
O
"'
I
)
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72
Chapter 3
Diode Rectifie11
= 211/. Thus
However, the frequency of the source is f = lfT and w
(3.13) The nos value of a periodic waveform is defined as I V"", = [ 7
lo(1vl.(t) dt ]'12
For a sinusoidal voltage of 1.1i(t) = Vm sin wt forO s
t
s
Tl2, therms value of the output voltage is
.v""' = [ Tl /.T/2 (V., sin wt)2 dt ]112 = O
I
""'
V
; = 0.5V.,
V,.,, 0.5V., =-=--
R
(3.14)
R
From Eq. (3.1), P.., = (0.318V.,)2JR, and from Eq. (3.2), P,. = (0.SV.,)2/R. a. From Eq. (3.3), thc efficiency ll • (0.318Vm)'i (0.5Vm)2 • 40.5%. b. From Eq. (3.5), lhc FF • 0.5V..,I0.3l8V., • 1.57 or 157%.
c. FromEq.(3.7),theRF = Vl.572 - I= 121 or 121%. d.. The rms vol1agc of lhc transformer secondary is
V. = [TI lo(
2
(V., sin wt) dt
]'12 = \11 V = 0.701V.,
(3.15)
The nos value of the transformer secondary current is the same as that of the load:
o.sv..
I, = -;f"' The volt-ampere rating (VA) of the transformer, VA= V, I, = 0.101V,., X 0.5V.,IR. From Eq. (3.8) TUF = P,./(V, I,) = 0.3182/(0.707 x 0.5) = 0.286. e.. The peak revene (or inverse) blocking voltage PIV = V.,. · L l,(p..•J = V.,!R and I,= 0.5V.,IR. The CF of the input current is CF= l,(peuyl , = 1/0.5 = 2. 11, The input PF for a resistive load can be found from P,. 0.52 PF • VA • 0.707 x 0.5 • 0.707
Note: lffUF = 1/0.286 = 3.496 signifies that the transformer must be 3.496 times larger than that when ii is used to deliver power from a pure ac voltage. This rectifier has a high ripple factor, 121%; a low efficiency, 40.5%; and a poor TUF, 0.286. In addition, the transformer has to carry a de current, and this results in a de saturation problem of the transformer core.
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3.3
I+ • -I 0
+
+
J
+ i
o,
v, = V111 sin6:iC
·• 11
I I I I
-
.•
..
01------+-~--~....-..,
.- +
.
I I I I
73
+ • R :•
Vo ~Dm
-
-
..
Performance Parameters
L •
~
D 1conducts
I
-v.. (b) Waveromu
(a) arcuit diagram
v
0 1 conducts
vo
I
Dm conducts
I
(c) Waveforms
FIGURE 3.3 HaJJ.wavc rectifier with RL load.
Let us consider the circuit of Figure 3.la with an RL load as shown in Figure 3.3a. Due to inductive load, the conduction period of diode D1 will extend beyond 180° until the current becomes zero at wr = 11 + u. The waveforms for the current and voltage arc shown in Figure 3.3b. It should be noted that the average vL of the inductor is zero. The average output voltage is
v
V"" = ~
1~•0
211 0
=
Vm 211
v
sin wt d(wt ) =~[-cos wt][• 0
[I - cos(11 + a))
211
(3.16)
The average load current is /d, = V,,,/R.
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74
Chapter 3
Diode Rectifiers
It can be noted from Eq. (3.16) that the average voltage (and current) can be increased by making u s 0, which is possible by adding a freewheeling diode Dm· as shown in Figure 3.3a with dashed lines. The effect of this diode is to prevent a negative voltage appearing across the load; and as a result, the magnetic stored energy is increased. At t = t1 = ,r/w, the current from [)., is transferred to D., and this process is called commutation of diodes and the waveforms arc shown in Figure 3.3c. Depending on the load time constant, the load current may be discontinuous. Load current i 0 is discontinuous with a resistive load and continuous with a very high inductive load. Toe continuity of the load current depends on its time constant T = wUR. If the output is connected to a battery, the rectifier can be used as a battery charger. This is shown in Figure 3.4a. For v, > E, diode D 1 conducts. The angle a when the diode stans conducting can be found from the condition V.,sina = E
R
n: 1
+
+
••
.. .
'
io
+
II
o,
+ "'::" E
v,
(a) Cin:uit
wl
.,,
v.. - E
-
R-
OL-1'----'~-->'-'-~~~~~-'-L-~~~"-~~
"
fJ
,r'
2,r
(b) Wa,•clorms
wt
FIGURE 3.4
Baucry charger.
Copyrighted m~lcria
3.3
Performance Parameters
75
which gives a = sin- 1 Diode Di is turned off when v,
<
Ji. v...
(3.17)
E at
l3= 1T -(l The charging current i L• which is shown in Figure 3.4b, can be found from
. •o
Example 3.2
=
v, - E R
=
Vm sin wt - E R
fora < wr < 13
Finding the Performance Parameters of a Battery Charger
The battery voltage in Figure 3.4a is £ = 12 V and its capacity is 100 Wh. The average charging currentshould be Id, = S A. The primary input voltage is v, • 120 V, 60 Hz. and the transfonner has a tum ratio of n = 2:1. Calculate (a) the conduction angle 6 of the diode, (b) the current-limiting resistance R , (c ) the power rating P• of R, (d) the charging time h0 in hours, (e) the rectifier efficiency 11, and (I) the PJV of the diode.
Solution E = 12 V, v, = 120 V, V,
= V,ln = 120/2 = 60 V, and VM = vi V, = vi x 60 = 84.85 V. a. FromEq.(3.17),a = sin- 1( 12/84.85) = 8.13°or0.1419radfl = 180 - 8.13 = 171.87°. The conduction angle is 6 = fl - a = 171.87 - 8.13 = 163.74°.
b. The average charging current /"' is
(3.18)
which gives I
R • - I (2V.., cos a + 2£a - 1r£) 21' d< =
2
""
1
x
5
(2
x 84.85 x cos 8.13° + 2 x
12
x 0.1419 -
.,,
x
12) = 4.26 n
c. The rrns battery current / ,,,. is ,
/.~(Vm
I 2"' •
lrms = -
sin w1 R
2
-
£)'
( )
d wt
• -•-[(v~ + £')(,r - 2a) + v~sin2a - 4V Ecosa ] 21rR2 = 67.4
or /,... •
v'67.4 •
2
2
m
(3.19) 8.2 A. The power rating of R is PR = 8.2 2 x 4.26 = 286.4 W.
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76
Chapter 3
Diode Rectifiers
cl. The power delivered Pde to the battery is
P.,, = El"'
= 12 x 5 = 60 W
100 100 h0 P.,, • 100 or h0 = P.,, = = 1.667 h
60
e. The rectifier efficiency Tl i.s Tl •
power delivered to the battery P.., 60 17 32 total input power • Pde + PR = 60 + 286.4 = · %
I. The peak inverse voltage PIV of the diode is
PIV = Vm + E = 84.85 + 12 = 96.85 V
Example 3.3 Finding the Fourier Series of the Output Voltage for a Half-Wave Rectifier The single-phase half-wave rectifier o! Figure 3.la is connected to a source of V, = 120 V, 60 Hz. Express the instantaneous output voltage 1.b(t) in Fourier series.
Solution The rectifier output voltage 1.1) may be described by a Fourier series as ~
1.b(I) •
v.., +
~ (a. sin nwt + b. cos nwt) 11• 1.2.. ••.
V. V"' = - I J.' •1.l) d(wt) = - I /.' V.. sin wt d(wt) = ...!!
2 ff
2 ff
O
a. • .!.f. ' '...., sinn ..1d(..1)
=
O
1r o
vm
ia -
1T
.!.f.'vmsin ..1sinn..1d(wt) 1T o
forn ;;;;: J
2
=0
for n = 2, 3, 4, 5, 6, .. .
b. = .!.J.'•"l) cosnwtd( ..1) = .!.J.·v.,,in wtcosnwtd(wt) "' o "' o Vm 1 + ( - 1)" =-:;;- _ Corn= 2, 4,6, ... 1
n'
=O
for n
~
1, 3, 5, ...
Substituting a. and b• . the instantaneous output voltage becomes V. , "'
1.1) (1) a -
Vm , 2
2V,. 2Vm 2Vm cos2wt - cos4wt - c016'"1 3,r 15"' 35..,
+ - smwt - -
(3.20)
where V., = ~ x 120 = 169.7 V and "' = 2,r X 60 • 3n rad/s.
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Single-Phase Full-Wave Rectifiers
77
Key Points or Section 3.3 • The performance of a half-wave rectifier that is measured by certain parameters is poor. Toe load current can be made continuous by adding an inductor and a freewheeling diode. The output voltage is discontinuous and contains harmonics at multiples of t!ie supply frequency. 3.4
SINGLE-PHASE FULL·WAVE RECTIFIERS
A full-wave rectifier circuit with a center-tapped transformer is shown in Figure 3.Sa. Each half of the transformer with its associated diode acts as a half-wave rectifier and the output of a full-wave rectifier is shown in Figure 3.5b. Because there is no de current flowing through the transformer, there is no de saturation problem of transformer core. Toe average output voltage is 2 {Tl2
Vck = T
Jo
2V
(321)
V,.sinwtdt = .,"' = 0.6366V,.
Instead of using a center-tapped transformer, we could use four diodes, as shown in Figure 3.6a. During the positive half-cycle of the input voltage, the power is supplied
'" I I I
-v.,, - . -- - --- ~-- -
r~/'\-+-/'\ '
Vm~
I+ •o, -I 0,
+ ••
••
0 0 "o
"i•
\
.. ....,,
2
2..
i,,
R
+ 1•• o,
••
+
I+ •o, -I (a) Cin:uit diagram
(b) Wavefomu
FIGURE 3.5
FulJ.wa,·c rcc:tifier • i th center-tapped transformer.
Copyrghtcd matcria
78
Chapter3
Diode Rectifien v,
". I I I
- v.. -------- ~---
'
i,,
+
••
+
II
o,
-+
o, R
v,
o,
••
o,
'·v:57\ 0
.
w·
...
~21r"'l
-v ---"'03, ltl)I ID
Vo1, Voz
(b) Wavc(orms
(a) Cucuit diag,am FIGURE 3.6
Full,wave bridge reclll~r.
to the load through diodes Di and D,_. During the negative cycle, diodes ~ and D4 con· duct. The waveform for the output voltage is shown in Figure 3.6b and is similar to that of Figure 3.Sb. Toe peak-inverse voltage of a diode is only Vm. This circuit is known as a bridge rectifier, and it is commonly used in industrial applications (1, 2).
Eumple 3,4 Finding the Perform1nce P1rameters of a Full-Wave Rectlfler with Center-~pped Transformer If the rectifier in Agure 3.Sa has a purely resistive load o! R,determine (a) the efficiency, (b) the FF, (c) the RF,(d) the 11JF, (e) the PIV of diode D1, and (f) the CF of the input current.
Solution From Eq. (3.21), the average output voltage is 2v..
v.,. - - ... -
0.6366V.,
and the average load current is Ide •
v.,. 0.6366V,. -R- • - -R- ~
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3.4
Single-Phase Full-Wave Rectifiers
79
The rms value of the output voltage ii
V.m, = [T2 Jo{rn (V., sin w1) 2 dt ]'
V,..,
l.,.,=R=
12
V = Ji = 0.707V,.
0.707V,. R
From Eq. (3.1) P« = (0.6366V,.)2/R, and from Eq. (3.2) P" = (0.707V,.) 2/R. L From Eq. (3.3), the efficiency -ri = (0.6366V,.)21(0.707V,.)2 = 81%. b. From Eq. (3.S), the form factor FF = 0.707V.,I0.6366V,. = I.II.
= v'
=
c. From Eq. (3.7), the ripple factor RF 1.112 - I 0.482 or 48.2%. cl. The rms voltage of the transformer secondary V, = V,,l\/2 = 0.707V,.. The rms value of transformer secondary current l, = O.SV.,IR. The volt-ampere rating (VA) of the transformer, VA = 2V,l, = 2 x 0.707V,. x 0.511,.,IR. From Eq. (3.8),
TUF •
0.63662 x 0.707 x O.S • 0.5732 • 57.32% 2
e. The peal: reverse blocking voltage, PIV = 2V,.. t 1,1,,..,) = V.JR and l , • 0.107V,./R. The CF of the input current is CF= l,tpc•tll, = 110.707 = \/2, s, The input PF for a resistive load can be found from P,.
PF =
0.7072
VA = 2 x 0.707 x o.s = 0·707
Note: 1/TUF = 1/0.5732 = 1.75 signifies that the input transformer, if present. must be 1.75 times larger than that when it is used to deliver power from a pure ac sinusoidal voltage. The rectifier has an RF of 48.2% and a rectification elficiency of 81%. Note:The perfonnance of a full-wave rectifier is significantly improved compared wilh lhat of a half-wave rectifier. Example 3.5 Finding the Fourier Series of the Output Voltage for a Full-Wave Rectifier The rectifier io Figure 3.5a has an RL load. Use the method of Fourier series to obtain cxprcs·
sions for output voltage t\l(r).
Solution The rectifier output.voltage may be described by a Fourier series (which is reviewed in Appeo dix E) as t\)(t) = V"' + • •
fi."' . ?• cos
nwt + b. sin nwt)
Copyrghlcd matcria
80
Chapter 3
Diode Rectifiers
where VJ,
£?•t\,(t) d(wt ) = -2 J.'o v., sin J(wt) J.'r. cos nwt d( 21.•V,"sin cos nwt d(wt )
I a:;-
2
-1T. o
a" = -I
Vo
'iT O
4V...
'ff
wt) • -
wt
1i' 0
~"'
= - .,,
-1
(11- ! )( n + I)
. . ........
for"= 2.4.6, .. .
=0
for n = I. 3, 5.... 1
b. =
.!.f. 'TT
2V = ___.!:!
wl
,;r
O
""" d ( w1) •
' t\l sin
!f.' lf
O
Vm sin wl sin""" d( "'') - 0
Substiluling 1hc values of a,1 and b,.. the expression for the output voltage is 2V,,, 4V,11 4V,lf 4Vm 't'Q(t) = - cos 2wt - - cos 4wt - - cos 6wt " J,r 15,r 35,,
(3.22)
Note: The output of a full-wave rectifier contains only even harmonics and the second harmonic is the mosl dominant one and its frequency is 2/( = 120 Hz). The output voltage in Eq. (3.22) can be derived by spectrum multiplication of switching function. and this is explained in AppendLx C. Example 3.6
Finding the Input Power Factor of a Full-Wave Rectifier
A singlc·phasc bridge rectifier that supplies a very high inductive load such as a de motor is sho\\'n in Figure 3.7a.The turns ratio of lhe trnnsfom1er is unily. The load is such that the motor
dra,,·s a ripplc·frcc armature current of /,, a.s sho,m in Figure 3.7b. Determine (a) the HF of input current. and (b) the input PF of the rectifier.
Solution Normally. a de motor is highly inducti,•c and acts like a filter in reducing the ripple current of the load.
a. The waveforms for the input current and input voltage of the rectifier are shown in Figure 3.7b. The input current can be expressed in a Fourier series as ~
i,(1) = /"'
+
L
(n,, cos""" + b,, sin 11w1)
lf• l-t . .. .
I.,, • -
J.
a. = -I
!."
1 1 1 ' i,(r) tl(wt) • 1, d(wt ) = 0 211' 0 2'r.' 0
1f
O
J.
10
21.· / cos
i,(t) cos m•lil(wt) = -
'ff
O
0
nwt d(wt) = 0
b" • .!.J.!1ri,(1)sinnwtd(wt) . !.J."1. sin11(Ntt/(wt) • Tro
1T o
41 " fl'l'l'
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Single-Phase Full-Wave Rectifiers
81
"·
.
i,
+
+
"•
II
Fundamental
ls. ~.,...........,,,. compontnt (,,----,. '
o,
o,
o,
o,
-
/ ' Of----~~----+-"'' /, 2ff
io = ·~
" ... _. ~· '
-•.
v,
(a) Circuit diagram
(b) Wa,'Cforms
....
FlGURE 3.7
Full-wa,·e bridge rectifier with de motor load.
Substituting the values of a11 and b,.. the expression for the inpul current is . ( ) 4/• (sin wt sin 3w1 + sin Swt ) I I = - -- + -- +· ·· '
,r
I
5
3
(3.23)
The rms value of the fundamental component of input current is 41.
,,, = irv'2 = 0.90/, Therms value of the input current is
,, = ,,~'·[ 1 +
(1Y (iY (~Y (iY rn= ,. +
+
+
+ ···
From Eq. (3.10), HF= THO = [ ( .~)' - I]'" = 0.4843 or 48.43% 0
b. The displacement angle 4> • 0 and OF • cos 4> • I. Flom Eq. (3.11). the PF (/, 111,) cos 4> = 0.90 (lagging). Key Points or Section 3.4 • There are two types o r single-phase rectifiers: center-tapped transformer and bridge. Their performances are almost identical, except the secondary current o r
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Chapter 3
Diode Rectifiers
the center-tapped transformer carries unidirectional (de) current and it requires larger VA rating. The center-tapped type is used in applications less than 100 W and the bridge rectifier is used in applications ranging from IOOW to 100 kW. The output voltage of the rectifiers contains harmonics whose frequencies are multi· pies of 2/(2 times the supply frequency). 3.5
SINGLE-PHASE FULL-WAVE RECTIFIER WITH RL LOAD
With a resistive load, the load current is identical in shape to the output voltage. ln practice. most loads are inductive to a certain extent and the load current depends on the values of load resistance Rand load inductance L. This is shown in Figure 3.8a. A battery of voltage E is added to develop generahzed equations. If v, = Vm sin wt = v2 V, sin wt is the input voltage. the load current i 0 can be found from
.
L dt + Rio + E = v2 V, SID wt for
i0
.!:
0
which has a solution of the form i0
a
v2V,
.E
Z
R
- - sin(wr - 6) + A,e-
(3.24)
where load impedance Z = [R 2 + (wL)2) 112• load impedance angle 6 = ta·n-1(wUR), and V, is therms value of the input voltage. Case I : continuous load current. This is shown in Figure 3.8b. The constant At in Eq. (:J.24) can be dctcm1incd fron'I the condition: at wt =
1r. ; 0
=
l o-
Substitution of A 1 in Eq. (:l.24) yields
v2V, . 1.0 = - s1D(w1 - U) + Z
( 1 + -E - -v2V, . ) eCRIL)(ff/w _1> - -E - s1D9 0 R
Z
R
Under a steady-state condition. i 0( wt = 0) = i0 (wt = 1r). That is. i0 (wt plying this condition. we get the value of 10 as lo =
v2V, .
""'z"" SID 9
1+
e -(R/L)(•lw)
E
I -
e - (R/L)(•lw)
R
(3.25)
= 0) = 10. Ap-
for 10 .!: 0
(3.26)
which. after substituting 10 in Eq. (3.25) and simpli~ication, gives
v2V,[ io = z- sin(wr - 6) + I _ for O s
(wt
- 6)
2
. - ( RIL)(,1..1
s 1r and i0
. (R/L)] smee' - -E R 2: 0
(3.27)
Copyrighted m~lcria
3.5
Single-Phase Full-Wave Rectifier with RL load
83
v.
'•
i,
+
o,
R
D3
+
v.
v,
o.
o,
...
~
,.,
L
E
I
0
9
(a) Circuit
2"
"
"+ 9
2,r
.,
(b) Wa•eforms ~
-Jo ----------
I
(c) Supply line current
v
"
0
a
a (d) Di1continuou1 current
FIGURE 3.8 Full-bridge rectifier with RL load.
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Chapter 3
Diode Rectifiers
The rms diode current can be found from Eq. (3.27) as
I,= [ ~ 1•{5 tl(wt)
2
rz
and· the rms output current can then be determined by combining the rms current of each diode as
Inn, = (/; + /;)Ill = -../2/, The average diode current can also be found from Eq. (3.27) as
1,1 =
I(' 211
Jo i0 ti( wt )
Case 2; discontinuous load current. This is shown in Figure 3.Sd. The load current !lows only during the period"' s wt s 13. Let us define .t = EIV,., = El viV, as the load battery (emf) constant, called the voltilge rmio. The diodes start to conduct at wt = a given by
At wt = a. i 0 (wr) = 0 and Eq. (3.24) gives
£ A,= [ R
- ~-../iv, sin (n
- 0) ] e l Ri L)(o lw}
which. after substituting in Eq. (3.24). yields the load current
i0 = At wr
-../iv, sm(wt . -"z
- 0)
[£
-../iv, sin (c, - 8) ] , .( llil.l(• lw-r) + R - -"z
-
E R
(3.28)
E=
O
(3.29)
= 13. the current falls to zero.and i0 ( wr = 13 ) = 0. That is. ...JiV, -../iv, - sin(l3 - 8) + [E - - - sin(c, - O)· ] e\ Rt L)( .. - n)tw Z R Z
Dividing Eq. (3.29) by V2V,IZ. and substituting RIZ sin(l3 - e) +
-
-
R
= cos Oand wLIR = tan O. we get
(-cos(O-<- ) - sin(c, - o))e~ - _ .._. = O cos(O)
(3.30)
f3 can be determined from this transcendental equation by an iterative (trial and error) method of solution. Start with 13 = O. and increase its value by a very small amount until the left-hand side of this equation becomes zero. As an example, Mathcad was used to find the value of 13 for e = 30°, 60°. and x = 0 to I. The results are shown in Table 3.l. Ask increases.13 decreases.At k = 1.0, the dio<.lcs do not conduct and no current n ows.
Copyrghlcd malcria
3.s· TABLE 3.1
Variations orAngle $ \\ilh the Vohage Ralio..t
VOitage Ratio.x ~ for.8
= 30'
~
e
forB
95
Single-Phase Full-Wave Rectifier with l!L Load
60'
0
0.1
0.2
0.3
OA
0.5
0.6
0.7
0.8
0.9
1.0
210 244
203
197 225
190 215
183
175 194
167 183
158
·~7 157
132
90
205
I.IS
90
23"
171
Therms diode current can be found from Eq. (3.28) as
lpi5
I, = [ 2I.,.. •
d( wt)
J'n
The average diode current can also be found fro m Eq. (3.28) as
16
Id = - I . i 0 d(wr) 21T " Boundary condirions:The condition for the discon1inuous current can be fou nd by SCI· ting / 0 in Eq. (3.26) to zero. V, v2 . [ I + e·(~H;l ] £ 0= • , - z- srn( O) 1 - e·t;>l: l R which can be solved for the voltage ralio
x = £ /( V2V,) as
I + e·<,;:·.;l]
x(O): = [ _ _._ sin( O) cos( O) l - e 1,.,••,1
(3.31 )
The plot of the voltage ratio x against the load impedance angle 6 is s hown in Figure 3.9. The load angle 6 cannot exceed n /2. The value of x is 63.67% at O = 1.5567 rad. 43.65% at e = 0.52308 rad (30' ) and 0% al a = O.
Example 3.7 Finding the Performance Parameters of a Full-Wave Rectifier with an RL Load The singlc-ph>se full-wave rec1ilier or Figure 3.8a has L = 6.5 mH. I! = 2.5 n. and£ = 10 V. The inpul voltage is V, = 120 Val 60 Hz. ( a) Dc1erminc (I) lhc slcady·slalc load currcnl /0 at wt G 0. (2) the avcrngc diode current /,1• (3) therms diode current I,. and (4) therms output cur· rent lrm...• (b) Use PSpic.c 10 plol the ins1an1ancous ou1pu1 current in. Assume diode parame ters JS • 2.22E - IS. ev = 1800 v.
Solution II is not known \\•hether 1he load curre nt is continuous or discontinuous. Assume that the load current is continuous and proceed \\'ilh 1hc solution. Ir the assumption is not corrccl. the load current is zero a nd then n1ovcs to the case fo r a discon1inuous curre nt.
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Chapter 3
Diod1t R1tctlflers . Boundary. Discontinuous/Continuous Region
:.-_-~::-J.,._-__--- -~ll-------,-----t-,------'.. ,·:. :--, I
I
.!!
I ,--·-b
I
!~ •(9) 0.4 ;>
I
!
I
,2
I
I
!
i
i
I
I I
I
•
- - t --·--·
!
I
--~-
---- ---+---1---L--t-·- -l------
0.2
1
I 0.2
1
I
'
I
0.4
0.6
I
i
0.8
1
1.2
1.4
8
Load Impedance Angle. radians FIGURE 3.9
Boundary of continuous and discontinuous regions for siaglc·phase rectifier. L
R = 2.5 n. Ls 6.5 mH, f = 60 Hz. w = 2,r x 60 = 3n rad/s, [R 2 + (wL)'J'n = 3.5 0 , and 9 • tan. 1(wUR) • 44.43°. (1)
v, = 120 v.
Z =
The steady-state load current al OJI = 0, / 0 • 32.8 A. Because / 0 > 0, the load current is continuous and the assumption is correct.
(2) The num~rical integration of 4, in Eq. (3.27) yields the average diode current as
Id= 19.61 A. (3) By numerical integration of i3between the limits wt • 0 and"• we get the rrns diode current as I, = 28.5 A. (4) The rrns output current I,.., = VZI, = V2 X 28.50 = 40.3 A.
Notes L i0 hasa minimum value of25.2A at WI = 25.5° and a maximum value of 51.46A at
wt = 125.25°. i0 becomes 27.41 A at"" = e and 48.2 A at"" = e + "· Therefore, the minimum value of;,, occurs approximately at"" = 9. 2. The switching action of diodes makes the equations for currents nonlinear. A numerical method of solution for the diode currents is more efficient than the classical tech· nique& A Mathcad program is used to solve for Io, Id. and I, by using numerical integration. Students arc encouraged to verify the results of this example and to appreciate the usefulness of numerical solution, especially in solving nonlinear equations of diode circuits.
Copyrighted m~lcria
3.6
o,
+
o.•
R
+
2
o,
o,
2.s n
s
•o
•• 0
87
;.,
3 i,
Multiphase Star Rectifle11
L
v,
6.SmH
6 lOV
FIGURE 3.10
Single-phase bridge rectifier
4
for PSpicc simulation.
b. The single-phase bridge recrifier for PSpice simulation is shown in Figure 3.10. The list of the circuit file is as follows.:
~ l e 3. 7 Single- Phase Bridge Rectifier with RL load VS L R
1 5 3
vx
6
01
2 02 4 03 0 04 4 VY 1 .MODEL .TRAN
0 6
SIN 10 6.SMH
5
2. 5 OC lOV
4 3 0 3
2
IM)D
169. 7V
60HZ)
Voltage source to measure the -Output current ; Diode model
r:MlO IM)O r:MlO 2
ooc
IM)O
DIIS=2 . 22E-15
lUS
32HS
.PROBE .ENO
BV=lBOOV)
16 . 667HS
Diode model parameters Transient analysis Graphics postprocessor
The PSpice plot of instantaneous output current i0 is shown in Figure 3.11. which gives / 0 = 31.83 A, compared with the expected value of 32.8 A. A Obreak diode was used in PSpice simulation to specify the diode parameters. Key Points of Section 3.5
• An inductive load can make the load current continuous. There is a critical value of the load impedance angle 9 for a given value of the load emfconstantx to keep the load current continuous. ' 3.6
MULTIPHASE STAR RECTIFIERS
We have seen in Eq. (3.21) the average output voltage that could be obtained from singlephase full-wave rectifiers is0.6366V.. and these rectifiers are used in applications up to a power level of 15 kW. For larger power output, three-phase and-multiphase rectifiers are
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88
Chapter 3
Diode Rectifiers
60A -1--,-- -+---+---+--,--+----+---4---+----4-
-r----------------~-~~l~-----------------------~-----~ 4-0A
:
:··-··········· ..... ....·-·····-····1 ~
I
I 1
20A+--'---+---+---+-~--4---+---...:....---+----1r;.• I (VX)
::-· j
ov -IOOV+--'---+---+---+-~-+----+---4.----+----1l6ms 18ms 20 ms 22 ms 24 ms 26 ms 28 ms 30 ms 32 ms • V (3.4) Cl • 22.747 m. 50.179 C2 = 16.667 m. 31.824 lime dif = 6.0800 m. 18.355
FIGURE 3.11
PSpicc plol for Example 3.7.
used. The Fourier series of lhe oulpul voltage given by Eq. (3.22) indicates that the output contains harmonics and the frequency of the fimdame111al compo11ent is two times lhe source frequency (2/). ln praclice, a filter is normally used to reduce the level of harmonics in the load; and the size or the filter decreases with the increase in frequency or the harmonics. In addition to the larger power output of multiphase rectifiers, the fundamen1al frequency of the harmonics is also increased and is q times the source frequency (qf). This rectifier is also known as a s1ar rectifier. The rectifier circuit of Figure 3.Sa can be exlended to multiple phases by having multiphase windings on the transformer secondary as shown in Figure 3.12a. This circuit may be considered as q single-phase half-wave rectifiers and can be considered as a half-wave type. The kth diode conducts during the period when the voltage of ktb phase is higher than lhat of other phases. The waveforms for the voltages and currents are shown in Figure 3.12b. The conduction period of each diode is 2n/q. It can be noticed from Figure 3.12b that the current flowing through the secondary winding is unidirectional and contains a de component. Only one secondary winding carries current at a particular time, and as a result the primary must be connected in delta to eliminate the de component in the input side of the transformer. This minimizes the harmonic content of the primary line current. Assuming a cosine wave from nlq to 2n/q, the average output voltage for a q-phase rectifier is given by (3.32)
Copyrghlcd malcria
3.6
oi= Vmsinwt
Multiphase Star Rectifiers
89
02
o,
3
+ I I
D,
4
'
••
R
I
' -- - - - - - - -- ---- - -- - --!>1--
o. (a) Circuit diagram
o,.on 0
q"
o,
Di on 4,,
2ff
q
o,
o, g,.
6,,-
q
q
Dq IOrr
q
q
2rr
wt
(b) Waveform\
FIGURE 3.12
l\.-tultiphasc rectifiers.
Vrms
2
1o•lq
2
-Tr q O V; cos wt •d(wt) 112 2 = V [.!!....(Tr+ !.sin .,)] m 2Tr q 2 q = [ ;;-1
]112
(3.33}
If the load is purely resislive, the peak currenl through a diode is Im = V,,/R and we can find the nns value of a diode current (or transformer secondary current) as
I, = [ ~ [ •\ ; cos2 wt d(wr )
2
[_!_(,r +.!.sin Z 1r)] 2ir q 2 q
r
12
112
= I m
= Vrms R
(3.34)
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90
Chapter 3
Diode Rectifiers
Example 3.8 Finding the Performance Parameters of a Three-Phase Star Rectifier A lhree·phasc star rectifier has a purely resistive load with R ohms. Determine (a) the efficiency. (b) the FF. (c} the RF. (d) the TUF factor, (c) the PIV of each diode, and (f) the peak current through a diode if the rectifier deliven /
Solution For a lhree·phasc rectifier q = 3 in Eqs. (3.32) to (3.34)
a. From Eq. (3.32).V"' = 0.821Vm and Id, = 0.821V.,JR. From Eq.(3.33), V.m, = 0.84068V., and lrm, = 0.84068V.,JR. From Eq. (3.1), P"' = (0.827V.. )21R; from Eq. (3.2), Pac • (0.84068V.,)21R; and from Eq. (3.3), the efficiency (0.827V.,}2 Tl = (0.84068V,,.)2 = 96.77%
b. From Eq. (3.5), the FF = 0.84068/0.827 = 1.0165 or 101.65%. <.
From Eq. (3.7), the RF =
\!1.0165' -
I = 0.1824 = 18.24%. = V.,Jvi = 0.707V,..FromEq.(3.34), therms current of the transformer scoondary,
d. Tbermsvoltageoflhetransformersecondary,V,
I,
m
0.48541., •
0.4854V.. R
The VA rating of the transformer for q = 3 is VA= 3V,/, = 3 x o.101v.. X
0.4854V., R
From Eq. (3.8),
TUF
a
0.8272 3 x 0.707 x 0.4854 • 0 ·6643
0.840682
PF • 3 X 0.707 x 0.4854 • 0 ·6844
e. The peak invel$C voltage of each diode is equal to the peak V31ue of the secondary linc·to·linc voltage. Tbrce•phailC circuits arc reviewed in Appendix A. The line-to-line voltage is v3 times the phase voltage and lhus PIV = v3 v_ L The average current through each diode is
/4 = -
2 /.~,,
2ir o
l.,coso>ld("'1)
I
.
11'
q
= /.,-sin~
(3.35)
For q • 3, 14 = 0.2757/,.. The average current through each diode is I, = 30/3 = 10 A and Ibis gives the peak current as I ., = 10/0.2757 = 36.27 A.
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3.6
Multipha,e Star Rectifier,
91
Example 3.9 Finding the Fourier Serles of a q·Phase Rectifier L Express the outptit voltage of a q·phase rectifier in Figure 3.12a in Fourier series. b. If q m 6, Vm = 170 V, and the supply frequency is f = 60 Hz, determine the rm., value
of the dominant harmonic and its frequency.
Solllllon a. The waveforms for q-pulses are shown in Figure 3.12b and the frequency of the output is q times the fundamental component (qfl. To find the constants of the Fourier series, we.integrate Crom - -rrlq to -rr/q and theconstanu are
b,. • O Dn
= - I,'fftq
1''•
Vm COS WI COS nwt d(wt)
-wlq
qVm{sin((n - 1)-rr/q) sin((n + 1)-rr/q)} + -~-~~ -rr n - 1 n + l
- =
..
qVm (n + I) sin[(n - 1)-rr/q) + (n - I) sin(('! + 1) -rr/q)
n2
I
-
After simplification and then using trigonometric relationships, we get sin(A + B) = sin A cos B + cos A sin B and sin( A - B) = sin A cos B - cos A sin B we get
a"=
2qVm ( ns1n-cos . mr . ") -1T - cos-n,r s1n,r(n2 - I} q q q q
(3.36}
For a rectifier with q pulses per cycle, the harmonies of the output voltage are: qth, 2qth, 3qth, and 4qtb, and Eq. {3.36) is valid for n = 0, lq, 2q, 3q. The term sin(,..,,/q) = sin" = 0 and Eq. (3.36) becomes On =
-2qvm (
-rr(n2 - I )
• ") cos -n,r sin-
q
q
Toe de component is found by letting n - 0 and is
ao
q .
11'
V,. = - = V. -sm2 ffl 1T q
(3.37)
which is the same as Eq. (3.32). The Fourier series of the output voltage 11o is expressed as
a 11o(1) = ;
I ... oo
+ ••
a. cos nwt
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Chapter 3
Diode Rectifiers
Substituting the value or a., we obtain 11> = Vm ! sin 2:. ( I 1T q
f - 2- 1 cos !!.!q cos nwl)
(3.38)
•·,i..... n2 -
b. fur q • 6, the output voltage is expressed a.,
t1,(t) = 0.9549V., ( I +
:s
cos 6"" -
!
13
cos J:Z.,, + · · ·)
(3.39)
The sixth harmonic is the dominant one. The rms value or a sinusoidal voltage is 11\12 times its peak magnitude, and therms or the sixth harmonic is V61> • 0.9549Vm x 2/(35 x \12) = 6.56 V and its frequency is / 6 = 6/ = 360 Hz. Key Points of Sedlon 3.6
• A multiphase rectifier increases the amount of de component and lowers the amount of the harmonic components. The output voltage of a p-phase rectifier contains harmonics whose frequencies arc multiples of p (p times the supply frequency),p/.
3.7
THREE-PHASE BRIDGE RECTIFIERS A three· phasc bridge rectifier is commonly used in high-power applications and it is shown in Figure 3.13. This is a full-wave rectifier. It can operate with or without a transformer and gives six-pulse ripples on the output voltage. The diodes are numbered in order or conduction sequences and each one conducts for 120°. The conduction sequence for diodes is D1 - Di, Di - Di, Di - D,, Ds - D,. Ds - D6, and Di - D,,. The pair of diodes which are connected between that pair of supply lines having the highest amount of instantaneous line·to·line voltage will conduct The line-to-line volt· age is V3 times the phase voltage of a three-phase Y-connected source. The waveforms and conduction times of diodes are shown in Figure 3.14 [4] . If Vm is the peak value of the phase voltage, then the instantaneous. phase voltages can be described by
v,. = Vm sin(wt)
Vt,,,=
V01 sin(wt - 120°)
Ven=
;,,,
;.
~
c
o, R
n -
Vao
b
c
-+
Secondary
Prinury
a
Vm sin(wt - 240°)
b
i,
+
"·
o,
FIGURE 3.13
Three-phase. bridge rectifier.
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3.7
0
2,r
ir
i, V3Vm - R-
4,r
':
T
3I
T
I I I
I I
Three-Phase Bridge Rectifiers
s..
2,r
T
93
wt
Une curre.nt
------.!..I
01--~~+-~ ~-!-~~--!e-~~+-~~-!-,--~-+~-,/, 2ir ~ 2ff a.it 3 T ,
- V3Vftl I I I ~R~ - - - ---~------~----- - ~-------- -
: : : Diodecurrr:=tj
·~~1
3"
7rr
"'
T
2ir wt
FIGURE 3.14 Wavlforms and conduction times of diodes.
Because the line-line voltage leads the phase voltage by 30°, the instantaneous line-line voltages can be described by
= \/3 Vm sin( wt + 30°) Vco = V3 Vm sin(wt - 210') v••
Vi,,
=
\/3 Vm sin( wr - 90' )
The average output voltage is found from
Vdc
= 211'216 1·16 V3Vmcoswtd(wt) 0
= J\/3Vm = 1T
1.654Vm
(3.40)
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Chapter 3
Diode Rectifiers
where V.,. is the peak phase voltage. The nns output voltage is
r/63V; cos
v.... = [ 2..216 lo
2
"9v'3)
3
+( 241'
a
-
wt d( wt)
]'n
112
...1 6554V.'"
V.
(3.41)
m
If the load is purely resistive, the peak current through a diode is/.., the nns value of the diode current is
= Y3 V,./R and
r/6
I, = [ 24,r lo I; cos2 wt d(wt) ]'n
1 21')]'12 -1 (1T - +-sin[ "' 1r 6 2 6
=I a
0.5518/.,.
(3.42)
and the rms value of the trnnsformer secondary current,
11
= [2~
1~
16 /~
cos2 wt d(wt )
rn
112
[! (.!
= I '"1r6 + .!.2 sin 261r) ] = 0.78041...
(3.43)
where /.,. is the peak secondary line current. For a three-phase rectifier q = 6, Eq. (3.38) gives the instantaneous output voltage as
1-1>(1) = 0.9549V.,. ( l +
ts
cos(6<.>r) -
!
1 3
cos(l2wr) + ·· · )
(3.44)
Example 3.10 Finding the Performance Parameters of a Three-Phase Bridge Rectifier A three-phase bridge rectifier bas a purely resistive load of R. Delcnnine (a) the efficiency, (b) lhe FF,(c) the RF,(d) theTUF,(e) the peak inverse (or reverse) vollage (PJV) of each diode.and (0 the peak current through a diode. The rectifier delivers J.,. - 60 A at an output voltage of V.,. • 280.7 V and the source frequency is 60 Hz.
Solution a. From Eq.(3.40), v.,. • l.654V.,. and I.,.• l.654V.,JR. From Eq.(3.41), V,,.. • l.6554V.,. and /..., = t.6554V.,JR. From Eq. (3.1), P.,. ~ (1.6S4V.,) 2/R, from Eq. (3.2), P., = ( l.6554V.,) 2/R, and from Eq. (33) the efficiency 11
=
(t.654V.,.)2 (t.6S54V.,)'
= 99.83%
Copyr ghtcd matcria
3.8
b, From Eq. (3.5), the FF
Thr~·Phase Bridge Rectifier with RL Load
= 1.6554/1.654 = 1.0008 =
V
95
100.08%.
2
c. From Eq. (3.6), the RF = 1.0008 - I = 0.04 = 4%. d. From Eq. (3.15), therms voltage of the transformer secondary, V, = 0.107V.,. From Eq. (3.43), the rms current of the transformer secondary, . r.
v..
,, • 0.18041., s 0.1804 x v3 "'ii The VA rating of the transformer,
VA = 3V,l, = 3
. r. v.. X 0.7804 X v 3 R
x 0.707V.,
From Eq. (3.8) , TUF =
3
l.6S4'
x v3 x 0.707 x 0.7804
c
0.9542
e. From Eq. (3.40), the peak line-to-neutral voltage is V., • 280.7/1.654 - 169.7 V. The peak inverse voltage of each diode is equal to the peak value of the secondary line-toline voltage, PlV • v3 V., • v3 X 169.7 • 293.9 V. t The average current through each diode is
,, = £''\ .. = 2:
cos wt d(wt)
The average current through each diode is 14 I., = 20/0.3183 = 62.83 A.
m
=,..
~sin
f
= 0.3183/.,
60/3 • 20 A; therefore, the peak current is
Nott: This rectifier bas considerably improved performances compared with those of the multiphase rectifier in Figure 3.12 with six pulses.
Key Points or Sedloa 3.6 • A three-phase bridge rectifier has considerably improved performances compared with those of single-phase rectifiers.
3.8
THREE·PHASE BRIDGE RECTIFIER WITH RL LOAD Equations that are derived in Section 3.5 can be applied to determine the load current of a three-phase rectifier with an RL load (similar to Figure 3.15). It can be noted from Figure 3.14 that the output voltage becomes
V,o =
vi V,o sin wl
for
11'
211' .
3
3
- :s wt :s -
where V,o is the line-to-line rms input voltage. The load current i0 can be.found from
"" . L dio dt +Rio+ E = •v2 V,osm wt
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96
Chaptl!r 3
Dlodl! Re
;.
4
i.
8
o,
o,
Ds
+
2.S n
R
6
+ ;.
••
2
1.SmH
L
3
o,
D,
;.
+
o,
v,
7
IOV
s FIGURE 3.15
'Ibrce~pbue bridge rcc1ifier for PSpicc 1imuladon.
which has a solution of the form
v'ivob sin(o>l - 8) + A,e- (RIL), - E Z R 2 112 where load impedance Z = (R + (wL)2} and load impedance angle i0 =
(3.45)
e = tan-•
(o>UR). The constant A 1 in Eq. (3.45) can be determined from the condition: at wt • .,,13, i0 • lo,
A = [1 + ! - v:.i, sin(f- 8) 1
0
]e
Substitution of A 1 in Eq. (3.45) yields
. v'iVob . [ / +----s10 E v'iVob . ('IT E •o---s10(w1-8)+ - - 8 ) ] e( RIL)(fflJ..- ' > - 0 R z · Z 3 R (3.46) Under a steady-state condition, i0 (w1 = 21'/3) = i0 (o>t = 'IT/3). That is, i0 (w1 21r/3) = 10• Applying this condition, we get the value of / 0 as
l0 -
viV.i, sin(21f/3 - 8) - sin(1r/3 - 0)e-(R/Ll (• l3w) E Z 1 - e-(R/L)(•l3w) R
for / 0
.,,
0
a
(3.47)
which, after substitution in Eq. (3.46) and simplification, gives .
ro a
V2Vo1,[ . ( -
2-
SID
wt -
)
8 +
sin(21f/3 - 8) - sin(1r/3 - 8)
1
_
e-
( R/L)(•l3w- r) )
e
. (3.48)
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Three-Phase Bridge Rectifier with RL load
3.8
97
Therms diode current can be found from Eq. (3.48) as
1,:·
13
I,= [ 2~
2
i3t1(w1) ]"
and the nns output current can then be determined by combining the rrns current of
each diode as
u: + l; + l ;)'n •
!,.... •
v'3 /,
The average diode current can also be found from Eq. (3.47) as
212•/3
Id= 2'If
iod(wt )
T/3
Boundary conditions: The condition for the discontinuous current can be found by setting / 0 in Eq. (3.47) to zero. • ""
v .
•
2 3
[sin ( "' -
e) -
sin (~ -
t -
e-{f)(~l
which can be solved for the vo ltage ratio x
sin ( ¥ x(e):= [
3
e) e-CflCi:ll _ E
= O
R
= El(V2V,. 8 ) as
e)- sin
(f- e)e-<;;;;.,jll
-c-·-i 1 - e .\IM(I)
cos(0)
(3.49)
e e=
The plot of the voltage ratio x against the load impedance angle is shown in Figure 3.16. The load angle 0 cannot exceed 'lf/2. The value o f xis 95.49% at 1.5598 rad, 95.03% at = 0.52308 (30°), and 86.68% at =
e
e o.
Example 3.11 Finding the Performance Parameters of a Three-Phase Bridge Rectifier with an RL Load The three-phase full-wave rectifier of Figure 3. 15 has a load of L = 1.5 mH, R = 2.5 n, and E • 10 V. The line-10-line inpul vollage is v.. = 208 V, 60 Hz. (a) Dclermine (1) 1he steadystatc load current /0 at w, • 'ffl3, (2) the average diode current l o, (3) the rms diode current /,, and (4) therms output current !,..,. (b) Use PSpicc 10 plot the instantaneous output current io, Assume diode parameters IS = 2.22E - 15, BV = 1800 V.
Solution L
R - 2.5 0, L - 1.5 mH, f = 60 Hz. w = 2tt x 60 = 377 rad/s, v.. = 20& V, Z = [R2 + (CdL)2] 112 = 2.56 n.and 8 = tan" 1(wUR) = 12.74°. L The steady-state load current at "" = -rr/3, / 0 = 105.77 A. 2. The numerical integration of i 0 in Eq. (3.48) yields the average diode current as 14 = 36.09 A. Because /0 > 0, the load current is continuous.
Copyrghtcd matcria
98
Chapter 3
Diode Rectifiers BouodMy: Discootiouous/Cootiouous Region
0.95 0 ·.::, ~
..: !'. !!
~ "O
•(0) 0.9
M
.3 0.8S
0.4
0.2
0.6
0.8
e
1.2
Load lmpedance·Angle, radians
2"
FIGURE 3.16
Boundary of continuous and disoontinuous rcpons for three-phase rectifier.
3. By numerical inlel!J'ation of i~ between the limits wt = .,13 and 2'ff/3, we get the nns diode current as I, = 62.733 A. 4. The nns output current I,.. = '1/31, a V3 X 62.53 = 108.31 A. b. The three-phase bridge rectifier for PSpice simulation is shown in Figure 3.15. The list of lhe circuit file is as follows:
~ l e 3.ll Vl\N 8 0 Vffi 2 0 la 3 0 L 6 7 R 4 6
vx
7
w
8 l
DI. D3 J:6 02 D4
2 3 5 5
5 1 4
Three-Phase Bridge Rectifier with RL load SIN (0 169.7V 60HZ) SIN (0 169.7V 60HZ O O 1200EG) SIN (0 169.7V 60HZ O O 2400EG) 1.SMH
2.5
DC lOV DC ov Ill)!)
4
lHlD
4
lH)O
3 1
lH)O
voltage source to measure the output current vol tage source to measure the input current , Diode model
IHlD
Copyrighted m lcria
3.8 06
S
2
.!DDEL
lM)O
.TRAN
lOOS
Three·Phase Bridge Rectifier with RL Load
99
lMlD
D (IS=2.22E-15 BV=lBOOV) 16.667MS lOOS 25MS
Diode m:xlel 1)4rameters Transient analysis
Graphics postprooessor .PROBE . options I'l'LS•O abstol • l.OOOn relt ol = . 01 vntol = l .OOOm .END
The PSpice plot of instantaneous output current i0 is shown in Figure 3.17, which gives / 0 = 104.89 A, compared with the expected value of 105.77 A. A Dbreak diode was used in PSpice simulation to include the specified diode parameters.
Example 3.12 Finding the Input Power Factor of a Three-Phase.Rectifier with a Hlghly Inductive Load The load current or a three-phase rectifier in Figure 3.13 is continuous, with a negligible ripple contenL Express the input current in Fourier series, and determine the HF or the in put current, the OF, and the input PF.
Solution The waveform or the line current is shown in Figure 3.14. The line current is symmetric at the angle (q = p/6) when the phase voltage becomes zero, not when the line-line voltage
v,•
108A 104A
+-- - + - - - -- - f - ---'i-- -+- --t----+---+----+ !!>; I (VX) JOOV+---+-- --h----+----.+- --+---t---;---;r-----+ 100 A
280V
260V 240V +-- -+------f----'i---+---+---+--'--+-- --+ 19ru ~ru Dru nms ~ru ~ru 16ms 17ms 18ms oV(4,7) Cl • 18.062 m, I0-1.885
w~ lime
C2 • 19.$92 m. 110.911 di! = -1.8300 m -6.0260
FIGURE 3.17
PSpice plot for Example 3.1 1.
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100
Chapter 3
Diode Rectifiers
+ 211) = f(x), the input current can
becomes zero. Thus, for satisfying the condition of f(x be d escribed by
11 s,, for - :s "" :s -
i,(1) • 1. i,(1)
=-
6
6
71f
1. for
6
:s"" :s
1111
6
which can be expres,ed in a Fourier series as ~
i,(1) = J.,_
+
~
L (a. cos(n"'1) + b. sin(n"'1)) = L c. sin(n"'1 + 4>.) 11•t n• I
where the coefficients arc
a.
b.
