Ee328lecture3 Diodes Uncontrolled Rectifiers

Diodes and Uncontrolled rectifiers EE328 Power Electronics Assoc. Prof. Dr. Dr. Mutlu BOZTEPE Ege University, Dept. of E&E

Outline of lecture 

S C I N O R T C E L E R E W O P 8 2 3 E E

 

Power semiconductor diodes Reverse recovery effect Power diode types

– General purpose diode – Fast-recovery diodes – Schottky diodes   

Series and parallel connection of diodes Performans parameters of rectifiers Single-phase half wave rectifiers

– Resistive load Resistive-inductive load – Resistive-inductive 

Single-phase full- wave rectifiers

– Resistive load Resistive-inductive load – Resistive-inductive  

Multiphase star rectifiers Three-phase bridge rectifiers

EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

2

1

Power semiconductor diodes 





A power diode diode is a two-terminal pn-junction pn-junction device Forward biasing: When the anode potential is positive with respects to the cathode, the diode conducts. It has a small voltage drop across it which depends on manufacturing process and junction temperature. Reverse biasing: When the cathode potential is positive with respect to the anode, the diode is cut-off. It has a small reverse current (leakage current) which increases slowly in magnitude with reverse voltage.


3

I-V characteristic of a power diode 

Schockley diode equation V  nV    1 I D  I S e     D

T

 





V D&I D= Diode voltage & current I S= leakage current, typically micro- or nanoampere nanoampere n= Empirical constant, diode factor, ranges from 1 to 2. V T = Thermal voltage V T 

  

kT q

k =1.3806x10 =1.3806x10-23 J/K, Boltzman’s constant, q=1.6022x10-19 C, Electron charge T = Absolute temperature in K °

Thermal voltage at 25 °C junction temperature

V T 

kT q



1.3806 x10

23

273  25

1.6022 x10

19

 25.8mV


4

2

Forward-biased region V D>0 

Until the treshold voltage the diode current ID is very small.



The diode conducts fully if the VD is higher than the treshold value.



Case of 0
Let’s assume V D=0.1, n=1, VT=25.8mV. Corresponding diode current is; .1  100.0258   I D  I S  e 1    I S 48.23  1  47.23I S   It is still small since  

I s is very small. 

Case of V treshold <
D

T

T

 I D  I S It is much larger than I s


5

Reverse-biased region 

If VD<-0.1, the exponential term in diode equation becomes negligibly small compared to unity.

 nV V  I D  I S  e  1   I S     D T



The diode current flows in the reverse direction



The reverse current is called as leakage current



In the real diodes, the leakage current increases with increasing the reverse voltage


6

3

Avalanche region 











The barrier of the diode breaks if the reverse voltage across it goes beyond a certain value. This voltage is known as breakdown voltage, V BR Reverse current increases rapidly with a small change in the reverse voltage beyond V BR The operation of this region is not destructive provided that the power dissipation is within the safe level that is specified in the manufacturer datasheet. However, it is often necessary to limit the reverse current in breakdown region. Zener diodes use this region! Keep the diode operating point away from this region in rectifiers! EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

7

Exercise 1 

The forward voltage drop of a power diode is V D=1.2V at ID=300A. Assuming that n=2 and VT=25.8mV, find the saturation current I S.

Solution: Using diode equation, V  nV   I D  I S e  1     .2  2 01.0258    30 0  I S e  1      I S  2.38371 x108 A D

T


8

4

Reverse recovery characteristics 



If the current in a fwd biased diode is reduced to zero, the diode continues to conduct due to minority carriers which remains stored in the PN-junction. It takes certain time to recombine and to be neutralized. This time is known as reverse recovery time, t rr .

t rr  t a  t b  



t a: is due to charge storage in the depletion region. t b: is due to charge storage in the bulk semiconductor material.

Peak reverse recovery current is

di

I RR  t a



dt Reverse recovery time depends on the temperature, forward current (I F ) and down-slope of the forward current (di/dt). EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

9

Reverse recovery charge, QRR 

QRR is the amount of charge carriers that flows through the diode in reverse direction during recovery process.



It can be determined by integrating the reverse current.