=;J. =;1. 1
h
,.
i, (t) cos(n"'1) d(wt) 2 • l,(t) cos(n"'1)
d(wt)
0
~
=;[II, •
cos(n"'1) d("'1) -
f'
}1f
I,
cos(n"'1) d ("'1)] = 0
=;1 [ J~r2t/ sin(nwt) d ("'1) - j,Tr"tI, sin(n"'1) d("'1) ] 4
•
•
whi~ after integration and simplification, gives b,._ as
-41, cos ( n1r )Stn . b,. • ~ b. = 0
(n") . (m') T sin 3
n = 2, 3, 4, 6, 8, 9, .. .
for
. ., c. = ,y I (a.)2 + (b.) 2 = -4/, n'ff cos(mr)Stn
4>. -' arctan (::) =
for n = I, 5, 7, 11, 13, ...
(n") . (=) 2 3 sin
0
Thus, the Fourier series of the input current is given by 1
'
=
f 4.v'3l, ••,
211
(sin(wt) _ sin(5"'1) _ sin(7wt) 1 5 7
+ sin(11"'1 ) + sin(l3"'1) _ sin(17"'1) _ .. . ) 11
13
17
(3.50)
The rms value of the nth harmonic input cuncnt is given by I
I 2~1. nir 2 2 112 -sin., .. = -vi (a" + b") = - n1r 3
(3.51)
The rms value of the fundamental current is
v'6
1, 1 = - I, 1f
= 0.7797 I,
Copyrghtcd matcria
3.10
Rectifier Circuit Design
101
The nns input current
[z: 1:·
16
I,
=
HF • [ (::,
112
I: d(b>I) ]
= 10
'3 =
r- (fr -
OF = cos cj,1
1 ]'" = [
= cos(O)
,,,
PF = - cos(O)
/,
1]
0.8165/•
112
= 0.3108
or 31.08%
= I
o.n97 a -- •
0.8165
0.9549
Nott: If we compare the PF with that of Example 3.10, where the load is purely resistive, we can notice that the input PF depends on the load angle. For a purely resistive load, PF = 0.8166.
Key Points or Section 3.8 • An inductive load can make the load current continuous. The critical value of the load electromotive force (emf) constant x for a given load impedance angle 6 is higher than that of a single-phase rectifier; that is. x = 86.68% at 6 = 0. • With a highly inductive load, the input current of a rectifier becomes an ac square wave. The input power factor of a three-phase rectifier is 0.955, which is higher than 0.9 for a single-phase rectifier. 3.9
COMPARISONS OF DIODE RECTIFIERS The goal of a rectifier is to yield a de output voltage at a given de output power. Therefore, it is more convenient to express the performance parameters in terms of V.sc and P.i.,. For example, the rating and turns ratio of the transformer in a rectifier circuit can easily be determined if therms input voltage.to the rectifier is in terms of the required output voltage Vi1c. The important parameters arc summarized in Table 3.2 (3). Due to their rel· ative merits, the single-phase and three-phase bridge rectifiers are commonly used. Key Points or Section 3.9
• The single-phase and three-phase bridge rectifiers. which have relative merits. are commonly used for dc-ac conversion.
3.10
RECTIFIER CIRCUIT DESIGN The design of a rectifier involves determining the ratings of semiconductor diodes. The ratings of diodes are normally specified in terms of average current, rms current, peak current, and peak inverse voltage. There arc no standard procedures for the design, but it is required to determine the shapes of the diode currents and voltages. We have noticed in Eqs. (3.20), (3.22), and (3.39) that the output of the rectifiers contain harmonics. Filters can be used to smooth out the de output voltage of
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Chapter 3
Diode Rectifiers
TABL£3.2 Pcrfonnanoe Parameters of Diode Rectifiers with a Resistive Load
Performance Pvametcn
Single-Phase Rectifier with CenterTapped 'll-ans!ormer
Bridse Rectifier
Six-Phase St.Ir Rectifier
Bridge Rectifier
3.14V..,
1.s1v...
2.09V..,
1.0SV"'
1.uv.. 0.5-01..,
l.llV,..
0.501"'
0.74V« 0.1671"
0.428V"' 03331"'
1.571.., 0.785/"'
1.57/"' 0.7851"'
6.281.., 0.409/"
3.141.., 0.5791.,.
l.S7 0.81
1.57 0.81 I.II 0.482 123P"' 123P"' 2/,
2.45 0.998
1.74 0.998
Peat repetitive revenc voltage. V«RN Rm, i.nput voltage per tnns(ormcr leg. V, Diode ave"'ge l,c•Vl
=~
Peak repetitive fort¥Vd c:unent, J,..,., Diode mu cunent, 11(«.USJ Form factor of diode current,
1,<•.,s{IF1,VJ Rectification ratio,~ Form factor, FF
Singlc,.Phlsc
I.II
Ripple l.lclor, RF
.0.482 1.23P.,. 1.1SP.., 2/,
Transformer rating primary,VA
'll-ans!ormcr rating rec:ondMy, VA Output ripple frequency,/,
Threc,.Pbase
1.0009
1.0009
0.042 l.28P.., I.SIP"'
0.042 1.osr.. 1.osr..
6/,
6/,
the rectifier and these are known as de filters. Toe de filters are usually of L. C, and LC
type, as shown in Figure 3.18. Due to rectification action, the input current of the rectifier contains harmonics also and an ac filter is used to filter out some of the harmonics
from the supply system. The ac filter is normally of LC type, as shown in Figure 3.19. L, +
+
+
v.
••
R
••
(a)
+
c,
••
+ R
+
c,
••
(b)
"ll
R
(c)
FIGURE 3.18 De ftltcrs.
a
+ v5 • V,n sino.,t
-
+
c,::
Rectifier
v0
-
FIGURE 3.19 Ac filtcn.
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3.10
Rectifier Circuit Design
103
Nonnally, the filter design requires determining the magnitudes and frequencies of the harmonics. The steps involved in designing rectifiers and filters are explained by examples.
Example 3.13 Finding the Diode Ratings from the Diode Currents A three-phase bridge rectifier supplies a highly inductive load such that the average load current is Id, = 60 A and the ripple content is negligible. Determine the ratings of the diodes if the l.ineto-neutral voltage oftbe Y-eonnected supply is 120V at 60 Hz.
Solution The currents through the diodes are shown in Figure 3.20. The average current of a diode IJ = 60/3 = 20 A. The rms current is ·
(11.•
I, • The PIV =
v'3 Vm
=
2...
la., d(WI)
, IJ
v'3 x V2 x
]'12 = ~ I = 34.64 A VJ
120 = 294 V.
t,
Ot----+----+-----+---1----+----+---~t
f 2:i '; 4:i '· -----r-----....--....---. I..,
2..
' Ot---~---+---+---1--- - -- ---wt I I
: •. ict3-----r-----~-----+---.... --...., I
I
I
I
I
Ot----,----+---~---1---- -+-- ----wt , I : .... ' '
------~-----1-----------4-------~ I
I
O>---~'-- -~'- - - - - - ~ - - + - - - - ~ - - wt
;..,
1,1 - - - - l
Ot-- - ~- - - + - - - - - - - - - - . a - - - - -- wl i.o6 l,1 - - - - - - - 1 o~------+---- - - - - - - - - - ~ - - w t FIGURE 3.20
~ent through diodes. ·
Note: The factor of V'2 is used to convert rms to peak value.
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Chapter 3
Diode Rectifiers
Example 3.14 Finding the Diode Average and rms Currents from the Waveforms The current tluough a diode is shown in Figun: 321. Determine (a) tho nns current, and (b) the average diode current if 11 100 µs, r2 350 µs , rJ 500 µs, f = 250 Hz, f, 5 kHz, I., = 450 A, and /, = 150 A.
=
=
=
=·
Solution· L
The rms value is defined as I, =
t /,,'' I! dJ ]'12 (/.,.sin"'•!)' dt + T [TI f.'' O
(3.52)
= (/;, + /;, )112 where w, = 2nf,
= 31.415.93 racl/s, 11 = 'fflw,
= 100 µs, and T • 1/f.
1 (''
/,1 = [T =
Jo (/., SID. w,1)2 di ]"' =
(F,
lmyT
(3.53)
50.31 A
and I,., =
(T11·,, ·
I, dt
)2= I, Vf(t, - r,)
(3.54)
= 29.05A Subsoituting Eq~ (3.53) and (3.54) in Eq. (3.52), the rm• value is
/~1,
+ ( 1.f(r, - '2)
(50.312
29.052 ) 112 =
I, = [ - 2
=
+
]'/2
(3.55)
58.09 A
b. The average current is found from
Id= [ T I = /do
(/.,sin w,1).d1 + T ,, J,dr J.'' O I / ''
]
+ l,n
FIGURE3.21 Currcn1 " 'llvcform.
Copyr ghtcd matcria
3.10
Rectifier Circuit Design
105
where
I,. = 142
1 ["
.
lmf
T Jo (/.., sin w,J) d1 = nf,
(3.56)
If'' / d1 = IJ (t, - 1,)
(3.57)
=-
0
T ,,
Therefore, the average current becomes
Id,
l,,J a ,
+ 1./(13 - tz)
a
7.16
+ 5.63 = Ji.79 A
1f1 '
Example 3.15 Finding the Load Inductance to Limit the Amount of Ripple Current
The single-phase bridge rectifier is supplied from a 120-v ,' 60.Hz source. The load resistance is R = 500 Calculate the value of a series inductor L that limits therms ripple current /"' to less than 5%of 1,,..
n.
Solution The load impedance Z = R + j(nwL) =
VR
2
+ (nwL)2 &
(3.58)
and 9
"
IJWL = tan- 1-
(3.59)
R
and the instantaneous current is ;0•(1)
a
1.,. -
[l-
4Vm
n Y R2 + (nwL)2 3
cos(2'»1 - &,) +
I
15
cos(4w, -
o,) .. .
l
(3.60)
where
Equation (3.60) gives the rrns value of the ripple current as 2
{2 = K
(1)
2
2
(I)'
(4V,.) (4V,.) -+ -+ 2n'[R' + (Z..L)'J 3 2n'[R2 + (4..L)2] 15 .
Considering only the lowest order harmonic (n = 2), we have
1 " =
'
4V..,
Vi"ff y R2 + (2wL) 2
(1) 3 Copyrighted m lcria
106
Chapter 3
Diode Rectifiers
Using the value of I.,. and after simplification, the ripple factor is
RF = I,. I.,.
=
= 0.05
0.4714
VI + (2wUR)
2
For R • 500 n and f = liO Hz, the inductance value is obtained as 0.4714 2 • 0.052 (! + (4 x 60 X 11US00)2] and this gives l = 6.22 H. We can notice from Eq. (3.60) that an inductance in the load o(fers a high impedance for the hannonic currents and acts like a filter in reducing the hannonics. However, this inductance introduces a time delay of the load current with respect to the input voltage; and in the case of the single-phase half-wave rectifier, a freewheeling diode is required to provide a path for this
inductive currenl.
Example 3.16 Finding the Filter capacitance to Limit the Amount of Output Ripple Voltage
•
A single-phase bridge-rectifier Ls supplied from a 120-V, 60-Hz source. The load resistance is R = 500 n. (a) Design a C filter so t.hat the ripple factor of the output voltage is less than 5%. (b) With the value of capacitor C in part (a),calculate the average load voltage V.,,.
Solution a. When the instantaneous voltage v, in Figure 3.22a is higher than the inst antaneous capacitor voltage u,. the diodes (D1 and D, or D:, and D,) conduct; and the capacitor is 1hen charged from the supply. If the ins tantaneous supply voltage v. falls below the in· stantaneous capacitor voltage v.., the diodes (D1 and D, or D:, and D,) arc reverse bi-
ased and the capacitor C, discharges through the load resistance R. The capacilor voltage ·vc VI.tries bct¥1ccn a minimum Vt(min) and maximum value Vc(mai1)· This is sh~ in Figure 3.22b. Let us assume that 11 is the charging time and that 12 is the discharging time of capacitor C,. The equivalent circuit during charging is shown in Figure 3.22c.The capacitor charges almost instantaneously to the supply voltage v,. The capacitor C, will be charged to the peak supply voltage v~, so that Vc(t = 11) = Vm. Figure 3.22d shows the equivalent circuit during discharging. The capacitor discharges exponentially through R. ~,Jio dt + Vc(t = 0) + Rio ~ 0 which, with an initial condition of uh = 0)
= Vm, gives the discharging current as
The output (or capacitor) voltage Vo d uring the discharging period can be found from
Vo(t) • Rio • The peak-to-peak ripple voltage V,(pp) •
v,..,-11RC,
V,(Pl'I can be
found from
Vo(I - ,,) - Vo(I • ,, ) - V,.. - V..,e"'l• c. -
v... p - ,-,,iRC,>
(3.61)
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3.10
Rectifier Circuit Design
107
D
v,
+
+
c,
v,
R
•o
wl
(•) Circuit model
II I I I
1 1
I I
.,,
v,
(b) Waveforms for full-w::tYC·rectifier
o, + >
c,
'>R >
wl
(c) Charging
(d) Discharging
FIGURE 3.22
Single-phase bridge rectifier ,,ith C filter.
Since,,-, "' I - .r, Eq. (3.61) can be simplified to Vmtz
12 )
Vm
V,cpp) • Vm ( I - I + RC, • RC, • 2/RC,
Therefore, the average load voltage v., is given by (assuming 12 = 1/2 /)
v...- = v -
"'
V,,,
V,( pp)
- - - =v. - - 2
"'
4/RC,
(3.62)
Copyrghtcd matcria
108
Chapter 3
Diode .Rectifiers
Thus, the nns output ripple voltage v.. can be found approximately from V,,,. 4V2/RC,
V. • V,tpp) •
"
2V2
The RF can be found from V,,. V., 4/RC, I RF = - = = ~~-"---Vole 4v'1.fRC, V.,(4/RC, - 1) v'1. (4/RC, - I}
(3.63)
which can be solved for C,: 1 C, • 4;R ( I + v'2 RF) = 4 x
~ x 500 ( I + v'2 ~ 0.05) = lZ6.2 µ.F
b. From Eq. (3.62), the average load voltage V"' is
v.., = 169.7 -
169.7 1'
~ =
x 60 x 500 x 126.2 x 10
169.7 - 11.21
= 158.49 V
Example 3.17 Finding the Values of an LC Output Filter to Umit the Amount of Output RIPPie Voltage An LC filter as shown in Figure 3.18c is used to reduce the ripple content of the output voltage for a single-phase full-wave rectifier. The load resistance is R = 40 0, load inductance is L = 10 m.H, and source (requency is 60 Hz (or 377 rad/a). (a) Determine 1hc vnluc:s of L~ and C~ w that the RF of the output voltage is 10%. (b) Use PSpicc to calculate Fourier components of
the output voltage 1.\)- Assume diode parameters IS
= 2-22E - 15, BV
=
1800 V.
Solution a. The. equivalent circuit for the hannonics is shown in Figure 3.23. To make it easier for the nth harmonic ripple current to pass through the filter capacitor, the load impedance must be much greater than that of the capacitor. That is.
VR2+ (nwL) 2 >> _1_ n..C, This condition is generally satisfied by the relation
VR2 + (no,L)'
R _
V ..(no,)
X,, • n~C,
c.
= __!Q__
(3.64)
n..C,
+ v .,.(no,)
L
FIGURE 3.23
Equivalent cil'"t'Uit for harmonics.
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3.10
Rectifier Circuit Design
109
and under this condition, the effect o r the load is negligible. Therms value or the nth harmonic com ponent appearing on the output can be found by using the voltage· divider rule and is expressed as
V00
.
~
I I
I
I
-1/(nwC,) -1 V,i,j = V".11 (nwl,) - 1/(nwC,) (11w)2L, C, - I
(3.65)
The total amount or ripple voltage due to all harmonics is
v. = ac
( L..
v2
),n
(3.66)
on
11•2,,4.6, . . .
For a specified value of V" and with the value ol C, from Eq. (3.64). the value or L, can be computed. We can simplify the computation by considering only the dominant harmonic. From Eq. (3.22) we find that the second harmonic is the dominant one and its rms value is = 4V..,1(3"11r) and the de value. V"' = 2V.,f,r. For n = .2, E qs. (3.65) and (3.66) give
v,,.
v • v02 - I
- I 1\/.2 (2w) 2L,C, - I •
"'
The value of the filter capacitor C, is calculated from
VR
2
+ (2wl )2
=
__!!!... 2..C,
or
C, =
10 41rfV R 2
+ (4,rfl)2
= 326µF
From Eq. (3.6) the RF is defined as RF =
v"
v"'
=
v., = v,. v.,
= "11
1
v"' (41rf)2L,C, -
I
t
3 ((4,rf) 2L,C, - I]
I = o. J
or (4-rrf) 2L,C, - I = 4.714 and L, = 30.83 mH. b. The single-phase bridge rectifier for PSpicc simulation is shown in Figure 3.24. The list or the circuit file is as follows: - l e 3.17 . vs 1 LE
)
CE
7
RX
8 5
L R
7
S ingle- Phae,e Bridge Ractifiot' ,..,ith LC Fil ter 0 SIN (0 169.7V 60HZ)
8 4
6
326UF SOM lOMH DC CN
7
vx
6
5 4
V'i
l 2
2 3
Dl
30.83MH
Used to converge the 'solution
40 DC
CN
r:t10D
Vol tage source to measure the oucput current Voltage source to measure the ifll'.)Ut current ; Diode rrodels
Copyrighted m~leria
110
Chapter3
Diode Rectifiers 0 3 2
03 04
4 0 4
.MODEL
IM)[)
D (IS=2. 22E-15 BV=1800V}
•TRAN .FOUR
lOUS
SOMs 33MS V(6 . 5 }
ll2
120HZ
. options
I'l'LS=O
sous
; Diode DK>del parameters
Transient analysis ; Fourier analysis of output voltage
abstol = 1. OOOU reltol = . 05 vntol = O. Olm
.END
The results of PSpice simulation for the output voltage V(6, 5) are as follows:
FOURIER CCMl'ONEm'S OF TRANSIENI' RESl'CfiSE V(6,5) . DC CCMPONEtn' = l.140973E+02 l'.l®IJNIC FIIEl:lUm:'i N:>Rlll\LIZEI> FOURIER N)
(HZ)
1 2 3 4 5 6
1.200E+02 2 . 4008+02 3.600E+02 4.8008+02 6.000E+02 7.2008+02 8. 4008+02 9 . 600E+02 1. 080E+03
7
8 9
TCfrAL HARMONIC
CCtlPCmm'
CCt!PONmI'
Pl!ASE
OORl!l\LIZED
(DB1}
PHl\SE (DID)
l.3048+01 1.000E+OO 1.038E+02 1.2368+02 6. 4968- 01 4.9818- 02 9.2268+01 2.277E-01 1. 7468-02 l.5668- 01 1.201E- 02 4 .8758+01 l.274E-01 9.7678-03 2 .2328+01 l.0208-01 7.8228- 03 8.358E+OO 1 . 9978+00 8.2728-02 6.3438- 03 -1.0612+00 6.982E-02 5.3548-03 6.0158- 02 4.61.28° 03 - 3. 4361!+00 DISTORTICN • 5.6360708+00 PERCENT
O.OOOE•OO 1.9888+01 - 1.1508+01 -5.501.E+Ol - 8.144E+Ol -9.540E+Ol - 1.0188+02 -1.048E+02
-1. 072E+02
which verifies t he design.
i..
3
v, ~
30.83mH
o,
ov
R,
7
80m0
R
03
+
2
••
v,
0
8
o,
o,
400
s
c,
3261
L
v,
IOmH 6 OV
4
FIGURE 3.24
Singlc,.phasc bridge rcct~cr for PSpice simullltion.
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3.1 0
111
Rectifier Circuit Design
Example 3.18 Finding the Values of an LC Input Filter to Limit the Amount of Input Ripple Current An LC input filter as shown in Figure 3.19 is used to reduce the input current harmonics in a single-phase full-wave rectifier of Figure 3.7a. The load current is ripple Cree and its average value is /,. Uthe supply frequency is f = 60 Hz (or 3n rad/s), determine the resonant frequency of the filter so that the total input harmonic current is reduced to 1% of the fundamen· tal component.
Solution The equivalent circuit for the nth harmonic component is shown in Figure 3.25. Therms value of th e nth harmonic current appearing in the supply is obtained by using the current-divider rule. ~ -
I
1/(nwC,)
(nwL1 - 11( nwC1)
I I
I
I ~ (nw )2 L;C1 - I
~=
(3.67)
where 1•• is the rrns value of the nth harmonic current. The total amoun t of harmonic current in the supply line is
and the harmonic factor of input current (with the ftlter) is
~ (/....!!)']'12 ,= -Is,.l =[n • L 2.,J.... f,1 From Eq. (3.23), / 1• = and (3.68) we get
41,lv'i."' and r•• =
,, =
f (/•I,,•)'
=
• •3.>.7. .. .
4/,f( v'i. mr) for n
f
I
• •JJ.7... .
(3.68)
= 3, S, 7 , .... From Eqs. (3.67)
(w' L,c, - 1)2
n 2[(nw)2L;C; - 1)2
j
(3.69)
This can be solved for the value of L1C1.To simplify the calculations, if we consider only the third hannonic, 3[(3 x 2 x"' x 60) 2L1C1 - 1V(w 2L,C1 - I} • 1/0.01 • IOOor L1C1 = 9.349 x 10-6 and the filter frequency is 1/VL;C, 327.04 rad/s, or 52.05 Hz. Assuming that C1 1000 µF, we obtain L 1 = 9.349 mH.
=
I,,
l
X,, • nwC.
=
C;
FIGURE 3.25
Equivalenc circuit for barmonic cunent.
Copyrghtcd matcria
112
Chapter 3
Diode Rectifiers
Nore: The ac filter is generally tuned to the harmonic frequency involved, but it requires a careful design to avoid the possibility of resonance with the power system. The resonant frequency of the third-harmonic current is 377 x 3 = 1131 rad/s. Key Points or Section 3.10 • The design of a rectifier requires determining the diode ratings and the ratings of filter components at the input and output side. Filters are used to smooth out the output voltage by a de filter and to reduce the amount of harmonic injection to the input supply by an ac filter.
3.11
OUTPUT VOLTAGE WITH LC FILTER
The equivalent circuit of a full-wave rectifier with an LC filter is shown in Figure 3.26a. Assume that the value of C, is very large, so that its voltage is ripple free with an average value of Vo(dc) · L , is the total inductance, including the source or line inductance, and is generally placed at the input side to act as an ac inductance instead of a de choke. If Vde is less than Vm, the current i 0 begins to flow at a, which is given by
vdc = v.. sin a This in tum gives
••
vm i,,
I«
v""
..
+ 0
v,
C, 0
"
2,r
v""
jL
.,, 0
(a) Equivalcnl cin:ui1
...
I I I I I I I I
+
a
...
IJ ,r+ a (b) Waveforms
FIGURE 3.26 Output voltage with LC filter.
Co.,yrghlcd matcria
3.11
Output Voltage with LC Filter
113
where x = Vd.fVm. The output current io is given by
dil
L~dt
c
Vm sin wt - Vdc
which can be solved for i0•
io
=
l"'(Vm sin wt - Vo.:) d(wr)
lL w ' •
= - vm
wLe
vd, (wt
(cos a - cos wt) - -
wLe
- a)
for wt ;;,: a
(3.70)
The critical value of wt = 13 = ,r + a at which the current i 0 falls to zero can be found from the condition i 0( wt = 13) = ,r + a = 0. The average current /de can be found from
11•+uio(t) d(wt)
Ide= -
,r •
which, after integration and simplification gives
Ide
(I - .!:.) ] 2
= vm [ ~ + x wL,
(3.71)
ir
For Vd, = 0, the peak current that can flow through the rectifier is lpt malizing /de with respect to / pk, we get k (x)
= -Ide = •y ,,::----Z 1 - r + x ( -2 ~·
= V,./wL,. Nor-
- -1t)
1T
(3.72)
2
Normalizing therms value I,... with respect to l pk• we get k,(x)
= -l,rru =
(3.73)
/pk
Because a depends on the voltage ratio x, Eqs. (3.71) and (3.72) are dependent on x. Tobie 3.3 shows the values of k(x) and k,(x) against the voltage ratiox. Because the average voltage of the rectifier is V"' = 2 V.,,/1t, the average current equal to
Thus,
2 - Vm = lo.: = lpt k(x) = -V.. [-y ,,:;----z 1 - x- + x (2 - - -"')] nR wL, 1T
2
Copyrghtcd matcria
114
Chapter 3
Diode Rectlfie11
which gives the critical value of the inductance L.,.( = L.} for a continuous current as
Le,=
:[~+x(i-¥)]
(3.74)
Thus, for a continuous current through the inductor, the value of L, must be larger than the value of Le,· That is,
L, > Lcr
=: [ ~ +
x(~ -¥)]
(3.75)
Discontinuous cue. The current is discontinuous if wt = p s (1r + a). The angle P at which the current is zero can be found by setting in Eq. (3.70) to zero. That is,
cos(a) - cos(P) - x(P - a)
=0
which in terms of x becomes
~Example 3.19 Load Current
x(P - arcsin(x))
=0
(3.76)
Finding t he Crit ical Value of Inductor for Continuous
The nns input voltage to the circuit in Figure 3.26a is 220 V. 60 Hz. (a) If the de o utput voltage is Vc1c • 100 V at 1.,, 2 10 A, determine the values of critical inductance L,, a, and I...,. (b) If 1.,, - 15 A and L, - 6.S mH, uoc lllblc 3.310 determine the values of Vc1c, a, ll, and/..,..
TABLE 3.3 Normalized Load Cunent x%
0
s
10 IS 20 25 30 35 40 45
so SS <,()
6S
70 72
72.5 73 73.07
,.;,,.% 100.0
95.2 90.16
84.86 79.30 73.47 67.37 <,0.98 54.28 47.26 39.89 32.14 23.95· 15.21 6.02 2.14 I.IS 0.15 0
l...JI,. %
a Deg,ees
flDevees
122.47 115.92 lO'J.1 102.01 94.66
0
180 182.97 185.74 188.63 191.54 194.48 197.46 200.49 203.SS 206.74 210.00 213.37 216.87 220.54 224.43 226.0S 226.47 226.89 226.95
87.04 79.18 71 .1 62.82 54.43
46.06 38.03 31.0S 26.58 26.75 28.38 28.92 29.51 29.<,0
2.87
S.14 8.63 11.54 14.48 17,46 20,49 23.SS 26.74 30.00 33.37
36.87 . 40.54 44.27 46.0S 46.47 46.89
46.95
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3.11
OUtput Voltage with LC Filter
115
Soludon "' = 21r x 60 = 377 rad/s, V, = 120 V, V.., = vi x 120 = 169.7 V. L Voltage ratio x • V.,/V,., a 100/169.7 • 58.93%; " • sin- 1(x) • 36.87o/o. Equation (3.72) gives the avenge current ratio k = l ,i,llpt = 25.75%. Thus, fpt. = l,i,}k = 10/0.2575 • 38.84 A. The critical value of inductance is
v...
169.7
L" = wl•• = 377 x 38.84 ~ 11·59 mH
Equation (3.73) gives the IDlS current ratio k, = 1.,../lpt = 32.4%. Tbus,
1,.., = k, lpt = 0.324 x
38.84
= 12.58 A.
b. L, • 65 mH, lpt. • V,.,l(wL,) • 169.7/(377 x 65 mH) • 69.25 A.
loc
1..
k • -
IS 69.25
• - - • 21.66%
Using linear interpolation, we get .%
v.,.
= ;rlt +
(x• • , - x.)(k - k.)
k,.., - k,i (65 - 60)(21.66 - 23.95) • 60 + 15.27 - 23.95 • 6132% = xv.., = 0.6132 x 169.7 = 104.06 V (a,,+ 1
a=n.+ •
36 87 · +
-
k
a.)(k - k.)
11-+-l
- k ti
(4054 - 36.87)(21.66 - 23.95) 37 840 1527 - 23.95 = ·
f3 • 13. + (13... - 13.)(k - k.) k,.+l - k,. (220.54 - 216.87)(21.66 - 23.95 ) - 216.87 + _ _ . • 211.ss• 15 27 23 95 I,... (k,<•• •> - k,<. J)(k - k.) k, = J a k,(•) + k - k pk
1t+ l
l'I
(2658 - 31.05)(21.66 - 23.95) = JI.OS + 1527 - 23.95
= 29 .87%
Thus, lmu • 0.2987 x lpt. = 0.2987 x 69.25 = 20.68 A.
Key Points of Section 3.11 • With a high value of output filter capacitance C, the output voltage remains almost constant. A minimum value of the filter inductance L, is required to maintain a continuous current. The inductor L, is generally placed at the input ~ide to act as an ac inductor instead of a de choke.
Copyrighted m lcria
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3.12
Chapter 3
Diode Rectifiers
EFFECTS OF SOURCE AND LOAD INDUCTANas
In the derivations of the output voltages and the performance criteria of rectifiers, it was assumed that the source has no inductances and resistances. However, i.n a practical transformer and supply, these are always present and the performances of rectifiers are slightly changed. The effect of the source inductance, which is more significant than that of resistance, can be explained with reference to Figure 3.27a. The diode with the most positive voltage conducts. Let us consider the point wt = "'where voltages v"' and v1,c arc equal as shown in Figure 3.27b. Toe current /,J,: is still flowing through diode D 1• Due to the inductance L1, the current cannot fall to zero immediately and the transfer of current cannot be on an instantaneous basis. The current i,n decreases. resulting in an induced voltage across L 1 of +1.1o1 and the output voltage becomes Vo = v.., + I.bi· At the same time the current through ~. i 43 increases from zero, inducing an equal voltage across Li of -tm and the output voltage becomes tin = 1/1>< - Vo:2· Toe result is that the anode voltages of diodes D 1 and ~ are equal;
c
L.,
..
- Vu ..
1.,
o,
o,
...,p,
t"'
i4.1
o, b
o,
o.
- "u+
(a) Circuit diagram •1.2
-Vm
·:r
I
..., . I
x
3"
x x=. ii11
2..
T
I
4,,
"H T Q
•
5,,
i111
2,r
T
(b) Wa,•crorms FIGURE 3.27
Three~phase bridge reclifier with source inductances.
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3.12
Effects of Source and Load Inductances
117
and both diodes conduct for a cenain period which is called commutation (or overlap) angle µ.. This transfer or current from one diode to another is called commutation. The reactance corresponding to the inductance is known as commutating reactance. The effect of this overlap is to reduce the average output voltage of conveners. The voltage across I,z is
Vu=
di
Lz -dt
(3.77)
Assuming a linear rise of current i from Oto I& (or a constant dildt = tiiltit), we can write Eq. (3.77) as
Vu tit = /.,z {ij
(3.78)
and this is repeated six times for a three-phase bridge rectifier. Using Eq. (3.78), the average voltage reduction due to the commutating inductances is
V,
=
1
7
2(vL1 +Vu+ v1..3)tit
= 2/(Lt + Lz + y
. ) ti,
= 2/(Lt + Lz + L3) /c1c
(3.79)
lf all the inductances are equal and L, = Lt = I,z = L 3, Eq. (3.79) becomes V, = 6/L,l""
(3.80)
where /is the supply frequency in hertz.
Example 3.20 Finding the Effect of Line Inductance on the Output Voltage of a Rectifier A three-phase bridge rectifier is supplied (rom a Y-connected 208-V 60-Hz supply. The average load current is 60 A and has negligible ripple. Calculate the percentage reduction or output volt· age due to commutation i( the line inductance per phase is 0.5 mH.
Solution L , = 0.5 mH, V, = 208/v':i = 120 V, f = 60 Hz, l0c = 60 A, and Vm = vi x 120 = 169.7 V. From Eq. (3.40). V"' = 1.654 x 169.7 = 280.7 V. Equation (3.80) gives the output voltage reduction,
_ 100 V, • 6 x 60 x 0.5 x 10 3 x 60 • 10.8 V or 10.8 x 'iiio.7 = 3.85% and the effective output voltage is (280.7 - 10.8) = 269.9 V.
Example 3.21 Finding the Effect of Diode Recovery Time on the Output Voltage of a Rectifier The diodes in the single-phase rull,wave rectifier in Figure 3.6a have a reverse recovery time or t,, = 50 fLS and therms input voltage i, V, = 120 V. Determine the effect or the reverse recovery time on the average output voltage if the supply frequency is (a)/, e 2 kHz. and (b) /, = 60 Hz.
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Chapter 3
Diode RectifieB
FIGURE 3.28
Effect of reverse recovery time on output voltage.
Solution The reverie recovery time would affect the output voltage of the rectifier. In the Cull-wave recti· fier of Figure 3.6a, the diode Di is not off at "" • 'IT; imtead, it continues to conduct until t = "'"' + t,r As aresult of the reverie recovery time, the average output voltage is reduced and the output voltage waveform is shown in Figure 3.28. If the input voltage is v • V"' sin u>t • v2 V, sin cJll, the average output voltage reduction is
If.'"
V. = V. sin Oll dt = ZVm [- ~]'" "To"' T a>o
v.,
= -:;;- (1 - cos""")
v_ - v'2 v. - v'2 x
120 - 169.7 v
(3.81)
Without any reveBO recovery time, Eq. (311) gives the average output voltage v., = 0.6366V., = 1()8.03 v. 1.
For 1,. = SO f1.S and/, - 2000 Hz. the reduction of the average output voltage is Vm
V,, = -(I - cos 2Tr/,t")
" = 0.061Vm a JO.J V or 9.Sl%otV.,.
b. For'" = 50 f1.S and/, • 60 Hz, the reduction of the output de voltage
v,, •
v"'(t - cos2,r/,1.,) = 5.65 x 10-sv., 'ff
= 9.6 x 10-3 v
or 8.88 x 10-1% of v..,
Note: The effect of 1,, is significant for high-frequency source and for the case of normal 60-Hz source, its effect can be considered negligible.
Key Points or Section 3.U • A practical supply has a source reactance. As a result, the transfer of current from · one diode to another one cannot be instantaneous. There is an overl!'P known as
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119
commutation angle, which Jowers the effective output voltage of the rectifier. The effect of the diode reverse time may be significant for a high-frequency source.
SUMMARY There are different types of rectifiers depending on the connections of diodes and input transformer. The performance parameters of rectifiers are defined and it bas been shown that the performances of rectifiers vary with their types. The rectifiers generate harmonics into the load and the supply line; and these harmonics can be reduced by filters. The performances of the rectifiers are also influenced by the source and load inductances.
REFERENCES (11 J. Schaefer, Rtctifior Circuits-Tlreory and Design, New York: John W~cy & Sons, 1975. (21 R. W. Lee, Power Convert
REVIEW QUESTIONS 3.1 3.2 3.3 3A 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.2.o 3.21 3.22 3.23
J..24
What is the turns ratio of a transformer? What is a rectifier? What is the difference between a rectifier and a converter? What is the blocking condition of a diode? What are the performance parnmeters of a rectifier? What is the significance of the fonn factor of a rectifier? What is the significance of the ripple factor of a rectifier? What is the efficiency of rectification? What is the significance of the transformer utilization factor? What is the displacement facior? What is the input power factor? What is the harmonic factor? What is the difference between a hall-wave and a fuU-wave rectifier? What is the de output voltage of a single-phase hall-wave rectifier? What is the de output voltage of a single-phase full-wave rectifier? What is the fundamental frequency of the output voltage of a single-phase full-wave rectifier? What are the advantages of a three-phase rectifier over a single-phase rectifier? What arc the disadvantages of a multi-phase hall-wave rectifier? What are the advantages of a three-phase bridge rectifier over a six-phase star rectifier? What arc the purposes of filters in rectifier circuits? What are the differences betwe.en ac and de filters? What are the effects of source inductances on the output voltage of a rectifier? What are the effects of load inductances on the rectifier output? What is a commutation of diodes? \Vhat is the commutation angle of a rectifier?
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Diode Rectifiers
PROBLEMS 3.1 A single-phase bridge rectifier bas a purely resistive load R • 10 n, the peak supply volt· age V,. = 170 V, and the supply frequency f = 60 Hz. Determine the average output voltage of the rectifier if the source inductance is negligible.
1.2 . Repeat Problem 3.1 if the source inductance per phase (including transformer leakage in· ductance) is L, = 0.S mH. 3.3 A six-phase sta r rectifier has a purely resistive load of R = 10 fl, the peak supply voltage V,. = 170 V, and the supply frequency f = 60 Hz. Determine the average o utput voltage o f the rectifier if the source inductance is neglibile. 3.4 Repeat Problem 33 if the source inductance per phase (including the transformer leakage inductance) is L, = 0.5 mH. 3.5 A three·phase bridge rectifier bas a purely resistive load of R • 100 0 and is supplied Crom a 280-V, 60-Hz supply. The primary and secondary o! the input transformer are connected in Y. Determine the average o utput voltage of the rectifier if the source induc· tanccs are negligible. 3.6 Repeat Problem 3.S if the source inductance per phase (including transformer leakage in· ductancc) is L, 0 O.S mH. 3.7 The single-phase bridge rectifier of Figure 3.6a is required to supply an average voltage of V"' = 400 V to a resistive load of R = 10 0 . Determine the voltage and current ratings of diodes and transformer. 3.8 A three-phase bridge rectifier is required to supply an average voltage of V"' = 750 Vat a ripple-free current of l0c = 9000 A. The primary and secondary of the transformer arc connected in Y. Determine the voltage and current ratings or diodes and transformer. 3.9 The single-phase rectifier or Figure 3.Sa has an RL load. Ir the peak input voltage is
VM • 170 V. thcsupply frequency f
3.10 3.11
3.12
• 60 Hz. and the load resisu1.nee R •
IS 0, determine
the load inductance L to limit the load current harmonic to 4% of the average value 1• .,. The three-phase star rectifier of Figure 3.12a has an RL load. If the secondary peak voltage per phase is Vm = 170 V at 60 Hz. and the load resistance is R = IS 0 , determine the load inductance L to limit the load current harmonics to 2%of the average value /dcThe battery voltage in Figure 3.4a is E = 20 V and its capacity is200Wh.Theaverage charging current should 10 A. The primary input voltage is v, = 120 V, 60 Hz. and the trans[ormer has a turns ratio of h = 2: I. Calculate(•) the conduction angle 6 o[ the diode, (b) the curre nt-limiting resistance R. (c) the power rating PR or R, (d) the charging time /1 in hours, (e) the rectifier efficiency '1, and (f) the peak in verse voltage PIV of the diode. The singlc·phasc full-wave rectifier of Figure 3.8a has L • 4.S mH, R a S 0, and E = 20 V. The input voltage is V, = 120 V at 60 Hz. (1) Determine (I) the steady-state load current /0 at wt = 0, (2) the average diode current 14 , (3) therms diode current /,, and (4) therms o utput current (b) Use PSpicc to plot the instantaneous o utput cur· rent io, Assume diode parameters IS = 2.22E - IS, BV = 1800 V. The three-phase full-wave rectifier or Figure 3.13a has a load of L a 2.5 mH. R a S 0 , and E = 20 V. The line-to-line input voltage is v,. = 208 V, 60 Hz. (1) Determine (I) the steady-state load current /0 at "'' = ..,13_ (2) the average diode current IJ. (3) the rrns diode current/,, and (4) the rrns o utput current 1..... (b) Use PSpicc to plot the instantaneous output curre nt i0• Assume diode parameters JS 2.22E - IS, BV 1800 V. A ,ingle-phase bridge rectifier is supplied from a 120· V. 60-Hz source. The load resis· tance is R = 200 n. (1) Design a C-filter so that the ripple factor of the output voltage is less than 5%. (b) With the value of capacitor C in part (a), calculate the average load voltage Vd<. R epeat Problem 3.14 for the single-phase half-wave rectifier.
be'•, -
!,,.,.
3.13
=
3.14
3.15
=
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121
3.16 The rms input voltage to the circuit in Figure 3.22a is 120 V, 60 Hz. (a) If the de output voltage is V"' = 48 Vat /
voltage harmonics. V~. 3.18 The LC filter shown in Figure 3.18c is used 10 reduce the ripple content of the output voltage for a six·phasc slar rectifier. The load resistance is R = 20 0. load inductance is L s 5 mH. and source frequency is 60 Hz. Determine the filler parameters L, and C, so lha11he ripple factor of the output voltage is 5%. 3.19 The three-phase bridge rectifier or Figure 3.3a has an RL load and is supplied from a Y
3.20
3.21
3.22
3.23 3.24
connected supply. (a) Use the method of Fourier series 10 obtain expressions for lhe oul· put voltage Vo(t) and load current i 0(1). (b) If peak phase voltage is Vm = 170 Val 60 Hz and the load rc-sislance is R = 200 !I, determine the load inductance L to limit the ripple current 10 2% of the average value I«· The single-phase half-wave rcc1irier of Figure 3.3a has a rrecwhecling diode and a ripple· free average load current of,•. (a) Draw the waveforms for the currents in D1• D.,. and the transformer primary; (b) express the primary current in Fourier series; and (c) determine the input PF and HF of the input current at the rectifier input Assume a transformer turns ratio of unily. The single-phase full-wave rectifier of Figure 3.Sa bas a ripple-free average load current of 1.- (a) Draw the waveforms for currenLli. in Di, Dz, and transformer' primary; (b) express the primary current in Fourier series: and (c) determine the input PF and HF of the input current at the rectifier input. Assume a transformer turns ratio of unity. The multiphase star rectifier or Figure 3.12a has three pulses and supplies a ripple-free average load current of/,. The primary and secondary or the transformer are connected in Y. Assume a transformer turns ratio of unity. (•) Draw the waveforms for currents in 0 1• Dz., D3• and transformer primary: (h) express the prin1ary current in Fourier scries:and (c) determine the input PF and HF of input current. Repeal Problem 3.22 if the J>rimary of the lransfom,er is connected in della and secondary in Y. The multiphase star rectifie r of Figure 3.12a has six pulses and supplies a ripple-free average load current of/~ The primary or the t ransformer is connected in delta and secondary in Y. Assume a transformer turns ratio of unity. (a) Draw the waveforms for currents in D 1, D,, and transformer primary: (b) express the primary current in Fourier series: and (c) determine the input PF and HF or the input current. The three-phase bridge rectifier of Figure 3.13a supplies a ripple-free load current of I,. The primary and secondary or the transformer arc connected in Y. Assume a transformer turns ratio or unity. (•) Draw the waveforms for currents in Di, /);, D, and the secondary phase current or lhc transformer; (b) express the secondary phase current in Fourier se· ries; and (c) determine the input PF and HF of the input current. Repeal Problem 3.25 if the primary of the transformer is connected in delta and secondary in Y. Repeal Problem 3.25 if the primary and secondary of the transformer arc connected in delta.
D,.
3.25
3.26 3.27
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4
Power Transistors The /taming objectives of this cliapter an as /ollow11: • To learn the characteristics of an ideal switch • To learn about different power transistors such as BJTs, MOSFETs, SITs, IGBTs, and COOLMOS. • To learn the limitations of transistors as switches • To understand the characteristics, gate control requirements, and models of power transistors
4.1
INTRODUCTION
Power transistors have controlled turn·on and tum-off characteristics. Toe transistors, which are used as switching elements, are operated in the saturation region, resulting in a low on-state voltage drop. The switching speed of modem transistors is much higher than that of thyristors and they are extensively employed in de-de and dc-ac convert· ers, with inverse parallel-connected diodes to provide bidirectional current flow. However, their voltage and current ratings are lower than those of thyristors and transistors are normally used in low- to medium-power applications. Toe power transistors can be classified broadly into five categories: · 1, Bipolar junction transistors (BJTs) 2. Metal oxide semiconductor field-effect transistors (MOSFETs} 3. Static induction transistors (SITs) 4. Insulated-gate bipolar tra.nsistors {IGBTs) S. COOLMOS
BJTs, MOSFETs, SITs, JGBTs, or COOLMOS can be assumed as ideal switches to explain the power conversion techniques. A transistor can be operated as a switch. However, the choice between a BJT and an MOSFET in the converter circuits is not obvious, but each of them can replace a switch, proyided that its voltage and current 122
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4.2 Bipolar Junction Transistors
123
ratings meet the output requirements of the converter. Practical transistors differ from ideal devices. The transistors have certain limitations and are restricted to some applications. The characteristics and ratings of each type should be examined to determine its suitability to a particular application.
4.2
BIPOLAR JUNCTION TRANSISTORS A bipolar transistor is formed by adding a second p- or n-region to a pn-junction diode. With two n-regions and one p-region, two junctions are formed and it is known as an NPN-transisror, as shown in Figure 4.Ja. With two p-regions and one n-region, it is called as a PNP-transistor, as shown in Figure 4.lb. The three terminals are named as collutor, tmittu, and bast. A bipolar transistor has two junctions, collector-base junction (CBJ) and base...:mitter junction (BEJ) (1- 5] NPN-transistors of various sizes are shown in Figure 4.2. There are two n+·regions for the emitter of NPN-type transistor shown in Figure 4.3a and two p+·regions for the emitter of the PNP-type transistor shown in Figure 4.3b. For an NPN-type, the emitter side n-layer is made wide, the p -base is narrow, and the collector side n-layer is narrow and heavily doped. For a PNP-type, Collector
CoUector
c n
c
le
le
le
le E
E
(a) NPN-1ranl istor
(b) PNP-tran,isror
FIGURE 4.1
Bipolar transis1ors.
FIGURE 4.2
NPN-cranslstors. (Courtesy or Powerex. Inc.)
Co yr ,hi
m
ri,1
124
Chapter 4
Power Transistors
Emiuer
Base
Collector
Base
I n'
I
I "'"""' '--"'
I ·rr,,rrrr
n•
p•
p
I
'7777">
I
p•
n
n
p
Collector
Emitter
(a) NPN·transisior
(b) PNP-transis1or
FIGURE 4.3
Cross sections ol BJTs.
the emitter side p·layer is made wide, the 11-base is narrow, and the collector side p-layer is narrow and heavily doped. Toe base and collector currents flow through two parallel paths, resulting in a low on-state collector-emitter resistance., RcE(ON)·
4 .2.1
Steady-State Characteristics Although there are three possible configurations--<:ommon collector, common base, and common emitter, the common-emitter configuration, which is shown in Figure 4.4a for an NPN-transistor, is generally used in switching applications. The typical input characteristics of base current 18 , against base-emitter voltage VnE, are shown in Figure 4.4b. Figure 4.4c shows the typical output characteristics of collector current le, against collector-eminer voltage Ve£· For a PNP-transistor, the polarities of all currents and voltages are reversed. There arc three operating regions of a transistor: cutoff, active. and saturation. In the cutoff region, the transistor is off or the base current is not enough to tum it on and both junctions are reverse biased. In the active region, the transistor acts as an ampli· lier, where the base current is amplified by a gain and the collector-emitter voltage decreases with the base current. The CBJ is reverse biased, and the BEJ is forward biased. In the saturation region, the base current is s ufficiently high so that the collector-emitter voltage is low, and the transistor acts as a switch. Both junctions (CBJ and BEJ) arc forward biased. The transfer characteristic, which is a plot of VeE against 18 , is shown in Figure 4.5. The model of an NPN-transistor is shown in Figure 4.6 under large-signal de operation. The equation relating the currents is I,:, = l e
+
18
(4.1)
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Bipolar Junction Transisto"
125
v
Vce1 CE2
. "~T"· +
Ra la Vo
Vni::
+
le
'------'-
O
(a) Circui1 diagram
VaE
(b) lnpul characteri.iics
le l1:1n >
------,,..
le1 > loo
FIGURE 4.4
Charac1eris1ics or 1\fPA'-transi:stors. VC'E
Cutoff '-Acli\'e--'Vee
Saturation
' I I
I I
I I I
I I
I I
V CE(o~
L-_-_-_-...;i~---_-- --_-_-_-_-_-_-_,!'--========- I8
,.,
o----,,--------+------~v6 E 0.5
FIGURE 4.5
Transfer characteristics.
The base current is effectively the input current and the collector current is the output current. The ratio of the collector current l e, to base current 18 , is known as the forward current gain, 13F: 13,.. =
Ji FE
le =. 18
(4.2)
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Chapter 4
Power Transistors
c le l cro
B
•• IE
FIGURE 4.6 E
Model ot NPN-transislon.
The collector current has two components: one due to the base current and the other is the leakage current of the CBJ. (4.3) where ICEo is the collector-to-emitter leakage current with base open circuit and can be considered negligible compared to PF/a, From ·Eqs. (4.1) and (4.3),
I,; = Ia(l + PF) + ICEo
(4.4)
"' ls(l + PF) I,;"' Because PF
/e (l +
(4.4a)
1 _!_) = le !3,· + PF PF
(4.5)
>> 1, the collector current can be expressed as (4.6)
le "' aFlc where the c9nstant aF is related to J3 by
(4.7) or (4.8)
Let us consider the circuit of Figure 4.7, where the transistor is operated as a switch.
Is
= Va -
VaE
Ve
= VCE = Vee -
(4.9)
Rs .
~-~+~
IeRc
= Vee
ppRe
- ~(Va - Va£l B
~·~
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4.2 Bipolar Junction Transistors
127
Re le
'• +
v.
Rn +
+
Vee
-+-
YcE
v••
le
FIGURe 4.7
Transistor S\\'llch.
or (4.11) Equation ('4.11 ) indicates that as long as VCE 2: V8£, the CBJ is reverse biased and the transistor is in the active region. The maximum collector current in the active region, which can be obtained by setting Vc 8 = 0 and V8 E = VcE, is Vee - VeE
le..,=
Vee - VaE
Re
Re
(4.12)
and the corresponding value of base current
I
_ lc.11
BM -
13F
(4.13)
If the base current is increased above / 8M, V8 E increases, the collector current increases, and the VCE falls below VaE· This continues until the CBJ is forward biased with V8c of about 0.4 to 0.5 V. The transistor then goes into saturation. The transistor saturation may be defined as the point above which any increase in the base current does not increase the collector current significantly. In the satura tion, the collector current remains almost constant. If the collector-emitter saturation voltage is Vc£1..,,, the collector current is' _ Vee - VeE(sa1) 1cs Re
(4.14)
and tJ1c corresponding v3lue of ba.sc c urrent is
Ics las= -
13F
(4.15)
Normally. the circuit is designed so that 18 is higher than I BS· The ratio of I 8 to I as is called the overdrive facto r (ODF): ODF=~
l as
(4.16)
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Chapter 4
Power Transistors
and the ratio of !cs to I 8 is called as forced 13, 13ror
13corccd
fcs
= ls
(4.17)
The total power loss in the two junctions is
Pr = VnEIB +
(4.18)
VCE/c
A high value of ODF cannot reduce the collector-emitter voltage significantly. However, Va£ increases due to increased base current, resulting in increased power loss in theBEJ.
Example 4.1
Finding the Saturation Parameters of a BJT
.The bipolar transistor in Figure 4.7 is specified to have llFin the range of 8 to 40. The load rcsis· tance is Re = 11 n. The de supply voltage is Vee = 200 V and the input voltage to the base circuit is V8 = 10 V. If Vc£(g t) = 1.0 V and VnE(u t) = 1.5 V, find (a) the value of Rs that results in saturation with an ODF of 5,(b) the f3tortt<1• and (c) the power loss Pr in the transistor.
Solution
Vee= 200V, 13.,.;n = 8, P,,.., = 40. Re= 11 n. ODF = 5, Vs a IOV. Vet(m) = 1.0V, and Vst(u•) = 15 V. From Eq. (4.14), fcs = (200 - 1.0)/11 = 18.1 A. From Eq. (4.15), i1,s = 18.1/!3,.10 = 18.118 = 2.2625 A. Equation (4.16) gives the base current for an overdrive factor
of 5, Is
=S x
2.2625
=
I l.3125 A
a. Equation (4.9) gives the required value of Rn, R = Vs - Vst (u•) = 10 - 1.5 = 07514 0 8
'•
11.3125
.
b. FromEq.(4.17).f3,.,..d m 18.1/11.3125 • 1.6. c. Equation (4.18) yields the total power loss as
Pr = 1.5 X
11.3125 + 1.0 x 18.I
G
16.97 + 18.1
G
35.07 w
Note: For an ODF of 10, ls = 22.265 A and the power loss is Pr = 1.5 x 22.265 + 18.1 = 51.5 W. Once lhe transistor is saturated, the collector-emitter voltage is not reduced in relation to the increase in base current. However, the power loss is in· creased. At a high value of ODF, the transistor may be damaged due to thermal runaway. On the other hand, if the transistor is underdriven (/ 8 < l c8 ), it may operate in the active region and Ve£ increases. resulting in increased power loss.
4.2.2
Switching Characteristics A forward-biased pn-junction exhibits two parallel capacitances: a d epletion-layer capacitance and a difrusion capacitance. On the· other hand, a reverse-biased p11-junc1ion has only depletion capacitance. U nder steady-slate conditions. these capacitances do
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4.2 Bipolar Junction Transistors
129
i\.
in
B
'"'
c \'b
c
c,,.
ro = 'a
B
'"'
+
c...,
v...,
T. oi· g,. •
i1: E (a) ~ 1odel with current gain
c
c,.
,, \'t,t
(b) ~1ode l with lransconductancc
FIGURE 4.8
Transient model of OJT.
not play any role. However, under transient conditions. they influence the tum-on and tum-off behavior of the transistor. The model of a transistor under transient conditions is shown in Figure 4.8, where C,,, and c,,.. are the effective capacitances of the C BJ and BEJ. respectively. The trtmsw11,i11c11mce, g,... of a BJT is defined as the ratio of Ille to AV8 c. These capacitances are dependent on junction voltages and physical construction of the transistor. C"' affects the input capacita nce significant!)' due to the Miller multiplication effect (6). lhe resistances of collector to emitter and base to emitter. are ' " and ,,,., respectively. Due to internal capacitances. the transistor does not tum on instantly. Figure 4.9 illustrates the waveforms and switching times. As the input voltage v8 rises from zero to V1 and the base current rises to I 81 . the collector current does not respond immediately. Titere is a delay, known as delay time. rd before any collector current flows. This delay is required to charge up the capacitance of the BEJ to the forward-bias voltage . V8 r. (approximately 0.7 V). After this delay; the collector current rises to the stcadystate value of fcs· The rise time 1, depends on the time constant determined by BEJ capacitance. The base current is normally more than that required to saturate the transistor. As a result, the excess minority carrier charge is s tored in t{e base region. The higher the ODF, the greater is the amount of extra charge stored in the base. This extra charge, which is called the sn111rn1i11g charge, is proportional to 1he excess base drive and the corresponding current I ,: I , = In -
l cs
13 =
ODF · lns - I /JS= lus( ODF - 1)
(4. 19)
and the saturating charge is given by
Q,
= TJ, = T, IBS(ODF -
I)
(4.20)
where Ts is kno,vn as the s torage ti,ne constant of the lransistor. When the input voltage is reversed from V1 to - V2 and the base current is also changed to -/82, the collector current does not change for a time 1,, called the .<1orage limc. lhe r, is required to remove the saturating charge from the base. Because u8 c is s till positive with approximately 0.7 V o nly, the base current rc,•crses its di rection due
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Chapter 4
Power Transistors
,. v, ----0 kT
-v,
( 1 - k)TI I
o~~~4-~~~~~~~~~,1-~~~~-~-'--'
- Iu!
.------+----------------- ·1---r/ O t
~· le~ _____
O.CJ1t'!i
I I·
J_____ I
- ------------r--- -
----- : ---I
I I
t I
I I
I
I I
0.11(0
I I
I I
I
I
-- -- -j- -- ~------ - ------~----:-l11
,,
t,
FIGURE 4.9
Switching limes of bipolar transistors.
10 !he change in !he polarily of v8 from Vi 10 - Vi. The reverse currenl, - /Bl, helps 10 discharge the base and remove !he extra charge from the base. Without - 182, the satu- · raling charge has lo be removed enlirely by recombination and the slorage lime would be longer. Once the extra charge is removed, !he BEJ capacitance charges 10 the input voltage - Vi , and !he base currenl falls 10 zero. The fall lime tr depends on lhe time constant, which is determined by the capacitance of the reverse-biased BEJ. Figure 4.lOa shows the extra storage charge in the base of a saturated transistor. During tum-off, this extra charge is removed first in 1ime·t, and !he charge profile is changed from a 10 c as shown in Figure 4. lOb. During fall lime, !he charge profile decreases from profile c unlil all charges are removed. The !Um-on lime 100 is !he sum of delay time Id and rise time.I,:
and the tum-off time 10 ,r is the sum of storage time 1, and fall time If fofJ ;:: fs
+
tr
Copyrghtcd matcria
4.2 Bipolar Junction Transistors Base
E mitter
131
C0 Ucc1or
(a) Charge SIOrage in ba.se
(b) Charge profile.during turn-off
FIGURE 4.10
Charge storage in saturated bipolar 1ransis1ors.
Example 4.2
Finding the Switching Loss of a BJT
The waveforms of the transistor switch in Figure 4.7 are shown in Figure 4.11. The parameters are Vee a 250 V, Vs, c~•> a 3 V. l o a 8 A. Vcs(..,) = 2 V, l cs = 100 A, '• = 0.5 )l,S, t , = I µs, ,., • 5 fl.S. ,1 e 3 µ.s. and f, e 10 kHz. The duty cycle is k = 50%. The collector-to-emitter leakage current is few = 3 mA. Determine !he power loss due 10 collector curren1 (a) during turnon '•• + '" (b) during conduciion period ,•• (c) during turn-off to1r = t, + tr, (d) during off-lime r.,. and (e) 10131 average power losses Pr,({) PI011he insianlaneous power due 10 collec· tor current P,.(r ). 0
,,
Solution T • llf, • 100.)1,S. k • 0.5. kT • t, + t, + '• • 50 µ.s. '• = 50 - 0.5 - 1 = 48.5 µs. ( I k )T = r, + r1 + 10 = 50 fl.S, and 10 = 50 - 5 - 3 = 42 fl.S.
a. During delay time. 0 :s t s ltt:
i,(t) = few Vce(t )
= Vee
The instantaneous po\ver due to the collector current is
P.,(t} = i,vc, = lcwVcc = 3 x 10-J x 250 • 0.75 W The average power loss during the delay time is Pd -
TI J['' o Pr(1),1t
- lct:0Vcc1df,
(4.21)
= 3 x 10- 3 x 250 X 0.5 x IO .. X IO X During rise time. 0
:G t
s
JO'= 3.75mW
t,:
lcs
,,
i,.(t ) = -
t
vc,(t . ) = Vee+
CVa,,., . 1-
I
Vee) -t,
Copyr ghtcd matcria
132
Chapter 4
Power Tran.sistors
Vee Vee
V cE(,.,1)
0
le,
,-,----~-- -~ - !\.
'c
I I
I I I I I
l cEO 0
,..
t--,.,,~
. f--1.,--t
•• ,,
,,
,,
I I
,,
'•
ie
1-- - - - --
T = 11~-------l
•I I
I I VU E(1>tt)
FIGURE 4.11
W• 1.·c(orms or ua n,inor switc.h.
P, (1) = i,vcr. = lcs !.. [vcc + (Vc£(mJ - Vee)!..] I,.
I,
(4.22)
The power Pc(t) is maximum v.·hcn t = t.,,. where c,Vcc r,.• ---'-=-2(Vcc - VcE (,.,,]
=
250 I x ( 0_ ) 2 25 2
(42.3)
= 0.504 µs
and Eq. (4.22) yields the peak power
Vl·clcs
p = -=.,-=:=--e ' 4(Vcc - Vcr.(u1)) _
,
100
(4.24) _
- 250 x 4(250 - 2) - 6300 w
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4.2 Bipolar Jun,tion Transistors I
P, = -
7
1·
[Vee + Vcco") - Vee]
P,(1)d1 = f ,lcsr, , _
0
= 10 x 10·' x 100 x 1 x
3
io-<>[ 2~0
+
2
-/
50
]
133
(4.25)
= 42.33 W
The total pO\\'Cr loss during the 1urn·on is
+ P, = 0.00375
Pon = I~,
(4.26)
+ 42.33 = 42.33 W
b. 1be conduction period. 0 :s; 1 :s; r,,: i,.(1} = lcs va(1}
= Vcl:(,,,)
P,(r) =
= \'cc(utllcs
it. lJCt:
=2X l00 =200 W P,,
1 ('•
= Tlo P,(1)<11 = Va
(4.27} 3
2 X 100 X 48.5 X 10-6 X 10 X 10
c:. The storage period, 0 s
t
•
97 \V
:s ,,:
i,(1) = lcs VC£(1) = Va(w1)
~.(,) = i/Vct: = Vr,:c ...11)/cs = 2 x 100 = 200 W
P, =
l be fall time, 0 :,;
f" P, (1)<11 = Vac,")lcs1J, = T1 lo
(4.28)
2 x 100 x 5 x io-<, x 10 x 10·' = IO W
1 :,;
'I'
i,(1)
=
/cs( I - t). neglecting lc,:o Vee
.
(4.29}
11c,(1) = - ,- r. neglecting fem I
P, (r) - i,.vc,: - l'cc-lcs [ ( I -
~) ~]
Tiiis power loss during fall cime is maximum when 1 = ,1n = 1.5 µs and Eq. (4.29)
gives the peak po "·cr.
Vcclcs Pi,,=-4-
100
= 250 X -
4
(4.30)
= 6250 W
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134
Chapter 4
Power Transistors I [''
P1 = Tlo P, (1)d1
= 250 x
=
Vc,·fcst1f ,
6
(4.3 1)
100 x 3 x 10"" x 10 x 10-' 6
= 125 W
The power loss during tum-off is
P.,, = P, +Pt = fcsf, (1,Vcc(.,,1 + Ve;'/)
(4.32)
= 10 + 125 = 135 W cl. Off-period, 0 :S t ::;; 1,,: i,(1) • l cco vcc(t) = Vee
P,(1) • i, vCE • f cEOVcc = 3 x 10-J x 250 = 0.75 W
(4.33)
I ['•
Po = Tlo P,(t )dt = l cEOVcctJ, ,,;. 3 x 10-J x 250 x 42 x 10"" x 10 x 10-' = 0.315 W e. The total power loss in the transistor due to collector current is
Pr
Pon + P,, + P0 r, + Pt1 = 42.33 + 97 + 135 + 0.315 = 274.65 W =
(4.34)
r. The plot of the instantaneous power is shown in Figure 4.12
-r' n2001
o.10~sf
I
A..
IL.. r l \ I------------------------------~--- l l
I I
I,) ' ,, I
'·
,,
.I
t lr
FIGURE 4.12
Plot or instantaneous power tor Example 4.2.
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4.2 Bipolar Junction Transistors
Example 4.3
135
Finding the Base Drive Loss of a BJT
For the parameters in Example 4.2. ca.lcuh:ue the averasc po,,•cr loss due to the base current.
Solution Vor.(,,..) = 3 v, ' • = 8 A . T s l lf, c 100 µs . k 9 0.5. kT = so µs. 1, = 0.5 µ.S. 1, = I µs. ,,, = so - l.S = 48.Sµ.s, '~ =- Sµ.s, ,, = 3 µ.s. , Cll\ = Id+ Ir = L.5 µ.s. and '"'' = ,, + ,, ~ 5 +3=8µ.s. During the period. 0 s I s (10 ,
+ r.,): i.(1)
= los
Vot{I) = V11c(,,.) The insrnntaneous power due to the base current is
P;(t ) = i,vor. = lnsV11sc,.,, =8X3=24W During the period. 0 s t S 10 Pn
=
(T - 100
l n.sVnF.(~1)(lon + '"
• 8
4.2.3
•
x3X
-
t., - t, - t;): P,(1) • 0. The average power loss is
+ t , + ti )/,
( 1.5 + 48.S + 5 + 3)
x 10-6 x
10
x
10·' • 13.92 W
(4.35)
Switching Limits Second breakdown (SB). The SB, which is a destructive phenomenon, results from the current flow to a small portion of the base, producing localized hot spots. If the energy in these hot spots is sufficient , the excessive localized heating may damage the transistor. Thus, secondary breakdown is caused by a localized thermal runaway, re· suiting from high current concentrations. 'Ille currcnl concentration may be caused by defects in the transistor structure. The SB occurs at certain combinations of voltage, current, and time. Because the time is involved, the secondal')' breakdown is basically an energy dependent phenomenon. Forward-biased safe operating area (FBSOA). During turn-on and on-state conditions. the average junction temperature and second breakdown limit the power· handling capability of a transistor. The manufaclurers us ually provide the FBSOA curves under specified test conditions. FBSOA indicates the i,-vCE l.imits of the transistor; and for reliable operation the 1ransistor must not be subjec1ed to greater power dissipation than that shown by the FBSOA curve. Reverse-biased safe operating area (RBSOA). During turn-off, a high current and high voltage must be sustained by the tra nsistor, in most cases with the basc-toemitter junction reverse biased. The collector-emitter voltage must be held to a safe level at, or below, a specified value of collector current The manufacturers pro vide lhc le- Vee limits during reverse-biased turn-off as RBSOA. Power derating. The lhcrmal equivalent circuit is.shown in Figure 4. 13. If the tolal average power loss is Pr, the case temperature is
Tc
= 7i
- P.r R1c
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136
Chapter 4
Power Transistors
..--.w~---v'vv~--.r, RK'
R<"<
P·r FIGURE 4.13 Thermal e,1uivalent circuit o( a trdnsi.sl(,r.
The sink temperature is
The ambient temperature is T,1 = Ts - PrRs,1
and (4.36)
where R1c = thermal resistance from junction to case. 0 CIW; Res = thermal resistance from case to sink, 0 C/W; RsA = thermal res istance from sink to ambient, 0 CIW. The maximum power dissipation
Pr is normally specified at Tc
= 25°C. If the
ambient tempera ture is increased to T,. = 7"1 == t50°C, the transis to r can dissipate
zero power. On the other hand, if the junction lemperature is Tc =
o•c. the device can
dissipate 1naximurn po,vcr and this is not practical. Therefore. the an1bicnt tcn1perature
and thermal resistances must be considered when interpreting the ratings of devices. Manufacturers show the dcrating curves for the thermal derating and second break· down derating. Breakdown voltages. A breakdoiv11 voltage is defined as the absolute maximum voltage bel\vecn two terminals with the third terminal open, shorted, or biased in ci · ther forward or reverse direction. At breakdown the voltage remains relatively con· slant, where lhe current rises rapidly. The following brcukdown-voltagcs arc quoted by the manufacturers: V,: 80 : the maximum voltage between the emitter temtinal and base terminal with collector terminal open circuited. Vc£v or Vccx : the maximum voltage between the collector terminal and emitter terminal a t a specified negative voltage applied between base and emitter. Vcro(sus): the maximum sustaining voltage between the collector temtinal and emitter terminal with the base open circuited. 'Ibis rating is specified at the maximum collector current and voltage, appearing simultaneously across the device with a specified value of load inductance.
Let us consider the circuit in Fi_gure 4.14a. When the switch SW is closed, the col· lcctor current increases, and after a transient. the stcady·state ~ol1ector current is
Copyrghlcd matcria
4.3 Power MOSFETs
Le
+
137
le
le,
c
ll D
''
Re
Pure / ',
rcsisti\'e load
V (X'
0
(a)Tcst circuil
''
''
''
A
Yee
Vc-·E VC'EfsuH
(b) Load lines
FIGURE 4.14
Turn-on and 1um-off load lines.
l es = (Vee - Vcc(,.,J)IRe. For an inductive load. the load line would be the path ABC shown in Figure 4.l 4b. If the switch is opened to remove the base current, the collector current begins to fall and a voltage of L(diltit) is induced across the inductor 10 oppose the current reduction. The transistor is subjected to a transient voltage. If this voltage reaches the sustaining voltage level, the collector voltage remains approximately constant and the collec1or curren1 falls. After a short time, 1he 1rnnsistor is in the off-state and the turn-off load line is shown in Figure 4.14b by the path CDA.
Finding the Case Temperature of a BJT
Example 4.4
The maxin1um junction ten1perature of a transisto r is T1 -=- 150°C and the ambient ten1peraturc is TA = 25°C. U the thermal impedances arc R,c = 0.4°C/W, Res = O.t•Cfw. and R.u = 0.5°C/\V, calculate (a) the n1axin1um p<>\\'e r dissipa tion. and (b) the case te1npera1ure.
Solution ·
a.
r, -
T,. "' P7 (R,c + Res+ Rs,.) "' P7 R1 A- R1 ,. = 0.4 + 0.1 + 0.5 = 1.0,
= I.OP, . ·which gives the mu,.imun1 po,,·c r dissipation as Pr = b. Tc "' T1 - PrR1e = 150 - 125 x 0.4 "' 100°C. and 150 - 25
4.3
125 \ V.
POWER MOSFETs
A BJT is a current-controlled device and requires base currcnl for current now in the collector. Because the collec1or current is dependent on the input (or base) currcn1. the current gain is highly dependent on the junction temperature. A power MOSFET is a voltage-conirolled device and requires only a small input current. The switching speed is very high and the switching times are of the order of nanoseconds. Power MOSFETs find increasing applications in low-power high· frequency converters. MOSFETs do not have the problems of second breakdown phenomena as do BJTs. However, MOSFETs have the problems of electrostatic discharge and require special care in handling. In addi1ion, it is relatively difficult to prolect them under short-circui1ed fault conditions.
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138
Chapter 4
Power Transistors
Metal s\lbstrate
Ri
p-type substrale
T'·· u +
lu
VG,
Basic structure (a) n-Channel deple1ion-type MOSFET
0
s
lo
Ro Voo
+
Symbol
lo D ,-J
G
--= ~Vo, +
s
p' p
.•
I n-type subiarntc
...,- 0,anncl
p'
I
,,.,.MelaI '> '> '>
-
- ~ Yoo -= +
O
lo
u Vo,
s
+ Basic structure (b) p-Channel depletion-lype MOSFET
+
Symbol
FIGURE 4.15
Depleiion-type MOSFET&
The two types of MOSFETs are (I) depletion MOSFETs, and (2) enhancement MOSFETs [HJ. An 11-channel depletion-type MOSFET is formed on a p-type silicon substrate as shown in Figure 4.15a, with two heavily doped 11+ silicon for low-resistance connections. The gate is isolated from the channel by a thin oxide layer. The three terminals are called gate, drai11, and source. The substrate is normally connected to the source. The gate-to-source voltage Yes could be either positive or negative. If Yes is negative. some of the electrons in the 11-channel area are repelled and a depletion region is created below the oxide layer, resulting in a narrower effective channel and a high resistance from the drain to source R 0 5. lf Ye s is made negative enough, the channel becomes completely depleted, offering a high value of Ros, and no current flows from the drain to source, Ios = 0. The value of Ye s when this happens is called pinchoff voltage Yr, On the other hand, Yes is made positive, the channel becomes wider, and / 05 increases due to reduction in Ros· With a p-channel depletion-type MOSFET, the polarities of Yos, los, and Yes arc reversed as shown in Figure 4.15b. An 11-channel enhancement-type MOSFET has no physical channel, as shown in Figure 4.16a. If YGs is positive, an induced voltage attracts the electrons from the
.
\
Copyrighlcd m~leria
4.3 Power MOSFETs
Met•I!bstrale
D Meta.I
RD
p-Type
0
139
substrate
+
D
Yoo
lo
s Oxide Basie structure (a) n-Channcl cnhancemenl-lypc MOSFET
D Mctnl
n-Type ,ub5tratc
0
Ll
s Oxide
Vn, +
s
Voo
+
Basic structure Symbol (b) p·Channcl cnhanccmenMypc MOSFET FIGURE 4.16
Enhanccmenl-lype MOSFETs.
p·subslrale and accumula1e them at the surface beneath the oxide layer. If VGs is greater than or equal to a value known as threshold voltage Vr, a sufficient number of electrons are accumulated 10 form a virtual n-<:hannel and the current flows from the drain to source. The polarities of V05 , 105, and Vcs are reversed for a p-channel enhancement-type MOSFET as shown in Figure 4.16b. Power MOSFETs of various sizes are shown in Figure 4.17. Because a depiction MOSFET remains on at zero gate voltage ,vhcrcas an en·
hancement type MOSFET remains off at zero gate voltage, the enhancement 1ype MOSFETS are generally used as switching devices in power e lectronics. The cross section of a power MOSFET known as a vertical (V) MOSFET is shown in Figure 4.18a. When lhe gate bas a su((icicntly positive voltage with respect lo the source, the effect of its electric field pulls electrons from the 11+ layer into the p layer. This opens a channel closest to the gate, which in turn allows the current to flow fro.rn the drain to lhe source. There is a silicon oxide (SiO) dielectric layer between the gate metal and the 11+ and p junction. MOSFET is heavily doped on rhe drain side to create an 11+ buffer below 1he 11-drift layer. 1l1is buffer prevents the depletion layer from reaching
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140
Chapter 4
Power Transistors
FIGURE 4.17
Po~'Cr MOSFET~ (Courtesy of International Rectifier.)
the metal, evens out the voltage stress across the n layer, and also reduces the forward voltage drop during conduction. Toe buffer layer also makes it an asymmetric device with rather low reverse voltage capability. MOSFETs require low gate energy, and have a very fast switching speed and low switching losses. The input resistance is very high, 109 to 1011 n. MOSFETs, however, suffer from the disadvantage of high forward on-state resistance as shown in Figure 4.18b, and hence high on-state losses, which makes them less attractive as power devices, but they are excellent as gate amplifying devices for thyristors (see Chapter 7).
Source
...
510
p
p D
Gate
SourceS()
Gate
-
epitaxy
-•
I/
...L\/ .-
I/
~
R..,
n- q,i
n• epit.uy n+,11h
Drain (a) Cross section o!V-MOSFET
R,.., \.)D
Drain (b) On-state $Cries resistances o!V-MOSFET
FIGURE 4.18
Cross seaions of MOSFETs. lRef. 10.G. Deboy)
Co yr h1 1
ma•eric1
4.3 Power MOSFETs 4.3.1
141
Steady-State Characteristi~ The MOSFETs arc voltage-controlled devices and have a very high input impedance. The gate draws a very small leakage current, on the order of nanoamperes. The current gain, which is the ratio of drain current 10 , to input gate current l a, is typically on the order of 109. However, the current gain is not an important parameter. The tra11sco11d11cta11ce, which is the ratio of drain current to gate voltage, defines the transfer characteristics and is a very important parameter. The transfer characteristics of 11-channel and p-cbannel MOSFETs are shown in Figure 4.19. Figure 4.20 shows the output characteristics of an 11-channel enhancement MOSFET. There are three regions of operation: (! ) cutoff region, where Vas :s V7 ; (2) pinch-off or saturation region, where V0 s .e: Vas - Vr; and (3) linear region, where Vos s Vas - V7 . The pinch-off occurs at Vos = Vas - V7 . In the linear region, the drain current / 0 varies in proportion to the drain-source voltage V0 5. Due to high drain current and low drain voltage, the power MOSFETs are operated in the linear region for switching actions. In the saturation region, the drain current remains almost constant for any increase in the value of Vos and the transistors are used in this region for voltage amplification. It should be noted that saturation has the opposite meaning to that for bipolar transistors. The steady-state model, which is the same for both depletion-type and enhancement-type MOSFETs, is shown in Figure 4.21. The transconductance g,,, is defined as !J./D Cm=--
I
(4.37)
AVcs vru•cocuLant
v, - - - -o : :-r----,~-v
v,
0
~ " " ' - - - - - " ' ' - - - Yo,
n-.channel
p-channcl
(a) Depiction-type MOSFET
v,. -----::..-i,o;:--vc.
lo
O VT _......;..L.....;...c_ ___
- lo
v(~
n-cht'lnnel
p-
(b) Enhanccmcnt-t)'p< MOSFET FIGURE 4.19
Thlnsfer choractcristics of MOSFETs.
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142
Chapter 4
Power Transistors
Linear
. rc-gion
lo
_,-,:._.-- .
Pinch-of£ region or saturation region Vo,.c
> Vou > Vou > Vo,1 > Vy
----- v,., FIGURE 4.20
Ou1put characteristics of enhancement-type: MOSFET.
The output resistance, r0
0
= Ros, which is defined as .I. Vos Ros=--
(4.38)
f>lo
is normally very high in the pinch-off region, typically on the order of megohms and is very small in the linear region, typically on the order of milliohms. For the depletion-type MOSFETs. the gate (or input) voltage could be either positive or negative. However, the enhancement-type MOSFETs respond to a positive gate voltage only. The power MOSFETs are generally of the enhancement type. However, depletion-type MOSFETs would be advantageous and simplify the logic design in some applications that require some form of logic-compatible de or ac switch that would remain on when the logic supply falls and Vcs becomes zero. The characteristics of depletion-type MOSFETs are not discussed further.
4.3.2
Sw itching Characterist ics Without any gate signal, an enhancement-type MOSFET may be· considered as two diodes connected back to back or as an NPN-transistor. The gate structure has parasitic
Ro
Ro
D
lo
D
~
...
+
VG
Ve,
+
+ Yo,
s
(a) Circuit diagram
s (b) Equivalent cirwit
FIGURE 4.21
Steady-slntc switching model of MOSFETs.
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Power MOSFETs
4.3 D
143
D
lo
lo
s
s
(a) Parasitjc biploar
(b) lnterlllll diode
RGURE 4.22
Parasitic model or enhancement or MOSFETs.
0 ~ 1-- + - --+-- ---+---<>D 10
C1:b
Cp
VP
rob
FIGURE 4.23
s v,
v,
Vo
Vo,
I
I
------~----j J---- -......------!-. r - - - - - - - - ----~
V 0sp - - - - -
Yr -----rI O
Switching model of MOSFET~
I
I
1
,
I
1+-t-,...
1'"' 1
ld(oa)
FIGURE 4.24
Swilching waveforms and limes.
capacitances to the source, C,,. and to the drain, C,J· The npn-transistor has a reverse-bias junction from the drain to lhc source and offers a capacitance, Cd,. Figure 4.22a shows the equivalent circuit of a parasitic bipolar transistor in parallel with a MOSFET. The base-to-emitter region of an NPN-transistor is shoned at the chip by metalizing the source terminal and the resistance from the base to emitter due to bulk resistance of 11and p-regions. Ri,,, is small. Hence, as MOSFET may be considered as having an internal diode and the equivalent circuit is shown in Figure 4.22b.The parasitic capacitances arc dependent on their respective voltages. The switching model of MOSFETs is shown in Figure 4.23. The typical switching waveforms and times are shown in Figure 4.24. The mm-on delay ,.,..,1 is the
Copyr ghtcd matcria
144
Chapter 4
Power Transistors
time that is required to charge the input capacitance to threshold voltage level. The rise/ me 1, is the gate-charging time from the threshold level to the full-gate voltage VcsP , which is required to drive the transistor into the linear region. Toe tum-off delay time IJ(off) is the time required for the input capacitance to discharge from the overdrive gate voltage Vi to the pinch-off region. Vcs must decrease significantly before V05 begins to rise. Toe fa/I time 11 is the time that is required for the input capacitance to discharge from the pinch-off region to threshold voltage. If Vos :S V7 , the transistor turns off. 4.4
COOLMOS
COOLMOS (9-11), which is a new technology for high voltage power MOSFETs, implements a compensation structure in the vertical drift region of a MOSFET to improve the on-state resistance. It has a lower on-state resistance for the same package compared with that of other MOSFETs. Toe conduction losses are at least five times less as compared with those of the conventional MOSFET technology. It is capable of handling two to three times more output power as compared with that of the conventional MOSFET in the same package. The active chip area of COOLMOS is approximately five times smaller than that of a standard MOSFET. Figure 4.25 shows the cross section of a COOLMOS. The device enhances the doping of the current conducting n-dopcd layer by roughly one order of magnitude without altering the blocking capability of the device. A high-blocking voltage V 8 R of Source
s
••
•• p
n-~i
n • nJb
FIGURE 4.25
Cross sc.·i·1ion or COOUiOS.
D
Drain
Copyrighted m~leria
4.5 SITs
145
the transistor requires a relative thick and low-doped epitaxial layer leading to the weU-known law (12) that relates the drain to source resistance to VaR by (4.39)
where kc is a constant between 2.4 and 2.6. 'This limitation is overcome by adding columns of the opposite doping type that are implemented into the drift region in a way that the doping integral along a line perpendicular to the current flow remains smaller than the material specific breakthrough charge, which for silicon is about 2 x 1012 cm·2• This concept requires a compensation of the additional charge in then region by adjacently situated p-doped regions. These charges create a lateral electric field, which does not contribute to the vertical field profile. In other words, the doping concentration is integrated along a line perpendicular to the interface created by the p- and n-regions. Majority carriers provide the electrical conductivity only. Because there is oo bipolar current contribution, the switching losses are equal to that of conventional MOSFETs. The doping of the voltage sustaining the layer is raised by roughly one order of magnitude; additional vertical p-stripes, which are inserted into the structure compensate for the surplus current conducting n-charge. The electric field inside the structure is fixed by the net charge of the two opposite doped columns. Thus, a nearly horizontal field distribution can be achieved if both regions counterbalance each other perfectly. The fabrication of adjacent pairs of p- and n-doped regions with practically zero net charge requires a precision manufacturing. Any charge imbalance impacts the blocking voltage or the device. For higher blocking voltages only the depth of the columns has to be increased without the necessity to alter the doping. This leads to a linear relationship (10) between blocking voltage and on-resistance as shown in Figure 4.26. The on-state resistance of a 600-V, 47-A COOLMOS is 70 mn. The COOLMOS has a linear v-i characteristic with a low-threshold voltage (10). The COOLMOS devices can be used in applications up to power range or 2 kVA such as power supplies for workstations and server, uninterruptible power supplies (UPS), high-voltage converters for microwave and medical systems, induction ovens, and welding equipmcnL These devices can replace conventional power MOSFETs in all applications in most cases without any circuit adaptation. At switching frequencies above 100 kHz, COOLMOS devices offer a superior current-handling capability such as smallest required chip area at a given current. The devices have the advantage or an intrinsic inverse diode. Any parasitic oscillations. which could cause negative under· shoots of the drain-source voltage, are clamped by the diode to a defined value.
4.5
SITS An SIT is a high-power, high-frequency device. Since the invention of the static induction devices in Japan by J. Nishizawa (17), the number of devices in this family is growing (19). It is =ntially the solid-state version of the triode vacuum tube. The silicon er= section of an SIT (15) and its symbol are shown in Figure 4.27. lt is a vertical structure device with short multichannels. Thus, it is not subject to area limitation and is suitable for high-speed, high-power operation. The gate electrodes are buried within the drain and source 11-epi layers. An SIT is identical to a JFET except for vertical and buried gate construction.
Copyr ghtcd matcria
146
Chapter 4
Power Transiston
20
16
< I:"'
12
x ~
Js.
8
4
0
600
200
800
1000
Breakdown voltage Vc»R)OSS (VJ AGURE 4.26
The linear relationship between blocking voltage and oo-resi>tanoe. [Rel 10. 0. Deboy)
Source
s
•
••
{b)Symbol
D Drain
(a) Cross section FIGURE 4.27
Cross section and symbol !or SJTs.
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4.G IGBTs los
0 - I -2
(mAI
-3
147
-4
600
200
200
400
600
800
FIGURE 4.28 'lyp«:al characteristics of SIT~ (Ref. 18.19(
which gives a lower channel resislance, causing a lower drop. An SIT has a short channel length, low gate series resistance, low gate-source capacitance, and small thermal resistance. It has a low noise, low distortion. and high audiofrcquency power capability. The tum-on and tum-off times arc very small, typically 0.25 µs. The on-state drop is high, typically 90 V for a 180-A device and 18 V for an 18-A device. An SIT normally is an on device, and a negative gate voltage bolds it oft The normally on-characteristic and the high on-s tate drop limit its applications for general power conversions. The typical characteristics of SITs are shown in Figure 4.28 [18). An electrostatically induced potential barrier controls the current in static induction devices. The SITscan operate with the power of 100 KVA at 100 kHz or lOVA at 10 GHz. The current rating of SlTs can be up to 1200 V, 300 A, and the switching speed can be as high as 100 kHz. It is most suitable for high-power, high-frequency applications (e.g., audio, VHF/UHF. and microwave amplifiers). 4.6
IGBTs
An JGBT combines the advantages of BJTs and MOSFETs. An IGBT has high· input impedance, like MOSFETs, and. low on-state conduction losses, like BJTs. However, there is no second breakdown problem. as with BJTs. By chip design and structure, the equivalent drain-to-source resistance Ros is controlled to behave like that of a BJT (13--14). The silicon cross section of an JGBT is shown in Figure 4.29a, which is identical to that of an MOSFET except the p• substrate. However, the performance of an !GBT is
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148
Chapter 4
Power Tran,ls1ors Collector
p• substrate n• • Buffer layer
. - epi
p• p
p
Gate
Emitter (1) Cross section
c
c
oo-J
• E (b) Equivalent
circuit
E (c) Simplified
circuit
FIGURE 4.29
Crog section and equivaleot circuit for IOBTs.
closer to that of a BJT than an MOSFET. This is due to the p• substrate, which is responsible for the minority carrier injection into the 11-region. The equivalent circuit is shown in Figure 4.29b, which can be simplified to Figure 4.29c. An IGBT is made of four alternate PNPN layers, and could latch lilr.e a thyristor given the necessary
Copyrghlcd matcria
149
4.6 IGBTs
condition: (a.,.. + Dp,,p) > 1. The n•-buffcr layer and the wide epi base reduce the gain of the NPN-terminal l;>y internal design, thereby avoiding latching. lGBTs have two structures of IGBTs: punch-through (PT) and nonpunch through (NPT). ln the PT IGBT structure, the switching time is reduced by use of a heavily doped n-buffer layer in the drift region near the collector. In the NPT structure, carrier lifetime is kept more than that of a PT structure, which causes conductivity modulation of the drift region and reduces the on-state voltage drop. An IGBT is a voltage-controlled device similar to a power MOSFET. Like an MOSFET, when the gate is made positive with respect to the eminer for tum-on, n carriers are drawn into the p-channel near the gate region; this results in a forward bias of the base of the npn-transistor, which thereby turns on. An IGBT is turned on by just applying a positive gate voltage to open the channel for n carriers and is turned off by removing the gate voltage to close the channel. It requires a very simple driver circuit. It has lower switching and conducting losses while sharing many of the appealing features of power MOSFETS, such as ease of gate drive, peak current, capability, and ruggedness. An IGBT is inherently faster than a BIT. However, the switching speed of IGBTs is inferior to that of MOSFETs. The symbol and circuit of an JGBT switch are shown in Figure 4.30. The three terminals are gate, collector, and emitter instead of gate, drain, and source for an MOSFET. The typical output characteristics of ic versus Vee are shown in Figure 4.3la for various gate-emitter voltage Vee, The typical transfer characteristic of ic versus Vee is
c
Oale signal
le Ro
, --- ------ I • I
I I
R.
G
I
I I
I I
V ee
E
R(;E
...
I
FIGURE 4.30
I I
I
Symbol and circuit for an IGBT.
L ••••••••••
ic
7
~ 6
<
-
Yoe= IOV
'::c s -
84-
E 2
9V
a 3i 2-
~
aa
SV
,
i
e
7V 6V
0
I
0
< ~
u
e
ic
3
I
.
I
.
I
Vcr
I
2 4 6 8 10 CoUector- emiuer ,·oltagc
12
I
0
VGE
0
2 4 6 Gate- emitter voltage
FIGURE 4.31
Typical output and tran.s£cr characteristics o{ IGBT$..
Copyrghtcd matcria
150
Chapter 4
Power Transistors
shown in Figure 4.31b. The parameters and their symbols are similar to that of MOSFETs, · except that the subscripts for source and drain are changed to emitter and collector, respectively. The current rating of a single lGBT can be up to 1200 V, 400 A, and the switching frequency can be up to 20 l
4.7
SERIES AND PARALLEL OPERATION lransistors may be operated in series to increase their voltage-handling capabiiity. lt is very important that the series-connected transistors arc turned on and off simultaneously. Otherwise, the slowest device at turn-on and the fastest device at turn-off may be subjected to the full voltage of the collector-emitter (or drain-source) circuit and that particular device may be destroyed due to a high voltage. The devices should be matched for gain, transconductance, threshold voltage, on-state voltage, turn-on time, and turn-off time. Even the gate or base drive characteristics should be identical. Voltage-sharing networks similar to diodes could be used. lransistors are connected in parallel if one device cannot handle the load current demand. For equal current sharings, the transistors should be matched for gain, transconductance, saturation voltage, and turn-on time and tum-off time. In practice, it is not always possible to meet these requirements. A reasonable amount of current sharing (45 to 55% with two transistors) can be obtained by connecting resistors in series with the emitter (or soun:e) terminals, as shown in Figure 4.32. The resistors in Figure 4.32 help current sharing under steady-state conditions. Current sharing under dynamic conditions can be accomplished by connecting coupled inductors as shown in Figure 4.33. If the current through Q1 rises, the L(dildt) across L 1 increases, and a corresponding voltage of opposite polarity is induced across inductor L 2• The result is a low-impedance path, and the current is shifted to Q 2• The inductors would generate voltage spikes and they may be expensive and bulky, especially at high currents. BJTs have a negative temperature coefficient. During current sharing, if one BJT carries more current, its on-state resistance decreases and its current increases further, whereas MOSFETs have a positive temperature coefficient and p_arallel operation is
1., FIGURE 4.32
+
R.,
Vee
Parallel connection of uans-lstors.
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4.8 dildt and dvldt Limitations
151
Ir
·~ ·
'Et
R.,
R.,
+
Y ee
11~z11~.
FIGURE 4.33
Dynamic currcn1 sharing.
rclati\'ely easy. Toe MOSFET that initially draws higher current beats up faster and its on-state resistance increases, resulting in current shifting to the other devices. IGBTs require special care to match the characteristics due to the variations of the temperature coefficients with the collector current. Finding the Current Sharing by Two Parallel MOSFETs
Example 4.5
Two MOSFETs that arc connected in parallel similar 10 Figure 4.32 carry a total current or I r = 20 A. The drain·IO·source yoltage ol MOSFET M1 is = 2.5 V and that of MOSFET M 2 is Von = 3 V. Determine the drain current or each transistor and difference in current shar· ing ii the current sharing series resistances arc {a) R,1 = 0.3 n and R12 = 0.2 n. and {b) R,, = R,2 = 0.5 0.
v,,,.
Solution a. I o,
+ Im
= Ir and Vos, + lo,Rs,
= V052 + 10 ,R..,,. = Von ... RnUr - l o,),
Von - Vos, + I r R,, Sim = ~~-=~-~ R,1 + R,2
= 3 - 2.5 + 20 x 0.2 = 9 A or 45%
(4.40)
0.3 + 0.2
I m • 20 - 9 • 11 A or 55% {),/ = 55 - 45 = 10%
b.
I
DI
=3 -
Z.S + ZO x 0.5 0.5 + 0.5
/ 02 = 20 - 10.5 = 9.5 A t:,,/ = 52.5 - 47.5 = 5%
4.8
= 10 5 A .
or 52.5%
or 47.5%
dVdt AND dvldt LIMITATIONS Transistors require certain turn-on and tuin-off times. Neglecting the delay time td and the storage time,,, the typical voltage and current waveforms of a BJT switch are
Copyr ghtcd matcria
152
Chapter 4
Power Transistors
01----i---..,__ _ _ _ __ _ __,..__..__ _ _ _
IL
I
I
:
I
I I I I I
I I I I I
~
-----~---~·- -- ------...
FIGURE 4.34
Voltage and current wa\•cforms.
shown in Figure 4.34. During tum-on, the collector current rises and the dildt is di h - =dt I,
= -le,I,
(4.41)
During turn-off, the collector-emitter voltage must rise in relation to the fall of the collector current, and d vldt is -dv = -v, = -v" dt
,, ,,
(4.42)
The conditions dildt and dvldt in Eqs. (4.41} and (4.42) are set by the transistor switching characteristics and must be satisfied during turn-on and turn-off. Protection circuits are normally required to keep the operating dildt and dvldt within the allowable limits of the transistor. A typical transistor switch with dildt and dvldt protection is shown in Figure 4.35a, with the operating waveforms in Figure 4.35b. The RC network across the transistor is known as the snubber circuit, or snubber, and limits the dvldt. Toe inductor L,, which limits the dildt, is sometimes called a series s,mbber. Let us assume that under steady-state conditions the load current h is free· wheeling through diode Dm, which has negligible reverse recovery time. When transistor Q1 is turned on, the collector current rises and current of diode Dm falls, because Dm behaves as short-circuited. The equivalent circuit during turn-on is shown in Figure 4.36a and turn-on dildt is di d(
-
G
-
v,
L,
(4.43)
Equating Eq. (4.41) to Eq. (4.43) gives the value of L,. (4.44)
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4.8 dildt and dvldt Limitations
153
+
o.. ir
L
v,
(b) Waveforms
(a) Protection circuits FIGURE 4.35
Transis.tor switch with dVdt and d ti/dt protection.
+
L.
-·L
+ \'
•
v.
0
(a) Mode I
(b) Mode 2
(c) Mode 3
FIGURE 4.36 Equ.ivaleat ciTCUilS.
During tum-off. the capacitor C, charges by the load current and the equivalent circuit is shown in Figure 4.36b. The capacitor voltage appears across the transistor and the dvldt is (4.45) Equating Eq. (4.42) to Eq. (4.45) gives the required value of capacitance,
h 11 c' ~v,
(4.46)
Once the capacitor is charged to V., the freewheeling diode turns on. Due to the energy stored in L,. there is a damped resonant circuit as shown in Figure 4.36c. The transient analysis of RLC circuit is discussed in Section 16.4. The RLC circuit is normally
Copyrghlcd matcria
154
Chapter 4
Power Tran5istors
I· ' • ' I
FIGUijE 4.37
Discharge current of snubber cap.1citor.
N 1+--
~ ..
- - T :s 1/Ct - - -..
made critically damped to avoid oscillations. For unity critical damping, 6 = I, and Eq. (18.11) yields
(4.47)
R, =2 ~
The capacitor C, has to discharge through the transistor and this increases the peak curre nt rating of the transistor. The discharge through the transistor can be avoided by placing resistor R, across C, instead of placing R, across D,. The discharge current is shown in Figure 4.37. When choosing the value of R,. the discharge time. R,C, = T, should also be considered. A discharge time of one-third the switching period T, is usually adequate.
3R,C, = 7~ = ; , or I
(4.48)
R, = 3/C J
Example 4.6 BJT Switch
'
Finding the Snubber Values for limiting dv!dt and dVdt Values of a
A bipolar transistor is operated as a chopper S\.\ilch at a frequency or f~= 10 kHz. lbe circuit
arrangement is shown in Figure 4.35a. The de voltage of the chopper is V, = 220 V and the load current is f l = 100 A. Vc£(ui) ::w O V . The switching times arc IJ = 0. t, • 3 µ.s, and 11 -= 1.2 µs. Determine the values of (a) l ,; (b) C,: (c) R, for critically damped condition; (d) RP if the discharge time is limited to one-third of switching period; (e) R,, if the peak discharge current is limited to 10% of load current; and (f) power loss due to R C snubber P,, neglecting the effect of inductor L1 on the voltage of the snubber capacitor C,4
Solution
= 220V,f, = IO kHz. r, = 3 µs, and 't = 1.2 µs. a. From Eq. (4.44), L, = V,1,111• = 220 x 31100 = 6.6 p.H. b. From Eq. (4.46), C, • /itl V, • 100 x 1.21220 • 0.55 p.F. c. From Eq. (4.47). R, = 2Vl/C, = 2v'6Mi35 = 6.93 n. d. From Eq. (4.48). R, = 1/(3[,C, ) = 10'1(3 x 10 x 055) • 60.6 n. •· V/R, = 0.1 X /Lor220IR, • 0.1 x IOOorR, • 220.
Ii= IOOA, V,
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4.9
SPICE Models
155
r. The snubber loss. neglecting the loss in diode D,, is P, '" 0.5C, V;f,
(4.49}
= 0.5 x 0.55 x 10... x 220' x 10 x 10' = 133.1 W
4.9
SPICf MODELS Due to the nonlinear behavior of power electronics circuits, the computer-aided simu· lation plays an important role in the design and analysis of power electronics circuits and systems. Device manufaclurers often provide SPICE models for power devices.
4.9.1
BJT SPICE Model The PSpice model. which is based on the integral charge-control model of Gummel and Poon (16), is shown in Figure 4.38a. The static (de) model that is generated by PSpice is shown in Figure 4.38b. If certain parameters are not specified, PSpice as· sumcs the simple model of Ebers-Moll as shown in Figure 4.38c. The model statement for NPN-transistors has the general torm . t-DOEL O,UJ
and the general form for PNP-transistors is .MJOEL QNAME PNP t Pl=Vl P2 =V2 P3 =V3 . . • PN=VNI
where QNAME is the name of the BJTmodel. NPN and PNP are the type symbols for NPN· and PNP-transistors, respectively. Pl , P2, . . and VI , V2, . .. are the parameters and their valucs, rcspectivcly. The parameters that affect the switching behavior of a BJT in power electronics arc JS, BF, CJE, CJC, TR, TF. The symbol for a BJT is Q, and its name must start with Q. The general form is Q NC NB NE NS ~'1E ( (a.rQa) value}
where NC. NB, NE, and NS are the collector, base, emitter, and substrate nodes, respectively. The substrate node is optional: If not specified, it defaults to ground. Positive current is the current that flows into a terminal. That is, 1he current flows from the collector node, through lhe device, to lhe emitter node for an NPN-BJT. The parameters thal significantly influence the swi1cbing behavior of a BJT are: IS BF CJE CJC TR TF
4.9.2
P-N saturation current Ideal maximum forward beta Base-emitter zero-bias p11 capacitance Base-collector zero-bias pn capacitance Ideal reverse transil time Ideal forwa rd transil lime
MOSFET SPICE Model The PSpice model (16) of an 11-channel MOSFET is shown in Figure 4.39a. The slalic (de) model lhat is generaled by PSpicc is shown in Figure 4.39b.111e model statcmenl
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156
Chapt er 4
Power Transistors C
Collector
c.., s -+-~~
+-~--<>--~-+---!f---o Subs1n1te
C;c
Cic
l t,c?