It is approximately equal to

1

QRR

1

Q RR  I RR (t a  t b )  I RR t rr 2 2 

Substituting the I RR into the equation yields

Q RR  

1 2

t a t rr

di dt

If t b is neglected, that is usually the case, t rr  t a  t b  t a , then

t rr 

2QRR

di dt

I RR 

2Q RR

di dt

QRR and di/dt are given in manufacturer datasheet!!


10

5

Measured reverse recovery time

Ref. “One More Way to increase the Recovery Softness of Fast High-Voltage Diodes”, By Chernikov A. A., Gubarev V. N., Stavtsev A. V., Surma A. M., and Vetrov I. Y., Proton, Electrotex ISC, www.powerguru.org EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

11

1N540x vs. UF540x





Slow recovery (Not suitable for high switching frequency applications. Use for line rectification purposes.!!) High peak surge current capability (IFSM=200A)

 



Fast recovery (50ns) Suitable for high frequency switching operations. Relatively low peak surge current capability (IFSM=150A)


12

6

Forward recovery time 









If a diode is in a reverse-biased condition, a leakage current flows due to the minority carriers. When a fwd. voltage is applied, the diode requires a certain time to be turned on, known as forward recovery (or turn-on) time. During recovery time, majority carriers are formed along the junction If the rate of rise of the forward current is high and forward current is concentrated to a small area of junction, the diode may fail. Thus the forward recovery time limits the rate of the rise of forward current and switching speed. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

13

Measured forward recovery time

Ref. “The Grey Area”, By Thomas Schneider, www.powerguru.org


14

7

Exercise 2 

The reverse recovery time of a power diode is t rr =3us and the rate of fall of the diode current di/dt =30A/dt. Determine; a)

The storage charge QRR ,

b)

The peak reverse current I RR .

Solution: a)

Q RR 

1 di 2 dt

2  t rr

b)

I RR  2Q RR

di dt

1 2

30 A /  s 3 x106   135 C 2

 2 135 x106  30 x106  90 A


15

Summary of forward and reverse recovery


16

8

Power diode types (1)  







Ideal diode have no reverse/forward recovery times. Reducing the reverse recovery time in a diode increases its manufacturing cost. In many applications, the effects of reverse recovery time is not significant, and therefore inexpensive diodes can be used. Depending on the recovery characteristics and manufacturing technics, the power diodes classified into three categories; Standard or general purpose diodes

– Relatively high reverse recovery time trr  25us – Used in low speed (up to 1kHz) where the recovery time is not critical, e.g. Line rectifiers etc. – from 1A to several kA, from 50V to around 5kV EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

17

Power diode types (2) 

Fast-recovery diodes

– – – – – 

Low recovery time t rr < 5us Used in high speed dc/dc and dc/ac converters from 1A to hundreds of A, from 50V to around 3kV Epitaxial diodes has trr<50ns Various market names; superfast diodes, ultrafast diodes etc.

Schottky diode

– Metal-semiconductor junction (instead of semiconductor-semiconductor) – Majority carrier device which results in no recovery effect due to minority carriers. But junction capacitance is large and causes reverse recovery effect which is much less than junction diodes.

– Maximum voltage is generally limited to 100V, and currents from 1 to 300A

– Ideal for high current and low voltage circuits. – Forward voltage drop is considerable lower than junction diodes. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

18

9

Reverse recovery behavior of different diode technologies 



Note that the forward voltage drop increases as the reverse recovery time decreases. SiC schottky has the smallest reverse recovery time and recovery charge QRR.


19

Effects of reverse recovery time 

  

 





SW is turned on at t=0 and remains on long enough until the current reaches steady-state of Io=Vs/R Diode Dm is reversed biased. SW is turned off at t=t1. Dm conducts suddenly and load circulates through Dm (freewheeling ) Now the SW is turned on again at t=t2. Dm behaves as a short-circuit for a certain time until reverse recovery process is finished. There is no device to limit the peak reverse current, therefore Dm can be failed due to excessive peak current In practice, switching speed of SW is reduced, i.e. the switch current rise time (di/dt) is extended, or series di/dt limiting inductor is added to the circuit.