. B
Ra
t.,.1113R
I
Base lbt1
(I...,, - lM)IK.,b
I.., 1113F
Ci,
E
Emitter
(•) Gummcl-A:>on model
c c 1,.
I
C1,c
arlE
l1,c1IJJR
.......?
Ra
(I,.1 - l"'1)1Kq,
I
B
Ro B
11,.,t/lF
lh(1
E (b) De model
le E (c) Ebcrs-Moll model
FIGURE 4.38
PSpic:c BJT model
Copyrghlcd m~leria
4.9 SPICE Models
D
Dntin Ro
c..,
D
c,,,
Ro
+ VJd -
- vbd +
- Vw +
G
'•
Ros
Oa1c
t+
v ..
1-
B G
Bulk
'•
Ros l
- vbl + +
157
VP -
- Vm
B
+
Rs
c,,.
Cp
s (b) De model
c,. S
Source
( a) SPICE model FIGURE 4.39
PSpice n-channel MOSFET model.
•
of n-channel MOSFETs has the general form . M:JDEL MNA.liE ?MJS (Pl • Vl P2=V2 P3 =V3 . . . PN•VN)
and the statement for p-channel MOSFETs has the form . M)OEL MW.ME PMOS (Pl=\/1 P2=V2 P3=V3 ••• PN=VNJ
where MNAME is the model name. NMOS and PMOS are the type symbols of n-channel and p-cbannel MOSFETs. respectively. The parameters that affect the switching behavior of an MOSFET in power electronics are L, W, vro,KP, IS, CGSO, and CGDO. The symbol for an MOSFET is M. The name of MOSFETs must start with M and it takes the general form M
~m I L•
(W:a)
•
(AI):: I {~:: J
+
( POa l (PS• } (NRD=J (NRS:} (NRG=t INRB=)
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158
Chapter 4
Power Transistors
where ND, NG. NS. and NB are the drain, gate, source, and bulk (or substrate) nodes, respectively. The parameters that significantly influence the switching behavior of an MOSFETare:
L
w
VTO IS CGSO CGDO
Channel length Channel width Zero-bias threshold voltage Bulk pn-saturation current Gate-source overlap capacitance and channel width Gate-drain overlap capacitance and channel width
For COOLMOS, SPICE does not support any models. However, the manufacturers provide models for COOLMOS [11J. 4.9.3
IGBT SPICE Model
The 11-channel IGBT consists of a p11p-bipolar transistor that is driven by an 11-channel MOSFET. Therefore, the IGBT behavior is determined by physics of the bipolar and MOSFET devices. Several effects dominate the static and dynamic device characteristics. The internal circuit of an IGBT is shown in Figure 4.40a. An IGBT circuit model (16), which relates the currents between terminal nodes as a nonlinear function of component variables and their rate of change, is shown in Figure 4.40b. The capacitance of the emitter-base junction Cro is implicitly defined by the emillcr-basc voltage as a function of base charge. l u b is the emitter-base capacitor current that defines the rate of change of the base charge. The current through the collecror-emiller redistribution capacitance l ,m is pan of the collector current, which in contrast to/'" depends on the rate of change of the base-emitter voltage./•,, is part of the base current that does not flow through C,• and does not depend on rate of change of base--collector voltage. There are two main ways to model IGBT in SPICE: (1) composite model and (2) equation model. The composite model connects the existing SPICE p11p-BJT and 11-channel MOSFET models. The equivalent circuit of the composite model is shown in Figure 4.4la. It connects the existing BIT and MOSFET models of PSpice in a Darlington configuration and uses the built-in equations of the two. 'The model computes quickly and reliably, but it does not model the behavior of the JGBT accurately. The equation model (22, 23] implements the physics-based equations and models the internal carrier and charge to simulate the circuit behavior of the IGBT accurately. This model is complicated, often unreliable, and computationally slow because the equations are derived from the complex semiconductor physics theory. Simulation times can be over 10 times longer than those for the composite model. There are numerous papers of SPICE modeling of IGBTs and Sheng (24} compares the merits and limitations of various models. Figure 4.41 b shows the equivalent circuit of Shcng's model (21 f that adds a current source from the drain to the gate. It has been found that the major inaccuracy in dynamic electrical propenics is associated with the modeling of the drain to gate capacitance of the 11-channel MOSFET. D uring high-voltage switching. the drain-to-gate capacitance C,8 changes by two orders of
Copyrghtcd m~leria
4.9 SPICE Models
159
Gate
l
Emitter
g
c,.. Emillcr
c
?
~Ic. .
c
'
c
Tc•.1
c~
C.i.i
td
I=
p - base
p•
c"'
Gale
I,
b---+-' d
t b
t I«« le,,
t
1,..
C..,
t
tl«b c..
e
'---+-' - - - - - ' n- drift
e
p• substrate
..
Collector
C0Uec1or (b) Circuil model
(a) Internal circuil n1odel FIGURE 4.40
IOBT model. !Rd. 16. K. Shenoi]
magnitude due to any changes in drain-to-gate vollage Vdg· This is, Cd, is expressed by Cdg
E.uCoxd = -;a===-'"----~--2t,;VJs
- -,.,- c.,d q1 B
+ Adg<,,
where A,18 is the area or the gate over the base; <,; is the dielectric constant of silicon; Coxd is the gate-drain overlap ox:idc capacitance; q is the electron charge; N8 is the base doping dcnsiiy.
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Chapter 4
Power Transistors
c
c
4
PNP
PNP
s
o,
o,
G 6
G
MOSFET
MOSFET
0
E (n) Composite model
E (b) Sheng !'Spice model
FIGURE 4.41
Equivalent circuits of IGBT SPICE model~ (Ref. 21. K. Sheng)
PSpice does nol incorporate a capacitance model involving the square root, which models the space charge layer variation for a step junction. PSpicc ntodcl can i1nple·
ment the equations describing the highly nonlinear gate-drain capacitance into the composite model by using the analog behavioral modeling function of PSpice.
4 .10
COMPARISONS OF TRANSISTORS Table 4.1 shows the comparisons of BJTs, MOSFETs, and IGBTs.
SUMMARY Power transistors are generally of five types: BJTs. MOSFETs, SITs, IGBTs, and COOLMOS. BJTs are current-controlled devices and their parameters are sensitive to junction temperature. BJTs suffer from second breakdown and require reverse base current during tum-off to reduce the storage time, but they have low on-state or saturation voltage. MOSFETs are voltage-controlled devices and require very low gating power and their parameters are less sensitive to junction temperature. There is no second breakdown problem and no need for negative gate voltage during turn-off. The conduction losses of COOLMOS devices is reduced by a factor of five as compared with those of the conventional technology. It is capable to handle two to three times more output power as compared with that of a standard MOSFET of the same package. IGBTs, which combine the advantages of BJTs and MOSFETs, are voltage-controlled devices and have low on-state voltage similar to BJTs. COOLMOS, which has very low onstate loss, is used in high-efficiency. low-power applications. IGBTs have no second
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TABLE 4.1
Comparisons or l\>wcrTronsistors 8 m1c/Ga1c
Charnc.tcrislic
Switching Frequency
Con1inuou.s
Mediun,
Control
Swi1ch l)'pc
Conuol Variable
BIT
Current
On,Statc
rvtax.
Vohngc
Voltage
l)TQ()
Rating VJ
Low
1.5 kV
l\1ax.. Current Rating I,
I kA
Advantage~
Limitations
Simple switch
Current controlled dc,·icc and rcq_uires a higher base
Low on.srnte drop
20kHz
S, -= V1 / , = 1.5 ~·IV,\
S., • V~I, • 1.5 MVA
1,tighcr off-state vollagc capability High swi1ching los..;
current 10 turn .o n and
sus1ain on-state cu.rrcnt
Dase drive power los's Charge recovery tin,c and slower SW'i tching s.pccd Secondary b reakdown
MOSFET
u
~
~
Continuous
Very high
High
I kV S, • \l,I~
V. 1,
• 0. 1 MVA
IOOA
COOLMOS
Vohagc
Conlinuous
Very hii;h
Low
l kV
IG BT
Vol1age
Continuous
High
Mcdiu1n
3.5 kV
2 kA
s, - v,,.
SJ • V,I,
SJT
--"'
150 A
s, •
• 0 .1 MVJ\
= 1.5 MVA
() 0
"<5 & 3 "'Q
Voh:i.gc
Voltage
Continuous
Very high
High
• 1.5 MVA
region High s"itching losses Unipo lar vullagc device tfighcr switching speed Hig.h on-state drop as Low switching loss high as IOV S i1nplc g:itc drh c circuh Lower off-stale vollagc capability Little gale po....·cr. Unipolar voltage device Ncg..11ive 1en1pera1urc coefficient on rain current and facilil::itcs paraHe-1opc-rution Low-powe r device Low gate drive Low voltage and current requirement and low ratings on,s1a1c power drop Lower ort-sta1c vo!tagc Low on-.s1:11e voltage capability Little g:uc pO\\'Cf Unipolar voltage device Higher on·slatc \'Olla ge dr()p High-,.·oltage ra1ing Lo"·er current ratings 1
162
Chapter 4
Power Transistors
breakdown phenomena. SITs are high-power. high-frequency devices. They are most suitable for audio, VHF/UHF, and microwave amplifiers. They have a' normally oncharacteristic and a high on-state drop. Transistors can be connected in series or parallel. Parallel operation usually requires current-sharing elements. Series operation requires matching of parameters, especially during tum-on and tum-off. To m aintain the voltage and current relationship of transistors during turn-on and turn-off, it is generally necessary to use snubber circuits to limit the dildt and dvldt. The gate signals can be isolated from the power circuit by pulse transformers or optocouplers. The pulse transformers are simple, but the leakage inductance should be very small. The transformers may be saturated at a low frequency and a long pulse. Optocouplers require separate power supply.
REFERENCES [1) B. J. Baliga. Power Semico11d11ctor Devices. Boston, MA: PWS Publishing. I996. [21 S. K. Ghandi. Semico11d11ctor Power Devices. New Yo rk: John Wiley & Sons. 1977. (3) S. M. Sze, Modem Scmico11tl11c1or De•ict Physics. New York: John Wiley & Sons 1998. (4) B. I. Baliga and D. Y. Chen. Power Trtmsist0n: De,•ice Design mu/ Appliwrio11s. New York:
IEEE Press 1984. (SI Westinghouse Electric, Silico,r Power Tra11sistor Ha11dbook. Pittsburgh: Westinghouse Electric Corp. 1967. [6) R. Severns and J. Armijos, MOSPOWER App/icarion Ha11tlbook. Santa Clara, CA: Sili· conix Corp. 1984. (71 S. Clemente and B. R. Pelly. "Understanding power MOSFET switching performance." Solid-Sratc Elwronics, Vol. 12. No. 12, 1982, pp. 1133-1141. (8) 0 . A. Grant and I. Gower, Power MOSFETs: Theory and Applicario11s. New York: John Wiley & Sons. 1988. [9] L. Lorenz, G. Deboy. A. Knapp, and M. Mar,. "COOLMOST>'-a new milestone in high vollage power MOS," Proc. /SPSD 99. Toronto. 1999. pp. 3-IO. (10) G. Deboy. M. Marz. J. P. Stcngl. H. Strack. J.1ilhanyi. and H. Weber. "A new generation or high voltage MOSFETs breaks the limit of silicon," Proc. I EDM 98, San Francisco, 1998, pp. 683-685. [U] Infineon Technologies. Coo/MOS"': Power Semicomlucrors. Germany: Siemen~ 2001. www.infincon.co [U] C. Hu, ··Optimum doping profile !or minimum ohmic resistance and high breakdown voll· age," /EEETta11sacrio11s 011 E/ectro11ic Devices. Vol. ED-26, No. 3, 1979. (131 B. J. Baliga, M. Cheng. P. Shafer, and M. W. Smith, ··Toe insulated gate transistor (IGT): a new power switching device:· IEEE Industry Applications Society Conference Record. 1983. pp. 354-,363. (14) B. J. Baliga. M. S. Adler, R. P. Love, P. V. Gray. and N. Zommer. ··Toe insulated gate transis· tor: a new three-terminal MOS controlled bipol:ir power dc,ice." I£££ Tra11sacrio11s Elcc1ro11 Dcl'ices, ED-31.1984. pp.821-828. (15] /GBT Desig11er's Ma1111al. El Segundo. CA. International Rectifier. l'l91. (161 K. Shcnai. Power Electro11ics Ha11dbook.cdited by M. H. Rashid. Los Angeles. CA: Academic Press. 2001, Chapter 7.
Copyrighted m lcria
Review Questions
163
(17) I. Nish izawa and K. Yamamoto . ''High·!rcquency high·powe r static induction transistor,"
IEEETra,r.sacrions on Electron Devices, Vol. E025. No. 3, 1978, pp. 314-322. (18) J. Nishizawa, T. Tcrasaki, and J. Shibata, "Fie ld·cffcct transistor versus analog transistor {static induction transistor),'' IEEE Transactions on Electro11 Devices, Vol. 22, No. 4, April 1975, pp. 185-197. (19) · B. M. Wilamowski. Power Electronics Handbook.edited by M. H. R ashid. Los Angeles. CA: Academic Press. 2001, Chapter 9. (20) M. H. Rashid. SPICE for Powu Electro11ics and Electric Power. Englewood Cliffs, NJ: Prentice-Hall. 1993. (21) K. Sheng, S. J. Finney. and B. \V. Williams. "Fast and accurate IGBT model for PSpice." Electro11ics letters, Vol. 32, No. 25, December 5. I 996, pp. 2294-2295. (22) A. G. M. Strollo. "A new IGBT circuit model for SPICE simulation,'' Power Electronics Specialists Confem,ce,June 1997, Vol. 1, pp.133- 138. (23) K. Sheng, S. J. Finney, and B. W. Williams. ''A new analytical IGBT model \\;th improved elec1rical characteristics." I £££ Transnctious on ro,ver £/ectro11ics. Vol. 14. No. t, January 1999, pp. 98- 107. [24) K. Sheng. B. W. Williams, and S. J. Finney. "A revie w of IGBT models." IEEE Tra,isac1io,,s on Power Elee1ronics, Vol. 15. No.6. November 2000, pp. 1250-1266. [251 A. R. Hefner, "An investigation of the drive circuit require ments fo r the power insulated gate bipolar transistor (1GBT)," IEEE Trnn.sac1io11s on Po,ver f;lectron;cs. Vol. 61 1991. pp. 208-219. [26) C. Licitra, S. Musumeci, A. Raciti. A. U. Galluzzo, and R . Lctor, "A new driving circuit for IGBT d evices." IEEE Transactions Power Electronics, Vol. 10, 1995. pp. 373-378. (27) H. G. Lee, Y. H . Lee, B. S. Suh. and J. \V. Lee, "A new intelligent gate control scheme to drive and protect high pO\\'Cr IGBTs." European Po iver Electronics Conftrtnct Recordr. 1997,
pp. l.400-1.405. (28) S. Bernet, "Recent developments of high power converters for industry a nd traction appli· cations." IEEE Tra,,sacliom on Power Electronics. Vol. 15, No.6. November 2000,pp. 1102-1117.
REVIEW QUESTIONS 4.1 4.2 4.3 4.4 4.S 4.6 4.7 4.3 4.9 4.10
What is a bipolar transistor (BJT)? What a re the types o f BJTs? What are the differences between NPN·trn11sistors and ?NP-transistors? \Vha1 are the input characceris1ics of /\'P1\f.1ransis1ors? What are the output characteristics of NPN·transistors? What are the three regions of operation for BJTs? What is a bctu ( 13) of BJTs? What is the difference between beta, 13, and forced beta, i3r of BJTs? Wl,at is a transductance of BJTs? What is an overdrive factor of BJTs'! 4.11 What is the switching model of BJTs? 4.U What is the cause of delay time in BJTs? 4.13 What is the cau.,;e of storage time in BJTs?
4.14 \Vhat is the cause of rise ti1ne in BJTs? 4.15 What is the cause of fall time in BJTs'/ 4.16 \Vhat is a saturation mode or BJTs'!
Copyrghtcd matcria
164
Chapter 4 4.17 4.18 4.19 4.20 4.ZI
4.22 4.23
4.24 4.25 4.26
4.27
4.28 4.29
4.30 ~.31 4.32 4.33 4.34 4.35
4.36 4.37
4.38 4.39 4.40 4.41 4.42 4.43 4.44 4.45
4.46 4.47 4.48 4.49 4.50
4.51 4.52
4.53
Power Transistors
What is a tum-on time of BJTs? What Ls a tum-off time of BJTs? What is a FBSOA of BJTs? What is a RBSOA of BJTs? Why is it necessary to reverse bias BJTs during tum-off? What is a second breakdown of BJTs? What are the advantages and disadvantages of BJTs? What is an MOSFET? What are the types of MOSFETs? What arc the differences between enhancement-type MOSFETs and depletion-type MOSFETs? What is a pinch-off voltage of MOSFETs? What.is a threshold volt.igc of MOSFETs? What is a transconductance of MOSFETs? What is the switching model of n-ehanncl MOSFETs? What are the transfer characteristics of MOSFETs? What are the output characteristics or MOSFETs? What are the advant.iges and disadvant.iges of MOSFETs? Why do the MOSFETs not require negative gate voltage during tum-off? Why docs the concept of saturation differ in BJTs and MOSFETs? What is a tum-on time of MOSFETs? What is a turn-off time of MOSFETs? What is an SIT? What are the advantages of SITs? What are the disadvantages of SJTs? What is an IGBTI What arc the transfer characteristics of IGBTs? What are the output characteristics of IGBTs? What arc the advantages and disadvantages of IGBTs? What are the main differences between MOSFETs and BJTs? Whal arc the problems of parallel operation of BJTs? What arc the problems of parallel operation of MOSFETs? What are the problems of parallel operation o f IGBTs? What are the problems of series operation of BJTs? What are the problems of series operations of MOSFETs? What are the problems of series operations of IGBTs? What are the purposes of shunt snubber in transistors? What is the purpose of series snubber in transistors?
PROBLEMS 4.1 The beta (13) of bipolar transistor in Figure 4.7 varies from 10 to 60. The load resistance is Re 5 n.The de supply voltage is Vee 100 V and the input voltage to the base circuit is V8 = 8 Y. If Ve« ..•> = 2.5 V and V8 £ (u•) • 1.75 V. find (a) the value of Rs that will result in·satura1ion with an overdrive fac1or of 20; (b) the forced ~. and (c) the power loss in
=
=
the transistor Pr. 4.Z The beta (13) of bipolar transistor in Figure 4.7 varies from 12 to 75. The load resistance is Re = l.S The de supply voltage is Vee a 40 V and the input voltage to the base circuit is V8 = 6 Y. If Ve£("'>= 1.2 Y, V8 £c.. ,J = 1.6 V, and Rs • 0.7 n. determine(•) the OOF, (b) the forced I). and (c) the power loss in the transistor PT.
n.
CopyrghCcd m~leria
Problems
165
4.3 A 1ransis1or is used as a swi1ch and lhe waveforms are shown in Figure 4.11. The parame· 1crs arc Vee = 200 V. V6 ,<.,•J= 3 V. / 8 = 8 A. Va(>atJ = 2 V. l cs = 100 A.,,,= 0.5 f',S, ,, = I µ.s, ,, = 5 µ.s . ,1 = 3 µ.s, and / , = 10 kHz. The duty cycle is k = 50%. The collcc1or-emincr Jcakage current is lcr.o = 3 rnA. Determine-1hc power loss due to thccolleclor currenl (a) during 1um-on '•• = 1, + t,; (b) during conduc1ion period In, (c) dur· ing 1urn·off 11.,ff a t, + 11. (d) during off.time 1 and (e) the total average power losses Pr , (0 Plot the ins1an1aneous power due lo 1he collector current P,(t ). 11,
4.4 The n1aximun1 junction temperature o f the bipola r lransistor in Problem 4.3 is T1 ~ t 50°C and 1hc ambicnl lcmperalurc is TA = 25°C. II lhc thermal rcsis1anccs arc Ru; = 0.4°C/W and Res = 0.05'CIW. calculale lhe thermal resistance or heat sink RsA · (flim: Neglec1 the
power loss due to base drive.) 4.5 For the parame1ers in Problem 4.3, calculate the average power loss due to lhe base cur·
rent P8 . 4.6 Repeal Problem 4.3 if VHt.(•u) = 2.3 V. 18 = 8 A. Va(~") = 1.4 V. '• = 0.1 µs. t, = 0.45 µs. ,, = 3.2 µ.<. and ,1 = I.I µs. 4.7 An MOSFET is used as a switch. The parameters are v,," = 40 V, '" = 35 A. Ros = 28 mn, VGs = 10 ' ' · '"(o,,) = 25 ns, t, = 60 ns, ,d(off) == 70 ns. 11 = 25 ns., and /, = 20 kHz.. TI,e drain source leakage current is I oss = 250 µA. The duty cycle is k a 60"A. Determine the power loss due to the drain current (a) during tum-on t0 "
= td(t1 ) +
tr; (b) during con-
duction period ,,.. ( c) during lurn-orr 10 ,r = '"<•") + 11. (d) during ofl·lime Im and (c) the total average power losses Pr. 4.8 The maximum junc1ion temperature of the MOSFET in Problem 4.7 is T1 = 1500C and the an1bient tc1npcraturc is T14 - 300C. I( the thcnnal resistances arc R1c - l KJ\V and Res = I K/\V, calculate the thermal resistance or 1he heal sink RsA· (Nore: K =
·c + 213.) 4.9 Two BJTs are connected in parallel similar to Figure 4.32. The total load current of Ir = 200 A. 1he collector-<:miner vollagc of transisor load current: and (0 po,vcr loss due to RC snubbe r P,.
neglecting 1hc effect of inductor L , on the vollagc of snubber capacilor C,. Assume 1ha1 Vcr:c,a1J = 0.
4.U An MOSFET is opcralcd as a chopper switch at a frequency of J; = 50 kHz. The circuil arrangement is shown in Figure 4.35a. The de input vollagc
or 1hc chopper is
V,
= 30 V
and the load current is / 1_ = 40 A. The switching times are,, = 60 ns and 11 = 25 ns. De· ,ermine the values of (•) L,; (b) C,.: (c) R, for criticall)' damped condi1ion; (d) R, if the dis· charge lime is limited 10 one-lhird of switching period: (c) R, if peak discharge currenl is lintilcd 10 5% of load current; and (0 power loss due 10 RC snubber P., negl~cting the effect o( inductor L, on the voltage o( snubber capacitor C, ....\ s..~ume that Vct.·c~i) = 0.
Copyr ghlcd malcria
CHAPTER
5
De-De Converters The leamlng object/vu of tliis diopter an as follows: • • • • • •
5.1
To learn the switching technique for do-
In many industrial applications, it is required to convert a fixed-voltage de source into a variable-voltage de source. A de-de converter converts directly from de to de and is simply known as a de converter. A de converter can be considered as de equivalent to an ac transformer with a continuously variable turns ratio. Like a transformer, it can be used to step down or step up a de voltage source. De converters are widely used for traction motor control in electric automobiles, trolley cars, marine hoists, forklift trucks, and mine haulers. They provide smooth acceleration control, high efficiency, and fast dynamic response. De converters can be used in regenerative braking of de motors to return energy back into the supply, and th.is feature results in energy savings for transportation systems with frequent stops. De converters are used in de voltage regulators; and also arc used. in conjunction with an inductor, to generate a de current source, especially for the current source inverter. 5.2
PRINCIPLE OF STEP-DOWN OPERATION The principle of operation can be explained by figure 5.1a. When switch SW, known as the chopper, is closed for a time 11, the input voltage V, appears across the load. If the switch remains off for a time 12, the voltage across the load is zero. The waveforms for the output voltage and load current are also shown in Figure 5.lb. The converter switch can be implemented by using a (1) power bipolar junction transistor (BJT), (2) power metal oxide semiconductor field-effect transistor (MOSFET), (3) gatc-- turn-off thyristor (GTO). or (4) insulated-gate bipolar transistor (IGBT). The
166 Copyrighted m lcria
5.2 Principle of Step-Down Operation
:~,.I
I+Converter v,. -1 i. +
I =
O' +
SW
v,
R
v,
(•) Circuil
:E·
,,
,.
,, kT (b) W3'·clorms
167
·[
I,
I
T
50
45
8c
!!
40
·!
35
;;
30
~
c..
·= -~ R,(k ) ZS u
t; ~
.,,';; -~ ;; E
20 15
"
0
z
10 5
00
10
20
30
50
60
70
80
90
100
k Duty cycle.%
(c) Effec1h•e inpu1 resistance again.st duty cycle FIGURE 5, 1
Step-down con\•cncr v.ith resistive load.
practical devices have a finite voltage drop ranging from 0.5 to 2 V, and for the sake of simplicity we shall neglect the voltage drops of these power semiconductor devices. 'The average output voltage is given by
11''
,
V="'· dt =-V=ftV = kVs "To-u T s 1 1
and the average load current. /0
(5.1)
= V,/R = kV/R. Copyrghtcd matcria
168
Chapter 5
De-De Converters
where Tis the chopping period; k = 11/T is the duty cycle of chopper; f is the chopping frequency. The rrns value of output voltage is found from
v. = (~
lkTvf,d,yn= VkV,
(5.2)
Assuming a lossless converter, the input power to the converter is the same as the output power and is given by
P;
l lkT l lkTtr. y2 =-To Voi di = ~ dt = k_! T o R R
(5.3)
The effective input resistance seen by the source is
V, V, R R· = - = - - = ' I, kV,IR k
(5.4)
which indicates that the converter makes the input resistance Ri as a variable resistance of Rik. The variation of the normalized input resistance against the duty cycle is shown in Figure 5.lc. It should be noted that the switch in Figure 5.1 could be implemented by a BJT, an MOSFET, an IGBT, or a GTO. The duty cycle k can be varied from Oto I by varying It , T, or f Therefore, the output voltage V0 can be varied from Oto V, by controlling k, and the power Dow can be controlled. 1. Constant-frequency operation: The converter, or switching, frequency f (or chopping period 7) is kept constant and the on-time It is varied. The width of the pulse is varied and this type of control is known as pulse-wid1Ji-mod11/a1ion (PWM) control. 2. Variable-freq11ency operation: The chopping, or switching, frequency f is varied. Either on-time It or off-time 12 is kept constant. This is called frequency modularion. The frequency has to be varied over a wide range to obtain the full output voltage range. This type of control would generate harmonics at unpredictable frequencies and the filter design would be difficult. Example 5.1 Finding the Performances of a De-De Converter The de conver,er in Figure 5.la has a resistive load of R m 10 0 and the input voltage is
v"'
V,
s 220 V. When the converter switch remains on, its voltage drop is = 2 V and the chop· ping frequency is f = I kHz. If the duty cycle is 50%, determine (a) the average output voltage V., (b) the rms output voltage V., (c) the converter efficiency, (d) the clfcctive input resistance R1 of the converter, and (e) therms value of the fundamental component of output harmonic voltage.
Solution V, = 220 V, k 1.
e
0.5. R • 10 0, and
v,• = 2 V.
From Eq. (5.1 ). v. = 0.5 x (220 - 2) = 109 V. • y'ij3 x (220 - 2) • 154.15 V.
b. From Eq. (5.2), V0
Copyrghlcd matcria
5.2 Principle of Step-Down Operation
169
c. The output power can be found from
I!."
l Rdt = -:;:
Vij
Pn =r
O
= 0.5 x
J.'r(V, -R v,.)2 dt =k~(\/, - Vc!,) 2 R- 0
2) 2
(220 -
10
(5.5)
= 2376.2 W
The input power to the converter can be found from
P;
=
T1 J.' O
1
- v"') dt = T1 J.'rv,(V, O R
•
v,, dt =
V,(V, - v..,)
k ~ -R-~
(5.6)
= 0.5 x 220 x
220 - 2 10
= 2398 W
The converter efficiency is
d. From Eq. (5.4). R, = 10/0.S = 20 11. e. The oulpUI voltage as shown in Figure S.lb can be expressed in a Fourier series a.,; v.(1) = kV,
v .,
+ -' L sin 2nr.k cos 211'rrft nn n• I
(5.7)
v .,
+ - ' L(l - cos 2mrk)sin 2mrft n-rr " '"'
The fundamental component (for n = I) or output voltage hannonic can be detcnnined from Eq. (5.7) as
v
v1(1) =...! (sin 2,rk cos 2,rft ,r
+ ( I - cos 21rk)sin 2r.ft) (5.8)
220X2 . • sm(2,r 1T
X
. 10001) = 140.06 sm(6283.2t)
and its root·mcan·square (rms) value is V, • 140.06/v'i • 99.04 V.
Note: The efficiency calculation, which includes the conduction loss of. the con· verter. does not take into account the switching loss due to tum-on and tum-off of practical converters. The efficiency of a practical convener v.aries between 92 and 99%.
Key Points or Section S.2 • A step-down chopper, or de converter, that acts as a variable resistance load can
produce an output voltage from Oto V5.
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170
Chapter 5
Dc-Oc Converters
• Although a de converter can be operated either at a fixed or variable frequency, it is usually operated at a fixed frequency with a variable duty cycle. • The output voltage contains harmonics and a de filter is needed to smooth out the ripples.
5.2.1
Generatlon of Duty Cycle The duty cycle k can be generated by comparing a de reference signal v, with a sawtooth carrier signal v.,,. This is shown in Figure 5.2, where V, is the peak value of v,, and V,, is the peak value of Ver, The reference signal v, is given by
v,
V, -T t
::i:
(5.9)
which must equal to the carrier signal v,, = V,, = kT. That is,
V.,, =
V,
7 kT
which gives the duty cycle duty k as
v.,,
k= - = M
(5.10)
v,
where Mis called the mod11/ation index. By varying the carrier signal v,, from Oto V"' the duty cycle k can be varied from O to I. The algorithm to generate the gating signal is as follows:
L Generate a triangular waveform of period T as the reference signal v, and. a de carrier signal v". 2. Compare these signals by a comparator to generate the difference v, - v., and then a hard limiter to obtain a square-wave gate pulse of width kT, which must be applied to the switching device through an isolating circuit. 3, Any variation in v,, varies linearly with the duty cycle k . y
v
v,
'
v"
v"
,.
0
v, 0
tT
T
FIGURE 5.2
Con1paring :1 reference signal with a carrier signal.
Copyrghlcd malcria
5.3 Step,Oown Converter with RL Load
5.3
171
STEP-DOWN CONVERTER WITH RL LOAD
A converter (1 Jwith nn RL load is shown in Figure 5.3. The operation of the converter can be divided into two modes. During mode l, the convener is switched on and the current nows from the supply to the load. During mode 2, the converter is switched off and the load current continues to now th rough freewheeling diode 0.,. The equivalent circuits for these modes are shown in Figure 5.4a . The load current and output voltage waveforms are shown in Figure 5.4b with the assumption that the load current rises tin· early. However, the current nowing through an RL load rises or falls exponentially with a time constant. The load time cons tant (1 = U R ) is generally much higher than the switching period T. Thus. the linear approximation is valid for many circuit conditions a nd simplified expressions can be derived within reasonable accuracies,. .,. The load current for mode 1 can be found from
V,
= Ri 1 +
L
dit dt +£
which with initial curre nt i 1(t = 0) = / 1gives the load current as i1(1)
= l1e-
£
R
(J -
.
,-,RIL )
(5.11)
This mode is valid O s t s 11 (= kT ); and at the end of this mode. the load current becomes .
= /2
i1 (1 = t1 = kT)
(5.12)
The load current for mode 2 can be found from
di,
0 = Ri2 + L -· + E dt
With initial current i2 (t = 0) ginning of mode 2, we have
= [z and redefining the time origin (i.e.,, = 0) at the be· (5.13)
This mode is valid for O :S rent becomes
t :S 12 [
= (I -
k )T). At 1he e nd of 1his mode, the load cur-
(5.14) Chopper
+
SW
t =O
+
L
v
'
+ _-
R
E
FIGURE 5.3
De conver1er with RL. loads.
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172
Chapter 5
De- De Converters Yo
v,
.--11
0
+
R
I
I
L
v,
1,
---------
1, ;,
I I I I I I
+
o------~-TE
1,
Continuous current I I
tT--!-(t - k)T-
Mode I
0
;,
I
T
I
I
I I
tT
T
,, i
L
Discontinuous current
T
Mode2 (b) Waveforms
(a) Equivalent circuits
FlGURE 5.4
Equivalent circuits and waveforms for RL loads.
At the end of mode 2, the converter is turned on again in the next cycle after time, T = 1/f = 11 + 12, Under steady-state conditions, / 1 = iJ. 'lbe peak-to-peak load ripple current can be determined from Eqs. (5.11) to (5.14). From Eqs. (5.11) and (5.12), / 2 is given by (5.15) From Eqs. (5.13) and (5.14), his given by 1 = /1 = 3
Solving for / 1 and fz, we get
f,t - (1 - k )TRIL _
f
e-( 1-k)TRIL)
R
_Vs(ek'-1) _E e' - I R
11 - R where z = T
E (1 _
(5.16)
(5.17)
is the ratio of the chopping or switching period to the load time constant.
Copyrghlcd malcria
5.3 Step-Down Converter with RL Load
(e-h - I) - -R
Ii-_ -V, R
E
e-: - I
173
(5.18)
The peak-to-peak ripple current is
which after simplifications becomes
61
V, I - e-k, + e-, 1 - e- •·
e -( l - k ),
=R
(5.19)
The condition for maximum ripple,
d( 6/) - - =0 dk
e-•, -
e-(•-•lz = 0 or -k gives ripple current (at k = 0.5) is
= - (I
(5.20)
- k) or k
V,
61...,. For 4/L
= 0.5. The maximum peak-to-peak R
= R tanh 4/L
(5.21)
> > R, tanh O "' 0 and the maximum ripple current can be approximated to 61.,.x
v, = 4fL
(5.22)
Note: Equations (5.11) to (5.22) are valid only for continuous current now. For a large off-time, particularly at low-frequency and low-output voltage. the load current may be discontinuous. The load current would be continuous if UR > > T or Lf >> R. In case of discontinuous load current, / 1 = 0 and Eq. (5.11) becomes i1(1) and Eq. (5.13) is valid for O :s t
E
R
(1 - e- •RIL)
:s 12 such that i 2(1 = 12
Because 1
= v:' -
=
L
R In
(
l
+
12) = / 3 = / 1 =
0, which gives
R/2) E
= kT, we get i1(1)
= 12 = V, ;
E
(1 - e-t,)
which after substituting for Ii becomes
Copyrghtcd matcria
174
Chapter 5
Dc-Oc Converters
Condition fo r continuous current: For / 1
-"
0, Eq. (5.17) gives
- 1 _ E) ( e-t' e' - 1 V,
"" 0
which gives the value of the load electromotive force (emf) ratio x = EIV, as E V,
x= - s
e-t'-1
(5.23)
e'- 1
Example 5.2 Finding the Currents of a De Converter with an RL Load A convener is feeding an RL load as shown in Figure S.3 with V, = 220 V, R = Sn , l = 7.S mH, / • 1 kHz. k • O.S. and E • 0 V. Calculate (a) the minimum instantaneous load cunent J,. {b) the peak instantaneous load current / 2, (c) the maximum peak-to-peak load ripple current, (d ) the average value of load current /~ (e) the rms load current /0 , ( [ ) the effective input resistance R1 seen by the source, (g) the nns chopper current JR , and (b) the critical value or the load inductance for continuous load current. Use PSpice to plot the load current, the supply current, and the freewheeling diode currenL
Solution V, = 220 V, R = S 0 , L = 7.S mH, E = 0 V, k = 0.5, and f = 1000 Hz. From Eq. (S.15), / 2 = 0.7165/1 + 12.473 and from Eq. (S.16), / 1 = 0.7165/2 + 0. Solving these two equations yields / 1 • 18.37 A. b. ,, • 25.63 A. c. A/ = / 2 - / 1 = 25.63 - 18.37 = 7.26 A. From Eq. (5.21), A/mu Eq. (S.22) gives the approximate value, Al ,..,. = 7.33 A .
L
= 7.26 A
and
d. The average load cuncnt is. approximately, I,= /2 + /1 = 25.63
2
+ 18.37 = 22 A 2
e. Assuming that the load current rises linearly from / 1 to I,. the instantaneous load cur·
rent can be expressed as .
I
'1 -
Alt I+
kT
for O <
t
< kT
The rrns value of load current can be found from
!."
1 o lo= ( kT
i} dt
)'n = [11 + (I2 -3
1
I I)
+ /1(/2 -
/1) ] "'
(5.24) =
f.
22.1 A
The average source current
I, • kl, • O.S x 22 • 11 A and the effective input resistance R1 = V,JI,
= 220111 = 20 n.
Copyrghlcd m~leria
5.3 Step-Dow n Converter wit h RL l oad
175
30A -,------- --------- --- ---- ----- ------------------ :- --I - -, J _J
,.. I
'
I I
SEL>> :
OA
--- -------------------- ----- --- -
i:_oJ l(R)
~5..155
,\I
-
9.50901,
,\2
-
901lXlni.
17.q
J,f
•
50S q2qu.
74948
30 A--,--------------------- ----- --- ----- ---------- --- -- --, I' ' I' I
OA
I I
o -l(V,)
30A -,- - - -------- -- - ----- -- - -- -- - -- ----- --- ----- -- ------- - ,
..l.D_ltLLD_b_h_ D_ D_D_lj Os
D
5 ms
IOms
l (D.,)
lime FIGURE 5.5
SPIC£ plots of load. input. and diode currents lor Example 5.i .
g. The rms converter current can be found from
1 J."·l ) 1. = ( T O ,, dt
= ,ffc1 = 0
~
112
2
= vii [ l 2I + (/, -3 1,) + J,(l, - I ,) ]'"
(5.25)
x 22.1 = 15.63 A
b. We can rewrite Eq. (5.23) as
) (•''e· --1 1
Vs - . -
= £
which. after iteration, gives. l = TR/L • 52.S and L • I ms x 5152.5 • 0.096 n1H. The SPICE simulation results [32) are shown in Figure 5.5. which shows the load cur· rent /( E), the supply current -I(V,), and the diode current l (D,.). We get 11 ~ 17.96 A and /2 = 25.46 A.
Example 5.3
Finding the Load Inductance to Limit the Load Ripple Current
The converter in Figure 5.3 has a load resistance R = 0.25 n,input voltage V, = 550 V. and bat· tcry voltage £ • 0 V. The a\1erage lo:id current /" = 200 A . and chopping (rcqucncy
Co;,ynghled m tcria
176
Chapter 5
De-De Converters
f = 250 Hz. Use the average output voltage to calculate the load inductance L , which would limit the maximum load ripple current to 10"/o of/, .
Solution V,c550V, R s 0.250, £ c 0V, fa250Hz, T s l/f • 0.004s, and lli • 200X0.I = 20 A. The average output voltage v. = kV, = RI,. The voltage across the inductor is given by di L dr • V, - RI, • V, - kV, = V,(I - k ) Ir the load current is assumed to rise linearly, dt = r1 = kT and di = lli: V (1 - k )
lli = '
L
kT
For the worst-case ripple cond~tion~
d( Ai) --=O
dk
This gives k = 0.5 and lli L = 20 x L = 550(1 - 0.5) x 0.5 x 0.004
and the required value of inductance is L • 27.5 mH.
Nore: For lll
= 20 A, Eq. (5.19) gives, = 0.036 and L = 27.194 mH.
Key Points or Section S.3
• An inductive load can make the load current continuous. However, the critical value of inductance, which is required for continuous current, is influenced by the load emf ratio. The peak-to-peak load current ripple becomes maximum at k = 0.5. 5.4
PRINCIPLE OF STEP-UP OPERATION
A converter can be used to step up a de voltage and an arrangement for s tep-up operation is shown in Figure 5.6a. When switch SW is closed for time l a, the inductor current rises and energy is stored in the inductor L. If the switch is opened for time 12, the energy stored in the inductor is transferred to load through diode D 1 and the inductor current falls. Assuming a continuous current now, the waveform for the inductor current is shown in Figure 5.6b. When the converter is turned on, the voltage across the inductor is di
Vt
= L dt
and this gives the peak-to-peak ripple current in the inductor as A/=
v, L''
(5.26)
Copyrghlcd m~leria
5.4 Principle of Step-Up Operation ic
+
L
o,
+
v,
177
-+
I I
I
VL
I I
Chopper
Load
T
v.
CL
I
• (a) S1cp-up arr.1ngcmcn1
6
s 3
,,
--- -----
6j
2
•1---if---lz-
0.2 (b) Cunent wavcronn
G.4
0.6 0.8
1.0
(c) Outpu1 voltage
FIGURE 5.6 Arrangement for step-up operation.
The average oulput voltage is Vo =
V, + L~ = lz
V ,(I + !!_) = t2
V,_!_ k I -
(5.27)
If a large capacitor Ci is connected across the load as shown by dashed lines in Figure 5.6a, the output voltage is continuous and v0 becomes the average value V0 • We can notice from Eq. (5.27) that the voltage across the load can be stepped up by varying the duty cycle k and the minimum output voltage is V, when k = 0. H owe ver, the converter cannot be switched on continuously such that k = 1. For values of k tending to unity, the output voltage becomes very large and is very sensitive to changes io k, as shown in Figure 5.6c. This principle can be applied 10 transfer energy from one voltage source to another as shown in Figure 5.7a. The equivalent circuits for the modes of operation are shown in Figure 5.7b and the current waveforms in Figure 5.7c. The inductor current for mode I is given by
,u,
V = L-
dt
s
and is expressed as i1(t )
v,
= T. ' + f,
(5.28)
Copyrghtcd matcria
178
Chapter 5
De-De Converters L
+
+
v,
~L
I
o,
Chopper
-0 (a) O rcuit diagram
E
T
L
+
i,
v,
o~~~~~~~~~~-
kT T (c) Current waveforms
Mode 2 (b) Equivalent drcuils FIGURE 5.7 Amtngcmcnt for transfer of energy.
where / 1 is the initial current for mode I. During mode I, the current must rise and the necessary condition,
-di, > 0 dt
or
V,> 0
The current for mode 2 is given by \t'. '
= L -di2 + dt
E
and is solved as
(5.2.9) where Ii is initial current for mode 2. for a stable system, the current must fall and the condition is
Copyrighted m lcria
5.5
Step-Up Converter with a Resistive Load
179
If this condition is not ~atisfied, the inductor current continues to rise and an unstable situation occurs. Therefore, the conditions for controllable power transfer are ~~
O
Equation (5.30) indicates that the source voltage v, must be less than the voltage E to permit transfer of power from a fixed (or variable) source to a fixed de voltage. In electric braking of de motors, where the motors operate as de generators, terminal voltage falls as the machine speed decreases. The converter permits transfer of power to a fixed de source or a rheostat. When the converter is turned on, the energy is transferred from the source V, to inductor L. If the converter is then turned off, a magnitude of the energy stored in the inductor is forced to battery E. Nore: Without the chopping action, v, must be greater than E for transferring power from V, to E.
Key Points or Section 5.4 • A step-up de converter can produce an output voltage that is higher than the input. The input current can be transferred to a voltage source higher than the input voltage. 5.5
STEP·UP CONVERTER W ITH A RESI STIVE LOAD A step-up converter with a resistive load is shown in Figure 5.8a. When switch S1 is closed, the current rises through Land the switch. The equivalent circuit during mode 1 is shown in Figure 5.8b and the current is described by
V '
=
L!!_i1 dt
which for an initial current of Ii gives (5.31)
L
L R
+
~ v,
L R
+
+
+
+
E-=-
-T
(a) Circuit
(b) Mode t
(c) Mode 2
FIGURE 5.8 Step.up convener with a reslsti\•e load.
Copyrghtcd matcria
180
Chapter 5
De- De Converters
which is valid for O s r s kT. At the end of mode 1 at t = kT,
v,
/2 = i1(t = kT) = - kT + 11 L
(5.32)
When switch S1 is opened, the inductor current flows through the RL load. The equivalent current is shown in Figure 5.8c and the current during mode 2 is described by
which for an initial current of / 2 gives
i2(t) = V, L-
E(1 - ··•) + [ze,•.
(5.33)
eT
which is valid for O s r s (1 - k)T. At the end of mode 2 at t = (1 - k )T.
Ii
= 12(1 = (1 -
k )T)
=V , L- E[1 -
e-(1-k)z
] + /2e-
(5.34)
where z = TRI L. Solving for / 1 and / 2 from Eqs. (5.32) and (5.34), we get I= 1
Vk, ,
,-<1- t ),
R 1-
I = V,kz 2 R 1-
V - E
e-{1-k)z
1
+~'--
(5.35)
V, __ - E + _,._
(5.36)
R
R
. - (1-k),
The ripple current is given by
v,
A/ = /2 - 11 = - kT L
(5.37)
These equations are valid for E s V,. If E ;,:, V, and the converter switch S1 is opened, the inductor transfers its stored energy through R to the source and the inductor current is discontinuous.
Example 5.4 Finding the Currents of a Step-up De Converter Thcstcp-upconvertcrinfigurc5.8a has V, • 10 V,f • I kHz. R • 5 n, L • 6.5 mH, E • 0 V, and k = 0.5. Find / 1, / 2, and /1/. Use SPICE to find these values and plot the load, diode, and switch current.
Solution Equations (5.35} and (5.36} give / 1 = 3.64 A (3.36 A from SPICE} and / 2 2 4.4 A (4.15 A from SPICE). The plots of the load current /(L}. the diode current /(D,.} and the switch current /C(Q1 ) are shown in Figure 5.9.
Copyrghlcd m~leria
5.6 Performance Parameters
181
S~A -r--------------------------------------- ---------= -:
!
r···········
I I
I
SEL>> 1
OA
I----------------------------CD! I (L)
5.0 A ~- - - - -- - ----- --- - ----- - -- --- - --
' I I
I I I
OA ..L
o IC(01)
~:}Il_GD_QD-~DQD_QD_QD_~D Os
15 ms
IOms
Sms
a l (D,.)
lime FIGURE 5.9
SPICE plots of load, input, and diode current for Example S.4.
Key Points of Section S.S
• With a resistive load. the load current and the voltage arc pulsating. An output filter is required to smooth the output voltage.
5.6
PERFORMANCE PARAMETERS The power semiconductor devices require a minimum time to turn on and turn off. Therefore, the duty cycle k can only be controlled between a minimum value km,n a nd a maximum value km,x• thereby limiting the minimum and maximum value of output · voltage. The switching frequency of the converter is also limited. It can be noticed from Eq. (5.22) that the load ripple current depends inversely on the chopping freq uency f.
Toe frequency should be as high as possible to reduce the load ripple current and to minimize the size of any additional series inductor in the load circuit. Toe performance parameters of the step·up and step·down converters are as follows: Ripple current of the inductor. !J. h: Ma,cimum switching frequency, fmax: Cond.ition for continuous or discontinuous inductor current: Minimum value of inductor to maintain continuous inductor current;
Co;.,ynghted ma!cri.i
182
Chapter 5
Dc-Oc Converters
Ripple content of the output voltage and output current, THD; Ripple content of the input current, THD. 5.7
·CONVERTER CLASSIFICATION
The step-down converter in Figure 5.la only allows power to flow from the supply to the load, and is referred to as first quadrant converter. Depending on the direi:tions of current and voltage flows. de converters can be classified into five types: 1. Fust quadrant converter 2. Second quadrant converter 3. First and second quadrant converter 4. Third and fourth quadrant converter S. Four-quadrant converter
first quadrant con~er1er. The load current flows into the load. Both the load voltage and the load current are positive, as shown in Figure 5.lOa. This is a singlequadrant converter and is said to be operated as a rectifier. Equations in Sect.ions 5.2 and 5.3 can be applied to evaluate the performance of a first quadrant converter. Second quadrant conwrter. The load current flows out of the load. The load voltage is positive, but the load current is negative, as shown in Figure 5.lOb. This is also
- - - + - - - VL
0
0
0
(a) First qu3drant
(b) Seoond quadrant
converter
converter
(<) First and Seoond quadrant converter VL
+Vt
0
- le
0
... i1. - vL
(d) Third ond Fourth quadrant converter
(e) Four,quadr1.nt converter
RGURE 5.10
De converter classification.