Freewheeling diode


20

10

Diodes in series 









In many high voltage applications (e.g. HVDC transmission lines) one commercially available diodes can not meet the required voltage rating. In this case diodes are connected in series in order increase the reverse blocking capability. However, in reality even two diodes of the same part number will not have the same characteristics due to tolerances in the production process. This gives rise to problems when diodes are connected in series, since the blocking voltages will differ slightly. In the figure, each diodes has to carry same leakage current, but blocking voltage would differ significantly. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

21

Diodes in series 





This problem is solved by forcing equal voltage sharing by connecting a resistor across each diode Due to the equal voltage sharing the leakage current of each diode would be different In this arrangement, the total leakage current must be shared by a diode and a resistor. Hence

I S  I S 1  I R1  I S 2  I R 2 



We know

I R1



V D1 R1

I R 2



V D 2 R2

Therefore we obtain,

I S 1 

V D1 R1

 I S 2 

V D 2 R 2


22

11

Diodes in series I S 1  



V D1 R1



R 2

If IR1>>IS1 and IR2>>IS2 then the equation becomes R1



V D 2

We need to get V D1=V D2

V D1



 I S 2 



V D 2 R2

If R=R1=R2 the two diode voltages would be slightly different depending on the dissimilarities of the to v-i curves By using R1R2 , equal diode voltages can be obtained. But it may not be preferred due to leakage current varies the temperature considerably. For transient conditions an RC can be used. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

23

Exercise 3 

Two diodes are connected in series to share a total voltage of 5kV. The reverse leakage currents od the diodes are IS1=30mA, IS2=35mA

a) Find the diode voltages if the voltage sharing resistances are equal R1=R2=R=100k  b) Find the resistance values R1 and R2, if the diodes voltages are equal Solution: a) V D1

I S 1 

R

 I S 2 

V D 2 R

b)

V D1 R1

 I S 2 

V D 2



R 2

 I s 2  I s1   2750V

V D1  5kV  2750V  2250V

V D 2  V D  V D1 I S 1 

V D1 

V D 2 R2

V D  V D1  V D 2

Assuming R1  100k R2 

V D1 R1 V D1  R1 ( I s 2  I s1 )


 125k 24

12

Diodes in paralel 









In parallel operation of diodes, current sharing depends on the magnitude of their forward voltage drops. Uniform current sharing can be achieved either by the use of equal inductances or by connecting current sharing resistors, the later option may not be practical due to power losses incurred by the resistive components. Dynamic current sharing is achieved with the use of coupled inductors If the current through diode D1 rises, then the voltage across inductor L1 (Recall V L = L di/dt ) increases, causing a voltage of opposite polarity to be induced across inductor L2. The disadvantage of using current sharing devices under dynamic conditions is that the inductors would generate voltage spikes and would be expensive and bulky. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

25

Rectifiers 



The purpose of a rectiﬁer may be to produce an output that is purely dc, but in practice it has a dc component with a certain ripple

In practice, the half-wave rectiﬁer is used most often in low-power applications because the average current in the supply will not be zero, and nonzero average current may cause saturation problems in transformer performance. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

26

13

Performans parameters of rectifiers 

The average value of output (load) voltage given as Vdc



The average value of output (load) current given by Idc



The output dc power given by Pdc = VdcIdc



The rms value of output voltage given as Vrms



The rms value of output current given as Irms



The output ac power given by Pac = VrmsIrms



The efficiency or rectification ratio of a rectifier is given by

  P dcac P



The output voltage consists of two components, such as ac component and dc component. The effective or (rms) value of the ac component of output voltage is given by V ac  V rms  V dc 2

2


27

Performance parameters of rectifiers 

The form factor which is a measure of the shape of the output voltage is given by FF 





V rms

The ripple factor which is a measure of the ripple content is given by V RF  ac V dc 2 2 or, V rms  V dc2  V rms  V ac   1  FF 2  1 RF     V dc V dc  V dc  The transformer utilization factor is defined as TUF 



V dc

P dc

V s I s where Vs and Is are the rms voltage and rms current of the transformer secondary respectively EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

28

14


vs is the sinusoidal input voltage



is is the instantaneous input current



is1 is the fundamental component of is





The displacement angle  is the angle between fundamental components of input current and voltage. The displacement factor (DF) or Displacement Power Factor (DPF) is defined as DF  cos


29


The harmonic factor (HF) also known as total harmonic distortion (THD) is a measure of the distortion of a waveform. The harmonic factor of the input current is given as HF 



I s2  I s21 I s21

2

 I    s   1  I s1 

The crest factor is a comparison of the peak input current to its rm s value. I s , peak CF  I s Ideal rectifier should have:

  100%, V ac  0 RF  0, TUF  1 HF  THD  0 PF  DPF  1 EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

30

15

Single phase half-wave rectifier

 







It is the simplest type of rectifiers. During the positive half cycle D1 conducts and the input voltage appears across the load. During the negative half cycle Diode D1 is blocking condition, and output voltage is zero. Note that there is a small forward voltage drop across the D1. Disadvantages of half wave rectifier

– DC output voltage is discontinues and contains harmonics. – Input current is not sinusoidal – Transformer output current has dc component (saturation problem) EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

31

Single phase half-wave rectifier



Average output voltage V dc 





1

T

1

T 2

 v (t )dt  T  V sin  t dt

T 0

L

m

0

 V m   T   1  cos  T  2 

 

2

T    

V dc 

V m

dt

 V rms 



Effective output voltage V rms 

1

T

v (t )dt  T  2 L

0

1

T 2

V sin  t  T 

2

m

0


V m 2 32

16

Exercise 4 



For half-wave rectifer, determine; a) efficiency, b) form factor, c) ripple factor, d) TUF, e) PIV of D1, f) CF of input current. Solution: V dc 

V m

I dc 

V dc

 0.318V m





R

0.318V m

P dc P ac

I rms 

R

P dc  V dc I dc 

a)  

V rms 

0.318V m 2

V m 2

V rms R

 0.5V m 

0.5V m

P ac  V rms I rms 

R

0.318V m   40.5% 0.5V m 2

R

0.5V m 2 R

2



Low efficiency

b) FF  V rms  0.5V m  1.57 or 157% V dc 0.318V m EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

33

Exercise 4 c)

RF  FF 2  1  1.57 2  1  1.21 or 121%

Very high ripple factor.

d)

– Rms voltage of transformer

V s 

V m

 0.707V m 2 – Rms value of transformer secondary current I s is equal to rms value of the load current secondary is

– The Volt-ampere rating (VA)

VA  V s I s  0.707V m

of transformer

– Transformer utilization factor TUF 

P dc V s I s



0.3182 0.3535

 0.286

CF 

I s , peak I s



0.5V m

R

0.5V m R



0.3535V m2 R

Transformer should be 1/0.286=3.496 times larger than that when it delivers pure ac power to a load.

e) The peak reverse blocking voltage f) Crest factor,



V m R 0.5V m R

PIV  V m

2


34

17

Exercise 5 

The source voltage is 120Vrms at frequency of 60Hz. The load resistor is 5 . Determine;

a) Average load current b) Average power absorbed by the load c)

Power factor of the circuit

Solution:


35

Half-wave rectifier with RL load 





Due to the inductive load the conduction period of diode D1 will extend beyond 180° until the current becomes to zero. Remember that the average inductor voltage is zero. The average output voltage is V dc 



V m 2 V m 2

 



sin  t d  t  

0

V m 2

 cos  t  0 

1  cos   

Note that, the average output voltage decreases with inductive load. 

The average load current is I dc 

V dc R


36

18

Adding a freewheeling diode 







Freewheeling diode prevents negative voltage appearing across the load. The magnetic energy stored in inductor increased. The current transferred from D1 to Dm this process is called as commutation. The load time constant is  



Freewheeling diode

L R

If the load time constant is large enough the load current can be continues. Note that, the average output voltage is equal to the case of the resistive load