Copyrighted m lcria
5.7 Converter Classification
183
:L~--~ ~ 1'.~ 0
L
•
(b) Load current
R
T
(I+ k)T
(c) Load voltage
(a) Circuit FIGURE 5.11
Second quadrant convener.
a single-quadrant converter, but operates in the second quadrant and is said.to be operated as an inverter. A second quadrant converter is shown in Figure 5.1 la, where the battery E is a part of the load and may be the back emf of a de motor. When switch S4 is turned on, the voltage E drives current through inductor Land load voltage vL becomes zero. The instanlaneous load voltage vL and load current iL are shown in Figure 5.llb and 5.llc, respectively. The current iL, which rises, is described by
which, with initial condition iL(t ; 0) ; /1, gives iL;
ft e-(RIL)t -
E (I -
e- (RIL)t )
R At t ;
for Ost s kT
(5.38)
It,
iL(t ;
11 ;
kT ) ; /2
(5.39)
When switch S4 is turned off, a magnitude of the energy stored in inductor L is returned to the supply V, via diode Di, The load current iL falls. Redefining the time ori· gin t s 0. the load c.u rrent h is described by
dh
V, ; L di + RiL + E which, with initial condition i(t ; 12) ; 12, gives ;
L
; I
, - (RIL), 2
+ V., - E (I _ R
e- (RJL),)
forOs,sr2
(S.40)
Copyrghlcd malcria
184
Chapter 5 where r2
De-De Converters
= (I
- k )T.At r
= 12,
iL(t = lz) = I , for steady-state continuous current =O for steady-state discontinuous current
{5.41)
Using the boundary conditions in Eqs. (5.39) and (5.41), we can solve for / 1 and
- Vs I, - R
[1 - e-11-kJ,]E e-• R
lz as
{5.42)
1-
(5.43) First and secoud quadrant converter. The load cur.ent is either positive or negative, as shown in Figure 5.lOc. The load voltage is always positive. This is known as a 1wo-q11adranr co11 vuter. The first and second quadrant converters can be combined to form this converter, as shown in Figure 5.12. S 1 and D4 operate as a first quadrant converter. 52 and D4 D1 operate as a second quadrant convener. Care must be taken to ensure that the two switches are not fired together; otherwise, the supply V, becomes short-circuited. This type of convener can operate either as a rectifier or as an inverter. Third and fourth quadrant converter. Toe circuit is shown in Figure 5.13. Toe load voltage is always negative. The load current is either positive or negative, as
R
+ D,
s,
E
FIGURE 5.12 fint and second quadrant converter.
L
+
R
o,
s,
o, . ,'
s,
E
FIGURE 5.13
Third and fourth quadrant conver1cr.
Copyrghlcd matcria
5. 7 Converter Classification
185
shown in Figure 5.lOd. S3 and Di operate to yield both a negative voltage and a load current. When SJ is closed, a negative current flows through the load. When S3 is opened, the load current freewheels through diode Di, S2 and D; operate to yield a negative voltage and a positive load current. When S2 is closed, a positive load current flows. When S2 is opened, the load current free wheels through diode D;. It is important to note that the polarity of£ must be reversed for this circuit to yield a negative voltage and a positive current. This is a negative two-quadrant converter. This converter can also operate as a rectifier or as an inverter. tour-quadrant converter (2]. The load current is either positive or negative, as shown in Figure 5.!0e. The load voltage is also either positive or negative. One first and second quadrant converter and one third and fourth quadrant converter can be combined to forrn the four-quadrant converter, as shown in Figure 5.14a. Toe polarities of the load voltage and load currents are shown in Figure 5.14b. The devices that are operative in different quadrants are shown in Figure 5.14c. For operation in the fourth quadrant, the direction of the battery E must be reversed. This converter forrns the basis for the single-phase full-bridge inverter in Section 6.4. For an inductive load with an emf(£) such as a de motor, the four-quadrant converter can control the power flow and the motor speed in the forward direction ( vL positive and i L positive), forward regenerative braking (vL positive and iL reverse), reverse direction (vL negative and iL reversing) and reverse regenerative braking (vL negative and iL negative).
s,
s,
o,
+
+
s,
E
L
ic
Vs
o,
v,. s,
D,
(a) Circuit
Inverting VI, +
il -
Vt \'c
VL -
Vt
jl, -
Vt
Rcccifyi.ng yl
+ v,
i1,. + v,
Rectifying
\
(b) Potari1ies
S1 ( modulating), S, (modul31ing), 0 2 S2 (continuously on) D 1, D i
53 (modula1ing),
s?. o,
S2 (modubting), D,
S, (continuoustyon) 0 3.
s,, o,
o,
(c) Conducting d evices
FIGURE 5.14
Four-quad.rJnl convener.
Copyrighted m
lcria
186
Chapter 5
De-De Converters
Key Points or Section
s.7
• With proper switch control, the four-quadrant converter can operate and control flow in any of the four quadrants. For operation in the third and fourth quadrants, the direction of the load emf E must be revers.ed internally. 5.8
SWITCHING-MODE REGULATOR5
De converters can be used as switching-mode regulators to convert a de voltage, normally unregulated, to a regulated de output voltage. The regulation is normally achieved by PWM at a fixed frequency and the switching device is normally BJT, MOSFET, or IGBT. The elements of switching-mode regulators are shown in Figure 5.15. We can notice from Figure 5.lb that the output of de converters with resistive load is discontinuous and contains harmonics. The ripple content is normally reduced by an LC filter. Switching regulators are commercially available as integrated circuits. The designer can select the switching frequency by choosing the values of R and C of frequency oscillator. As a rule of thumb, to maximize efficiency, the minimum oscillator period should be about 100 times longer than the transistor switching time; for example, if a transistor has a switching time of 0.5 µs, the oscillator period would be 50 µs, which gives the maximum oscillator frequency of 20 kHz. This limitation is due to a switching loss in the transistor. The transistor switching loss increases with the switching frequency and as a result the efficiency decreases. In addition, the core loss of inductors limits the high-frequency operation. Control voltage 11, is obtained by comparing the output voltage with its desired value. The 110 can be compared with a sawtooth voltage v, to generate the PWM control signal for the de converter. There are four basic topologies of switching regulators [33, 34[: 1•. Buck regulators 2. . Boost regulators 3. Buck- boost regulators 4. Ciik regulators 5.8.1
Bu ck Regulators In a buck regulator, the average output voltage v•. is less than the input voltage, V,hence the name "buck." a very popular regulator [6, 7). The circuit diagram of a buck
Input
De
chopper
+
'• L..fLr
v,
Output
+
v,
v, FIGURE 5.15
Control
Elements or switching.mode rcgulalon.
Copyrighted m~leria
5.8 Switching-Mode Regulators
.
187
o,
+
.,.1.
+
iL, IL
Dm
+
-~ c
v, I Conuol
io, I,
i,:. I(
:1 11.o•d
:
(a) Circuit diagram
v,
Vo
,,
,,
0 iL 1, 1,. 1, 0 ~
f'VVV"
+ is •' iL 0
L
v,
~r=f' +
1~
1, 0
kT
T
kT
T
------
I, I
kT
~
Load
-0
Mode I
L
jL
+ v,
Dm
io .., Ia
v, 0
;.
kT
T
I, Modc2
(b) Equivolcnt circuits
0
(c) Waveforms
FIGURE 5.16
Buck regulator with continuous ;L.·
regulator using a power BJT is shown in Figure 5.1 6a, and this is like a step-down con· verier. The circuit operation can be divided into two modes. Mode 1 begins when Iran· sistor Q, is switched on a l t = 0. The input current, which rises, flows through filter inductor L, filter capacitor C, and load resistor R. Mode 2 begins when transistor Q 1 is switched off at t = 11• The freewheeling diode Dm conducts due to energy stored in the
Copyr ghlcd matcria
188
Chapter 5
Oc-Oc Converters
inductor; and the inductor current continues to flow through L, C, load, and diode D,,,. The inductor current falls until transistor Q1 is switched on again in the next cycle. The equivalent circuits for the modes of operation are shown in Figure 5.16b. The waveforms for the voltages and currents are shown in Figure 5.16c for a continuous current Dow in the inductor L. It is assumed that the current rises and falls linearly. In practical circuits, the switch bas a finite, nonlinear resistance. Its effect can generally be negligible in most applications. Depending on the switching frequency, filter inductance, and capacitance, the inductor current could be discontinuous. The voltage across the inductor L is, in general, eL
di = L dt
Assuming that the inductor current rises linearly from / 1 to / 2 in time ti, \l'.-\l'.=L s
I2 - I 1 ti
a
•t
=L -'-'-
(5.44)
ti
or t:.I L 11
(5.45)
= \/'. - \/'. • s
•
and the inductor current falls linearly from 12 to 11 in time 12 • t:. l -\/'. = - L •
(5.46)
12
or 12
t:.I L
(5.47)
= --
v.
where t:.I = Ii - Ii is the peak-to-peak ripple current of the inductor L. Equating the value of t:.I in Eqs. (5.44) and (5.46) gives t:.I = (V, - v.)11 = v.ii
L
L
Substituting r1 = kT and 12 = (1 - k )T yields the average output voltage as It
\/'.a = \/'.s T - = k\l'.J
(5.48)
Assuming a lossless circuit, V, I, = V0 / 0 = kV, / 0 and the average input current
(5.49)
I; = k/0 The switching period T can be expressed as T
1
t:./ L
t:.I L
= -f = 11 + 12 = v, - v. + - v.-
Al LV,
= ---~
v.(v, - v.)
(5.50)
Copyrghlcd matcria
5.8 Switching-Mode Regulators
189
which gives the peak-to-peak ripple current as
v.(v, - v.) ti.I=
(5.51)
/LV,
or V,k( l - k)
(552)
/L
ti. I =
Using Kirchhofrs current law, we can write the inductor current i L as
If we assume that the load ripple current ti.i0 is very small· and negligible, t,.;L The average capacitor current, which nows into for ri/2 + r,f2 = T/2, is
= ti.i,.
I= ~ c 4 The capacitor voltage is expressed as
I/·
v,=c
1,dr+v,(r=O)
and the peak-to-peak ripple voltage of the capacitor is
ti. \I = v - v (1 = 0) = .!_lm~ dt = ti.I T = ti.I c c c C o 4 8C 8/C
(5.53)
Substituting the value of ti.I from Eq. (5.51) or (5.52) in Eq. (5.53) yields
v.(v, - v.) ti. V,
=
ti. \Ic
-
(5.54)
8LC/2V,
or _ V,k( I - k)
SLC/2
(5.55)
Condition for continuous inductor current and capacitor voltage. If h is the average inductor current, the i.nductor ripple current Ii /
-= 2 /L ·
Using Eqs. (5.48) and (5.52) , we get V5 (1 - k)k
fL
=
2h
2kVs
= 21.
=R
which gives the critical value of the inductor L, as L,
=L =
(1 - k)R Zf
(5.56)
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Chapter 5
Dc-Oc Converters
U V, is the average capacitor voltage, the capacitor ripple voltage ,iV, Eqs. (5.48) and (5.55), we get
_V~ s (_l _- _k=")_ k 8LCf2
= 2V,. Using
= 2V = 2kY:s .•
which gives the critical value of the capacitor C, as
1- k C=C= - c 16Lf2
(5.57)
The buck regulator requires only one transistor, is simple, and has high efficiency greater than 90%. The dildt of the load current is limited by inductor L . However, the input current is discontinuous and a smoothing input filter is normally required. It provides one polarity of output voltage and unidirectional output current. It requires a protection circuit in case of possible short circuit across the diode path.
Example 5.5 Finding the Values of LC Fiiter for the Buck Regulator The buck regulator in Figure 5.16a has an input voltage of V. = 12 V. The required average out· put voltage is V. = 5 Vat R = 500 {land the pealc-to-peak output ripple voltage'is 20 mV. The switching frequency is 25 kHz. If the peak-to-peak ripple current of inductor is limited to 0.8 A, determine (a) the duty cycle k. (b) the filter inductance L. and (c) the filter capacitor C. and (d) the critical values of Land C. Solution V. = 12 v. d V, • 20 mV, IJ./ - 0.8 A.f = 25 kHz. and v. = 5 V.
From Eq. (5.48), V0 b. From Eq. (5.51). L
= kV, and k = V.,IV, = 5/12 = 0.4167 = 41.67%. 5(12 - 5)
L = 0.8 x 25,000 x 12 = 145·83 µH
c. From Eq. (5.53),
C•
o.3
8 x 20 x 10-J x 25,000
d. From Eq. (5.56). we get L, =
(I - k)R 21 I - k
=
From Eq. (5.57), we get C, = 16Lf' = 16
5.8.2
= 200 11F
(I - 0.4167) X 500 Al
x
2X25Xhr I - 0.4167 5.83 x 1oi x (25
= 5.83 mH
x
IO-' )' = 0.4 µP
Boost Regulators In a boost regulator [8, 9) the output voltage is greater than the input voltage-hence the name "boosL" A boost regulator using a power MOSFET is shown in Figure 5.17a. Toe circuit operation can be divided into two modes. Mode 1 begins when transistor
Copyrghlcd matcria
5.8 Switching-Mo.de Regulators
+
L
+
+
v,
\'
'
191
I,. I,
c
(a) Circuit di.ngr:sm v,
v, ---- - -----
1,
,,
kT
~
kT
------
1, 0
v,
lz - ••
Mode 1
Dm +
"· it
v,
L
i,
+
;. v, c
------
T
0 I,
T I
v, V
I I
:-::"'A'v:",;;,;;.:.:;.:__ • ----- : ---· !' -
- ~I -
·AV - -
01-------.---+----~ k,
t
I
t, 1-----' ____._______ Mode2 ( b) Equivalent circuits
01--- - - - - - - - - - - ~ (c) Waveforms
FIGURE 5.17
Boost regulator . . .i th continuous it.·
M, is switched on at t = 0. The input current, which rises, flows through inductor Land transistor Q,. Mode 2 begins when transistor M1 is swilched off at t = 11• The current that was flowing through the transistor would now flow through L, C, load, and diode Dm. The inductor current falls until transistor M 1 is turned on again in the next cycle. The energy stored in inductor L is transferred to the load. The equivalent circuits for the modes of operation are shown in Figure 5.l 7b. The waveforms for vollages and currents are shown in Figure 5.l 7c for continuous load current, assuming that the current rises or falls linearly.
Copyrghtcd matcria
192
Chapter 5
Dc-Oc Converters
Assuming that the inductor current rises linearly from / 1 to / 2 in time ti,
V, = L 12 - 11 = L !ii 11
(5.58)
11
or
AIL V,
(5.59)
/1 a - -
and the inductor current falls linearly from Ii to / 1 in time 12,
V - V =-LA/ I
lz
•
(5.60)
or
AIL
12 = V _ V. ·•
where !i/ = / 2 and (5.60),
-
(5.61)
I
the peak-to-peak ripple current of inductor L. From Eqs. (5.58)
/ 1 is
V,11 (V, - V,)12 A/ = -L- = -'--"-L--=~
Substituting 11 = kT and 12 = (1 - k)T yields the average output voltage,
V. =V.T= ___!L • '12 1-k
(5.62)
Vs (1 - k) = -
(5.63)
which gives
v.
Substituting k =
rr =
11
1J into
Eq. (5.63) yields
-11 =
V, - Vs V.J
(5.64)
Assuming a lossless circuit, V,/1 = V,I, = V,1.,1(1 - k) and the average input current is
1. l,=1-k
(5.65)
The switching period T can be found from
T
1 &;: -
f
a
11
+ t2 s
!ilL + V,
--
AIL AILV, = V, V,(V, - V,)
v. -
(5.66)
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5.8 Switching-Mode Regulators
193
and this gives the peak-to-peak ripple current: A/=
V,(V. - V,) fLV.
(5.67)
or
V,k A/=fL
(5.68)
When the transistor is on, the capacitor supplies the load current for I a 11. The average capacitor current during time 11 is I, = I, and the peak-to-peak ripple voltage of the capacitor is
v - v (1 = 0) ,=. ,'
AV.
1''
11''
l = -Co' I di = I = -I,11 Co° C
(5.69)
Substituting 11 = (V. - V,)l(V.f) from Eq. (5.64) gives
AV,=
1.(v. - V.) V.JC
(5.70)
or !,)<
AV, = fC
(5.71)
Condldoo for continuous iodudor Clllftat and capadtor YO!tage. If his the average inductor current, the inductor ripple current A/ = 2h. Using Eqs. (5.62) and (5.68), we get
-kVs = 2/L = 2/ fL
O
2Vs (1 - k)R
m ---"-
which gives the critical value of the inductor L, as L, • L •
k(l - k)R
21
If V, is the average capacitor voltage, the capacitor ripple vpltagc AV. Eq. (5.71), we get
(5.72)
= 2V,. Using
which gives the critical value of the capacitor C, as
C,
k
= C = 2/R
(5.73)
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Chapter 5
Dc-Oc Converters
A boost regulator can step up the output voltage without a transformer. Due to a single transistor, it has a high efficiency. The input current is continuous. However, a high-peak current bas to flow through the power transistor. The output voltage is very sensitive to changes in duty cycle k and it might be difficult to stabilize the regulator. The average output current is less than the average inductor current by a factor of (1 - k ), and a much higher rms current would flow through the filter capacitor, resulting in the use of a larger filter capacitor and a larger inductor than those of a buck regulator.
Example 5.6 Finding the Currents and Voltage in the Boost Regulator A boost regulator in Figure 5.17a has an input voltage of V, = 5 V. The average output voltage V, = 15 V and the average load current I , = 05 A. The switching frequency is 25 kHz. If L • 150 µHand C = 220 µF. determine (a) the duty cycle k. (b) the ripple current of inductor 6 / , (c) the peak current of inductor /2, (d) the ripple voltage offilter capacitor .iv,. and (e) the critical values of L and C.
Sol11//on V, = 5 V, V. = 15 V, f
•
25 kHz. L • 150 µH , and C = 220 µF.
a. From Eq. (5.62), 15 = 51( 1 - k) or k = 213 = 0.6667 = 66.67%. b. From Eq. (5.67), 61 •
5X(l5-5) 25,000 x 150 x 10·• x 15
=0.89A
c. From Eq. (5.65), I, = 051( 1 - 0.667) = 1.5 A and peak inductor current, / 2 = I,
+
61
2
= 1.5
+
0.89
2
= 1.945 A
d. From Eq. (5.71),
6 V. = 05 x 0.6667 = 60.61 mV ' 25.000 x 220 x 10"6
e. R = v, = ~ = 30fi I,
0.5
From Eq. (5.72), we get L, =
(I - k )kR
21
(I - 0.6667) =
x 0.6667 x 30 "'
2 X 25 X hr
- 133 µH
k 0.6667 From Eq. (5.73), we get C, = 2/R = 2 x 25 x 10' x 30 = 0.44 µF
5.8.3
Buck-Boost Regulators A buck- boost regulator provides an output voltage that may be less than or greater than the input voltage- hence the name "buck-boost"; the output voltage polarity is opposite to that of the input voltage. This regulator is also known as an inverting regulator. The circuit arrangement of a buck- boost regulator is shown in Figure 5:18a.
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5.8 Switching-Mode Regulators
195
+
v,
L
+
;.,.
,.
(a) Circuit diagram
' 'o
-------,, --------
- v.
iL
,, ;, ,, -------0
kT
T
kT
T
1,
+
~
jL
v,
0
;.
c
L
;,
i0
= i1 lz - Ia 0
--------
---------
T
- 1,
Mode I
v,
Dm
_ ave- --
- v,
;,
0
c ic
;., i. = i,
Model (b) Equivalent c.ircuits
••0 (c) Wovcfom:LS
FIGURE 5.18
Buck-boost regulator with continuous ii..
The circuit operation can be divided into two modes. During mode 1, transistor Q1 is turned on and diode D., is reversed biased. The inpul current, which rises, flows through inductor L and transistor Q1. During mode 2, transistor Q1 is switched off and the current, which was flowing through inductor L, would flow through L , C, D.,, and the load . The energy stored in inductor L would be transferred to the load and
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196
Chapter 5
De-De Converters
the inductor current would fall until transistor Q1 is switched on again in the next cycle. The equivalent circuits for the modes are shown in Figure 5.18b. The waveforms for steady-state voltages and currents of the buck-boost regulator are shown in Figure 5.18c for a continuous load current. Assuming that the inductor current rises linearly from / 1 to Ii in time ti,
V, = L 12 - 11 = L Al
'•
or
(5.74)
'•
,, =b.lL --
(5.75)
v,
and the inductor current falls linearly from lz to / 1 in time 12, V
•
= - LA/ 12
(5.76)
or -b.IL
(5.77)
12=--
v.
where b./ = /2 and (5.76),
-
'1 is the peak-to-peak ripple current of inductor L. From Eqs. (5.74) V,11 -v.12 A/=-=-L L
Substituting r1 = kT and 12
= (1
- k)T, the average output voltage is V,k v• - --1-k
(5.78)
Substituting r1 = kT and 12 = (1 - k )Tinto Eq. (5.78) yields
(1 - k) Substituting 12
= (1
- Vs Vs
= v. -
(5.79)
- k)T, and (1 - k) from Eq. (5.79) into Eq. (5.78) yields 11
v. = (V. - Vs)f
(5.80)
Assuming a lossless circuit, V,J, = -v.1. = V,l.kl(l - k) and the average input current I , is related to the average output current / 0 by I0k I,= 1 - k
(5.81)
Copyrghtcd matcria
5.8 Switching·Mode Regulators
1,1
The switching period T can be found from T
1
= -f
a
11
+
12
111 L
111 L
= -+ -- = v, v.
Al L(V. - V,)
(5.82)
v,v.
and this gives the peak·to·peak ripple current,
Al
O
v,v. fL(V0
(5.83)
V, )
-
or
V,k Al=fL
(5.84)
When transistor Q 1 is on, the filter capacitor supplies the load current fort • 11• The average discharging current of the capacitor is le = /0 and the peak-to·peak ripple voltage of the capacitor is .1 V.,
Substituting 11
= VJ[(V, -
1 f'' 1 ['' It = Clo le dt = Clo I, dt = ;/
(5.85)
V,)f) from Eq. (5.80) becomes
1.v.
(5.86)
AV,= (V, - V,)fC or
(5.87) Condition for continuous inductor current ud capadtor voltage. If JL is the average inductor current, the inductor ripple current .1/ = 2/L. Using Eqs. (5.78) and (5.84), we get kV, 2kV, fL = 2/L = 21. = (1 - k )R which gives the critical value of the inductor L , as
Le = L
=
{l - k)R 21
If V. is the average capacitor voltage, the capacitor ripple voltage 11V, Eq. (5.87), we get
l.k
Cf
(5.88) e
2y•. Using
= 2V, = 2l R 0
which gives the critical value of the capacitor C, as k
c, = C = 2/R
(5.89)
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198
Chapter 5
Dc-Oc Converters
A buck-boost regulator provides output voltage polarity reversal without a transformer. It has high efficiency. Under a fault condition of the transistor. the dildt of the fault current is limited by the inductor Land will be V/L. Output short-circuit protection would be easy to implement. However, the input current is discontinuous and a high peak current Oows through transistor Q 1• Example 5.7 Finding the Currents and Voltage In the Buck-Boost Regulator
The buck-boost regulator in Figure 5.18a has an input voltage or V, = 12 V. The duly cycle k = 0.25 and the switching frequency is 25 kHz. The inductance L • 150 µ.Hand filler capacitance C = 220 µ.F. The average load current I , = 1.25 A. Dclcrminc (a) the average output voltage, V.,: (b) the peak-to·peak output voltage ripple, o.V,; (c) the peak-lo·peak ripple current of inductor, A/; (d) the peak current of the transistor. 1, : and (c) the critical vnlucs of Land C.
Solulion V, = 12 V, k = 0.25. /0 = 1.25 A, / = 25 kHz, L = 150 µ.H , and C = 220 µ.F. a. From Eq.(5.78), V, a - 12 X 0.25/(1 - 0.25) = - 4 V. b. From Eq. (5.87), the peak-to-peak output ripple ,•oltagc is AV, •
1.25 X 0.25 - 56.8 mV 25,000 x 220 x 10· •
e. From Eq. (5.84), the peak-to-peak inductor ripple is Iii •
12X0.25 • O.SA 25,000 x 150 x 10...
d. From Eq. (5.81). /, = 1.25 x 0.25/(1 - 0.25) = 0.4167 A. Because I, is the average of duration kT. the peak-lo-peak current of the transistor,
-v
4
e. R= - '= = 3.2fi I, 1.25 From Eq. (5.88), we get L,
=
( 1 - k)R Zf
k
From Eq. (5.89), we gel C, • Z[R •
5.8.4
=
( I - 0.25) X 31
"' = 450 µ.H.
2X25Xhr
0.25
"' = 1.56 µ.F. 2 X 25 X hr x 3.2
Cuk Regulators
The circuit arrangement of the Ctlk regulator [10) using a power bipolar junction transistor (BIT) is shown in Figure 5.19a. Similar to the buck-boost regulator, the Cuk regulator provides an output voltage that is less than or greater than the input voltage. but the output voltage polarity is opposite to that of the input voltage. It is named after its inventor [I]. When the input voltage is turned on and transistor Q1 is switched off,
Copyr ghtcd matcria
. 5.8 Switching-Mode Regulators
iL1,i.
+
+ eL -
C1 ;,1
+ +
L,
GJo,
v,
v.,I-=-
iu
+
+
L,
v•~
VT
199
Dm vcl
t
c,
V0 , Vi.
"'
(a) Circuit diagram db
\', - L1dt
•r
--------
0 -vd,n
kT
T
v., ill
kT
I Lll
I, ILII
0
lu, IL2
lu 1
+Ye~ iu +
L,
ill
;.,, c, +
0 L,
c,
v,.,
v,
~,
Loact
;.
~
1
1u
Vy
T
I
0 • c1
0
L,
-
~I
iw
+
v,
I
- v,
Model
~. c,
kT
i,2
v,., i, Modc2 (b) Equivalen1 drcuilS
c,
;,,
,.
0 ;o
Lood
~-·.,
l,
0 (c) Waveforms
FIGURE 5.19 Olk. rcgula1or.
Copyrghtcd matcria
200
Chapter 5
De-De Converters
diode D,,. is forward biased and capacitor Ct is charged through Li, Dm, and the input supply V,. The circuit operation can be divided into two modes. Mode 1 begins when transistor Q, is turned on at t = O. The current through inductor L1 rises. At the same time, the voltage of capacitor C 1 reverse biases diode D,. and turns it off. The capacitor C 1 discharges its energy to the circuit formed by Ci, Ci. the load, and Li- Mode 2 begins when transistor Q 1 is turned off at t = 11• The capacitor C 1 is charged from the input supply and the energy stored in the inductor Li is transferred to the load. The diode D,,. and transistor Q 1 provide a synchronous switching action. The capacitor Ct is the medium for transferring energy from the source to the load. The equivalent circuits for the modes are shown in Figure 5.19b and the waveforms for steady-state voltages and currents are shown in Figure 5.19c for a continuous load current. Assuming that the current of inductor L 1 rises linearly from I w to h 12 in time Ii, V,
= L, h12 - h11 = Li t:./1 11
(5.90)
11
or
tJ.fiL1
(5.91)
ti=--
v,
and due to the charged capacitor Ci, the current of inductor L 1 falls linearly from to h 11 in time t 2, tl./1
V, - V,1 = -L1 -
hn
(5.92)
12
or (5.93) where Vc1 is lhe average voltage of capacitor Ci, and Eqs. (5.90) and (5.92).
S~bstituting 11
a
kT and 12
a
tl./1
= h 12 -
fi 11 • From
( 1 - k )T , the average voltage of capacitor C 1 is (5.94)
Assuming that the current of filter inductor L,. rises linearly from l w to fi22 in time 11, (5.95)
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5.8 Switching-Mode Regulators
201
or
(5.96) and the current of inductor L2 falls linearly from
fi22 to / l2l
in time r2, (5.97)
or
(5.98) where !1/2 = l w - l u 1. From Eqs. (5.95) and (5.97). !1/2
Substituting It
= kT and 12 =
=
(Vc1 + v.)11 L2
v.12
= - -Li_
( I - k )T. the average voltage of capacitor Ct is
v.
v() = -k
(5.99)
Equating Eq. (5.94) to Eq. (5.99), we can find the average output voltage as kV
v" = - · I - k
(5.100)
v. = --"-v,, -: \Is
(5.101)
which gives
k
\Is
1 - k = -...C....Vs - V0
(5.102)
Assuming a lossless circuit, V.,l , = - v.1. = V,l.kl(l - k ) and the average:. input current,
I,
=I
k l. - k
(5.103)
The switching period Tcan be found from Eqs. (5.91) and (5.93): (5.104) which gives the peak-to-peak ripple current of inductor Lt as !11, -
- V,(V, - Vc1} fL V
(5.105)
t cl
Copyrghtcd matcria
202
Chapter 5
De-De Converters
or
V,k IJ./1 = -
(5.106)
fL,
The switching period T can also be found from Eqs. (5.96) arid (5.98): T
= -1 = 11 + 12 = f
t::.'2L2
v" + v.
t::.lz~ -- = - t::.lz~Vc1 v. v.cv,, + v.)
(5.107)
and this gives the peak-to-peak ripple current of inductor L 2 as
- V0 (V,1 + V0 ) l>lz =
fL
(5.108)
V.
2 cl
or
t::. '2 = -
V0 (l - k} f~
kV,
(5.109)
= -
fL2
When transistor Q, is off. the energy transfer cap'acitor C 1 is charged by the input cur· rent for time t = 12 • The average charging current for C 1 is 1, 1 = I, and the peak-topeak ripple voltage of the capacitor C1 is I
t::,. V.:i = C1
(''
lo
1' ' dt =
l
(''
c,lo 1' tit =
l,t
c.2
(5.110)
Equation (5.102) gives 12 = V,/l(V, - V,)fl and Eq. (5.110} becomes (5.111) or (5.112) If we assume that the load current ripple t,./0 is negligible, l>iu = M a . The average charging current of C 2, which nows for time T/2, is 1, 2 = l>lz/4 and the peak-to-peak ripple voltage of capacitor C2 is · l>Vc2 = -
I
1TIZ fc2 dt = - I 1Tf2-!!/,· cit = -t::,.fi-
C2 o
Cz o
4
8fC2
(5.113)
or (5.114)
Copyr ghted matcria
5.8 Switching-Mode Regulators
203
Condition for continuous inductor current and capacitor voltage. If h, is the average current of inductor Li, the inductor ripple current 1H 1 = 2/u, Using Eqs. (5.103) and (5.106). we get kVs -fL = 2fi1 = 2/s = 1
2k/0 - 1- k
=2
(
k
2
) Vs 1- k R
--
which gives the critical value of the inductor Ltt as ·
Ltt
=
L,
If hz is_the average current of inductor Using Eqs. (5.100) and (5.109), we get kVs
-f lz - =
2ft2
=
(l - k )1R 2kf
Lz, the
(5.115)
inductor ripple current /J./1 = 2lu.
2v. 2kVs = 21. = -R- = (1 - k)R
which gives the critical value of the inductor L a as
L,2 = L2 =
( 1 - k)R / 2
(5.116)
U V. 1 is the average capacitor voltage, the capacitor ripple voltage 6 V, 1 = 2V0 • Using 6 Vtt = 2 V0 into Eq. (5.112), we get I ( I - k)
s
JC,
= 2V = 21 R
•
•
which, after substituting for Is, gives the critical value of the capacitor Cc1 as
k Cr1 = C, = 2/R
(5.117)
If V,2 is the average capacitor voltage, the capacitor ripple voltage 6 Vc2 = 2V0 • Using Eq. (5.100) and (5.114), we get kVs 8C2 Lif2
which, arter substituting for C,2 as
=
zv.
=
0
2kVs 1- k
Lz from Eq. (5 .11 6).gives the critical value of the capacitor Ca = Ci= -
I
8/R
(5.118)
The Ctik regulator is based on the capacitor energy transfer. As a result, the input current is continuous. The circuit has low switching losses and has high efficiency. When transistor Q1 is turned on, it has to carry the currents of inductors L 1 and Lz. As a result a high peak current flows through transistor Q 1• Because the capacitor provides
Copynghlcd m~leria
204
Chapter 5
Dc-Oc Converters
the energy transfer, the ripple current of the capacitor C 1 is also high. This ci.r cuit also requires an additional capacitor and inductor. ·
Example 5.8 Finding the Currents and Voltages In the Cuk Regulator The inp111 vollage or a C~k convcrlcr in Figure S.19a, V, a 12 V. lbe du1y cycle k = 0.25 and 1he . switching frequency is 25 kHt. The filter inductance is L:, = 150 µHand filter capaci1ance is Ci = 220 µF. lbe energy iransfer capaciiancc is C 1 = 200 µF and inductance L 1 a 180 µH. The average load current is I. = 1.25 A. Determine (a) the average output voltage V,; (b) the average input currcnl I,; (c) the peak·to-peak ripple current of inductor Li, /J./1; (d) the peak-lo· peak ripple voltage or capacitor C 1, d v,,; (e) the peak-to-peak ripple current or inductor Li. dfi; (!) the pcak-10-pcak ripple voltage of capacitor C1, d V, 1; and (g) the peak current or the lransistor Ip·
Solution V, = 12 V, k = 0.25. /, C, = 220 µ.F.
= 1.25 A .f = 25 kHz. L
1
= 180 µH, C1 = 200 µF. L1
= 150 µ.H,
and
From Eq. (5.100). V0 • - 0.25 X 12/(1 - 0.25) = - 4 V. From Eq. (5.103). /, = 1.25 X 0.25/( 1 - 0.25) a 0.42 A. From Eq. (5.106). d/ 1 e 12 X 0.25/(25.000 X 180 X 10...) = 0.67 A. From Eq. (5.112).dV,i a 0.42 x (I - 0.25)/(25.000 x 200 x 10-<>) = 63mV. c. From Eq. (5. 109), d/2 • 0.25 x 12/(25.000 x 150 x 10... ) = 0.8 A. f. From E
1 V.1,., = -kV,., = -Yak - k == V,,
(5.119)
For a lossless circui1. / ,.1VJ," :;;; V11 / ,, and 1he average value of the curren1 in inductor Lz.is
,.. v..
,, , = • ,, .. VJ.,, •
(5.120)
= 1.25 A
· Therefore.the peak current or transislor is , ,, = , . +
5.8.5
di,
2
+ ,,., +
d/2
2
0.67
0.8
= 0.42 + - 2- + 1.25 + T = 2.405 A
Limitations of Single-Stage Conversion The fo ur regulators use only one transistor. employing o nly o ne stage conversion. and require inductors o r capacitors for energy transfer. Due to the current-handling limitation of a single transistor, the output power of these regulators is small. typically tens of wans. At a higher current, the size of these components increases. with increased component losses, 3nd the efficiency decreases. In addition, there is no isolation between the input and output voltage, ,vhich is a highly desirable criterion in n,ost applications. For high-power applications. multistage conversions arc used. where a de voltage is
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5.9 Comparison of Regulators
205
converted to ac by an inverter. The ac output is isolated by a transformer and then converted to de by rectifiers. The multistage conversions are discussed in Section 14-4. Key Points of Section 5.8
• A de regulator can produce a de output voltage, which is higher or lower than the de supply voltage. LC filters are used to reduce the ripple content of the output voltage. Depending on the type of the regulator, the polarity of the output voltage can be opposite of the input voltage. 5.9
COMPARISON OF REGULATORS
When a current flows through an inductor, a magnetic field is set up. Any change in this current changes this field and an emf is induced. This emf acts in such a direction as to maintain the flux at its original density. This effect is known as the self-induction. An inductor limits the rise and fall of its currents and tries to maintain the ripple current ·low. There is no change in the position of the main switch Q1 for the buck and buck-boost regulators. Switch Q1 is connected to the de supply line. Similarly, there is no change in the position of the main switch Q1 for the boost and Olk regulators. Switch Q1 is connected between the two supply lines. When the switch is closed, the supply is shorted through an inductor L, which limits the rate of rise of the supply current. In Section 5.8, we derive the voltage gain of the regulators with the assumptions that there were no resistances associated with the inductors and capacitors. However, such resistances though small may reduce the gain significantly (11, 12]. Table 5.1 summarizes the voltage gains of the regulators. Inductors and capacitors act as energy storage elements in switched-mode regulators, and as filter elements to smootb out the current harmonics. We can notice from Eqs. (B.17) and (B.18) in Appendix B that the magnetic loss increases with the square cif frequency. On the other hand, a higher frequency reduces the size of inductors for
TABLE 5.1 SummariesofRegula1orOains!Ret II) VollageGain,G(k ) • VJVs
Regulalor Buck
with Negligible Values of,,. and re k
Voltage Gain,G(k) = V,!Vs 1'i1h Fini1e Va.Juts of''- and re kR
R+
'L
I
I - k
-k I - k
- k
I - k
(I - k)' R
[
(I - k )'R
[
+ ' < + k{I -
k)(,;:\{)
2
( I - k) R
]
]
( I - k)2R + ' < + k (I - k)c: c:R)
Copyrghicd malcria
206
Chapter 5
De-De Converters
the same value of ripple current and filtering requirement. The design of de-de converters requires a compromise among switching frequcqcy, inductor sizes, capacitor sizes, and switching losses. S.10
MULTIOUTPUT BOOST CONVERTER
For a digital signal processor, high-speed computation requires a high supply voltage V, for fast switching. Because power consumption is proportional to the square of V,. it is advisable to lower V, when lower computation speed is needed. A boost converter can be used to power high-speed processor cores with a very low supply voltage. A singleinductor dual-output (SIDO) boost converter topology [12) is shown in Figure 5.20. The .two outputs v.. and Voi, share the inductor Land the switch S1• Figure 5.21 shows the timing of the converter. It works with two complementary phases 'Pa and lfJb· During lfJo = 1, Sb is opened and no current flows into V00, whereas 5 1 is closed first. -!be inductor current fi increases until the time k 10T expires (determined by the output of an error amplifier), where Tis the switching period of the converter. During the time k2,T. 5 1 is opened and 50 is closed to divert the inductor current into the output V... A zero current detector senses the inductor current, and when it goes to zero. the converter enters the time k 3aT, and s. is opened again. The inductor current stays at zero until 'Pb = 1. lbus, k 1,, k2o, and k'Ju must satisfy the requirements that k,. k,.
+ k 2o s 0.5
+ k2o + k3.
(5.121)
(5.122)
= 1
Du ring 'Po = I , the controllc,· multiplexes the inductor current into the output V00 during !fJo = 1. Similarly, the controller multiplexes the inductor current into the output V"" during 'Pb = 1. The controller regulates the two outputs, alternately. D ue to the presence of k3oT and k3.T, the converter operates into the discontinuous conduction mode (DCM), essentially isolating the control of the two outputs such that load variation in one output does not affect the other. Therefore, the problem of cross regulation is alleviated. Another advantage of DCM control is simple compensation of the system
v.,
S,
L
v,-
s.
+
s,
l v~ c.
C,
FIGURE 5.20
Single-inductor dut1l-0utput boosl ronverle-r. (Re( 12, D. Ma)
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5.10
Multioutput Boost Converter
207
9,_J ~
t ,.T -
s,
n...____ 9•1 ..___. .
s.----..........,1i- j'.-
L - k,.T-
n.___
FIGURE 5.21
Timing d.i.agram ror the dual-output boost converter.
because there is only one LHP pole in the transfer function of the loop gain of each of the output(l3). With similar time multiplexing control, the dual-output converter can easily be extended to have N outpu1s as shown in Figure 5.22, if N nonoverlapping phases are assigned to the corresponding outputs accordingly. Ry employing time multiplexing (TM) control, a single controller is shared by all the outputs. Synchronous rectification, in the sense that the transistor in replacing the diode is switched off when the inductor current tends to go negative, is employed, thus eliminating diode drops and enhancing efficiency. All power switches and the controller can be fabricated on-chip (14, 15) and with only one inductor for .ill outputs, off-<:hip components arc minimized.
Key Points or Sedlon 5.10
• The boost converter can be extended to yield multiple outputs by using only a single inductor. By employing TM control, a single controller is shared by all the outputs. All power switches and the controller can be fabricated on-chip and with only one inductor for all outputs, off-chip components are minimized. This converter could find applications such as a power supply for high-speed digital signal I processors.
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208
Chapter 5
Oc-Oc Converters L ~ -rYV-,1"\.._
_
_ _.ef
s, o----------,v.,
t-- - e f o-- - - - - - , Vos - 1
-'
.
SN-I
v
s.
FIGURE 5.22
Topology or boost oon,•erter \\ith N outputs.
5.11
DIODE RECTIFIER-FED BOOST CONVERTER
Diode rectifiers are the most commonly used circuits for applications where the input is the ac supply (e.g., in computers, telecommunications., fluorescent lighting, and air· conditioning). The power factor of diode rectifiers with a resistive load can be as high as 0.9, and it is lower with a reactive load. With the aid of a modern control technique, the input current of the rectifiers can be nladc si nusoidal and in phase ,vith the input
voltage, thereby having an input PF of approximate unity. A unity PF circuit that combines a full-bridge rectifier and a boost converter is shown in Figure 5.23a. The input current of the convener is controlled to follow the full-rectified waveform of the sinusoidal input voltage by PWM control (16-23]. The PWM control signals can be generated by using the bang- bang hysteresis (BBH) technique, similar 10 the delta modulatiory in Figure 6.26. This technique, which is shown in Figure 5.23b, has the ad· vantage of yielding instantaneous current control, resulting in a fast response. However, the switching frequency is not constant and varies over a wide range during each half-cycle of the ac input voltage. The frequency is also sensitive to the values of the circuit components. The switching frequency can be maintained constant by using the reference current l,.r and feedback current li b averaged over the per-swit.c hing period. This is shown in Figure 5.23c. 1,.1 is compared with /fb, If 1,.1 > /fb, the duty cycle is more than 50%. For l,.r = /fb, the duty cycle is 50%. For l,c1 < /fb, the duty cycle is less than 50%. The error is forced lo remain between the maximum and the minimum of the triangular waveform and the inductor current follows the reference sine wave, which is superimposed with a triangular waveform. The reference current l r,.r is generated from the error voltage V.( = V0 } and the input voltage Vin 10 the boost converter. The boost converter can also be used for the power factor (PF) correction of three-phase diode rectifiers with capacitive output filters (19, 29) as shown in Figure 5.24. The ~ost converter is operated under DCM of the inductor current mode to achieve a sinusoidal input current shaping. This circuit uses only one active switch,
V.., -
Copyrighted m lcria
5.11
+
Diode Rectifier-Fed Boost Converter
L
209
D.,
+ Gate
Ac
supply
Vin
Dri\'C circuil
u
M,
v.
V/1
Controller
COD\'Crlcr
K,
c,
Load
(a) Circuit arrangement
Hysteresis window
Sine reference current
/
(b) Bang- oong hysteresis current control gate signals J,d 1ine "''ave
Reference wave
supcrimJ>O$Cd Yiith triangular v.-avc
(c) Current control FIGURE 5.23
Power factor conditioning or diode rectifiers.
Copyrghlcd malcria
210
Chapter 5
Dc-Oc Converters
Dm
+
v. v,
L
vb
L
Js,
CL
Rt
v,
L
FIGURE 5.24
Thrcc·phuc rcctificr·fcd boost convener. [Ref: 29, C. Mufioz)
with no active control of the current. The drawbacks of the simple converter are excessive output voltage and the presence of fifth harmonics in the line current. This kind of converter is commonly used in industrial and commercial applications requiring a high input power factor because their input-current waveform automatically follows the input-vollage waveform. Also, the drcuit has an extremely high efficiency. However, if the circuit is implemented with the conventional constant-frequency, low-bandwidth, output-voltage feedback control, which keeps the duty cycle of the switch constant during a rectified line period, the rectifier input current exhibits a relatively large fifth-order harmonic. As a result, at power levels above 5 kW, the fifthorder harmonic imposes severe design, performance, and cost trade-offs to meet the maximum permissible harmonic current levels defined by the IEC555-2 document [30). Advanced control methods such as the harmonic injection method [31) can reduce the fifth-order harmonic of the input current so that the power level at which the input current harmonic content still meets the IEC555-2 standard is extended. Figure 5.25 shows the block diagram of the robust, harmonic injection technique introduced in [3-5). A voltage signal that is proportional to the inverted ac component of the {ectified, three-phase, line-to-line input voltages is injected into the output-voltage feedback loop. The injected signal varies the duty cycle of the rectifier within a line cycle to reduce the fifth-order harmonic and improve the THD of the rectifier input currents.
Key Points of Section S.ll • The full-bridge rectifier can be combined with a boost converter to form a unity power factor circuit. By controlling the current of the boost inductor with the aid of feedback control technique, the input current of the rectifier can be made sinusoidal and in phase with the input voltage, thereby having an input PF of approximate unity.
Copyrghlcd matcria
5.12 Chopper Circuit Design
21 1
,------Boost : I
i,
v,
v.
;.
EM!
filter
4
v.
: inductor L
I I
I I I
+
, ,,
I =-=;i I I I
c
l - - --7
,,
I I I
,
I
::.=.::..=..=:
I
!._ _____ _ I,
Senfingand scaling circuit OutpUl·VOltagc
dh•ider
,-- -, : R1
:
I
I I I I
f----r--+
High-pass
I I
lilter
:~ :
I
A
I
•
FIGURE 5.25
Three-phase DCM boost rectifier with a hannonic injection method. (Re( 31. Y. Jang)
• 5.12
CHOPPER CIRCUIT DESIGN
We can notice from Eq. (5.7) that the output voltage contains hannonics.An output fit. ter of C, LC, L type may be connected to the output to reduce the output harmonics (24, 25]. The techniques for filter design are similar to that of Examples 3.17 and 10.15. A converter with a highly inductive load is shown in Figure 5.26a. The load current ripple is negligible (A/ = 0). If the average load current is I,, the peak load current is Im = 1. + A/ = / 0 • The input current, which is of pulsed shape as shown in Figure 5.26b, contains bannonics and can be expressed in Fourier series as i.h(r)
= k/ +
0
+ .!.!._
I
co
f
nn n• l
sin 2n-rrk cos 2n-rr/r
(5.123)
--!. ~(1 - cos 2mrk) sin 2n-rrfr n1r n• l
The fundamental component ( n "' 1) of the converter-generated hannonic current at the input side is given by
. ()
llh I
1. . 21rk cos 21rf1 + -r. (1 = -sm 1r 1r
cos 2,rk
) SID . 21rf1
{5.124)
Copyr ghtcd matcria
212
Chapter 5
De- De Converters
1, 1 - - -
o~-~~-~--~~
kT
(a) Circuit diagram
T
(b) Qopper current
FIGURE 5.26
Input cunent • •.1.,·e.form of con\'encr. Con ..·ener
+
I,
c.
v,
D.,
FIGURE 5.27
Converter with input filler.
,,.
c.
FIGURE 5.28
Equivalcnl circuit for harmonic currents.
•
In practice, an input filter as shown in Figure 5.27 is normally connected to rnter out the converter-generated harmonics from the supply line. The equivalent circuit for the converter-generated harmonic currents is shown in Figure 5.28, and the rms value of the nth harmonic component in the supply can be calculated from
1 '"
1
1
= 1 + (2mrf)2L,C, 1"" = 1 + (nflf0 ) 2 1••
5 25 ( .1 )
Vlh)
where f is the chopping frequency and fo a 1,(2,r is the filter resonant frequency. If (flfo) >> l. which is generally the case, the nth harmonic current in the supply becomes
I,, = I,•(~;y
(5.126)
A high chopping frequency reduces the sizes of input filter elemenls, but the frequencies of converter-generated harmonics in the supply line are also increased; and this may cause interference problems with control and communication signals. If the source has some inductances, L,, and the converter switch as in Figure 5. la is turned on, an amount of energy can be stored in the source inductance. If an attempt
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5.12 Chopper Circuit Design
213
is made to tum off the converter switch, the power semiconductor devices might be damaged due to an induced voltage resulting from this stored energy. The LC input fiJ. ter provides a low-impedance source for the converter action.
Example 5.9 Fi nding t he Harmonic Input Curr.elit of a De Converter A highly inductive load is supplied by a converter as shown in Figure 5.26a. 11,e average load current is /0 c H)O A and the load ripple current can be considered negligible ( /J./ a 0). A sim· pie LC input fill.e r with L, = 0.3 mH and C, = 4500 µFis used. If the converter is operated al a frequency of 350 Hz and a duty cycle of 0.5, determine the maximum rms value of the funda· mental component of converter-generated harmonic current in the supply line.
Solution For I. = JOO A,/ = 350 Hz, k = 0.50, C, = 4500 µF, and L, 136.98 Hz. Equation (5.124) can be wri1ten as
= 0.3 mli, / 0 = l/(21T v'f.4) =
1,.(1 ) = A, cos 21T/I + B, sin 21r/1 where A 1 = (//-rr) sin 21Tk and 8 1 = (//,r)(l - cos 2,rk). The peak magnitude of this current is calculated from
lp1, = (Al + B/) 112 =
¥2/, (1 - cos21Tk) 112
(5.127)
1T
Therms value of this current is
/ 1•
= !! (1 - cos 2-:rk) 112 = 1T
45.02 A
and this becomes ma:
=
45.02
I + (350/136.98)2
= 5 98 A •
I, the harmonic current in the supply becomes approximately
,,, = ,,.
7 ('•)'
Key Points of Section S.U • The design of a de-de converter circuit requires (a) determining the converter topology, (b) finding the voltage and currents of the switching devices, (c) finding the values and ratings of passive elements such as capacitors and inductors, and (d) choosing the control strategy and the gating algorithm in order to obtain the desired output.