37

Exercise 6 

In the circuit the battery voltage is 12V and its capacity is 100Wh. The average current should be Idc=5A. The primary input voltage Vp=120V, 60Hz, and the transformer turns ratio is n=2:1. Calculate;

a) the conduction angle of diode  b) the current-limiting resistor R c)

the power rating P R

d) the charging time in hours e) the rectifier efficiency  f)

the peak inverse voltage PIV of the diode


38

19

Solution 

E= 12 V



Vp= 120 V



Vm =2 Vs= 84.85 V

a) for Vs > E the diode D1 conducts  E    V m   12   sin 1    8.13  84.85 

V m sin   E    sin 1 

  180    180  8.13  171.87

The conduction angle       171.87  8.13  163.74


39

Solution b) The current limiting resistance R is I dc



1



2 

V m

sin  t  E R





1 2 R

2V

m

 t 

d

    

cos   2 E    E 

which gives R 



1 2 I dc

2V m cos   2 E    E 

1

284.85 cos 8.13  2120.1419   12 2 5  4.26


40

20

Solution c)

The power rating of R is 2 P R  I rms R

The rms current Irms is 2 rms

I



1





V m sin  t  E 2 R 2

2 

d wt 

2  V m2  V m 2      E  2  sin 2  4 V E cos         m 2  2 R  2 2    67.4

1

I rms



67.4

 8.2 A

The power rating is P R  67.44.26  286.4W EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

41

Solution d) The power delivered to the battery is P dc

 EI dc  12 x5  60W

Charging time is

hour 

En er gy 10 0Wh   1.66 7h P dc 60W

e) Rectifier efficiency P dc 60W     17.32% P dc  P R 60W  28 6.4W f)

PIV =Vm+E=84.85+12=96.85V


42

21

Psim verification of Ex.6

Average current=4.995A RMS current=8.204 A =8.13 °

See exercise2.psimsch file on web page. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

43

Exercise 7

In the circuit, R=2, Vm=100V, and the frequency is 60Hz. Determine ; a) the average load voltage and current b) the power absorbed by the resistor in the circuit. c)

Verify solution via PSIM simulation (Homework)


44

22

Solution a) Average load voltage V dc 

V m





100 

Note that, the output voltage is not negative due to the freewheeling diode

 31.8V

b) Average load current I dc 

c)

V dc R



31.8 2

 15.9 A

We need Irms to find power absorbed by the resistor. We can calculate it by using integrals, but fourier series is another method.


45

Solution c)

The fourier series of half-wave rectified sinewave

V 0  V dc 

1

an 

 n 1, 2...

an sin n t  bn cos n t 

2



an 



V m

 V sin  t sin n t  d wt 

bn 

m

0

for n  1

bn 

V 0







V m

2



sin  t 



2

 V sin  t cos n t  d wt  m

0

V m 1  (1) n

for n  2,4,6...  1  n 2 bn  0 for n  1,3,5...

2 an  0 for n  2,3,4... V m

1



2V m

2  1) n  2 , 4 , 6.. . ( n

cos n t


46

23

Solution (cont.) c)

Lets calculate the rms current for first 5 terms of f ourier series V m

15

cos 4 t 

2V m

 31.8  50 sin  t  21.2 cos 2 t  4.24 cos 4 t  1.82 cos 6 t

3

cos 2 t 

2V m

V 0

2

sin  t 

2V m







V m

V 0

35

cos 6 t  ...

the current for each term can be found using load impedance

I n



V n Z n

V n



R

2

2



n L

Resulting rms current I rms

The resistor power



15.9

2

5.19 

2

2

1.12 

2

2

2

2

0.11 

2

0.03 

2

 16.34 A

2 P R  I rms R  16.342 x 2  534W


47

Single-phase center tap full-wave rectifier







Each half of transformer constitutes a half-wave rectifier. There is no dc current flowing through the transformer (no dc saturation prob.) The average output voltage is V dc 



2 T

T 2



V m sin  t dt 

0

2V m 

PIV of the diodes is 2Vm EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

48

24

Single-phase full-wave bridge rectifier



Full utilization of transformer



Requires four diodes



PIV of the diodes is Vm



The average output voltage is V dc 

2

T 2

V sin  t dt  T  m

0

2V m 


49

Exercise 8 



A full wave center tap rectifier has a purely resistive load of R. Determine; a) Efficiency, b) Form factor, c) Ripple factor, d)Transformer utilization factor, e) Peak inverse voltage of the diode, f) Crest factor of input current Solution: V V rms  m  0.707V m 2V m 2 V dc   0.6366V m 

I dc 

V dc R



0.6366V m R

P dc  V dc I dc 

a)  

P dc P ac



I rms 

0.6366V m 2 R

V rms R



0.707V m


R

0.707V m 2 R

0.6366V m 2  81% 0.707V m 2

b) V 0.707V m FF  rms   1.11or 111% V dc 0.6366V m EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University, 2016

50

25

Exercise 8 (cont.) c) RF  FF 2  1  1.112  1  0.482 or 48.2% d)

– Rms voltage of transformer

V m

 0.707V m 2 – Rms value of transformer secondary current I s is equal to rms value of the load current secondary is

– The Volt-ampere rating (VA) of transformer

V s 

TUF 

V s I s



 0.573 (57.3%)

0.707

Trans. input current

I s , peak I s



0.5V m R



0.707 V m2 R

PIV  2V m

e) The peak reverse blocking voltage f) Crest factor, CF 

R

Transformer should be 1/0.573=1.74 times larger than that when it delivers pure ac power to a load.

2

0.6366

0.5V m

VA  2V s I s  2  0.707V m

– Transformer utilization factor P dc



V m R 0.707V m R

 1.4142


51

Comparison of single phase rectifiers

Rectifier (R Load)

Full-Wave Transformer Rectifier (R Load)

Performance Parameter

Half-Wave

Center-Tapped

Bridge

Efficiency ()

40.5%

81%

81%

Form Factor (FF)

157%

111%

111%

Ripple Factor (RF)

121%

48.2%

48.2%

Transformer Utilization Factor (TUF)

28.6%

57.32%

81.1%

Vm

2Vm

2

1.414

Peak Inverse Voltage (PIV) Crest Factor (CF)


Vm 1.414 52

26

Exercise 9



A single-phase bridge rectifier supplies power to a highly inductive load such as a DC motor. The motor current is ripplefree. Transformer turns ratio is 1:1. Determine;

a) Harmonics factor (HF), b) Input power factor (PF) of rectifier.


53

Solution 

The input current can be expressed in a fourier series as 

 a

i1 (t )  I dc 

n

cos n t  bn sin n t 

n 1, 3,...

I dc 

an  bn 

2

1 2

1

 i (t )dt  0 1

0

2

 i (t ) co s n t d  t   0

 0

1

1

2

4 I a

 i (t ) sin n t d  t   n  1

0

i1 (t ) 

4 I a  sin  t sin 3 t sin 5 t     ...    1 3 5 


54

27

Solution i1 (t ) 



4 I a  sin  t sin 3 t sin 5 t     ...    1 3 5 

RMS value of fundamental component I s1 



4 I a  2

 0.90 I a

RMS value of input current 2

2

2

2

 1   1   1   1  I s  1              ...  I a  2  3   5   7   9  4 I a

a) Harmonic factor 2

 1  HF  THD     1  0.4843 or 48.43%  0.90 

b) Displacement angle =0 and cos=1. Then Power factor, PF 


I s1 I s

cos   0.90 lagging

55

Single-phase Full-wave rectifier with RL load 







In practice, most loads are inductive A battery is added to develop generalized equations. Input ac voltage source v s  V m sin  t  2V s The load current iL can be found from

L 

di L

 Ri L  E  2V s sin  t

dt

E

The differential equation has a solution of the form

i L 

2V s Z

R

sin  t     A1e

Z  R 2  ( L) 2

  tan 1

 t L



E R

 L R


56

28

Case 1: continues load current 

The constant A1 can be determined from condition: at t=, i L=I1

 E  RL   2V s A1   I 1   sin   e   R Z   

Substituting of A1 to diff. equation yields,

 E  R L     t   E 2V s i L   sin  t      I 1   sin  e   Z R Z R   Under steady-state condition i L(t=0) = iL(t=) = I 1. Applying this 2V s



condition,

I 1 

2V s Z

1 e

sin 

1 e

R  L 



R   L 



E

for I 1  0

R


57

Case 1: continues load current (cont.) 