Copyr ghicd matcria
214
Chapter 5
De-De Converters
Example 5.10 A buck converter is shown in Figure 5.29. The input voltage is I', = 110 V, the average load voltage is V., • 60 V, and the average load current is 10 • 20 A. The chopping frequency is f = 20 kHz. The peak-to-peak ripples are 2.5% for load voltage, 5% for load current, and 10% for filter L , current. (a) Determine the values or L,, L, and C,. Use PSpicc {b) to verify the re· suits by plotting the instantaneous capacitor voltage Ve, and instantaneous load current 11,.; and (e) to calculate the Fourier coefficients and the input current is. The SPICE model parameters of the transistor are IS • 6.734{, BF • 416.4, BR • 0.7371, CJC & 3.638P, cJE a 4.493P, TR = 239.SN, TF = 3012P, and tha t or the diode are IS = 2.2E-15, BV = 1800V, TT • O.
Solution V,
= llOV, v. = 60V, /• = 20A. AV, • O.Q25 x v. v. 60
• 0.025 x 60 •
1.5 V
R = -=-=30 1. 20
From Eq . (5.48). k
v.
60
= -v, = -110 = 0.5455
From Eq. (5.49),
= 0.5455 x 20 = 10.91 A
I,= k/0 ll.fc - 0.05
A/
= 0.1
x
I, - 0.05
x 20 - I A
x I, = 0.1 x 20 = 2 A
•· From Eq. (5.51). we get the value of L,:
v.(v, - v.J L, =
AlfV,
a
60 x (110 - 60) 2 x 20 kHz x 110 • 6Sl.S2 ILH
From Eq. (5.53) we get the value or C,:
C,
A/ X 8/ = 1.5
= AV,
,<
2 8 X 20kHz
;,
= S.JJ l'-f L,
L
4
+
v,
+ ltOV
o..
c.
v,
R
FIGURE 5.29 Buc.k con\ er1cr. 1
Copyrghlcd matcria
215 .
5.1 2 Chopper Circuit Design
v,
o,
2
L,
3
4
681.82 µH
ov
L 40.9lµH
6
v,
Ra
IIOV
R
2500
c,
Dm
7
v,
v, 0
8.33µF
8 30
s
ov
(a) O rcuit
20~, so,...
27.28µ,
0
(b) Control voltage
AGURE 5.30
Buck chopper for PSpice simulation.
Assuming a linearrise of load currentiL during the time from t • 0 to,, • kT, we can write approximately
6h
Ah
L - • L - • 6Vc 1, kT
which gives the approximate value of L: kTt,,V, kt,,V, L = ~ = t,,/J {5.128)
• 0.5454 x 1.5 • 40.91 H IX20kHz
µ
a 0.5455,/ = 20 kHz. T = 1/f = 50 µs, and /00 a k X T = 27.28 µs. The buck chopper for PSpicc simulation is shown in Figure 5.30a. The control voltage v, is shown in Figure 5.30b. The list of the circuit file is as follows:
b. k
EX.alrl)le S. 10
Buck converter
vs
l
0
VY Vg
l 7
2 l
oc oc
RB
7 l
6
250
4
681. 82UH
LE
uov ov
PULSE I OV
; Volt.age source to meo.sure input current 20V
0
O.lNS
O. lNS
27 . 28\IS SOUS)
, Transistor base resisUUlee
Copyrghtcd matcria
216
Chapter 5
De-De Converters
CE
4
0
8.llUP
L R VX
4 8 S
8 5 O
40.911JH 3
a<
0
J
tMlD
.!«lDEL 01
CH)ll
2
6
o::
IC=60V
:
initial voltage
; Volt.age source to measure load current.
OJ
ll(IS=2.2E-1S BV=lBOOV Tl'•O)
; Freewheeling diode ; Diode model par..,.ter•
3
; SJT ...it.ch
(Ja)
.l!Ot>EL Qllll) NPN IIS=6.734P 8F=416.4 BR=.7371 CJC=3 . 638P • C,Es4. 493P TR• 239.SN ~30l.2P) an model po.ro.meters • TRAN
lUS 1. 6MS 1. SMS lUS
; Transient analysis
UIC
. PROBE
; Graphics postprocessor .options abetol • l .OOn reltol • 0.01 vntol • 0 .1 ITLS• SOOOO ; convergence .FOUR
20KHZ
t{VY)
: Fourier analysis
. END
The PSpice plots arc shown in Figure 5.31, where l (VX ) = load current. /(l•) = inductor l, current, and V(4) s capacitor voltage. Using the PSpice cursor in Figure 5.31 gives v, s v, = 59.462 v . .i.v, = 1.782 v. '1./ • 2.029 A.1,,.1 • 19.813 A , IJ.h = 0.3278 A , and/,= 19.8249 A . Thisverilics the design;howcver, IJ.fi gives a better result than expected.
Enmple5.IO
A Buck COn\'eMer
Temperature: 27.0 80.0V +-- - - - -+-- - - - -,r --
- - --1.- - - - --t--- ---i-
40.0V-+------+-----+------+------+------+ D V (4) 20.0A+-- - - - - t - - - - -- ; - - - - - - t - - - - - - ; - - - ----;-
19.6A +------+------+------+------,,------+ o I (VX) 40.0A + - - - - - - t - - - - - - ; - - - - - - t - - - - - + - - - - - - t
20.0Aia-- - - - - - - - - - - - - - - - - - - - - - - + O.OA +--
----+-- ----+------+-----if------+
I.SO ms o I (Le)
1.52 ms
1.54 ms
1.56 ms
1.58 ms
l.60m,
Time
FIGURE 5.31 PSpicc plots for Example S.tO.
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5.13 State-Space Analysis of Regulators
217
c. The Fourier coefficients of the input current are FOURIER COMPONOll'S OF TRA.'-:SIENT RESPONSE I (VY}
DC' COMPOMlNr • l .079535E+Ol HARMCltflC F ~lCY =,UER IJ:)
l 2 3 4 6 7
2 .OOOE•04 4 .000&+04 6.000E•04 8. 000£+04 l.OOOE+OS 1.200£•05 1.400£•05
8
l . 600E•OS
s
== l. 251E•Ol
(HZ)
1. 769E+OO 3 .8 48E+OO 1. 686£•00 l. 939E•OO l. 577E+OO 1. 014£•00 l.435E•OO
1.800£+05 4. 385£- 01 9 TCYI'AL HARMJNIC DISl'ORTION =
NORK>.LIZED
PH>.SE
!IORMl\LIZED
CQ'.l'Ot
(llEX:l)
PHASE l llEX:l) O.OOOE+OO 9 .163E•Ol
l.OOOE+OO -l.195E+OI l. 415E- 01 7 . 969E+Ol -3 . 13 1.E•Ol 3.076E-01 1. 348£-01 5.500£•01 l. SSlE-01 -S . 187E•01 1.261£-01 3.347£+01 8.107£-02 - 7 . 328E+Ol 1.147£-01 l .271E•Ol 3. 506£- 02 -9. 751E+Ol 4.401661E•Ol PERCENI'
-1. 937E•Ol
6 . 695£•01 -3.992E•Ol 4 . 542£•01 - 6 . 13lE•Ol 2 .466E•Ol - 8 . 556E•Ol
Key Points of Section S.12 • The design of a de-de converter circuit requires (1) determining the converter topology, (2) finding the voltage and currents of the switching devices, (3) finding the values and ratings of passive elements such as capacitors and inductors, and
(4) choosing the control strategy a.nd the gating algorithm 10 obtain the desired output.
5.13
STATE-SPACE ANALYSIS OF REGULATORS Any 11th order linear or nonlinear differential equation in one time-dependent variable can be written (26) as n first-order differential equation in II time-dependent variables x, through x •. Let us consider, for example, the following third-order equation: (5.129) where y' is the first derivative of y, y' = (dldt) y . Let y be x 1• Then Eq. (5.129) can be represented by the three equations
= X2
(5.130)
,t2" :;; .t3
(5.131)
.t3~ :=: - a ox1 - a1,t2 - a3x3
(5.132)
X1'
In eacb case, n initial conditi 10 and for a given set of input
sources.
Copynghlcd m~leria
• 218
Chapter S
Dc-Oc Converters
x,
s,
L
-i
+
u,
s,
c
+
x,
R
(a) Con,·c rt circui1
-
x,
x,
L
+
u,
c
+
i. +
x,
, • Sz
R
(b) &juivalenl circuit !or mode I
C ::~ Xz
> '>R '>
•
(c) &juivalcnt circuit !or mode 2
FIGURE 5.32
Buck converter with state variables.
All state variables are subscribed x's and all sources are subscribed r,'s. Let us consider the basic buck converter of Figure S.16a, which is redrawn in Figure 5.32a. The de source V, is replaced with the more general source r, 1•
Mode L Switch S1 is on and switch S2 is off. The equivalent circuit is shown in Figure S.32b.Applying Kirchoffs voltage law (KVL), we get u1 =
CX2 '
Lx1' + x 2
= X1
-
1 x2
-
R
which can be rearranged to (S.133)
(5.134) These equations can be written in the universal formal:
Copyr ghtcd matcria
5.13 State-Space Analysis of Regulators
+ B1u 1
x' = A 1x
where
x
219
(5.135)
= state vector = ( ::)
1) -1
RC
u 1 = source vector
8 1 = source coefficient matrix = (
i)
Mode 2, Switch 5 1 is off and switch 52 is on. The equivalent circuit is shown in Figure 5·32c. Applying KVL, we get
which can be rearranged to Xt' c
-1 L
- xz - 1
x 2'
=C
(5.136)
I Xz + RC Xz
(5.137)
These equations can be written in the universal format:
x' where
x
= A 2x +
Bzu1
(5.138)
= state vector = (;:)
A, • '"" oo,ffici,o< =W • ( ;
:1) -1
RC Ut = source vector = 0
Bz = source coefficient matrix = (~) In feedback systems, the duty cycle k is a function of x and may be a function of u as well. Thus, the total solution can be obtained by state-space averaging, that is, by summing the terms for each analysis the switched linear mode. Using the universal
Copyrghtcd matcria
220
Chapter 5
Dc-Oc Converters
format, we get A
= A1k +
A2(l - k)
(5.139)
B
= B1k
8 2(1 - k)
(5.140)
Substituting for A., A 2, B., and
+
Bi, we can find
A=(:
~1)
(5.141)
C RC 8
= (;)
(5.142)
which in turn lead to the following state equations: k
- 1
T
(5.143)
Xz + L "•
x ,'
=
.r2'
= 7:x2 + RC.r2
-1
I
(5.144)
A continuous but nonlinear circuil thal is described by Eqs. (5.143) and (5.144) is shown in Figure 5.33. It is a nonlinear circuit because k in general can be a function of x l, x2. and u 1. Toe state-space averaging is an approximation technique that, for high enough switching frequencies, allows a con1inuous-1ime signal frequency analysis to be carried out separately from the switching frequency analysis. Although the original system is linear for any given switch condition. the resulting system (i.e., in Figure 5.33) gen· crally is nonlinear. Therefore, small-signal approximations have 10 be employed to obtain the linearized small-signal behavior before other techniques [27, 28J, such as Laplace transforms and Bode plols, can be applied.
-
x,
L
... ku 1
c
+
x,
R
FIGURE 5.33
Continuous equivalent circuil or the Buck converter with state ,·ariables.
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References
221
Key Points of S«tlon S.13
• The state-space averaging is an approximate technique that can be applied to describe the input and output relations of a switching converter, having clifferent switching modes of operation. Although the original system is linear for any given switch condition, the resulting system generally is nonlinear. Therefore, small-signal approximations have to be employed to obtain the linearized small· signal behavior before other techniques can be applied.
SUMMARY A de converter can be used as a de transformer to step up or step down a fixed de; voltage. The converter can also be used for switching-mode voltage regulators and for transferring energy between two de sources. However, harmonics arc generated at the input and load side of the converter, and these harmonics can be reduced by input and output filters. A converter can operate on either fixed frequency or variable frequency. A variable-frequency converter generates harmonics of variable frequencies and a filter design becomes difficult. A fixed-frequency converter is normally used. To reduce the sizes of filters and to lower the load ripple current, the chopping frequency should be high. The state- space averaging technique can be applied to describe the input and output relations of a switching converter, having different switching modes of operation.
REFERENCES
[JJ J. A . M. Bleijs, and J. A. Gow, "Fast maximum power point control or current-fed DC- DC converter for photo voltaic arrays," Elcc1ro11ics u11us, Vol. 37, No. 1, January 2001, pp. 5-6. (21 A. 1. Forsyth, and S. V. Mollov. " Modeling and control or DC-DC converters," Power Engineering Journal, Vol. 12, No. 5, 1998, pp. 229- 236. [3[ A. L. Baranovski. A. Mogel, W. Schwarz. ·a nd O. \Voywode, '"Chaotic control or a DCDC-Converter," Proc. IEEE ltuer11a1ional Sy11rp0Ji1un on Circuits a11d Systt:.fllS, Gcncva,Swit.zcrland, 2000, Vol. 2, pp.11-108-ll-111.
(4) H. Matsuo. F. Kurokawa. H. Etou, Y. lshizuka, and C Chen. Cbangfeng, "Design oriented analysis or 1hc dig.ilally controlled de-de converter," Proc. IEEE Pol1''tr Elecsronics Specialists
Co11fertncc. Galway. U.K., 2000, pp. 401-407. [SJ J. L. Rodriguez Marrero, R. Sanlos Bueno, and G. C Vcrghc.c. "Analysis and control of chaotic DC-DC switching power con,•crtcrs." Proc. IEEE lntemational Symposi11m on Circuits and Sysrems, Orlando, FL. Vol. 5, 1999, pp. V-287-V-292. (61
O . Ioannidis. A . Kundinnis, and S. N. l\.1anias. "'Novel control design fo r the buck converter,"'
IEE Proceedings: Electric Powtr Applications, Vol. 145, No. 1, January 1998, pp. 39-47. (7) R . Oruganti.and M. Palaniapan. "Inductor voltage control or buck-type single-phase ac- dc converte r,'"/£££ Tranwctlons 011 Power Eltetro11/cs, Vol. 15, No. 2, 2000, pp. 4U-417. (8) V. J. ll1ottuvelil, and G. C. Verghese, "Analysis and control des ign of paralleled DC/DC coa-
vcrtcrs ,vilh current sharing." IEEE Transactions On Po1ver Eltctronics, Vol. 13, No. 4, 1998, pp. 635-644. (9) Y. Berkovich, and A . loinovici, "Dynamic model or PWM zero-voltage-transilion DC- DC boost convcrte-r... Proc. IEEE /11t~r,u11ional Sy,nposiwn 011 Circuits ancf'Systmzs. Orlando. FL. Vol.5, 1999, pp. V-254- \1-25.
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Chapter 5
Dc--Oc Converters
(10) S. Cllk and R. D. Middlebrook, "Advances in switched mode power conversion," IEEE Transactions on Industrial Electronics, Vol. lE30, No. l, 1983, pp. 10-29. [11) K. Kit Sum, Switch Mode Power Ccnvenion-Basic Theory and Design. New York: Marcel Dekker. 1984, Chapter I. [U). D. Ma, W. H. Ki, CY. Tsui, and P. K. T. Mok,"A 1.8-V single-inductor dual-output switching convener for power reduction techniques," Symposium on VLSI Circuits. 2001,pp. 137-140. (13) R. D. Middlebrook and S. Cuk, "A general unified approach to modeling de-to-de convert· ers in discontinuous conduction mode,'' IEEE Power Electronics Specialist Conference, 1977, pp.36-57. (14) H. S. H. Chung, "Design and analysis of a switched-capacitor-based step-up DC/DC con· vener with continuous input current," IEEE Tronsactions on Circuits and Systems I: Fundamental Theory and Applications, Vol 46,No. 6, 1999, pp. ~730. [15) H. S. H. Chung, S. Y. R. Hui, and S. C Tung. "Development of low-profile DC/DC converter using switched-capacitor circuilS and coreless PCB gate drive," Proc. IEEE Power Electronics Specialists Conference, Charleston, SC, Vol. 1, 1999, pp. 48-53. (16) M. Kazerani, P. D. Ziogas, and O. Ioos. "A novel active current wave shaping technique for solid-state input power factor conditionen," IEEE Transactions on Industrial Electronics, Vol. lE38, No. 1, 1991,pp. 7.2-78. [17) B. I. Thkahashi, "Power factor improvements of a diode rectifier circuit by dither signals," Conference Proc. IEEE-US Annual Muting, Seattle, WA, October 1990, pp. 1279-1294. [18) A. R. Prasad and P. D. Ziogas. "An active power factor correction technique for three phase diode rectilien,'" IEEE Transactions on Power Eltctronics, Vol. 6, No. I, 1991, pp. 83-92. [19) A. R. Prasad, P. D.Ziogas, and S. Manias. "A passive current wave shaping method for three phase diode rectifiers," Proc. IEEE APEC-91 Conference Record, 1991, pp. 319-330. (20) M. S. Dawandc, and 0. K. Dubey, "Programmable input power factor correction method for switch-mode rectifiers," IEEE Tronsactions on Powtr Electronics, Vol. 2, No. 4, 1996, pp. 585-591. [11) M. S. Dawande, V. R. Kanetkar, and O. K. Dubey, "Three-phase switch mode rectifier with hysteresis current control," IEEE Transaccions on Powtr Eltctronics, Vol. 2, No. 3, 1996, pp. 466-471. (22) E. L. M. Mehl. and I. Barbi. "An improved higb·power factor and low-east three-phase rectifier," IEEE Transactions on Industry ApplicJJtions, Vol. 33, No. 2, 1997, pp. 485-492. [23) F. Danie~ R. Chaffai, and K. Al-Haddad, "'l bree·phase diode rectifier with low harmonic distortion to feed capacitive loads," IEEE APEC Conference Proc. , 1996, pp. 932,-938. [24) M. Aorez-1..izarraga and A. F. Witulski, Input filter design for multiple-module DC power systems." IEEE Transactions on Power Electronics. Vol. 2, No. 3, 1996, pp. 472-479. (25) M. Alfayyoumi, A. H. Nayfch, and D. Borojcvic, " Input 6Jter interactions in DC-DC switching regulaton," Proc. IEEE Power E/,ctronics Specialists Conftrence, Charleston, SC, Vol. 2, 1999, pp. 926-932. (26) D. M. Mitchell, DC-DC Switching Rtgulator. New York: McGraw-Hill. 1988, Chapters 2 and 4. (27) B. Lehman and R. M. Bass. "Extensions of averaging theory for power electronic systems." IEEE Transactions on Power Electronics, Vol. 2, No. 4, 1996, pp. .542-553. (28). H. Bevrani, M. Abrishamchian, and N. Safari-shad," Nonlinear and linear robust control or switching power converters,·· Proc. IEEE International Conference on Centro/ Applications, Vof. I, 1999,pp. 808-813.
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Review Questions
223
(29) C. A. Mu!ioz, and I. Barbi. "A new high·power,facior three-phase ac-dc convener. analysis, design, and expcrin1cntation,'' IEE E Transactions on Power Electronics. Vol. 14. No. l. January 1999, pp. 90-97. (30( IEC P11blicatio11 555: Disturbances in supply systems caused by ho usehold ~ppliances and sin'lil~r equipment: Part 2: Harmonics. (31) Y. Jang, and M. M. Jovanovic, ''A new input,voltage feed forward harmonic-injection technique wi1h nonlinear gain control for single-s,.,tch. three-phase. DCM boost rectifiers." IEE:E Transacriom 011 Power Elecrro11ics. Vol. 28, No. I. March 2000. pp. 268- 277. (32) M. H. Rashid, SPICE for Power E:lcctro11ics Usi11g PSpicc. Englewood Cliffs. N.J.: PrcnticcHall. 1993. Chap1ers 10 and II . (33) P. Wood. Switching Power Co11vcrrers. NewYork:Van Nostrand Reinhold. 1981. (34] R. P. Severns and G. E. Bloom. Modem DC-10-D C Swilcl, Mode f'ower Com•errtr Circuits. New York: Van Nostrand Reinhold. 1983.
REVIEW QUESTIONS S.l What is a de choppcr,or
5.7 Whal is the effect of load inductance on the load ripple current'! S.8 \Vhat is the effect of chopping frequency on the load ripple current"/ 5.9 What arc the constraints for controllable transfer of energy between two de voltage sources? S.10 What is the algori1hm for generating the Ju1y cycle of a co nvener? 5.11 Whal is 1he modulation index for a PWM con1ro l'? S.U \Vha1 is a first and second t1uadrant co n"crter? S.13 What is a third and fourth quadrant co nverter? 5.14 \\lhat is a four-quadrant converter'! S.15 · \Vhat arc the pcrforn1ancc para1nctcrs of a converter? 5.16 Whal is a swi1ching-mode regulator? 5.17 What are 1he four basic types of switching-mode regulators? 5.18 Whn1 ore the advantages and disad"antagcs of a buck regulator? 5.19 What are the advantages and disad"anrngcs or a boost regulator? 5.20 Whal arc the ad\'antagcs and disadvan1agcs of a buck- boost regulator? S.21 Whal arc the ad"antages and disadvan1ages of a Ci!k regulator"/ Al whai du1y cycle docs 1he load ripple current become maximum'! 5.2.l What are the effects of chopping frequency o n filter sizes? 5.24 \\'hal is the discontinuous ntodc of operation of a regula1or? S.25 \\That is a multioulput boost converter? 5.26 Why n1us1 the nu11tiou11, ut boost convener be operated with tin1c: ntultiplcxing control? 5..27 Why ,nust the mul1iou1pu1 boost convcr1cr be opcra1cd in discontinuous ntode? S.28 Ho,v can the inpul curren1 of the rcclificr-fcd boost converter be n1adc sinusoidal and in phase ,vith the input voha.i;c? S.29 \Vhat 1s a state-space averaging technique'!
s:n
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224
Chapter 5
De-De Converteri
PROBLEMS 5.1 The de converter in Figure 5.la has a resistive load. R = 20 n and input voltage, V, = 220 V. When the converter remains on. its voltage drop is v.. a 1.5 V and chopping frequency is f = 10 kHz. If the duty cycle is 80%, determine (a) the average output voltage v•. (b) the rms output voltage v.. (c) the converter e!ficiency, (d) the effective input resistance R1, and (e) the rrns value of the fundamental component of harmonics on the output voltage. S.2 A converter is feeding an RL load as shown in Figure 5.3 with V, • 220 V. R • 10 L = 15.5 mH, f = 5 kHz, and E = 20 V. Calculate (1) the minimum instantaneous load current 11, (b) the peak instantaneous load current Ji, (c) the maximum pcalr.-to-pcak ripple current in the load, (d) the average load current 10 , (e) the rrns load current 10 , (f) the effective input resistance R1, and (g) therms value of converter current IR. 5.3 The converter in Figure 5.3 has load resistance, R • 0.2 0 : input voltage V. • 220 V; and battery voltage, E = 10 V. The average load current, 1. = 200 A, and the chopping frequency is f = 200 Hz (T • 5 ms). Use the average oulpul voltage 10 calculate the value of load inductance L, which would limit the maximum load ripple current to 5% of Ir 5.4 The de converter shown in Figure 5.7a is used to control power Row from a de voltage, V, • 110 V lo a battery voltage, E = 220 V. The power transferred lo the battery is 30 kW. The current ripple of the inductor is negligible. Determine (1) the duty cycle K , (b) the effective load resistance R"'. and (c) the average input current 1,. S.5 For Problem 5.4, plot the instantaneous inductor current and current through the battery E if inductor Lhasa finite value of, L = 7.5 mH. f = 2SO Hz, and k = 0.5. 5.6 An RL load as shown in Figure 5.3 is controlled by a converter. If load resistance R = 0.25 n,inductance L = 20 mH. supply voltage V. = 600, battery voltage E = ISO V, and chopping frequency f • 250 Hz. determine the minimum and maximum load current.
n.
the peak·tO··peak. load ripple current, and ave-rage load cunent tor k
S.7
5.8
S.9
S.10
S.11
S.12
= 0.1
10 0.9
with a
step or 0.1. Determine the m3Jlimum pcalt-to-pealt ripple current of Problem 5.6 by using Eqs. (521) and (5.22), and compare the results. . The step-up converter in Figure 5.8a has R • 10 L • 6.S mH, E • 5 V. and k • 0.5. Find· I ,. 12, and 111. U~ SPICE to find these values and plot the load, diode, and switch curre.nt. The buck regulator in Figure S.16a has an input voltage, V, • 15 V. The required average output voltage v. = 5 V ond the peak-to-peak output ripple voltage is 10 mV. The sv.itch· ing frequency is 20 kHz. The pcak-to-pc.a k ripple current of inductor is limited to 0.5 A. Determine (a) the duty cycle k, (b ) the filter inductance L , (e) the filter capacitor C, and (d) the critical values of L and C. The boost regulator in Figure S.17a has an input voltago, V, = 6 V. The avenge output voltage, v. = IS V and average load current, I, = 0.5 A. The switching frequency is 20 kHz. If L = 250 1,
n.
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Problems
225
Gltcr capacitance is C2 = 220 µF. The energy translcr capacitance is C 1 = 400 µF and in· ductance is L 1 = 250 µH. The average load current is 1. = 1.25 A. Determine (a) the average output voltage Va, (b) the average input current /H (c) the peak·to· peak ripple
current of inductor L1, IJ./1• (d) the peak-to-peak ripple voltage of capacitor C1• AV, 1• (e) the peak·lo·peak ripple current of inductor L,. A/2, (!) the peak·lo·peak ripple voltage of capacitor C2• AV, 2, and (g) the peak current of the transistor I,,. S.13 In Problem 5.12 for the COk regulator, find the critical values of L1, C1, L,. and C2• S.14 The buck converter in Figure 5.29 has a de input voltage VJ • 110 V. average load voltage V, = 80 V, and average load current I, = 20 A . The chopping frequency is f = 10 kHz. The peak-to-peak ripples arc 5% for load vo ltage. 2.5% for load current. and 10% for filler . L , current. (1) Determine the values of L,. l, and C,. Use .PSpice (b) 10 verify the results by plotting the instantaneous capacito r voltage vc and instantaneous load currcnl ii, and (c) to calculate the Fourier coefficients of the input current i,. Use SPICE model parame· ters of Example 5.10. S.15 The boost converter in Figure 5.l 7a has a de input voltage \.'., = 5 V. The load resistance R is 100 n. The inductance is L c 150 µH. and the filler capacitance is C a 220 µF. The chopping frequency is f = 20 kHz and the duty cycle of the converter is k = 60%. Use PSpicc (•) to plot the output voltage Ve. the input current i,. and the MOSFET voltage v,·; and (b) 10 calcula1e the Fourier coefficients of the input current i,. The SPICE model pa· ramcters of the MOSFET arc l = 2U. W = 0.3. VTO = 2.831. KP = 20.53U. IS ~ 194E- 18, CGSO = 9.027N. CGDO = 1.679N. S.1 6 A de regulator is operated at a duty cycle of k • 0.4. The load rcsis1ancc is R = 150 the inductor resistance is,,. = I 0. and the resistance of the filter capacitor is,, = 0.2 0. D etermine the voltage gain for the (a) buck converter, (b) boost converter and
n.
(c) buck- boost converter.
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C HAPTER
6
Pulse-Width-Modulated Inverters The learning objectives of this chapter arr as follol<'s:
• To learn the switching technique for dc-ac converters known as inverters and the types o( inverters • To study the operation of inverters • lb understand the performance parameters of inverters • To learn the different types or modulation techniques to obtain a near sinusoidal output waveform and the techniques to eliminate certain harmonics from the output • To learn the techniques for analyzing and designing inverters and for simulating inverters by using SPICE • To study the effe.cts of load impedance on the load current
6.1
INTRODUCTION
Dc-to-ac converters are known as inverters. 11,e function of an inverter is to change a de input vollagc to a symmetric ac output vollage or desired magnitude and frequency [!]. The output vollage could be fixed or variable at a fixed or variable frequency. A variable output voltage can be obtained by varying the input de voltage and maintaining the gain of the inverter constant. On the other hand, if the de input vollage is fixed and it is not controllable, a variable output voltage can be obtained by varying the gain of the inverter, which is normally accomplished by pulse-width-modulation (PWM) control within the inverter. The inverter gain may be defined as the ratio of the ac output vollage to de input vollage. The output voltage waveforms of ideal inverters should be sinusoidal. However, the waveforms of practical inverters are nonsinusoidal and contain certain harmonics. For low- and medium-power applicalions. square-wave or quasi-square-wave vollages may be acceptable; and for high-power applications, low disiorted sinusoidal waveforms are req uired. With the availability ol high-speed power semiconductor devices, 226 Copyrighlcd m lcria
6.2 Principle of Operation
227
the harmonic contents of output voltage can be minimized or reduced significantly by switching techniques. Inverters are widely used in industrial appl ications (e.g., variable-speed ac motor drives. induction healing, standby power supplies. and uninlerruptible power supplies). The input may be a battery, fuel cell, solar cell, or other de source. The typical singlephase outputs are (1) 120 Val 60 Hz. (2) 220 Vat 50 Hz, and (3) 115 Vat 400 Hz. For high-power three-phase systems. typical outputs are (I) 220 to 380 Vat 50 Hz, (2) 120 to 208 Vat 60 Hz, and (3) 115 10 200 Vat 400 Hz. lnver1ers can be broadly classified into two types:(!) single-phase inverlers, and (2) three-phase inverters. Each type can use controlled tum-on and tum-off devices (e.g., bipolar junction transistors IBJTs]. metal oxide semiconductor field-effect transistors [MOSFETs], insulated-gate bipolar transistors [lGBTs], metal oxide semiconductor-controlled thyristors IMCTs.J, static induction transistors, [S!Ts], and gate-tum-off thyristors (GTOs)). These inver1ers generally use PWM control signals for producing an ac output voltage. An inverter is called a vo/111ge-fed inverter (VFI) if the input voltage remains constant, a current-fed inverter (CFI) if the input current is maintained constant, and a variable de li11ked i11ver1er if the input voltage is controllable. If the output voltage or current of the invcrler is forced to pass through zero by creating an LC resonant circuit, this type of invcrt~r is called resona111-p11/se inverter and it has wide applications in power e lectronics. Chapter 8 is devoted to resonantpulsc inver1ers only. 6.2
PRINCIPLE OF OPERATION
The principle of single-phase inverters II] can be explained with Figure 6.la. The inverter circuit consists of two choppers. When only transistor Q, is turned on for a time Tof2, the instantaneous voltage across the load ·' ll is V/ 2. If transistor Q 2 only is turned on for a time TrJ2. - V,12 appears across the load. The logic circuit should be designed such that Q1 and Q2 are not turned on at the same lime. Figure 6.lb shows the waveforms for the output voltage and transistor currents with a resistive load. This inverter requires a three-wire de source, and when a transistor is off, its reverse voltage is V, instead of V,12. 1ltis inverter is known as a half-bridge inverter. The root-mean-square ( rms) output voltage can be found from
V •
= (~ ( ofl V; d1)112 = V, To Jo
4
2
(6.1)
The instantaneous output voltage can be expressed in Fourier series as
v0
a
oo
= _Q + L 2
(a,, cos( nwt) + b. s in(nwt))
""" l
Due to the quarter-wave symmetry along the x-axis, both a0 and a,, are zero. We get b,, as
b. • -1 [ 'IT
ho 2
-- V, d ( wt) +
::l
'
11\f2,
- tl(wt )
0
]
= -4V, /171
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228
Chapter 6
Pulse-Width-Modulated Inverters Fundamental
current. io1
v, +
T +
c,
D, R
v, 0 v, +
;.,
;,
t
2R 0
T.
v, ;,
;,
-voo • Yo +. a
T _ c,
v, ;,
t
o,
T
2R 0
T.
T
(a) Circuit
(b) Wa•,ctorms with re-.sistivc load
v, 4IL 0
(c) Load current \\ith ~ighly inductive load FIGURE 6.1
Single-phase half-bridge mver1cr.
which gives the instantaneous output voltage v0 as 00
2V, . -sm nwt
Vo=
n• l~ .S .. •. ll1T
=0
for 11 = 2, 4, ...
(6.2)
where w = 21'fo is the frequency of o utput voltage in rads per second. Due to the quarter-wave symmetry of the output voltage along the x-axis, the even harmonics voltages are absent. For n = I, Eq. (6.2) gives therms value of fundamental component as
V01
2V,
= .v21' "' = 0.45V,
(6.3)
For an inductive load, the load current cannot change immediately with the output voltage. If Q 1 is turned off at 1 = Tr/2, the load current would continue to Oow through Di, load, and lower half of the de source until the current fafls to zero. Similary, when Q2 is turned off at t = To, the load current flows through Di , load, and upper half of the de source. When diode D1 or Di conducts, energy is fed back to the de source and these diodes are known as feedback diodes. Figure 6.lc shows the load current and conduction intervals of devices for a purely inductive load. It can be noticed that for a purely
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6.2 Principle of Operation
229
inductive load, a transistor conducts only for Tif2 (or 90°). Depending on the load impedance angle, the conduction period of a transistor would vary from 90° to 180°. Any switching devices can replace the transistors. U 1011 is the tum-off time of a device, there must be a minimum delay time of td( = ,. rr) between the outgoing device and triggering or the next incoming device. Otherwise, short-circuit condition would result through the two devices. Therefore, the mnimum conduction time of a device would be 100 = T0 /2 - 'd· All practical devices require a certain turn-on and tum-off time. For successful operation of inverters. the logic circuit should take these into account. • For an RL load, the instantaneous load current io can be found by dividing the instantaneous output voltage by the load impedance Z = R + jnwL. Thus, we get 00
io =
">': •=l~....
2V
nnVR
2
'
+ (nwL)2
a.)
sin(nwr -
(6.4)
where a. = tan- 1(nwL/R). U / 01 is therms fundamental load current, the fundamental output power (for n = 1) is (6.5) (6.5a)·
Note: In most applications (e.g.. electric motor drives) the output power due to the fundamental current is generally the useful power, and the power due to harmonic currents is dissipated as heat and increases the load temperature. De supply current. Assuming a lossless inverter, the average power absorbed by the load must be equal to the average power supplied by the de source. Thus, we can write i \ ,(r)i,(r)dr
=i
\ .(1)i0 (1)d1
where Tis the period of the ac output voltage. For an inductive load and a relatively high switching frequency, the load current i0 is 'nearly sinusoidal; therefore, only the fundamental component of the ac output voltage provides power to the load. Because
the de supply voltage remains constant v,( r)
f\(t )dt
~
a
V,, we can write
= Vl (T V2V01 sin(wt)V2/0 sin("'1 '~
.
61)dr
= I,
where V01 is the fundamental rms output voltage; / is therms load current; 0 01 is the load angle at the fundamental frequency.
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230
Chapter 6
Pulse-Width-Modulated Inverters
Thus, the de supply current I, can be simplified to
{6.6) Gating sequence. The gating sequence for the switching. devices is as follows: 1. Generate a square-wave gating-signal v81 at an output frequency / 0 and a 50% duty cycle. The gating-signal v12 should be a logic invert of v81 • 2. Signal v81 will drive switch Qt through a gate-isolating circuit, and v82 can drive Q2 without any isolating circuit.
Key Point~ of Section 6.2 • An ac output voltage can he obtained by alternatively connecting the positive and negative terminals of the de source across the load by turning on and off the switching devices accordingly.1be rrns fundamenta.l component V01 of the output voltage is 0.45 V,. · • Feedback diodes are required to transfer the energy stored in the load inductance back to the de source.
6.3
PERFORMANCE PARAMETERS The output of practical inverters contain harmonics and the quality of an inverter is normally evaluated in tcrrns of the following performance parameters. Harmonic fa~or of nth harmonic (HF. ). The harmonic factor (of the nth harmonic), which is a measure or indh1dual harmonic contribution, is defined as
HF. n
= v••
for11>l
Vol
(6.7)
where V1 is therms value of the fundamental component and V0 ., is the rms value of the ,ith harmonic component. · Total harmonic distortion (THO). The total harmonic distortion, which is a measure of closeness in shape between a waveform and its fundamental component, is defined as
I(
THO= -V
ol
oo ~ ~
n • 2,J •. . .
2 V on
)'12
(6.8)
D istortion factor (D f), THO gives the total harmonic content, but it does not indicate the level or each harmonic component. If a filter is used at the output of inverters, the higher order harmonics would be attenuated more effectively. Therefore, a knowledge of both the frequency and magnitude of each harmonic is important. The OF indicates the amount of HD that remains in a particular wavefom1 aftcr,the liarmonics or 1ha1 ,vavcfo rn1 have been subjected to a second-order atlenuation (i.e.,
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6.3 Performance Parameters
231
divided by 112). Thus, OF is a measure of effectiveness in reducing unwanted hat monics without having to specify the values or a second-order load filter and is defined as
OF = - I [
(V
...!!!! )2 ]112
w
v.1 n=fi.... ,,z
(6.9)
The OF of an individual (or 11th) harmonic component is defined as OF,,
Von v0 1n
= - -2
for n > 1
(6.10)
Lowest order harmonlc ( LOH). The LOH is rhar harmonic component whose frequency is closest to the fundamental one, and its amplitude is greater than or equal to 3% of the fundamental component. Example 6.1
Finding the Paramet ers of the Single-Phase Half-Bridge Inverter
The single-phase half-bridge invener in Figure 6. la has a rcsi.srive load of R ~ 2.4 ll and lhe de input voltage is V, = 48 V. Delcrmine (a) therms output vollage al the fundamcnlal frequency V.,,. (b) the output power P•. (c) the average and peak cunen1s or each transis1or, (d) the peak reverse blocking voltage V88 of each transistor. (c) lhc THD. (f) the DF. and (g) 1he HF and DF ofthe LOH.
Solution V, = 48 V and R = 2.4 ll .
a. From Eq . (6.3), V0 1 = 0.45 x 48 = 21.6 V. b . From Eq. (6.1), V0 = V,12 = 4812 = 24 V. 1bc output power, P0 o V:,JR o 242/2.4 = 240W. c. The peak 1ransis1or curren1 I , = 24/2.4 = 10 A . Because each 1ransis1or conducrs for a 50% duty cycle, the average current or each rrnnsistor is lo = 0.5 x 10 = 5 A. d. The peak reverse blocking vol1age V8 • • 2 X 24 • 48 V. e. From Eq. (6.3). V01 = 0.45V, and therms harmonic vollagc v. VA =
(
fr,~ " "'
V 2<>"
)112 = ( V2 0
V'ol )112 = 0 · 2176Vs
.1....
From Eq. (6.8), THD = (0.2176V,)/(0.45V, ) = 48.34%.
r. From Eq. (6.2), we can find V.., and !hen find,
From Eq. (6.9). DF ~ 0.024V;J(0.45 V,) = 5.382%. g. The LOH is the 1hird, V03 • V0 113. From Eq. (6.7) . HF3 • V0 .,JV0 1 • 1/3 • 33.33%, and from Eq . (6.10), DF., = (V03 132 )IV01 = 1/27 = 3.704%. Because V03 /V0 1 = 33.33%, which is grca1cr lhan 3%, LOH = V.,,.
Key Points or Section 6.3
• The pe rformance parame1ers, which measure the quali1y of the inverlcr ourpul voltage, arc HF, THO, OF, and LOH.
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232
6.4
Chapter 6
Pulse-Width-Modclated Inverters
SINGLE-PHASE BRIDGE INVERTERS
A singlc·pbasc bridge voltage source inverter (VSI) is shown in Figure 6.2a. It consists of four choppers. When transistors Q 1 and Q2 arc turned on simultaneously, the input voltage V, appears across the load. If transistors Q3 and Q4 are turned on at the same time, the voltage across the load is reversed and is -V,. The waveform for the output voltage is shown in Figure 6.2b. Table 6.1 shows the five switch states. Transistors Qi, Q4 in Figure 6.2a act as the switching devices S1• S4 • respectively. If two switches: one upper and one lower conduct at the same time such that the output voltage is :I: V,, the switch state is 1, whereas if these switches arc off at the same time, t.h e switch state is 0. The rms output voltage can be found from
f2v 2 r1, )1/2 = v vo = (-To21T, o ' '
(6.11)
Equation (6.2) can be extended to express the instantaneous output voltage in a Fourier series as 00
v ,,
4V, .
_L -11 sin nwt
=
11 • 1J.S. . ..
(6.12)
r.
v,.
v, 20
~·: :] -~:+ ~
.I
-v
Q,
o,
O
T.
T
v,b
o,
o,
D,
~0
t
o,
v..,
Fundamental
v,
L
0
To (b) Wavclorms
(• ) Circuit
v, ;. 41c
01-~ --,,.~~._~.......~~~~.... I I I I
I
I
o, o, ! o, o, ! D; D, Io, Q, I on
on
on
Qn
(c) Load current with highly inductive lood FIGURE 6.2 Single-phase full-bridge inverter.
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6.4 Single·Pha.se Bridge Inverters
233
TABLE 6.1 Switch Statc-s for a Sing.le-Phase FulJ.Bridg.e Volt.age-Source Inverter (VSI)
·~ ...
Switch Staie•
State No.
State
••
Components Conducting S1 and S: if i0 > 0 D1 and Di if iu < 0 D4 and~ if i" > O
S1 and S2 are on ands_. and SJ arc oH
I
JO
Vs/2
-v.,n
Vs
S,. and SJ are on and S1 and S2 are ort
2
01
-v,12
Vs/2
-Vs
S1 ands, are on and S,. and~ are ore
3
11 ·
Vs/2
Vs/2
0
4
00
- Vs/2
- vs12
0
5
olr
-v,12
Vs/2 -Vs/2
-Vs
s. and S2 are on and 5 1 ands, are off s,. S2, S3. and S1. arc all off
Vs/2
S4 andS>iJi(} < 0 S1 and D, if;. > 0 0,nndS,ili. 0 S, and D, iii, < 0
D4 andD;iii0 > 0 D,andD:ifi0 < 0
v,
• I It an upper 1111·i1Ch is on and Oita lower switch is on.
and for 11 = I, Eq. (6.12) gives the nns value of fundamental component as 4V,
Vi
= V2'1T = 0.90V,
(6.13)
Using Eq. (6.4), the instantaneous load current i 0 for an RL load becomes 00
4V
~
i0 =
n•l.3.5.. ..
wrry'
R2
'
+ (11wL)2
sin(11w1 - 90 )
(6.14)
where e. = tan-1(11wL/R). When diodes D1 and D, conduct, the energy is fed back to the de source; thus, they are known as feedback diodes. Figure 6.2c shows the waveform of load current for an inductive load. De supply current. Neglecting any losses, the instantaneous power balance gives, v,(r)i,(t)
= v (1)i (1) 0
0
For inductive load and relatively high-switching frequencies, the load current i., and the output voltage may be assumed sinusoidal. Because the de supply voltage remains constant v,(r) = V,, we get i,(r)
V2V01 sin(wr)V2/0 sin(w1 - 61)
= ;
'
which can be simplified to find the de supply current as
.() V./ v., cos(Oi) - V. v., / cos(2wr - 6 )
r, r =
0
s
0
1
(6. l5)
'
where V01 is the fundamental rms output voltage: / 0 is the rrns load current: 61 is the load impedance angle at the fundamental frequency. Equation (6.15) indicates the presence of a second-order harmonic of the same order of magnitude as the de supply current. This harmonic is injected back into the de
Copyr ghtcd matcria
234
Chapter 6
Pulse-Width-Mod u lated Inverters
voltage source. Thus, the design should consider this to guarantee a nearly constant de link voltage. A large capacitor is normally connected across the de voltage source a.nd such a capacitor is costly and demands space; both features are undesirable, especially in medium to high power supplies.
Example 6.2
Finding the Parameters of the Single-Phase Full-Bridge Inverter
Repeat Example 6.1 for a single-phase bridge inverter in Figure 6.2a.
Solution V, = 48 V and R = 2.4 0.
a. From Eq. (6.13), Vt = 0.90 x 48 = 43.2 V. b. From Eq. (6.11), V0 = V, = 48 V. The oulpul power is P0 = V;IR = 48212.4 = 960 W . c. The peak 1ransis1or currenl is I , = 48/2.4 = 20 A. Because each transistor conducts for a 50% duty cycle, the average current or each transistor is la • 0.5 X 20 = 10 A. d. The peak reverse blocking voltage is V8 • • 48 V. e. From Eq. (6.13), Vt • 0.9V,. The rrns harmonic voltage v. is V• •
(._Ji..
V;. )"'
= (V5 -
V;t)
112
= 0.4352V,
From Eq. (6.8), THO • 0.4359V,l(0.9V,) • 48.34%.
t. [
f
(V-;)']"' = 0.048V,
11• 3.S.1.. . .
g.
ll
From Eq. (6.9), DF • 0.048V,l(0.9V,) • S.382%. The LOH is the third, V3 = Vt/3. From Eq. (6.7), HF3 = V0 3'V0 t = 113 • 33.33% and from Eq. (6.10), OF, = (VJ132)1V0 1 = 1'27 = 3.704%. .
Note:Thc peak reverse blocking voltage of each lransislor and !he quality of o ulpul voltage for half-bridge and full-bridge inverters are the same. However, for fullbridge inverters. the output power is fo ur times higher and !be fundamental component is twice that of half-bridge inverters. Example 6.3 Finding the Output Voltage and Current of a Single-Phase Full-Bridge Inverter with an RLC load The bridge invencr in Figure 6.2a has an RLC load with R = 10 0 , L = 31.5 mH, and C • 112 µF. The inverter freque ncy is fo m 60 Hz and de inpul voltage is V, = 220 V. (a) Express 1he inslantanco us load current in Fourier series. Calculate (b} the rrns load current at the fundamcn1al frequency /0 1; (c) the THO or lhc load current; (d) lhe power absorbed by the load Po and the fundamenlal power P01 : (c) 1hc average currenl or dcsupply /,:and ([) the rrns and peak currcnl of each 1ransiS1or. (g) Draw the waveform of fundamental load current and show the conduction inlcrvals or lransistors and diodes. Calcula1e the conduction time o r (h) the transistors. and (i) the diodes.
So/111/on
V, = 220 V, fo = 60 Hz. R = 10 n. L = 31.5 mH. C = 112 µF. and w = 2-rr x 60 1bc inductive rcnctancc for the ,11h harmonic voltage is X 1• = j.wL • flwrr x 60 x 31.5 x 10- 3
•
jl l.87n
= 377 rad/s.
n
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6.4 Single-Phase Bridge Inverters
235
The capacitive rcactance for the 11th harmonic ,•ohagc is jl06
j
x< = - - == - 2m, X 60 X nwC
- j23.68
112
a
11
n
The impedance for the 11th harmonic voltage is
IZnl
J
R2 + ( nwl -
=
,.:..C)' =
IJ02 + ( I l.87n - 23.68/n)'l'n
and the load impedance angle for the 11th harmonic voltage is
011 = tan
_1 11.8711 :.. 23.68/n 10
= ,an_,
(i.
2.368) 187 n - - II
a. From Eq. (6.12) , the instantaneous output voltage can be expressed as v0 (1) = 280.1 sio(3n1) + 93.4 sio(3 x 3771 ) + 56.02 sin (5 x 3n1)
+ 40.02 sin(7 X
3771)
+ 31.12 sin(9 X
3771)
+ · ··
Dividing the output voltage by the load impedance and considering the appropriate delay due to the load impedance angles, we can obtain the instantaneous load cur· rent as
i0 (1) = 18.1 sin(3n1 + 49.72') + 3.17 sin(3 x 3771 - 70.17') + sin(5 X 3771 - 79.63' ) + 0.5 sin(7 X 3771 - 82.85' ) + 0.3 sin(9 x 3771 - 84.52') + · · ·
b. The peak fw1damcntal load curre nt is Im, = IS.I A . llte rms load current al funda-
c.
mental frequency is /01 = l8.llv'2 = 12.8 A. Considering up to the nin1h harmonic, the peak load current.