Substituting I1 and simplification gives



2V s sin  t     i L  Z 

R 



sin  e

1  e L  RMS diode current can be found





R

2

I r

 t L

 E   R 

for 0   t   and iL  0



1

i d  t  2 



2

L

0



And the RMS output current can then be determined by combining the RMS current of each diode I rms





2

I r

 I r 2  2 I r

Average diode current  1 I d  i L d  t  2 0




58

29

Case 2: discontinues load current 

The load current flows only during the period   t  . The diodes start to conduct at t=  given by   sin



1

E V m

The constant A1 can be determined from condition at t= , iL(t)= 0

 E  RL   2V s   A1   R  Z sin     e   

Substituting it to diff. equation

 E  R L     t   2V s i L  sin t       sin   e   Z Z  R  2V s


59

Case 2: discontinues load current (cont.) 

At t= , the current falls zero, and iL(t=)= 0. That is,

 E  RL         2V s 0 sin         sin   e   Z Z  R  2V s



 an be found from this transcendental equation by using an iterative (trial and error) method.



After  found, the RMS diode current can be calculated as I r



1



i d  t  2  2

L





The average diode current I d



1

I dc must be calculated by using differential equation as like on the left.



i d  t  2  L



DO NOT USE

I dc




V dc

 E

R

!!!!!! 60

30

Multiphase star rectifiers 

Used in high power P>15kW



Less filter requirement



q single phase rectifier









Conduction angle of each diode is 2/q Secondary winding current is unidirectional and contains dc component. Therefore primary must be connected in delta to eliminate dc component in the input side of T.F. This method minimizes the harmonics content.


61

Multiphase star rectifiers (cont.) 

Output dc voltage V dc 

m

q 

0

sin

 q

Rms output voltage

V rms  

 V cos t d  t 

2 q

 V m 

 q

2

 q

2 2 q

 V

2 m

cos2  t d  t   V m

0

q  

1 2    sin  2  q 2 q 

If the load is purely resistive, the rms current of a diode

I s 

 q

2 2



I m2 cos 2  t d  t   I m

0

I m



V m R

q  

1 2  V   sin   rms 2  q 2 q  R


62

31

Exercise 10 



A three phase star rectifier has a purely resistive load with R Ohms. Determine; a) Efficiency, b) Form factor, c) Ripple factor, d)Transformer utilization factor, e) Peak inverse voltage of the diode, f) the peak current through a diode if rectifier delivers Idc=30A at an output voltage of Vdc=140V. Solution: q  V dc  V m sin  0.827V m 

V dc

I dc 

q



0.827V m

I rms 

R 0.827V m 2 P dc  V dc I dc  R

a) b)

 

R

V rms  V m

1 2    sin   0.841V m 2  q 2 q 

V rms R



0.841V m


R 0.841V m 2 R

0.827V m 2   99.77% 0.841V m 2

P dc P ac

FF 

q  

V rms V dc



0.841V m 0.827V m

 1.0165or 101.65%


63

Exercise 10 (cont.) c) RF  FF 2  1  1.01652  1  0.1824 or 18.24% d) Rms voltage of transformer secondary is V V s  m  0.707V m 2 Rms value of transformer secondary current is equal to rms value of the diode current I s  I m

q  

1 2  0.4854V m   sin   0.4854 I m  2  q 2 q  R

The Volt-ampere rating (VA) of transformer VA  3V s I s  30.707V m 

0.4854V m R

Transformer utilization factor TUF 

P dc 3V s I s



0.827 2

30.7070.4854

 0.6643 (66.43%)


64

32

Exercise 10 (cont.) e) The peak reverse blocking voltage

PIV  3V m f)The average current through each diode is

2 I d  2



6

 I

m

cos t d  t   I m

0

1 

sin



q

For q=3, I d = 0.2757 I m. The average current through each diode is 30  10 A 3 and this gives the peak current as

I d 

I m 

10  36 .27 A 0.2757


65

Three phase bridge rectifiers 









Commonly used in high power It is a full wave rectifier, and can operate with or without a transformer Diodes are numbered in order of conduction sequences and each one conducts for 120 °

Diode conductance sequence is 12,23,34,45,56,61,… Line-to-line voltage is 3 times of the line-neutral phase voltages.