Im = ( 18.12 + 3.172 + 1.02 + 0.5 2 + 0.3 2)'12 = 18.41 A The nns harmonic lond current is I h = ( / rtl 2 - 1' )'12 = 18.41' - 18.1' = 23789 A ,,. , \/2 . Using Eq. (6.8), the T HO or the load currenc. THO =
2 112 2 12 ( 2 18·41 / m - / mt ) = [( - ) - I ] ' =. 18.59%
Im,
18.l
d. The rrns load current is /0 . . J.,fvi = 18.4 1/v'2 = 13.02 A. and the load power is P0 • 13.02 2 x 10 = 1695 W. Using Eq. (6.5), thc fundamental output power is
Copyr ghtcd matcria
236
Chapter 6
Pulse-Width-Modulated Inverters i(I)
25 21.14
1<,(I)
20
IS
Fundamcn1al
10 5
0
-s
16.667 ms
- IO
i. • 2.639ms
- IS - 20
Q1 on
- 25
o,
on
0200
o, on
FIGURE 6.3
Waverorms for Example 6.3.
e. The average supply current I, = P.,IV, = 16951220 = 7.7 A. f. The peak transistor current Ip ;; /.,,. ~ 18.41 A. The maximum pcrmissibJc rms cur· rent of each transistor is /Q(mu) = l.,l'V2 = 1,12 = 18.41/2 = 9.2 A. g. The waveform for fundamental load current i 1(r) is shown in Figure 6.3. b. From Fig. (6.3). the conduction time of each transistor is found approximately from Wlo = 180 - 49.72 = 130.28° Orio= 130.28 x ,r/( 180 x 377) e 6031 µ.s. i. The conduction time of e.nch diode is approximately
,,
,, = (180 - 130.28) x 180 x 377 • 2302 µs
Notes:
1. To calculate the exact values of the peak current, the conduction time of transistors and diodes, the instantaneous load nrrent i0 (1) should be plotted as shown in Fi.g ure 6.3. The conduction time of a t ransistor must satisfy the condition i 0 (1 = 10) = 0, and a plot of i 0 (1 ) by .a computer program gives Ip= 21.14 A , 10 = 5694 µs , and 14 = 2639 µs. 2. This example can be repeated to evaluate the performance of an inverter with R, RL, or RLC load with an appropriate change in load impedance ZL and load angle 0n· Gating sequence. The gating sequence for the switching devices is as follows:
1. G enerate two square-wave gating-signals v, 1 and v12 at an output frequency [0 and a 50% duty cycle. The gating-signals v83 and Vg• should be the logic.invert of
v8 2 and v8 ,, respectively.
Copyrghlcd matcria
6.5 Three-Phase Inverters
237
2. Signals vg 1 and v13 drive Q1 and Q3, respec1ively, through gate isolation circuits.
Signals v,2 and v,~ can drive Q 2 and Q 4, respectively, without any isolation circuits.
Key Points or Sedion 6.4 • The full-bridge inverter requires four switching devices and four diodes. The output voltage switches between +V, and Therms fundamental component V. of the output voltage is 0.9V,. • The design of an inverter requires the determination of the average, rms, and peak currents of the switching devices and diodes.
- v,.
6.5
THREE-PHASE INVERTERS
Three-phase inverters are normally used for high-power· applications. Three singlephase half (or full)-bridge inverters can be connected in parallel as shown in Figure 6.4a to form the configuration of a three-phase inverter. The gating signals of single-phase inverters should be advanced or delayed by 120° with respect to each other to obtain three-phase balanced (fundamental) voltages. The transformer primary windings must be isolated from each other, whereas the secondary windings may be connected in Y or delta. The transformer secondary is normally connected in delta to eliminate triplen harmonics (11 = 3, 6, 9, ...) appearing on the output voltages and the circuit arrangement is shown in Figure 6.4b. This arrangement rcquir~ three single-phase transform· ers, 12 transistors. and 12 diodes. If the output voltages of single-phase inverters arc not perfectly balanced in magnitudes and phases, the three-phase output voltages are unbalanced. A three-phase output can be obtained from a configuration of six transistors and six diodes as shown in Figure 6.5a. Two types of control signals can be applied to the transis1ors: 180° conduction or 120° conduction. The 180° conduction has better utilization of the switches and is the preferred method. 6.5.1
180-Degree Conduction
Each transistor conducts for 180°. Three transistors remain on a t any instant of time. When transistor Q1 is switched on, terminal a is connected to the positive terminal of the de input voltage. When transistor Q4 is switched on. terminal a is brought to the negative te rminal of the de source. There are six modes of ope ration in a cycle and the duration of each mode is 60°. The transistors arc numbered in the sequence of gating the transistors {e.g., 123, 234, 345, 456, 561. and 612). Toe gating signals shown in Figure 6.5b are shifted from each other by 60° lo obtain three-phase balanced (fundamental) voltages. Toe load may be connected in Y or delta as shown in Figure 6.6. ll1e switches of any leg of the inverter (S1 and S4, S3 and S6, or S5 and S2) cannot be switched on simul taneously; this would result in a short circuit across 1he de link voltage supply. Similarly, to avoid undefined states and thus undefined ac output I.i nc voltages, the switches of any leg of the inverter cannot be switched off simultaneously; this can result in voltages that depend on the respective line current polarity.
Copyrighted m~leria
238
Chapter 6
Pulse-Width-Modulated Inverters
A
+
>-
Inverter 1
.
VAD
-
D
II
·-~
l
l
B >-
Inverter 2
,
+
+ VB£
-
E
c Inverter 3
-
'
+
• t'F
F I
b
II ....-
+
~
•
+
J >
II
c
v.,,
-
n
(a) Schem3tic:
..
+
~r
.J
- -
v,
I
0
+
~
2
A
•
8
c
!=,,
,J
D :J a
•
o,
F
R R
o,
o,·
• n
D
E
L
o;
o;
L
Y-conncctcd load
n R
(b) arcui t diagram
FIGURE 6.4
Three-phase inverter formed by three single-phase inverters.
Copyr ghlcd malcria
6.5 Three-Phase Inverters
+
'·r·~ g,
v,
0
-
v ~ c. -+ -t o,
~
D, i,
D,
a
~
o, ib
o.
~
239
o, i,
b
~
c
o,
(a) Circuit g, 0
2,r
g, 0
8.J 0 g,
, m.l
""
...
...
I
: 2,rlJ
s,0
wl
0
wl
0
wl
S6
••• v,
0 Vix
. v,
I I I I
"'
2,r
wl
-- T-I I
0
211
:'
wl
I I
v" v, -- J. ------~-0
: I
I
;,
,.. t
211
(b) Wa\'c(orms for 1so•conduc1ion FIGURE 6.5
Thrce-pha.s.e bridge in\'erter.
Table 6.2 shows eight valid switch states. Transistors QI, Q6 in Figure 6.6a act as the switching devices SJ, S6, respectively. If two switches: one upper and one lower conduct al the same time such that the output voltage is ± V,, the switch state is I, whereas if these switches are off at the same time, the switch state is 0. States 1 to 6 produce nonzero output voltages. States 7 and 8 produce zero line voltages and the line currents freewheel through either the upper or the lower frc~wheeling diodes. To
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240
Chapter6
Pul~·Width-Modulated Inverters
Switch States forThrcc·Phasc Voltagc·Sourcc Inverter (VSI)
TABLE6.2
State No.
State
S1• S1 , and S, arc on ands,, Ss, and .s3 are off S1• S3• and S1 arc on and Ss, $6• and $ 4 arc ofJ
S.,. S.,, and S2 are on and S6,, S1• and Ss arc off S.,, S5• and S1 arc on and S1, S2, and S6 are off Ss, S6• and 5 4 arc on and S1, $3 , and S1 arc off St,, S1. and Ss urc on and SJ, S,. and 52 arc orr S1, SJ, and Ss arc on and S.i, and S1 arc off 54• $6, and S2 arc on and S1, and S5 are off
s~ s,.
Switch States
....
v,.
v,.
Space Vector
100
Vs
0
-v,
v,
2
110
0
- Vs
v, • jl.155 • 21,3 L.90'
3
"•
010
- Vs
Vs
0
4
011
-Vs
0
Vs
V,
5
001
0
- Vs
Vs
v, • - jl.155 • 21,3 L.270'
6
IOI
v,
-Vs
0
v,
=1-
7
Ill
0
0
0
v,
=0
8
000
0
0
0
v.-
s
1
+ f0.5n a
+ ;o.sn
• 21,tJ ,:1so•
-I - j0.5n
= 21,3 L.210'
v, • - 1 a
21,tJ ,:30•
;o.sn
= 21,3
L.330'
0
generate a given vollage waveform, the inver1er moves from one stale 10 another. Thus, the resulling ac ou1pu1 line vollages are built up of discrete values of vollages of v,, 0, and - V,. To generate the given waveform, the selection of the states is usually done by a modulating technique that should assure the use of only the valid states.
For a della-connec1ed load, the phase currents can be obtained direclly from the line-to-line vollages. Once the phase currents are known, lhe line currenl\ can be determined. For a Y-connec1ed load, the line-10-neulral voltages must be determined 10 find the line (or phase) currents. There are three modes of operation in a half-cycle and the equivalent circuits are shown in Figure 6.7a for a Y-connected load. During mode l for O s wt < ,r/3, transistors Q., Qs, and Q6 conduct
R
3R
2
2
R a R + -•cq
. v, 21', ,.=-=Rcq 3R
•o---- R n R
R
b
co-------~
FIGURE 6.6
Dclt.i· and Y-conncctcd lo;id.
(a) Ocha conneclcd
(b) Y-c:onncclcd
Copyrghtcd matcria
6.5 Three-Pha$e Inverters
;,
...
R
a
;, R
v,
R
b
c
R
b
R
c
n
v,
+
•
;,
Mode I
n
+
R
b
R
c
R
v,
n
(a) Equivalent circuilS
...
~
3
R
Mode3
Mode 2
V 2V/3
•
241
.,,
0
~ 30 !---+-+---
-+-- +----+-+-- .,,
v,
--r ~ 30 l--+---+---1,---+-- t - --
+-,.- .,,
_ 2V, 3 (b) Phase vohagc$ for 180° conduction FIGURE 6.7
Equivalent circuits ror Y~ nncc1ed rcsisli\'e load.
i 1R
V,
Va,. =v,n=2=3 Vbn
= - i,R =
T-2V
During mode 2 for -rr/3 s wt < 2-rr/3, transistors Q,, Q 2 , and Q6 conduct
R
cq
= R + !!_2 = 3R 2
. V, 2V, 'z = - = -
3R
R cq
.
21', 3 - i 2R = - 2- =
van =12R = 'VIN,
During mode 3 for 2-rr/3 s wt <
= 1T,
V,n
- v, 3
lransistors Q,, Q2• and Q3 conduct
R
3R
R,q= R +2=2
Copyrghlcd malcria
242
Chapter 6
Pulse-Width-Modulated lnverten
.
13= -
v,
Rcq
v
••
vtn
2\1, 3R
= -
i3R V, = 2 3 -2V = i3 R = - -' 3
=v
b«
=-
The line-to-neutral voltages are shown in Figure 6.7b. The instantaneous line-to-line voltage vub in Figure 6.5b can be expressed in a Fourier series, Do
V.o = -
2
oo L (a. cos(nwl) + bn sin(nwt)) n =l
+
Due to the quarter-wave symmetry along the x-axis, both a0 and a. are zero. Assuming symmetry along the y-axis at wt = 1rl6, we ean write bn as 1 b. = :;;
[1
-,,H, - V,d(wt ) - S•/6
+
4\1, . mr . n1r f."16V,d(wt ) ] = -;;:; (2 )sm (3 ) • l6
SID
which, recognizing that Vab is phase shifted by 'fll6 and the even harmonics are zero, gives the instantaneous line-to-line voltage v.., (for a Y-connected load) as 00
vob =
4V, . -n,1 s1•n" ( - sm 3
•... n1r
n • I~
wl
+ -1r) 6
(6.J 6a)
Both 11/>t and v,. ean be found from Eq. (6.16a) by phase shifting V.o by 120"·and 240°, respectively, 00
v1,c=
L
1r) -4\1, sin -n,r sinn ( wr - n• t.3.S....t•1r 3 2 00
v,0 =
>.
. -n1r sin . n ( wt - -71r) -4V, s,n 3 6
•• ,°'.Js•...n1r
(6.16b) (6.16c)
We ean notice from Eqs. (6.16a) to (6.16c) that the triplen harmonics (n = 3, 9, 15, ...) would be zero in the line-to-line voltages. The line-to-line rms voltage can be found from 2 VL = [ 1r 2
,2.13V ; tl(wt )]'/2 = VJ12 V, = 0.8165V,
Jo
(6.17).
From Eq. (6.16a), the rms nth component of the line voltage is VLn = ~
4V,
v2111r
. n1r sm -3
(6.18)
Copyrghlcd m~leria
6.5 Three-Phase Inverters
which, for n
243
= I, gives therms fundamental line voltage. vL,
=
411, sin 60° • "' = o.7797V, v21r
(6.19)
The nns value of line-to-neutral voltages can be found from the line voltage, VL
VP = V3
V2 V,
=-
3
-
= 0.471411,
(6.20)
With resistive loads, the diodes across the transistors have no functions. If the load is inductive, the current in each arm of the inverter would ·be delayed to its voltage as shown in Figure 6.8. When transistor Q 4 in Figure 6.5a is off, the only path for the neg· ativc line current;. is through D1• Hence. the load terminal a is connected to the de source through D 1 until the load current reverses its polarity at l = 11• During the period for O s t s Ii, transistor Q 1 cannot conduct. Similarly, transistor Q, only starts to conduct at r = 12• The transistors must be continuously gated, because the conduction time of transistors and diodes depends on the load power factor. For a Y-conncctcd load, the phase ,•oltage is v•• = v.,J'l/3 with a dclayof30° with respect to ·vab. Therefore. the instantaneous phase voltages (for a Y-connected load) are
~ 4V, sm . (11"') . v.,, = ~V3n,r 3 sm(nwr)
for 11 = 1, 3, 5. . . .
(6.21a)
for 11 = I , 3, 5, . . .
(6.2lb)
00
vbN
4V,s i n (11'1') = L~ sin11 ( wr - -21r)
3
3
V311'1r sin(!!!)sin11(w1 3
~ ) 3
n•I V
vbN
311,r
= f ~ n= 1
Dividing the instantaneous phase voltage
for II
= 1, 3, 5,
.,.
(6.2Jc)
v.,., by the load impedance,
Z =. R + j11wL
2V,13 V,/3
A " v_
r-r-,-,
----+c--• 1
1-----~.~ ----tT1-
0
FIGURE 6.8 Three-phase in,·c rtcr wilh RI. load.
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Chapter 6
Pulse-Width-Modulated Inverters
Using Eq. (6.21a), the line current i 0 for an RL load is given by
i0 =
f [ v3[111tVRl4V.+
.. - 1.J.s....
(ll'!JL)l]
sin""] sin(11w1 - On) 3
(6.22)
where 0,. = tan- 1(11wUR). Note: For a delta-connected load, the phase voltages (v.N, vbN• and v,N) are the same as the line-to-line voltages (vab, vbc. and v,.) as shown in Figure 6.6b and as described by Eq. (6.16). De supply current. Neglecting losses, the instantaneous power balance gives
v,(c)i,(t) = Vab(t)i0 (1) + vbc(t )ib(t ) + Vca(t)ic(t) where i0 (1). i6( 1). and i, (I) arc the phase currents in a delta-connected load. Assuming that the ac output voltages arc sinusoidal and the de supply voltage is constant v,(t ) = V,. we get the de supply current
i,(1) =
V2V01 sin(U)() x vi/0 sin(wt - 91) V + viV0 1 sin(wt - 120°) x vit0 sin{wr ' + v'iV01 sin(wt - 240°) x V2/0 sin(w1 1
{
}
120° - 61) 24-0° - Oi)
The de supply current can be simplified to
v.1 /
I, = 3 V
'
0
• ,.
v., Ii cos(0 )
.
cos(0 1) = v 3 V
(6.23)
1
where h = 010 is the rrns load line current; V01 is the fundamental rms output line voltage; / 0 is the rrns load phase current: 6 1 is the load impedance angle at the fundamental frequency. Thus. if the load voltages are harmonic free. the de supply current becomes harmonic free. However, because the load I.inc voltages contain harmonics, the de supply current also contains harmonics. Gating sequence. The gating sequence for switching devices is as follows 1. Generate three square-wave gating-signals v1 i, v13 , and v15 at an output frequency lo and a 50% duty cycle. Signals v 14 , v16, and v12 should be logjc invert signals of Vgi, ·vgJ• and Vgs, respectively. Each signal is shifted from the other by 60°. 2. Signals v1 i, v 13 , and v 15 drive Qi, Q 3, and Q5, respectively, through gate-isolating circuits. Signals v 12, v 14, and v 16 can drive Q2, Q,, and Q6, respectively, without any isolating circuits. Example 6.4 Finding the Output Voltage a nd Current of a Three-Phase Full-Bridge Inve rter with an RL load The three-phase inverter in Figure 6.Sa bas a Y-connccted load of R • 5 n and L • 23 mH. The inverter frequency Ls fo = 60 Hz and the de input voltage is V, = 220 V. (a) Express the instan· tancous line-to-line voltage vab(t) and line current i3 (1) in a Fourier series. Determine (b) the rms line voltage v,.: (c) therms phase voltage VP: (d) therms line voltage vL, at the fundamental
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6.5 Three-Phase Inverters
245
frequcncy;(e) the 11115 phase voltage at the fundamental frequency v,1; (f) the THD;(g) the DF; (h} the HF and DF ol the LOH; (i) the load power P0 ; (j) the average transistor current IQ(••l; and (k) therms transistor current / Q(,m,)·
Solution
V. •
VJJV. R • 50,L • 23mH./0 = 60Hz,andw = 2n
x 60= 3nrad/s.
a. Using Eq. (6.16a), the instantaneous line·to-line voltage v.i,(1) can be written as v,.(1) = 242.58 sin(3n1 + 30°) - 48.52 sin 5(3771
+ 30") - 34.66 sin 7(3n1 + 30°) + 22.05 sin 11(3771 + 30°) + 18.66 sin 13(3n1 + 30") - 14.27 sin 17(3n1 + 30") +
Z1• =
\fR' + (nwL)2/ tan- 1(nwUR) = \!52 + (8.67n)2 /tan-1(8.67n15 )
Using Eq. (6.22). the instantaneous line (or phase) current is given by
i, 1,1 = 14 sin(3771 - 60°) - 0.64 sin(5 X 3771 - 83.4°) - 0.33sin(7 X 3771 -85.3°) + 0.13sin(ll x 3771 - 87°) + 0.l0sin(l3 x 3771 - 87.5°) - 0.06sin(l7 x 3771 - 88°) -
b. From Eq.(6.17), Vi= 0.8165 x 220 = 179.63 V. c. From Eq. (6.20), v, = 0.4714 x 220 = 103.7 V. d. From Eq. (6.19), V1,1 = 0.7797 x 220 = 171.53 V. = Vd\13 = 99.03 V. f, From Eq. (6.19), vll = 0.7797Vs
e.
v,,
(
~ ~
... ,.1.1 1. ...
From Eq. (6.8). THO
vjA)'" = cvl. - vi,) 112 = 0.24236V.
= 0.24236V,/(0.7797V,) = 31.08%. Therms harmonic line volt-
age is
g. vlh •
L.~i. . {~i; YJ".
0.00941 v,
From Eq.(6.9),DF = 0.00941V/(0.7797V,) = 1.211 %. h. The LOH is the fifth, v,., = Vd5. From Eq. (6.7), HF, = v1.,wL, = 115 = 20%. and from Eq. (6.10). DF5 = (V1..,t5 2)/V1.1 = 1/125 = 0.8%. L For Y-connected loads, the line current i.s the same as the phase current and the rms line current,
ft =
(142
+ 0.64 2 + 033 2 + 0.13 2 + O.!o' + 0.062) 112
v1
= 9.91 A
The load power Po = 3/l,R = 3 x 9.912 ?< 5 = 1473 W. j. The average supply current I, = Pfl20 = 14731220 = 6.7 A and the average transi.,tor current 10..., = 6.7/3 = 2.23 A. k. Because the line current is shared by three transistors. the rms value of n transistor current is IQ,- , = 11,1\13 = 9.91/\/3 = 5.72 A .
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6.5.2
Chapter 6
Pulse-Width-Modulated Inverters
120-Degree Conduction
In this type of control. each transistor conducts for 120°. Only two transistors remain on at any instant of time. The gating signals are shown in Figure 6.9. The conduction sequence or transistors is 61, 12, 23, 34,45, 56, 61. There are three modesof operation in one half-cycle and the equivalent circuits for a Y-connected load are shown in Figure 6.10. During mode I for s wt s -rr/3, transistors I and 6 conduct.
v,
v,
·von = - ·
Vi,,,= - -
2
2
I
WI
... r- ... wl
-----,
wt
wl
0
v,
vl!n
v,
T
-T
--------
t' ....
0
-
v,
-T
----------------v,
"''
2 FIGURE 6.9
On.ting sign:t.ls for 120, conduction.
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6.5 Three-Phase Inverters 1,
v.
•
R
b
R
i2 3
R
b
R
v,
n
R
c
R
;, b
R
c
R
v,
R
c
(a) Mode I
n
•
2.47
n
(e) Mode 3
(b) Mode Z
FIGURE 6.10
Equivalent circuits for Y..connected resistive load.
During mode 2 for -rrf3 :s wt s · 2-rrf3, lransistors 1 and 2 conduct .
v,
2
Van=
v '"
v,
= -2
During mode 3 for 2-rr/3 :s wt :s 3,r/3, transistors 2 and 3 conduct
v.,,
v,
=0
v
The line-to-neutral voltages that are shown in Figure 6.9 can be expressed in Fourier series as Van =
f
n•l.l.5•.. 00
vbn=
_ZV_, Sin n 1r sin n (we + .°!: ) 3 6 111' 2V, .
-
•=• .s.... 11,r
f
•
sm11
(
'.l
~
Ven =
n,r
sm -
wt - -,,.) 2
ZV, sin~ sin 11 (wt -
n• I.J.5•..
3
111T
(6.24a)
?,r) 6
(6.24b) (6.24c)
The line a-to-b voltage is va1, = VJ v,. wilh a phase advance of 30°.Therefore, the in· slantaneous line-to-line voltages (for a Y-<:onnected load) are
va1,
=
f
n• t
ZV3Vs sin (" ,,.) sin 11 IITi
3
L -2V3Vs --sin (",,.) sin r1•l 3 riff
for 11
= l, 3, 5. . . .
(6.25a)
( wt - -,,.)
for "
= 1. 3, 5. . . .
(6.25b)
for II = }, 3, 5, , ..
(6.25c)
·
00
v,,.. =
(wr + ~3~1
II
3
00
v,0
= L 2V3Vs sin (".,,) sin n (wt 3 n=I
nTr
- -rr)
There is a delay of 'IT/6 between the turning off Q1 and turning on Q4 • Thus, there should be no short circuit of the de supply through one upper and one lower transis· tors. At any time, two load terminals arc connected to the de supply and the third one remains open. The potential of this open terminal depends on the load characteristics and would be unpredictable. Because one transistor conducts for 120°, the 1ransistors
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Chapter 6
Pulse-Width-Modulated Inverters
are less utilized as compared with those of 180° conduction for the same load condition. Thus, the 180° conduction is -preferred and it is generally used in three-phase inverters.
Key Points of Section 6.5 • The three-phase bridge inverter requires six switching devices and six diodes. The rms fundamental component VLI of the output line voltage is 0.7798V, and that for phase voltage is v. 1 = VLll,13 = 0.4511, for 180° conduction. For 120° conduction, VPI 0.389811, and Vu ,13 Vpt 0.6753V,. The 180° conduction is the preferred control method. • The design of an inverter requires the determination of the average. rms. and peak currents of the switching devices and diodes.
=
6.6
=
=
VOLTAGE CONTROL OF SINGLE·PHASE INVERTERS
In many industrial applications. to control of the output voltage of inverters is often necessary (i) to cope with the variations·of de input voltage, (2) to regulate voltage of inverters. and (3) to satisfy the constant volts and frequency control requirement. There are various 1echniques to vary the inverter gain. The most efficient method of controlling the gain (and output voltage) is to incorporate PWM control within the inverters. The commonly used techniques are: l , Single-pulse-width modulation.
2. Multiple-pulse-width modulation. 3. Sinusoidal pulse-width modulation. 4. Modified sinusoidal pulse-width modulation. S. Phase-displacement control.
' 6.6.1
Sipgle-Pulse-Width Modulation In single-pulse-width modulation control, there is only one pulse per half-cycle and the width of the pulse is varied to control the inverter output voltage. Figure 6.11 shows the generation of gating signals and o utput voltage of single-phase full-bridge inverters. The gating signals are generated by comparing a rectangular reference signal of ampli· tude A , with a triangular carrier wave of amplitude A,. The frequency of the reference signal determines the fundamental frequency of output vo ltage. The instantaneous output voltage is v0 = V,(g1 - g,). The ratio of A, to A, is the control variable and defined as the amplitude mod11/ario11 illdex. 1l1e amplitude modulation index, or simply modulation index
A,
M =A ,.
(6.26)
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6.6 Voltage Control of Single-Phase Inverters e
A,
249
/ c,rricr s:ignnl
-- - ---
Reference
signal
01---..;.....1....::.i~-..;...--+--.--::,,,,:--.--+-~ .., 01
«2
I
I
81
I
I
I I I
I I I
I
I
I
L . - - 6 - : Gate signal for11r:insistor 0 1 I
0
1T6
g.,
2-2
rr6
2
J+2
Gate signal for
0
v, '
rr
o~
0
"
2rr
r.
2rr
"
2rr
.,, .,,
0
- v,
..,
y" - z6
I"
1! +! 2 2
lzt 2
FIGURE 6.11
Single pul.sc·wid1h modul,iion.
The rms output voltage can be found from
V0
21(.+6)/2
= [-
2-rr
V ; d ( wt )
]'12
{&
• V, /-;
(r. - 6)12
(6.27)
\ "
By varying A, from O 10 A ,. lhc pulse width 6 can be modified from 0° 10 180° and the rms output voltage V0 • from Oto V,. Th
v0 (1)
= L ,i c
4V, . 116 .
-
I.Jp~... . ll'IT
sm -
2
sm11wl
(6.28)
Due to the symmetry of the output voltage along the x-axis. !he even harmonics (for 11 = 2. 4, 6• . .. ) arc absent. A computer program is developed to evalua te the performance or single-p ulse modula tion for single-phase full-bridge inve rters. Figure 6.12 shows the harmonic profile with the variation of modulation index M. The dominant harmonic is the third. and the OF increases significantly at a low output voltage. The time and angles or intersections can be round from 11
a, Ts = - = (1 - M ) w 2 "'2
tz = :;- = ( l + M )
Ts
2
(6.29a)
(6.29b)
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250
Chapter 6
Pulse-Width-Modulated Inverters
v, v,
DF(%) 9
1.0 8
7
0.8
6 0.6
5 4
0.4
3 2
0.2
0.6 0.4 0.2 Modulation index. M
0.8
0
FIGURE 6.1 2
Harmonic profile of single-pulse-\.\'id1h modulation.
which gives the pulse width ti (or pulse angle 6) as d
6
= ;;; = 12 -
11
= MT,
(6.29c)
where T, = T12. Gating sequence. The algorithm for generating the gating signals are as follows: I . Generate a triangular carrier signal v,., of switching period T, = T/2. Compare v" with a de reference signal v, to produce the difference v, = u" - v,, which must pass through a gain limiter lo produce a square wave of width d al a switching pe· riod T,. 2. To produce the gating signal g,. multiply the resultant square wave by a unity sig· nal u,, which must be a unity pulse of 50% duty cycle at a period of T. 3. To produce the gat\ng signal g2, multiply the square wave by a logic-invert signal of v,.
6.6.2
Multiple-Pulse-Width Modulation 'The harmonic content can be reduced by using several pulses in each half-cycle of out· pul vollage. The generation of gating signals (in Figurc,6.13b) for turning on and off of ·transistors is shown in Figure 6.13a by comparing a reference signal with a triangular carrier wave. The gate signals are shown in Figure 6.13b. The frequency of reference
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6.6 Voltage Control of Single-Phase Inverters
251
Carrier signal
e
'
I t I
I I
I I
t.
I :
I
I
I I
I
I
(a) Oatc sign~l g~ncratioo I
·JH D DDD • ~j,___ _____.__..ri___.__._n '
21' I wt
---1-L-D_._.__D-L-L--t-D w,
(b) Gate signals
- v,
a..,, + " (c) Output \'Oltagc
FIGURE 6.13 .
Multiple-pulse-width modulation.
signal sets the o utput frequency / 0 , and the carrier frequency/, determines the number of pulses per half-cycle p. The modulation index controls the output voltage.1bis type . of modulation is also known as uniform pulse-width modulation (UPWM). The number of pulses per half-cycle is found from
f,
p= -
21.
m,
= -
2
(6.30)
where m 1 = J/ J0 is defined as the frequency m od11/atio11 ratio. The instanlaneous outpul vollagc is v0 = V,(g, - g,). Tite output voltage for single-phase bridge inverters is shown in Figure 6. I3c for UPWM.
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Chapter 6
Pulse-Width-Modulated Inverters
If 6 is the width of each pulse, the rms output voltage can be found from 2p1(•lp +6)/2 ]'/2 = v, ~ v. = [ -21r v~ d(wc) -1f (• lp-6)12
(6.31)
·Toe variation of the modulation index M from O to I varies the pulse width d from Oto Tr2p (0 to 1r/ p) and the rms output voltage V0 from Oto V,. The general form of a Fourier series for the instantaneous output voltage is 00
L
v0 (1) =
B. sin nwc
(6.32)
n= I.J..S. ...
The coefficient B. in Eq. (6.32) can be determined by considering a pair of pulses such that the positive pulse of duration 6 starts at wt = a and the negative one of the same width starts at wt = 1r + a. This is shown in Figure 6.13c. The effects of all pulses can be combined together to obtain the effective output voltage. If the positive pulse of mth pair starts at wr = Oro and ends at wr = cx111 + 6, the Fourier coefficient for a pair of pulses is
1··+& sin 4V, . 4 n6 [ . ( = ;;;;
2 [ bn = 1T
J."'"-·"'2 sin 36) - . ( + + 4
11wt d (wt ) -
o,...+612
sin
sin n
a,,.
sin 11
,r
] 6)] +4
11w1 d(wt)
1r+o.,.
am
(6.33)
The coefficient B. of Eq. (6.32) can be found by adding the effects or all pulses,
. -116[Sin. 11 (0, L• -411, SIO m=l mr 4 2
Bn
=
11
36) - sin. n (.orr +
+-
4
4
(6.34)
A computer program is used to evaluate the performance of multiple pulse modulation. Figure 6.14 shows the ham1onic profile against the variation of modulation index for five pulses per half-cycle. lbe order or harmonics is the same as that of singlcpulse modulation. Toe distortion factor is reduced significantly compared with that of single-pulse modulation. However, due to larger number or swit.ching on and off processes or power transis tors, the switching losses would increase. With larger values of p, the amplitudes of LOH would be lower, but the amplitudes of some higher order harmonics would increase. However, such higher order harmonics produce negligible ripple or can easily be filtered out. Due to the symmetry of the output voltage along the .t ·axis, A. = 0 and the even harmonics (for tr = 2, 4,. 6, .•. ) are absent. The mth time 1., and angle a., of intersection can be determined from t I
m
(I(,. T, = -W = (m - M) -2
o,n
m
= -
W
=
(111 -
T 1 + M ) ..2.
2
for
111
= 1, 3, ... , 2p
(6.35a)
for
,,, = 2 . 4 , ... . 2p
(6.35b)
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6.6 Voltage Control of Single-Phase Inverters
v,
1.0
v.
253
DF(%)
.'
5.0
p• S DF
0.8
4.0
0.6
3.0
0.4
2.0
v, 0.2
·-·-v7 ,,
-~·- -------- --:..-=-.:.'=':.: -
1.0
o-1-----~- - - --.-----1-0 . 1.0
0.8
0.4 0.2 0.6 Modulation index. M
0
FIGURE 6. 14
Harmonic profile of muhiplc-pulse-"idlh moduJ:uion.
Because all widths are the same, we get the pulse width d (or pulse angle 6) as 6
ti = - =
"'
l,n -,. 1 - Im
= M~
{6.35c)
where T, = Tl2p. The algorithm for generating the gating signals is the same as that for single-pulse modulation, except the switching period T, o f the triangular carrier signal v,, is Tl2p instead of 1'12. 6.6.3
Sinusoidal Pulse-Width Modulation
Ins1ead of maintaining the width of all pulses the same as in the case of multiple-pulse modulation, the width of each pulse is varied in proportion to the amplitude of a sine wave evaluated at the center of the same pulse (2]. The OF and LOH are reduced sig· nificantly. The gating signals as shown in Figure 6.15a are generated by comparing a si· nusoidal reference signal with a triangular carrier wave of frequency f,.This sinusoidal pulse-width modulation (SPWM) is commonly used in industrial applications. The frequency of reference signal f, determines the inverter output frequency[,;. and its peak amplitude A, controls the modulation index M. and then in turn the nns output volt· age v.. Comparing the bidirectional carrier signal v.,, with two sinusoidal reference sig· nals v, and -v, shown in Figure 6.1 Sa produces gating signals g 1 and g,, respectively, as shown in Figure 6.15b.111e output voltage is v0 = V,(g1 - g,). However, g1 and g, cannot be released at the same time. The number of pulses per half-cycle depe nds on the carrier frequency. Within the constraint that two transistors of the same arm {Q1 and
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254
Chapter 6
Pulse-Width-Modulated Inverters Carrier signal
v
v,
A.,
A, (a)
2,r
I I I I
I
'
I
,.,
g,
..1
f+-1.-l I,
0
..1
1T
1w
(b)
..1
0 v.
6m
v,
(c)
0
2,r
..,
- v,
v
FIGURE 6.15
Sinusoidal pulse·width moduJation.
Q4 ) cannot conduct at the same time, the instantaneou.s output voltage is shown in Figure 6.15c. The same gating signals can be generated by using unidirectional triangular carrier wave as shown in Figure 6.15d. It is easier to implement this method and is preferable. The algorithm for generating the gating signals is similar to that for the uniform PWM in Section 6.6.2, except the rererence signal is a sine wave v, = V, sin wt, instead or a de signal. The output voltage is v0 : V,(g, - g.). Therms output voltage can be varied by varying the modulation index M. It can be observed that the area or each pulse corresponds awoxirnatcly to the area under the sine wave between the adjacent midpoints of off periods on the gating signals. If 6m is the width or mth pulse, Eq. (6.31) can be extended to find therms output voltage
V.=V. O
J
(f
r, ~
m=1 1T
)'12
(6.36)
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6.6 Voltage Control of Single-Phase Inverters
255
Equation (6.34) can also be applied to determine the Fourier coefficient of output voltage as
Bn
= ~~
4V, sin n8m [ sin II ( am
;;;
1
4
36.,) - sin II ( 1T + a., + 8.,)] +4 4 for n • I, 3, 5, . . .
(6.37)
A computer program is developed to determine the width of pulses and to evalua te the harmonic profile of sinusoidal modulation. The harmonic profile is shown in Figure 6.16 for five pulses per half-cycle. The DF is significantly reduced compared with that of multiple-pulse modulation. This type of modulation eliminates all harmonics less than or equal to 2p - 1. For p = 5, the LOH is ninth. The mth time Im and angle am of intersection can be determined from I
am T, == tx + mw 2
m
(6.38a)
where 1, can be solved from
21= M sin [w (t, + -mT,)] 2 21 = M sin [w( + mT,)] T,
1 - T,
1,
2
for m
= 1, 3, .. . , 2p
(6.38b)
for m
= 2, 4,
(6.38c)
... , 2p
where T, e T!2(p + 1). The width of the mth pulse dm (or pulse angle 6.. ) can be found from (6.38d)
v,
DF (%
v,
0.8
1.0
0.8 0.6 0.6 0.4
0.4
-------------·----- ---
0.2
---
·--. . ...._
v9 ... v 15
0--1,-- ----------~ -
l.0
0.8
· -· 0
0.6 0.4 0.2 Modulo.tion index. M
0
Q2
FIGURE 6. 16
Harmonic profile o! sinusoidal pulse-width modulation.
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Chapter 6
Pulse-Width-Modulated Inverters
The output voltage of an inverter contains harmonics. 'The PWM pushes the harmonics into a high-frequency range around the switching frequency fc and its multi· pies, that is, around harmonics m1, 2m1, 3m1, and so on. The frequencies at which the voltage harmonics occur can be related by
(6.39)
fn = (jml ± k)fc
where the nth harmonic equals the kth sideband of jth times the frequency to modulation ratio m1.
n = ;m1 ± k = 2jp ± k
for j = 1, 2, 3, .. . and k = 1, 3, 5, ...
(6.40)
The peak fundamental output voltage for PWM and SPWM control can be found approximately from
v... = dV,
forOsdsl.O
(6.41)
Ford = 1, Eq. (6.41) gives the maximum peak amplitude of the fundamental output voltage as V., 1(mu) = V,. According to Eq. (6.12), V., 1cmas) could be as high as 4V,l-rr = 1.27311, for a square-wave output. To increase the fundamental output voltage, d must be increased beyond 1.0.The operation beyond d = 1.0 is called overmodulation. The value of d at which V.,11.,...1 equals 1.273V, is dependent on the number of pulses per half-cycle p and is approximately 3 for p = 7, as shown in Figure 6.17. Overmodulation basically leads to a square-wave operation and adds more harmonics as compared with operation in the linear range (with d s 1.0). Overmodulation is normally
avoided in applications requiring low distortion (e.g., uninterruptible power supplies
(UPSs)).
-----------------::;--;;-=-- ,
FIGURE 6.17
Pc,ak fundamental output volt.age ~cnus modulation index M.
0
2
3
M
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6.6 Voltage
257
6.6.4 ·Modified Sinusoidal Pulse-Width Modulation Figure 6.1Sc indicates that the widths of pulses nearer the peak of the sine wave do not change significantly with the variation of modulation index. This is due to the characteristics of a sine wave, and the SPWM technique can be modified so that the carrier wave is applied during the first and last 60° intervals per half-cycle (e.g., 0° to 60° and 120° to 180"). This modified sinusoidal pulse-width modulation (MSPWM) is shown in Figure 6.18. The fundamental component is increased and its harmonic characteristics are improved. It reduces the number of switching of power devices and also reduces switching losses. The mth time 1., and angle a,. of intersection can be determined from 1 m
am
= -
w
T, 2
= r +mx
for m = 1, 2, 3, . .. , p
(6.42a)
where 1, can be solved from
21 1 - T,
. [ w ( r, + mT,) ] = M sm 2
~ =Msin [ w(1,+m;,)]
= 1,3, ... ,p
(6.42b)
for m : 2, 4, .. . , p
(6.42c)
for m
The time intersections during the last 60° intervals can be found from am +t lm + l ;;;;; ---;-- :;;;;
2T -
for m
l2p-,n
•
= p, p +
I . . . , 2p - 1
(6.42d)
Carrier signal
Reference signal
s,
. FIGURE '6.18
Modified sinusoida1 pulic~wid1h modulation.
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Chapter 6
Pulse-Width-Modulated Inverters
v, v,
1.0
II_____,I ___
__
j__l___
:; --i---~ _: s 0.7 0.6
v0,!'
'I l
i
I
I__
.l--
i
-+ I
DF%
:
10
:
J DF 1
0.5 0.4
_ _. . _
0.3
0.2
2
0.1
I
FIGURE 6.19
HarmonJc profile of modified sinusoidal
pulse·widlh nKXlulalion.
OI
0.9 0.8 0.7 0.6 0.5 0.4 0.3
Modulation index. M
where T, - T/6(p + 1). The width of the mth pulse dm (or pulse angle 6m) can be found from ·
(6.42c) A compuler program w.as used lo determine ihe pulse widths and 10 evaluate the performance of modified SPWM. The harmonic profile is shown in Figure 6.19 for five pulses per half-cycle. The number of pulses q in the 60' period is normally related to the frequency ratio, particularly in three-phase inverters, by
(6.43) The instantaneous output voltage is v0 ;= V,(g, - g.). The algorithm for generating the gating signals is similar to that for sinusoidal PWM -in Section 6.6.3, excepl the ref· erence signal is a sine wave from 60• to 120° only.
6.6.5
Phase-Displacement Control Voltage control can be obtained by using multiple inverters and summing the output voltages of individual inverters. A single-phase full-bridge inverter in Figure 6.2a can be perceived as the sum of two half-bridge inverters in Figure 6.la. A tSO• ·phase
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6.6 Voltage Control of Single-Phase Inverters
259
••• v, (a)
(b)
2
...v, 0
wl
180'
2
•••v, (c)
(d)
(e)
0
...v,
...
360'
180' Q
T
0
- v.
...
180'
0
0
"''
180"
360'
••• v, (!)
...
,so·
0
180 + a
...
_a_)JQ=...._~ I
FIGURE 6.20
Phase-displacement control.
displacement produces an output voltage as shown in Figure 6.20c, whereas a delay (or displacement) angle of ex produces an output as shown in Figure 6.20e. For example, the gate signal g1 for the half-bridge inverter can be delayed by angle ex lo produce the gate signal gz. Therms outpul voltage,
V.= V.I y; ~ O
(6.44)
If V00
"' L
=
2V. _ s sin nwt
11 • 1..1..1.... n,r
then v,,.
=
2~ L - sin n(wt n• t.3.5... nn 00
ex)
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260
Chapter 6
Pulse·Width·Modulated Inverter,
The instantaneous output voltage, 00
= v.
v,b
vbo
0 -
n•l.3S... . n11
which, after using sin A - sin B tied to
= 2 sin[(A -
00
vab
2V.
= L -' (sin nwt -
= •• ,
~
sin n(wt - a))
B)/2) cos[(A + B)/2), can be simpli·
. na -4V, sm - cos n ( wt 2
, .. . mr
a)
-
2
(6.45)
The rms value of the fundamental output voltage is 4V, . a v.,c • -V2 sm.2
(6.46)
Equation (6.46) indicates that the output voltage can be varied by changing the delay angle. This type of control is especially useful for high-power applications, requiring a large number of switching devices in parallel. Uthe gate signals g1 and 82 are delayed by angles a 1 = a and a 2(= n - a), the output voltage v,b has a quarter-wave symmetry at ,r/2 as shown in Figure 6.20f. Thus, we get .. 2V.
v..
= n•Ll -n,r' sin(n(wr -
V1>o
= n•l L -' sin[n(wl nTr
v.b
= v00
00
a))
2V.
for 11 = I, 3, 5, . ..
+ a)]
for n
= l , 3, 5, ...
4V L -' cos(na) sin(nwt) n1T
for11
= 1,3,5
1r
00
6.7
-
vbo
=
n=I
(6.47)
ADVANCED MODULATION TECHNIQUES
The SPWM, which is most commonly used, suffer.; from drawbacks (e.g., low funda· · mental output voltage). The other techniques that offer improved performances are: Trapezoidal modulation Staircase modulation Stepped modulation Harmonic injection modulation Delta modulation For the sake of simplicity, we shall show the output voltage v00, for a half-bridge inverter in Figure 6.la. For a full-bridge inverter, v0 = v.0 - v1,o, where v00 is the inverse ofv..,. Trapezoidal modulation. The gating signals are generated by comparing a triangular carrier wave with a modulating trapezoidal wave (3) a s shown in Figur<: 6.21.
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6.7 Advanced Modulation Techniques
261
----- 7 I
'I
'
\
' \\
\
'
\
A, wt
I I ' ,,' '
\
,,
I
I~
I I
(a) Gntc signnl generation : I
' ''
I
v, 2
0
- v,
T
-- - - - - r - - - - ;,
-
~
~
~
~
u~
I
- - - -
2r.
WI
FIGURE 6..21
(b) Output voltage
Trapezoidal n1odulation.
The trapezoidal wave can be obtained from a triangular wave by limiting its magnitude to ±A,, which is related to the peak value A,(nwx) by
A,
=
( J A , (ma~)
where IT fs called the rriang11/ar factor, because 'the waveform becomes a triangular wave when" = 1. The modulation index Mis A, uA,(max) M= - = for0:$Af:$J A, A,
(6.48)
The angle of the nat portion of the trapezoidal wave is given by
24> = (! -
(6.49)
For fixed ,•alues of A,(max) and A., M that varies with the output vollage can be varied by changing the triangular factor"· This type of modulation increases the peak funda· mental output vollage up to 1.05V,, but the output contains LOHs. Staircase modulation. The modulating signal is a staircase wave. as shown in Figure 6.22. The staircase is not a sampled approximation lo the sine wave. The levels of the stairs are calculated to eliminate specific harmonics. The modulation frequency
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Chapter 6
Pulse-Width-Modulated Inverters
v
A,
..
A,
(a) Gate signal generation v,o
v,
T
-
l;,
0
-v.
T
-
2.....
L FIGURE 6.22
(b) Output vollagc
Staircase modulation.
ratio m1 and the number of steps are chosen to obtain the desired quality of output voltage. This is an optimized PWM and is not recommended for fewer than 15 pulses in one cycle. It has been shown (4] that for high fundamental output voltage and low OF, the optimum number of pulses in one cycle is 15 for two levels, 21 for three levels, and 27 for four levels. This type of control provides a high-quality output voltage with a fundamental value of up to 0.!/4V,.
Stepped modulation. The modulating signal is a stepped wave (4, 5] as shown in Figure 6.23. The stepped wave is not a sampled approximation to the sine wave. It is di· vided into specified intervals, say 20°, with each interval controlled individually to con· trol the magnitude of the fundamental component and to eliminate specific harmonics. This type of control gives low distortion, but a higher fundamental amplitude com· pared with that of normal PWM control. Harmonic injected modulation. The modulating signal is generated by injecting selected harmonics to the sine wave. This results in flat-topped waveform and reduces the amount of overmodulation. It provides a higher fundamental amplitude and low distortion of the output voltage. The modulating signal (6, 7) is generally composed of
·v,
=
1.15 sin wt + 0.27 sin 3wr - 0.029 sin 9wt
(6.50)
The modulating signal with third and ninth harmonic injections is shown in Figure 6.24. It should be noted that the injection of 3nth harmonics does not affect the quality of the output voltage, because the output of a three-phas\! inverter does not contain triplen harmonics. If o nly the third harmonic is injecled, v, is given by
v, = 1.15 sin wt + 0.19 sin 3wr
(6.51)
'
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6.7 Advanced Modulation Techniques
263
v,
v
v,
A. A,
'
0
'
...
/
(a) Gate signal generation
v,.
-
'
I I
-
-
0
~
......
.....
-
"'' FIGURE 6.23
(b) Output \'Ollagc:
Stepped modulation.
v
v,
..
(a) Gate signal generation
v
v• 2
I
-
- - -I
~
•I
~
0
-V
2 •
~
-
-
(b) Output voltage:
...... ...... ......
-
... FIGURE 6.24
Selected ha.rmonjc: injcetion modulation.
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264
Chapter 6
Pulse-Width-Modulated -Inverters v
v,
0 g,
8,,
I
3
omornrnnn nnoom g:t nonnooon no FIGURE 6.25 Harmonic injeclion modulalion.
"'' WI
wt
-v,
T
The modulating signal (8) can be generated from 2n/3 segments of a sine wave as shown in Figure 6.25. This is the same as injecting 311th harmonics to a sine wave. The line·toline voltage is sinusoidal PWM and the amplitude of the fundamental component is ap· proximately 15% more than that of a normal sinusoidal PWM. Because each arm is switched off for one-third of the period, the heating of 1he switching devices is reduced. Delta modulation. In delta modulation (9), a triangular wave is allowed to os· ciliate within a defined window t. V above and below the reference sine wave v,. The inverter switching function, which is idenlical to the output voltage v0 is generated from the vertices of the triangular wave v, as shown in Figure 6.26. -It is also known as hysteresis mod11/atio11. If the frequency of the modulating wave is changed keeping the slope of the triangular wave constant, the number of pulses and pulses widths of the modulated wave would change. The fundamental output voltage can be up 10 IV, and is dependent on the peak am· plitude A, and frequency f, of the reference voltage. The delta modulation can control the ratio of voltage to frequency, which is a desirable feature, especially in ac motor control.