66

33

Three phase bridge rectifiers




 

12 2 3 3 



 6

0

3V m cos  t d  t 

V m  1.654V m

Rms output voltage 12



 6

2 

V rms 

0

3 2



9 3 4

3V m2 cos 2  t d  t  V m  1.6554V m


67

Three phase bridge rectifiers






12

2 

 6

0

3 3 

3V m cos  t d  t 

V m  1.654V m

Rms output voltage 12

2 

V rms 



 6

3V m cos  t d  t  2

0

3 2



9 3 4

2

V m  1.6554V m


68

34

Three phase bridge rectifiers 

If load is purely resistive

I m



3 

V m

Rms value of diode current I r 



R

4

2 

1  

1 2    sin   0.5518I m 6    6 2

 6

I m2 cos 2  t d  t   I m

0

Rms value of transformer secondary current 8

I s 

 6

2



0

I m2 cos 2  t d  t   I m

2  

1 2    sin   0.7804I m 6    6 2


69

Exercise 11 



a) b)

A three phase bridge rectifier has a purely resistive load of R. Determine; a) Efficiency, b) Form factor, c) Ripple factor, d)Transformer utilization factor, e) Peak inverse voltage of the diode, f) the peak current through a diode if rectifier delivers Idc=60A at an output voltage of Vdc=280.7V. The source frequency is 60Hz. Solution: V dc 

3 3

I dc 

V dc





1.654V m

R 1.654V m 2 P dc  V dc I dc  R  

R

V m  1.654V m

P dc



P ac V rms

FF 

V dc

V rms 

3

I rms 

V rms

2 R

9 3



4



V m  1.6554V m

1.6554V m


R 1.6554V m 2 R

1.654V m 2  99.83% 1.6554V m 2 

1.6554V m 1.654V m

 1.0008 or 100.08%


70

35

Exercise 10 (cont.) c)

RF  FF 2  1  1.00082  1  0.04 or 4%

d) Rms voltage of transformer secondary is V V s  m  0.707V m 2 Rms value of transformer secondary current is I s  I m

2  

1 2  V   sin   0.7804 I m  0.7804 3 m   6 2 6  R

The Volt-ampere rating (VA) of transformer VA  3V s I s  30.707V m 0.7804 3

V m R

Transformer utilization factor TUF 

P dc



3V s I s

1.654 2 3 3 0.707 0.7804

 0.9542 (95.42%)


71

Exercise 10 (cont.) e) The peak reverse blocking voltage V 280.7  169.7V V m  dc  PIV  3V m  3 169.7   293.9V 1.654 1.654 f)The average current through each diode is

4 I d  2



6

 I

m

co s t d   t   I m

0

2 

sin



6

 0.3183I m

The average current through each diode is 60  20 A 3 and this gives the peak current as

I d 

I m 

20  62 .83 A 0.3183

Note that this rectifier has considerably improved performances compared to that of multiphase rectifier


72

36

Comparison of rectifier performances Performance Parameter

Single Phase FullSingle Phase Half- Wave CenterWave Rectifier Tapped Transformer (R Load) Rectifier (R Load)

Single Phase Full-Wave Bridge Rectifier (R Load)

Efficiency (h)

40.5%

81%

81%

99.83%

Form Factor (FF)

157%

111%

111%

100.08%

Ripple Factor (RF) 121%

48.2%

48.2%

4%

Transformer Utilization Factor (TUF)

28.6%

57.32%

81.1%

95.42%

Peak Inverse Voltage (PIV)

Vm

2Vm

Crest Factor (CF)

2

1.414

Three-Phase Bridge Rectifier (R Load)

Vm

3Vm

1.414

1.047


73

12-pulse rectifier 

In order to obtain equal secondary voltages, the number of turns of the two secondary windings must be in a ratio of 1:√3 (Turns ratio=4/7 or 7/12)

VRS(t)

VRDSD(t)


74

37

Ee328lecture3 Diodes Uncontrolled Rectifiers

Recommend Documents