6.8
VOLTAGE CONTROL OF THREE-PHASE INVERTERS A lhree-phase inverler may be considered as three single-phase inverters and the out-
put of each si ngle-phase inverter is shifted by 120°. The voltage control lechniques
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6.8 Voltage Control of Three-Phase Inverters
265
y
o f-:f-~~~~....,.. \ ~~.~ ~~~~~--,~r--.,,_~ w +t
' /'',,
Lower band
'
limit
v..,
v, 2 Ol-l---4--4~-+-l--~'--~.......-~-I--J.-'-I.-~•
- v, T
2~
wt
FIGURE 6.26
Delta modulation.
discussed in Section 6.6 are applicable to three-phase inverters. However, the following techniques arc most commonly used for three-phase inverters. Sinusoidal PWM Third-harmonic PWM 60° PWM Space vector modulation 6.8.1
Sinusoidal PWM The generations of gating signals with sinusoidal PWM are shown in Figure 6.27a.There are three sinusoidal reference waves ('v,., v,•• and ·v,.) each shifted by 120°. A carrier wave is compared with the reference signal corresponding to a phase to generate the gating signals for that phase [10[. Comparing the carrier signal v,, with tile reference phases v,.. v,b, and v,, produces g1, g3, and ~ . respectively. as shown in Figure 6.27b.The instantaneous line-to-line output voltage is v00 ~ V,(g1 - 8.J). The output voltage as shown in Figure 6.27c, is generated by eliminating the condition that two switching devices in the same arm cannot conduct at the same time. The normalized carrier freq ucncy 1111 should be odd multiple of three. Thus, all phase-voltage (v.N, vbN, and Vc,v) are identical, but 120° out of phase without even harmonics; moreover, harmonics at frequencies multiple of three are identical in
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Chapter 6
Pulse·Width·Modulated Inverters
(a)
(b)
(c)
FIGURE 6.27
Sinusoidal pulsc·\\*idtb moduJation for thrcc-pba.sc in\•crter.
amplitude and phase in all phases. For instance, if the ninth harmonic voltage in phase a is VaN9(1) = ~ sin(9wl)
(6.52)
the corresponding ninth harmonic in phase b will be, VbN
= VgSin(9(w1 - 120° )) = tl<)Sin{9wl - 1080°)} = i'!,Sin(9wt)
(6.53)
Thus, the ac output line voltage v.b = v 0 N - vbN does not contain the ninth harmonic. Therefore, for odd multiples of thr.c c times the normalized carrier. frequency m1, the harmonics in the ac output voltage appear al normalized frequencies fh centered around m1 and its multiples, specifically, at n = jm1 ± k
(6.54)
where j = 1,3,5, . .. fork = 2, 4,6, . .. ; and j = 2, 4, ... fork= 1,5, 7, ... , such that II is not a multiple of 1hrce. Therefore, the harmonics arc at m1 ± 2, m1 ± 4 .. .• 2mr ± l ,2m r ± 5, . .. , 3m1 ± 2, 3m1 ± 4, ... , 4m1 ± l , 4m1 ± 5, .... For nearly sinusoidal ac load current, the harmonics in the de link current are at frequencies given by 11
= jmf
±k ±
I
{6.55)
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CHAPTER
9
Multilevel Inverters Tht /taming objtclivu of this chapttr an as follows: • • • • •
9.1
To learn the switching technique for multilevel inverters and their types To study the operation and features of multilevel inverters To understand the advantages and disadvantages of multilevel inverters To learn about the control strategy to address capacitor voltage unbalancing To learn the potential applications of multilevel inverters
INTRODUCTION
The voltage source inverters produce an output voltage or a current with levels either O or ±Vc1c. They are known as the two-level inverter. To obtain a quality output voltage or a curren t waveform with a minimum amount of ripple content, they require high-switching frequency along with various pulse-width modulation (PWM) strategies. In high-power and high-voltage applications, these two-level inverters, however, have some limitations in operating at high frequency mainly due to switching losses and constraints of device ratings. Moreover, the semiconductor switc~ing devices should be used in such a manner as to avoid problems associated with their series-parallel combinations that are necessary to obtain capability of handling high voltages and currents. The multilevel inverters have drawn tremendous interest in the power industry. They present a new set of features that are well suited for use in reactive power compensation. It may be easier to produce a high-power, high-voltage inverter with the multilevel structure because of the way in which device voltage stresses are controlled in the structure. Increasing the number of voltage levels in the inverter without requiring higher ratings on individual devices can increase the power rating. The unique structure of multilevel voltage source inverters' allows them to reach high voltages with low harmonics without the use of transformers or series-connected synchronizedswitching devices. As the number of voltage levels increases, the harmonic content of the output voltage waveform decreases significantly (1, 2]. 406 Cooyr
9.2 Multilevel Concept
9.2
407
M ULTILEVEL CONCEPT
Let us consider a three-phase inverter system (4], as shown in Figure 9.la, with a de voltage Vdc· Series-connected capacitors constitute the energy tank for the inverter, providing some nodes to which the multilevel inverter can be connected. Each capacitor has the same voltage Em, which is given by
vd,
(9.1)
Em= m- 1
whcre m denotes the number of levels. Toe term level is referred to as the number of nodes to which the inverter can be accessible. An m-level inverter needs (m - 1) capacitors.
Im
v. V m- 1 lm- 1
vdc
~
V3
vdc
+· E.. :..L
2
+
E..
•• +
DOAC Pbwer
13
v••.
vb
Processinj:
System Vz
12
v,
,,
Ve
(a) Three-phase multilevel power processing system
Vs +
Em V4
Em
+
c,
+ C2
Vs
(b) Schematic of single pole of mullilcvcl inverter by a switch
FIGURE 9.1 General topology of multilevel inverters.
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Chapter 9
Multilevel Inverters
vtlJ ~......,...-...,,... _, ---, ...,·· ···..... ·.../
Fundamental WavcofVc - ab
V3 V2
V 1 -+J~-'-~--'-~....1-~--'-~--'-~.......__....~.---~,-~~ ~ ....--~-,--,...,.. ~t - V2
- V3
-v. Yot,
···...
..
. ·
FIGURE 9.2 1ypical output voltage of a five-level multilevel inverter.
Output phase voltages can be defined as voltages across output terminals of the inverter and the ground point denoted by o as shown in Figure 9. la. Moreover, input node voltages and currents can be referred to input tenninal voltages of the inverter with reference to ground point and the corresponding currents from each node of the capacitors to the inverter, respectively. For example, input node (de) voltages are de-
signated by Vi, V2, etc. and the input node (de) currents by 11, h, etc., as shown in Figure 9.la. V,,, Vb, and Yc, are the root-mean-square (rms) values of the line load v6ltages; 14 , Ii,, and le are the nns values of the line load currents. Figure 9.lb shows the schematic of a pole in a multilevel inverter where v. indicates an output phase voltage that can assume any voltage level depending on the St!lection of node (de) voltage Vi, V2 , etc. Thus, a pole in a multilevel inverter can be regarded as a single~pole, multiple-throw switch. By connecting the switch to one node at a time, one can obtain the desired output. Figure 9.2 shows the typical output voltage of a five-level inverter. The actual realization of the switch requires bidirectional switching devices for each node. The topological structure of multilevel inverter must (1) have less switching devices as far as possible, (2) be capable of withstanding very high input voltage for highpower applications, and (3) have lower switching frequency for each switching device. 9.3
TYPES OF MULTILEVEL INVERTERS
. The general structure of the multilevel converter is to synthesize a near sinusoidal voltage from several levels of de voltages, typically obtained from capacitor voltage sources. As the number of levels increases, the synthesized output waveform has more steps, which produce a staircase wave that approaches a desired waveform. Also, as more steps are added to the waveform, the harmonic distortion of the output wave decreases, approaching zero as the number of levels increases. As the number of le~els increases, the voltage that can be spanned by summing multiple voltage levels also
Cooyr
9.4
Diode-Clamped Multilevel Inverter
409
increases. The ouJput voltage during the positive half-cycle can be fo und from V 00
'" £ ,,SF" =L
(9.2)
n• l
where SF,, is the switching or control function of nth node and it ta kes a value of O or 1. Generally, the capacitor terminal voltages E h £,,, ... all have the same value Em· Thus, the peak o utput voltage is v . o( pc.,k) = (m - l )Em = Vdc- To generate an output voltage with both positive and negative values, the circuit topology has anothe r switch to produce the negative part v00 so tha t Vab = v00 + Vob = v0 0 - vbo. The multilevel inverters can be classified into three types [5]. Diode-clamped multilevel inverter; Flying-capacitors multilevel inverter; Cascade multilevel inverte r. 9.4
DIODE-CLAMPED MULTILEVEL INVERTER
A diode-clamped multilevel (m-level) inverter (D CMLI) typicaUy consists of (m - 1) capacitors on the de bus and produces m levels on the phase voltage. Figure 9.3a shows Vs
+ Es
DC
s.1
c, 0.1
S.i
V4
D,2
Es ~
vd<
o~,
Sa3
D2
s••
D1 Vo
s;I
s~
c.i D~ V2
Es
o:u c,
03
Vs- ab
L.
o_,
Con verier
s.1
V5
s.2
c,
s••
~
i.
V 3 V,x
Ve-ab
s~,
c.i
s;z 0 .3 0.4
+
V4
D.2
s;..,
s~
input side
Sa3
Oa3
V3
Es
04
Load side
V2
s~
D bl
s;4
0
c. v,
-
v, ~o (a) O ne leg of a bridge
(b) Single-phase bridge
FIGURE 9.3
Diode-clamped fivc-lc,•cl bridge multilevel inverter. [Ref. 4)
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ted
410
Chapter 9
Multilevel Inverters
one leg and Figure 9.3b shows a full-bridge five-level diode-clamped converter. The numbering order of the switches is S0 1, S112, S03, So4, S~i, S:O, and s;... The de bus consists of four capacitors, Ci, C2, C3, and C,. For a de bus voltage Vele, the voltage across each capacitor is Vdef4, and each device voltage stress is limited to one capacitor voltage level Vdcl4 through clamping diodes. An m-level inverter leg requires ( m - 1) capacitors,2( m - 1) switching devices and (m - l )(m - 2) clamping diodes.
s:a.
9.4.1
Principle of Operation
To produce a staircase-output voltage, let us consider only one leg of the five-level in· verter, as shown in Figure 9.3a, as an example. A single-phase bridge with two legs is shown in Figure 93b. The d e rail O is the reference point of the output phase voltage. The steps to synthesize the five-level voltages are as follows: 1. For an output voltage level v00 = Vc1c, tum on all upper-half switches S01 through Sa4·
2. For an output voltage level v00 = 3Vdcf4, tum on three upper switches S02 through So4 and one lower switch S~1• 3. For an output voltage level v00 = V.iJZ, turn on two upper switches Sa3 through Sa4 and two lower switches S~1 and 2 . 4. For .an output voltage level v00 = VM4, turn on one. upper switch So4 and three lower switches s~·1 through S:U. 5. For an output voltage level 1100 = 0, turn on all lower half switches S~1 through
s:
s~,.
Table 9.1 shows the voltage levels and their corresponding switch states. State condi· tion 1 means the switch is on, and state Omeans the switch is off. It should be noticed that each switch is turned on only once per cycle and there are four complementary switch pairs in each phase. These pairs for one leg of the inverter are (S0 1, S~1), (S.a, Sa), (Sa3, S~3 ) , and (S04, s:i.i). Thus, if one of the complementary switch pairs is turned on, the other of the same pair must be off. Four switches are always turned on at the same time. Figure 9.4 shows the phase voltage waveform of the five-level inverter. The line voltage consists of the positive phase-leg voltage of terminal a and the negative phase-leg voltage of terminal b. Each phase-leg voltage tracks one-half of the sinusoidal wave. The ·
TABLE 9.1 Diode-Oamped Vollage Levels and Their Switch States
Switch State Output Vo11
s.1
So1
v, = v.,.
I
I
V, • 3V...J4
0 0 0 0
0 0 0
v, - v.,.n V2 • V...,14 ~ 0
v.
s""'
I
l
0 0
s..
S~1
sd
s:.i
s:..
l I 1 1
0
0 0
0 0 0 1 l
0 0 0 0
0
1 l l 1
1 I I
l
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9.4 Diode-Clamped Multilevel Inverter
411
v.o
..
Vs
..•..
Fundamen1al WaveofVc-,b
../
V4 V3 V2
v , .............__~....___,__,__,__.__,__,..__.....,......__,__,__,__,__,__,__,--,--,--,--..... o 1 o2 o 3 a 4
o5
06
o1
08 •
...
w•
....·
FIGURE 9.4
Phase and fundamental voltage waveforms or a five-level inverter.
resulting line voltage is a nine-level staircase wave. This implies that an m-level converter has an m-level o utput phase-leg voltage and a (2m - 1 )-level o utput line voltage. 9.4.2
Features of Diode-Clamped Inverter
The main fea tures arc as follows:
1. High-voltage rating for blocking diodes:Although each switching device is only required to block a voltage level of V
=
m - 1- k l Vdc m-
(9.3)
where m is the numbe r of levels: k goes from 1 to (m - 2); Vdc is the total de Link voltage.
lf the blocking voltage rating of each diode is the same as that of the switching device. the num ber of diodes required for each phase is N0 = (m - 1) X (m - 2). This number represents a quadratic increase in m . Thus. for 111 = S, N0 = (5 - 1) X (5 - 2) = 12. When m is sufficien tl y high, the num ber of diodes makes the system impractical to implement, which in effoct Limits the number of levels.
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Chapter 9
Multilevel Inverters
2. Unequal switching device rating: We can notice from Table 9.1 that switch S01 conducts only during v00 = V<1c, whereas switch Sa4 conducts over the entire cycle except during the interval when v..., = 0. Such an unequal conduction duty requires different current ratings for the switching devices. Therefore, if the inverter design uses the average duty cycle to find the device ratings, the upper switches may be oversized, and the lower switches may be undersized. If the design uses the worst-case condition, then each phase has 2 x (m - 2) upper devices oversized. 3. Capacitor voltage unbalance: Because the voltage levels at the capacitor terminals are different, the currents supplied by the capacitors are also different. When operating at unity power factor, the discharging time for inverter operation (or charging time for rectifier operation) for each capacitor is different. Such a capacitor charging profile repeats every half-cycle, and the result is unbalanced capacitor voltages between different levels. This voltage unbalance problem in a multilevel converter can be resolved by using appro~ches such as replacing capacitors by a controlled constant de voltage source, PWM voltage regulators, or batteries. The major advantages of the diode-clamped inverter can be summarized as follows: • When the number of levels is high enough, the harmonic content is low enough to avoid the need for filters. • Inverter efficiency is high because all devices are switched at the fundamental · frequency. • The control m e thod is simple.
The major disadvantages of the diode-clamped inverter can be summarized as follows: • Excessive clamping diodes are required when the number of levels is high. • It is difficult to control the real power flow of the individual converter in multiconverter systems. 9.4.3
Improved Diode-Clamped Inverter
The problem of multiple blocking voltages of the clamping diodes can be addressed by connecting an appropriate number of diodes in series, as shown in Figure 9.5. However, due to mismatches of the diode characteristics, the voltage sharing is not equal. An improved version of the diode-clamped inverter [6] is shown in Figure 9.6 for five levels. The numbering order of the switches is S 1, S2, S3, S4 , S1, Si, S3, and S 4. There 11re a total of eight switches and 12 diodes of equal voltage rating, which are the same as the diode-clamping inverter with series-connected diodes. This pyramid architecture is extensible to any level, unless otherwise practically limited. A five-level inverter leg requires (m - 1 =) 4 capacitors, (2(m - 1) =) 8 switches and ((m - l ){m - 2) =) 12 clamping diodes. Principle of operation. The modified diode-clamped inverter can be decomposed into two-level switching cells. For an m-level inverter, there are (m - 1) switching cells. Thus, for m = 5, there are 4 cells: In cell 1, S2 , S 3, and S 4 arc always on whereas S1 and Sj are switched alternatively to produce an output voltage V
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9.4 Diode-Clamped Multilevel Inverter
413
Vs
~
c, o,
v,
011 V3
o,,
~
Oa
01
D2
;J
D,;i
03
Ds
;J
Os4
~
0 ;1
;J
o:i
~
o::,
~
o;..
0 09
o,
Ds
D 10
D 12
C3
A
V2 D6
c,
V1
FIGURE 9.5 Diode-clamped multilevel in verier with diodes in series. !"Re(. 6]
Voof4, respectively. Similarly, in cell 2, S3, S4 , and Si are always on whereas S2 and S2are switched alternatively to produce an output voltage VdJ4 and 0, respectively. In cell 3, S4 , S\ , and S2are always on whereas S3 and Sj are switched alternatively to produce an output voltage O and - VdJ2, respectively. In final cell 4, Sj, S 2• and S 3are a lways on whereas S4 and S 4are switched alternatively to produce an output voltage - Voof4 and - VdJ2, respective ly. Each switching cell works actually as a normal two-level inverter, except that each forward or freewheeling path in the cell involves ( m - 1) devices instead of only one. Taking cell 2 as an example, the forward path of the up-arm involves Di, S2 , S2, and S4, whereas the freewheeling path of the. up-arm involves Si, D 12, Ds, and Dz, connecting the inverter output to Vd.,14 level for either positive or negative current flow. The forward path of the down-arm involves S\, S 2, D 10, and D4, whereas the freewheeling path of the down-arm involves Dy,, D,, S 3, and S4, connecting the inverter output to
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Chapter 9
Multilevel Inverters
Vs
~
c,
;J
D,
v.
V3
02
07
03
Oa
On
0
o,
09
Os
010
012
~
Os3
;J
O,c A
:J OSi
I
V2
06
C4
v,
~
o::i
~
o:..
FIGURE 9.6
Modified diode-damed inverter wilh distributed clamping diodes. (Rel 6]
zero level for either positive or negative current flow. The following rules govern the switching of an m-level inverter:
L At any moment, there must be (m - 1) neighboring switches that are on. 2. For each two neighboring switches, the outer switch can only be turned on when the inner switch is on. 3. For each two neighboring switches, the inner switch can only be turned off when the outer switch is off. 9.5
FLYING-CAPACITORS MULTILEVEL INVERTER
Figure 9.7 shows a single-phase, full-bridge, five-level converter based on a flyingcapacitors multilevel inverter (FCMLI) [51, The numbering order of the switches is
9.5 Flying-Capacitors Multilevel Inverter Output AC side
Input De side
Converter
s••
v, ~ sb1 Ci
+ S.i
- - Ca3 +
S4.3
+ i,.
Sa4 a
+ VC- •b
s~ s;_,
4 15
c••
cb3
sbJ
Ciu
V4
V2
S1,2
V3
Cii2
Ci
sb4
Ca3
CM
Ct.3 V3 vdc
b
C,a
st,4
c..,
V4
Sb3
Ci,2
C,
Ct.3 V2
s~
Si,2
s;,
Si,,
C4
0
v, - -
FIGURE 9.7
Circuit diagram or a five-le,·el, flying-capacitors, single-phase inverter. (Re( 5]
S0 1, S0 2, S0 3, S0 4, S~4 , S~3, S~. and S~1. Note that the order is numbered differently from that of the diode-clamped inverter in Figure 9.3. The numbering is immaterial as along as the switches are turned on and off in the right sequence to produce the desired output waveform. Each phase leg has an identical structure. Assuming that each capacitor has the same voltage rating, the series connection of the capacitors indicates the voltage level between the clamping points. Three inner-loop balancing capacitors (C0 1, C,a, and C03) for phase-leg a are independent from those for phase-leg b. All phase legs share the same de-link capacitors, C1 through C4 • The voltage level for the flying-capacitors converter is similar to that of the diode-clamped type of converter. That is, the phase voltage v00 of an m-level converter has m levels (including the reference level), and the line voltage v0 b has (2m - 1) levels. Assuming that each capacitor has the same voltage rating as the switching device, the de bus needs (m - 1) capacitors for an m-level converter. The number of capacitors required for each phase is Ne = }:j. 1(m - j). Thus, form = 5, Ne = 10. 9.5.1
Principle of Operation
To produce a staircase-output voltage, let us consider the one leg of the five-level inverter shown in Figure 9.7 as an example. The de rail O is the reference point of
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Chapter 9
Multilevel Inverters
the output phase voltage. The steps to synthesize the five-level voltages are as follows: L For an output voltage level v"° = Vdc, tum on all upper-half switches Sat through Sa4. 2. For an output voltage level v00 = 3VdJ4, there are four combinations: a. v00 = Vdc - Vdt.14 by turning on devices Sat, S112, S113, and s:.i. b. Vao = 3Vdd4 by turning on devices So2, Sa3, Sa4, and S~1• c. v00 = Vite - 3Vdd'4 + V&/2 by turning on devices Sat, S43 , So4, and s:a. cl. v00 = Vc1c - Vdd2 + Vdcf4 by turning on devices Sai, Sa2, S44, and S~. 3. For an output voltage level v/J() = Vdcf2, there are six combinations: a. Vao = vdc - VciJ2 by turning on devices Sat, Sa2, s:u. ands;... b. v00 = V&!2 by turning on devices S113, Sa4, S~i, and c. v00 = Vc1c - 3V.icf4 + V
s:.z.
s:a.
s~.
s;....
There are many possible switch combinations to generate the five-level output voltage. Table 9.2, however, lists a possible combination of the voltage levels and their corresponding switch states. Using such a switch combination requires each device to be switched only once per cycle. It can be noticed from Table 9.2 that the switching devices have unequal tum-on time. Like the diode-clamped inverter, the line voltage consists of the positive phase-leg voltage of terminal a and the negative phase-leg voltage of terminal b. The resulting line voltage is a nine-level staircase wave. This implies that an m-level converter has an m-level output phase-leg voltage and a (2m - 1 )-level output line voltage.
TABLE 9.2 One Possible Switch Combination of the Flying-Capaciton Invert.er
Switch State Output V.o
Vs,. V4 = V3 = V1 =
v.;
Vele 3Vo!4 Vct,:12 Vocl4 0
s••
s.,,
S.3
Sa4
s;;(
l 1 I l 0
l l
l l 0 0 0
I
0
0 0 0 0
1
I
0 0
s~
s~.
1 I
0 0 1 l
0 0 0 l
0 0 0 0
l
1
1
1
s~>
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9.6 Cascaded Multilevel Inverter
9.5.2
417
Features of Flying-Capacitors Inverter
The main features are as follows:
L Large number of capacitors: The inverter requires a large number of storage capacitors. Assuming that the voltage rating of each capacitor is the same as that of a switch.ing device, an m-level converter requires a total of ( m - 1) x (m - 2)/2 auxiliary capacitors per phase leg in addition to (m - 1) main de bus capacitors. On the contrary, an m-level diode-clamp inverter only requires (m - l} capacitors of the same voltage rating. Thus, for m =·5, Ne = 4 x 3/2 + 4 = 10 compared with Ne = 4 for the diode-clamped ·type. 2. Balancing capacitor voltages: Unlike the diode-clamped inverter, the FCMLI has redundancy at its inner voltage levels. A voltage level is redundant if two or more valid ·switch combinations can synthesize it. The availability of voltage redundancies allows controlling the individual capacitor voltages. In producing the same output voltage, the inverter can involve different combinations of capacitors allowing preferential charging or discharging of individual capacitors. This flexibility makes it easier to manipulate the capacitor voltages and keep them at their proper values. It is possible to employ two or more switch combinations for middle voltage levels (i.e., 3VcJ.14, V
CASCADED MULTILEVEL INVERTER
A cascaded multilevel inverter consists of a series of H -bridge (single-phase, full. bridge) inverter units. The general function of this multilevel inverter is to synthesize a
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418
Chapter 9
Multilevel Inverters 4Vdc . r-,...
·•
~
v.
Vc1c
+ V•l( m- 1)12)
+
H-Bridge (m - 1)/2
+
~
. - 4Vdc .
~
Va2
H-Bridge2 vdc
+ n
!'.J
H-Bridge l v dc
~
'\ '·· ··7
\ .•• A
fud
al
v••
,;
\.
'
I,
!!:;:
"'' .,,
•
Va l
wt
V.z
c»t
Va3
wt
•
+
~ lla1 -
.
" J.
•
+ lla4
~
tvc1c
wt
{b) Ou1pu1 waveform of 9-level phase voltage
(a) Circuit diagram
FIGURE 9.8 Single-phase multilevel cascaded H-bridge inverter. [Ref. 7)
desired voltage from several separate de sources (SD CSs), which may be obtained from batteries, fuel cells, or solar cells. Figure 9.8a shows the basic structure of a singlephase cascaded inverter with SDCSs (7). Each SOCS is connected to an H-bridge inverter. The ac terminal voltages of different level inverters are connected in series. Unlike the diode-clamp or flying-capacitors inverter, the cascaded inverter does not require·any voltage-clamping diodes or voltage-balancing capacitors. 9.6.1
Principle of Operation Figure 9.8b shows the synthesized phase voltage waveform of a five-level cascadecl inverter with four SDCSs. The phase output voltage is synthesized by the sum of four inverter outputs, v011 = v01 + Vo2 + V0 3 + v04 • Each inverter level can generate three different voltage outputs,+ Vde, 0, and - V& , by connecting the de source to the ac output side by different combinations of the four switches, Si, S2 , S3, and S4 • Using the top yields Vo,.1 = + Vdc· Turning on S2 and S3 level as the example, turning on S1 and yields v"4 = -V11e- Turning off all switches yields v 4 = 0. Similarly, the ac output voltage at each level can be obtained in the same manner. If Ns is the number of de sources, the output phase voltage level is m = Ns + 1. Thus, a five-level cascaded inverter needs four SDCSs and four full bridges. Controlling the conductirtg angles at different inverter levels can minimize the harmonic distortion oft.h e output voltage. The output voltage of the inverter is almost sinusoidal, and it has less than·5% total harmonic distribution (THD) with each of the H-bridges switching only at
s.
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9.6 Cascaded Multilevel Inverter +
lv•i
v dc
j
Q.11
0
- vdc 0
+
"•i
Vc1c
O.;p
j G.;p
a.in
G.;.
211
'3trfl
1f
419
", I I
0
1fl2 .,. -:- er; I l
11-1-a; ,
.,.
I I I
2,r - a ·I
0
Q.ll
I
0
0
1f
wt
Gaip• G.; 0 is 1 if an upper switch is on and Oif a lower switch is on
(a) One H-bridge
(b) Switching time
FIGURE.9.9
Generation of quasi-square waveform. [Ref. 7]
fundamental frequency. If the phase current i 0 , as shown in Figure 9.8b, is sinusoidal and leads or lags the phase voltage Von by 90°, the average charge to each de capacitor is equal to zero over one cycle.Therefore, all SOCS capacitor voltages can be balanced. Each H-bridge unit generates a quasi-square wavefonn by phase shifting its positive and negative phase-leg-switching timings. Figure 9.9 shows the switching timings to generate a quasi-square waveform of an H-bridge. It should be noted that ea.ch switching device always conducts for 180° (or half-cycle), regardless of the pulse width of the quasi-square wave. This switching method makes all of the switching device current stresses equal. 9.6.2
Features of Cascaded Inverter The main features are as follows: • For real power conversions from ac to de and then de to ac, the cascaded inverters need separate de sources. The structure of separate de sources is well suited for various renewable energy sources such as fuel cell, photovoltaic, and biomass. • Connecting de sources between two converters in a back-to-back fashion is not possible because a short circuit can be introduced when two back-to-back converters are not switching synchronously. The major advantages of the cascaded inverter can be summarized as follows: • Compared with the mode-clamped and flying-capacitors inverters, it requires the least number of components to achieve the same number of voltage levels. • Optimized circuit layout and packaging are possible because each level has the same structure and there are no extra clamping diodes or voltage-balancing capacitors. • Soft-switching techniques can be used to reduce switching losses and device SIT~es.
The major disadvantage of the cascaded inverter are as follows: • It needs separate de sources for real power conversions, thereby limiting its applications.
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Chapter 9
Multilevel Inverters
Example 9.1 Fi nding Switching Al\gles to Eliminate Specific Harmonics
The phase voltage waveform for a cascaded inverter is shown in Figure 9.10 form = 6 (including Olevel). (a) Fmd tbe generalized Fourier series or the phase voltage. (b) Fmd the switching angles to eliminate 5th, 7th, 11th, and 13th harmonics if the peak fundamental phase voltage is 80% ofits maximum value. (c) Fmd the fundamental component 81, THO, and the distortion factor (OF).
Solution L
For a cascaded inverter with m levels (including 0) per half-phase, the output voltage per leg is '•
(9.4)
Due to the quarter-wave symmetry along the x-axis, both Fourier coefficients Ao and An are zero. We get Bn as 411.
Bn = ~
[1,.,i
1f
+
1.
B. = -4Vc1c n,r
,...
::.. ,,,,. ~
2,r
0
""'
-
"'
-
..
-
,1
I
.
I
I
..
,. ...
I
'
.
• I •
I
•
I
-
.... t
•• I
"'
I
"'
• I I'
••
' I
..
"'
. '
I
i\
r-1'
.,
••• I
"
. . •
"'
..
~
••
,..v•• 0
-
..
P'1
I
\
~
-
_.va2
'L,
I
I
J,rf2
,I Va,
0
(9.6)
/ •I
,4
f"o
_. va3
(9.5)
["'I-t cos(na;)]
~
,..v••
+
1
"·-·.
,r(l
,,
sin(nwr)d(wr)
O:
n sin(nwt)d(wr )]
t~-n
0
1. n
OJ
0 .. -
SVc1c
+
sin( nwt)d (wt)
~
I
..
I
PS
wt
.
wt
• •
wt
I •
•
I
•
I
..
FIGURE 9.10 Switching pattern swapping or the cascade invener for balancing battery charge. [Ref. 7]
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9.7 Applications
421
which gives the inslanlaneous phase voltage v,. as
i
1
l
v,.(w1) = -4V.ie [ "'- cos(11a1) sin(m"I) tiff , ••
(9.7)
b. If the peak output phase voltage V,n(,,..•l must equal to the carrier phase voltage, V,,(pat) = (m - l)V.ie. Thus, the modulation index becomes
(9.8) The conduccing angles a 1, o 2,
• .. •
o,,. _1 can be chosen such 1hat the tolal harmonic dis-
tortion of the phase voltage is minimized. These angles are normally chosen so as to cancel some predominant lower frequency harmonics. Thus, to eliminate 5th, 7th, 11th, and 13t.h harmonics provided that the peak fundamental phase voltage is 80%, of its maximum value, we must solve the following equations for modulation index M a 0.8. cos(5a 1) + cos(5a 2) + cos(5a3 ) + cos(Sa,) + cos(Sa,) = 0 cos(7a 1) + cos(7a 2) + cos(7a 3) + cos(7a,) + cos(7n5 ) = 0 cos(lla 1) + cos(lln2) + cos(lla3) + cos(lla,) + cos(lln,) = 0 cos(l3a 1) + cos( l3oz) + cos(l3a3) + cos(13a,) + cos(l3o5 ) = 0 cos(a 1) + cos(a 2 ) + cos(a 3 ) + cos(a,) + cos(a, ) = (m - l)M e 5 X 0.8 • 4
(9.9)
This set of nonlinear transcendental equations can be solved by an iterative method such as the Ne\\1on- Raphson method. Using Mathcad. we get ClJ
=
6.57°, 0 ) = 18.94°, O J = 27.)8°, U, = 45.15°, and a s = 62.24°
Thus. if the in,•erter output is symmetrically switched during the positive half-cycle of the fundamcntnl voltage to +V"' at 6.57°. +2V"' at 18.94°. +3Vd, at 27.18°. +4V"' at 45.15°, and +SV"' at 62.24° and similarly in the negative half-cycle to -v., at 186.57°. - 2V« at 198.94°, - 3V"' at 207.18°, - 4V"' at 225.15°, and - sv.., at 242.24°. the output voltage cannot contain the 5th, 7th, I I th, and 13th harmonics. c. Using Mat head, we get 8 1 = 5.0'.13%, THO = 5.975%, and OF = 0.08%
Note:The duty cycle for each of the voltage levels is different. This means that the level-I de source discharges much sooner than the level·5 de source. However, by using a switching pattern-swapping scheme among the various levels every half-cycle, as shown in Figure 9.10, all batteries can be equally used (discharged) or charged [7) . For example, if the first pulse sequence is P1, P2 , •• •• P,, then the next sequence is P2 , P3 , P,, Ps, P1, and so on. 9.7
APPLICATIONS There is considerable interest in applying voltage source inverters in high-power applications such as in utility systems for controlled sources of reactive power. In the steady-state operation, an inverter can produce a controlled reactive current and
Copyrghlcd matcria
422
Chapter 9
Multilevel lnverter1
operates as a static volt-ampere reactive (VAR)-compensator (STATCON). Also, these inverters can reduce tlie physical size of the compensator and improve its performance during power system contingencies. The use of a high-voltage invener makes possible direct connection to the high-voltage (e.g., 13-kV) distribution system, eliminating the distribution transformer and reducing system cost. In addition, the harmonic content of the inverter waveform can be reduced with appropriate control techniques and thus the efficiency of the system can be improved. The most common applications of multilevel conveners include (1) reactive power compensation, (2) back-to-back intenie, and (3) variable speed drives. 9.7.1
Reactive Power Compensation
An inverter converts a de voltage to an ac voltage; with a phase shift of 180°, the inverter can be operated as a dc-ac converter, that is, a controlled rectifier. With a purely capacitive load, the inverter operating as a dc-ac converter can draw reactive current from the ac supply. Figure 9.11 shows the circuit diagram of a multilevel converter directly connected to a power system for reactive power compensation. The load side is connected to the ac supply and the de side is open, not connected to any de voltage. For the control of the reactive power flow, the inverter gate control is phase shifted by 180°. The de side capacitors act as the load. When a multilevel converter draws pure reactive power, the phase voltage and current are 90° apart, and the capacitor charge and discharge can be balanced. Such a converter, when serving for reactive power compensation, is called a static-VAR generator (SVG).AII three multilevel converters can be used in reactive power compensation without having the voltage unbalance problem. The relationship of the source voltage vector Vs and the converter voltage vector Vc is simply Vs = Ve + ilcXs, where le is the converter current vector, and Xs is I, AC lnpu, Side
+ Reactive
v,
Load
~~-'-c='-"--'-'--"-'-'--'---'-~ ~~ 0c Load side
v, FIGURE 9.11
A multilc\'cl convenerconnected to a P<>"'e-rsystem ror reactive po"'er compensation. (Re(. SJ
Copyrghlcd matcria
9.7 Applications
423
v,
i.
l
!.----.- - - - ·
v,
v,
jl,X,
v, --============·~·====:; (a) Leading current
'
(b) logging currenl
FIGURE 9.12 Phasor diagrams of the source and 1he convener voltages for react,ive po\lo'er
compensation.
the reac1ance of the inductor ls. Figure 9.12a illustrates that the converter voltage is in phase with the source voltage with a leading reactive current, whereas Figure 9.12b illustrates a lagging reactive current. The polarity and the magnitude of the reactive current are controlled by the magnitude of the converter voltage Ve, which is a function of the de bus voltage and the voltage modulation index, as expressed by Eqs. (9.7) and (9.8).
9.7.2
Back-to-Back lntertie Figure 9.13 shows two diode-clamped multilevel converters that are interconnected with a de capacitor link. The left-hand side converter serves as the rectifier for utility interface. and the right-hand side converter serves as the inverter to supply the ac load. Each switch remains on once per fund~mental cycle. The voltage across each capacitor remains well balanced, while maintaining the staircase voltage wave, because the unbalance capacitor voltages on both sides tend to compensate each other. Such a de capacitor link is categorized as lhe back-to-back inte rtie. The back-to-back intcrtic that connects two asynchronous systems can be regarded as (I) a frequency changer. (2) a phase shifter. or (3) a power flow controller. The power flow between two systems can be controlled bidirectionally. Figure 9.14 shows the phasor diagram for real power transmission from the source end to the load end. This diagram indicates that the source current can be leading or lagging the source voltage. The converter voltage is phase shifted from the source voltage with a power angle, 6. If the source voltage is constant. then the current or power flow can be Rc.-c1irier Opcra1ion L,;
V s-hn lsb
Vs- (n I&
L,;
4
DCUnk
ln\'cr1cr Opcralion
V e-an
V e-be
\ r C- tt,
.p
.p
v,. .an
Le
\ ' 1- bft
LL
v,-a
LL
lu, v t.- ti"' 11.,c VL- 1:n
FIGURE 9.13
Rack-lo-back intertic sys1em using ,,vo <1iode-cl:1mpcd n1ul1ilC\'CI convcr1ers, IR1:t SJ
Copyrghtcd matcria
424
Chapter 9
Multilevel Inverters
'·
v,
~
il,X,
-
.
I,
V
~
jl,X,
v,
v, (a) Leading power factor
(b) Unity power facto r
(c) Lagging power factor
FIGURE 9.14
Phasor diagram or the source ,·ohage. converter voltage, and cunent showing real power
convcnions.
controlled by the converter voltage. For 6 = 0, the current is either 90° leading or lagging, meaning that only reactive power is generated. 9.7.3
Adjustable Speed Drives The back-to-back intertie can be applied to a utility compatible adjustable speed drive (ASD) where the input is the constant frequency ac source from the utility supply and the output is the variable frequency ac load. For an ideal utility compatible system, it requires unity power factor, negligible harmonics. no electromagnetic interference (EMI), and high efficiency. The major differences, when using the same structure for ASDs and for back-to-back inte rties. are the control design and the size of the capacitor. Because the ASD needs to operate at different frequencies, the de-link capacitor needs to be well sized to avoid a large voltage swing under dynamic conditions.
9.8
SWITCHING DEVICE CURRENTS
Let us take a three-level hair-bridge inverter, as shown in Figure 9.15a, where V0 and / 0 indicate the rms load vollage and current, respectively. Assuming that the load inductance is sufficiently large and the capacitors maintain their voltages so that the output current is sinusoidal as given by
i0 = /,,.sin(wl - ct,)
(9.10)
where I,,. is the peak value of the load current. and is the load impedance angle. Figure 9.15b shows a typical current waveform of each switching device with a simple stepped control of output phase voltage. The most inner switches such as S4 and Sj carry more current than the most outer switches such as S1 and S4. Each input node current can be expressed as a function of the switching function SF,, as given by for
11
= l, 2, . .. , m
(9.11)
Because the single-pole multiple-throw switch multilevel inverter, shown in Figure 9.lb, is always connected to o ne and only one input node at every instant, the output load curre nt could be drawn from one and only one input node. That is, (9.12)
Copyrghlcd matcria
9.9 DC-link Capacitor Voltage Balancing
v,
vi
+
v,
-i.
ES
+
13
ES+
_c, 1,
v, ES +
v, -f•
o,
0
o,
0
0
0
0;1Si
••
D,J
v i
s, s, s,
o.,
_c,
v, ":'
c,
s,
o,,
D3
0
0 0
i.,
o;, s; o;_, s; s;
••
~
w l (• )
>-,
~.
.,,..:;,
(b) I
(c)
,.. s,
(d) (c) (f)
-
.....
0
425
Sj
,
......
s;
-
0
0
1s;
~
(g) (h) (i)
s;
(b) Cum:nt wa,·cfomu
(a) Fivc-lC\'CI inverter circuit FIGURE 9.1 5
HaU·bridgc thrcc·levcl diode-clamped inverter. [Ref. 4]
and the rms value o f each current is expressed as m
/ ~(,nu)
(9.13)
= L i ~(mis) n
where /.,(nns) is the.rrns current of the nth node given by
l n(mi•) =
/ ' Jor2~S F.,i~ d (wt)
', i'll'
(9.14)
forn = 1, 2, . . . , m
For balanced switching with respect to Jhe ground level, we get ·2 ·2 d ·2 ' 1(rrns) c: 'S(mi~)· an l2(m,s} -
·2 '"(rms)
(9.15)
It should be noted that by structure, the currents through the opposite switches.such as
S\, ... , S'., would have the same rms current through S, , .. . , S1 respectively. 9.9
DC-LINK CAPACITOR VOLTAGE BALANCING
The voltage balancing of capacitors acting as an ene rgy ta nk is very important fo r the mult ilevel inver ter to work satisfactorily. Figure 9.16a shows the schematic of a halfbridge inverter with five levels and Figure 9. 16b illustrates the stepped-output voltage and the sinusoidal load current i0 = I .. sin (wt - 4>) ..
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426
Chapter 9
Multilevel Inverters
+
C1 I
Ve ,
v , ~ zl <"tl + V
l1<"1l V
• v:
CL(1v&)
C, t_Vc,
..L ~"""""v~,-+-~ ~~~-<>v, o
+
V2
~
I
-
v, C4
+
_ Ve.
v, ~~~~~~~~ (a) Schematic of a 3-lcvcJ
(b) Distribution of load currcnl
half-bridge inverter
FIGURE 9.16
Charge distribution of capaciton. (Rct 4]
The average value cif lhe input node current i 1 is given by l i(•vg)
= -2I
1•-•,
i.d{WI)
= 71
-'rf
1f OJ
1w-o,
!,. sin(wr - c!>)d(wr)
O:
(9.16) [,.
= - cos cj, COS <>2 1r
Similarly, the average value o f lhe input node current i 2 is given by l 2( avg)
= 1 2
1"'
'rt'
<111
iotJ(wl) = I 2
fl'
1"'
lm sin(wt .- c!>)d(WI)
a 1
(9.17)
Im
= -
1r
cos (cos a 1
-
cos a 2 )
By symmetry. / 3(a,g) = 0, /,1.,, > = - / 2(.,g)• and / s(avg) = - / 1(a,g)· Thus, each capacitor voltage should be regulated so that each capacitor supplies the average current per cycle as follows;
(9.18) (9.19)
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9.10 Features of Multilevel Inverters
427
Thcrefore, /c1(avg) < lci(avg) for a, < a2, This results in the capacitor charge unbalanc· ing, and more change flows from the inner capacitor C2 (or C1) than that of the outer capacitor Ct (or C 4). Thus, each capacitor voltage should be regulated to supply the appropriate amount of average current; otherwise, its voltage Ve2 .(or Vc1) goes to the ground level as times goes. Equations (9.18) and (9.19) can be extended to the nth capacitor of a multilevel converter as given by
,..
lcn(avg) = -:; cos cos
a.
(9.20)
Equations (9.18) and (9.19) give cos az / cz(ng) = cos a, l c,(a,g)
(9.21)
which can be generalized for the nth and (n - 1)th capacitors COS On
=
fc11(a\•g)
(9.22)
which means that the capaci1or charge unbalancing exists rcgardlc.ss of the load condi· tion and it depends on the control strategy such as ai, a 2 •. • .• a •. Applying control strategy that forces the energy transfer from the outer capacitors to the inner capaci· tors can solve this unbalancing problem [9- 11 ).
9.10
FEATURES Of MULTILEVEL INVERTERS A multilevel inverter can eliminate the need for the step-up transformer ancl reduce the harmonics produced by the inverter. Although the multilevel inverter structure was initially introduced as a means of reducing the output waveform harmonic content, it was found (1) that the de bus voltage could be increased beyond the voltage rating of an individual power device by the use of a voltage clamping network consisting of diodes. A multilevel structure with more than three levels can significantly reduce lhe harmonic content (2, 3). By using voltage-clamping techniques, the system KV rating can be extended beyond the limits of an individual device. The intriguing feature of the multilevel inverter structures is their ability to scale up the kilovolt-ampere {kVA)· rating and also to improve the harmonic performance greatly without having to resort to PWM techniques. The key features of a multilevel structure follow: • The output voltage and power increase with number of levels. Adding a voltage level involves adding a main switching device to each phase. • The harmonic content decreases as the number of levels increases and filtering requirements are reduced. • With additional voltage levels. the voltage waveform has more free-switching angles, which can be preselected for harmonic elimination. • In the absence of any PWM techniques, the switching losses can be avoided. In· creasing output voltage and power docs not require an increase in rating of individual device.
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428
Chapter 9
Multilevel Inverters
TABLE 9.3 Comparisont o r Component Requirements per Log of Three Multilevel Convenera (Rel 5) Converter 1ypc
Main switching devices Main diodes Camping diodes
De: bus capacitors Balancing c.apacitors
F1ying C..paciton
Cascaded Inverters
(m - I) x 2
(m - I ) x 2
(m - I) X 2 (m - I) X ( m - 2) (m - I) 0
(m - I) x 2 0 (m - I ) (m - I ) X (m - 2)n
(m - l) X 2 {m- l) X 2 0 (m - 1)/2 0
DiodcOamp
• Stalic and dynamic voltage sharing among lhe switching devices is built i.n to the structure through either clamping diodes or capacitors. • The switching devices do not encounter any voltage-sharing problems. For this reason, multilevel inverters can easily be applied for high-power applications such as large motor drives and utility supplies. • The fundamental output voltage of the inverter is set by the cic bus voltage Vc1c, which can be controlled through a variable de link.
9.11
COMPARISONS OF MU LTILEVEL CONVERTERS The multilevel converters (SJ can replace the existing systems thal use traditional multipulse converters without the need for transformers. For a three-phase system, the relationship between the number of levels m, and the number of pulses p, can be formulated by p = 6 x (m - 1). All three converters have the potential for applica· · lions in high-voltage, high-power systems such as an SVG without voltage unbalance problems because the SVG does not draw real power. The diode-clamped converter is most suitable for the back-to-back intertie system operating as a unified power flow controller. The other two types may also be suitable for the back-to-back intertie, but they would require more switching per cycle and more advanced control techniques to balance the voltage. Toe multilevel inverters can find potential applications in adjustable speed drives where the use of multilevel converters can not only solve harmonics and EMT problems but also avoid possible high-frequency switching dv/d1-induced motor failures. Table 9.3 compares the component requirements per phase leg among the three multilevel converters. All devices are assumed to have the same voltage rating, but not necessarily the same current rating. The cascaded inverter uses a full bridge in each level as compared with the half-bridge version for the other two types. The cascaded inverter requires the least number of components and has the potential for utility inter· face applications because of its capabilities for applying modulation and soft-switching techniques.
SUMMARY Multilevel converters can be applied to utility interface systems and motor drives. These converters offer a low output voltage THD, and a high efficiency and power
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References
429
factor. There are three types of multilevel converters: (1) diode clamped, (2) flying capacitors, and (3) cascaded. The main advantages of multilevel converters include the following: • They are suitable for high-voltage and high-current applications. • They have higher efficiency because the devices can be switched at a low frequency. • Power factor is close to unity for multilevel inverters used as rectifiers to convert ac to de. • No EMI problem exists. • No charge unbalance problem results when the converters are in either charge mode (rectification) or drive mode (inversion). The multilevel converters require balancing the voltage across the series-connected dcbus capacitors. Capacitors tend to overcharge or completely discharge, at which condition the multilevel converter reverts to a three-level converter unless an explicit control is devised to balance the capacitor charge. The voltage-balancing technique must be applied to the capacitor during the operations of the rectifier and the inverter. Thus, the real power flow into a capacitor must be the same as the real power flow out of the capacitor, and the net charge on the capacitor over one cycle remains the same.
REFERENCES (1) A. Nabae, I. Takahashi, and H . Akagi. '"A new neutral-point clamped PWM inverter," IEEE Transactions on Industry Applications, Vol. IA-17, No. 5, September/October 1981, pp. 518-523. (2) P. M. Bhagwat and V. R. Stefanovic, "Generalized structure of a mullilevel PWM inverter," IEEE Transactions on Industry Applications, Vol. 19, No. 6, November/December 1983, pp. 1057-1069. (3) M. Carpita and S. Teconi, "A novel multile vel structure for voltage source inverter," Proc. European ?ower Electronics, 1991, pp. 90-94. (4) N. S. Choi, L. G. Cho, and G. H . Cho, "A general circuit topology of multilevel inverter," IEEE Power Electronics Specialist Conference, 1991, pp. 96-103. [SJ J.-S. Lai, and F. Peng, "Multilevel converters-a new breed of power converters," IEEE Transactions on Industry Applications,Vol. 32, No. 3, May/June 1996, pp. 509-517. (6) X. Yuan and I. Barbi, "Fundamentals of a new diode clamping multilevel inverter," IEEE Transactions on Power Electronics, Vol. 15, No. 4, July 2000, pp. 711-718. (7) L. M. Tolbert, F. Z. Peng, and T. G. Habetler, " Multilevel converters for large electric drives," IEEE Transactions on Industry Applications, Vol. JS, No. 1, January/February 1999, pp.36-44. (8) C. Hochgraf, R I. Asseter, D. Divan, and T. A. Lipo, "Comparison of multilevel inverters for static-var compensation," IEEF-IAS Annual Meeting R ecord, 1994, pp. 921-928. (9) L. M. Tolbert and T. G. Habetler, "Novel multilevel inverter carrier-based P WM method," IEEE Transactions on Industry Applications, Vol. 35, No. 5, September/October 1999, pp.1098-1107. [10) L. M. Tolbert, F..Z . Peng, and T. G. Habetler, "Multilevel PWM methods at low modulation indices," IEEE Transactions on Power Electronics, Vol. 15, No. 4, July 2000, pp. 719-725.
z.
Cooyr
I
at
I
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Multilevel Inverters
Chapter 9
(11] J. H. Seo, C. H. Choi, and D. S. Hyun, "A new simplified space-vector PWM method for three-level inverters," 'IEEE TransactioflJ on Power Electronics, Vol. 16, No. 4, July 2001. pp. 545-550.
REVIEW QUESTIONS 9.1 Wh at is a multilevel converter? 9.2 What is the basic concept o r multilevel converters? 9.3 What are the features of a multilevel c-0nverter? 9.4 What are the types or multilevel converters? 9.5 What is a diode-clamped multilevel inverter? 9,6 What are the advantages or a diode-clamped multilevel inverter? 9.7 What arc the disadv:intagcs or a diode-clamped multilevel inverter? 9.8 What are the advantages of a modified diode-clamped multilevel inverter? 9.9 What is a flying-capacitors multilevel inverter? 9.10 What are the advantages of a flying-<:apacitors multilevel inverter? 9.11 What are the disadv!lJltages of a flying-capacitors multilevel inve rter? 9.12 What is a cascaded mu.ltilevel inverter? 9.13 What are th e advant!lges of a cascaded multilevel inve rter? 9.14 What are the disadvantages of a cascaded multilevel inverte r? 9.15 What is a back-to-back intertie system? 9.16 What does the capacitor voltage unbalancing mean? 9.17 What arc the possible applications of multilevel inverters?
PROBLEMS 9.1 A si ngle-phase diode-clamped inverter has m 9.2 9.3
= 5. find the generalized Fourier series and
T HO or the phase voltage. A single-phase diode-clamped inverter has 111 = 5. Find the peak voltage and current rat· ings of diodes and swi tching devices if V
=
=
9,4
9.S 9.6
9.7 9.8
9.9 9.10
=
=
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led
