tiristores_caracteristicas

Thyristors and controlled rectifiers EE328 Power Electronics Assoc. Prof. Dr. Dr. Mutlu BOZTEPE 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

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Thyristors

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Unijunction transistors

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Principal of phase control

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Single Phase controlled rectifier – Single Phase Half-wave Rectifiers – Single Phase Semiconverters – Single Phase Full converters – Single Phase Dual converters

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Three Phase controlled rectifier – Three Phase Half-wave Rectifiers – Three Phase Semiconverters – Three Phase Full converters – Three Phase Dual converters

EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

2

Thyristor 

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Most important type of power semiconductor device. They can operate as a bistable switch that can be turned on by a gate signal, but can not be turned off by the gate.

Proposed in 1950s at Bell Lab. The first thyristor was developed in 1957 by General Electric (GE) Have highest power handling capability up to ~4500 Volt, up to ~5000 Amperes Amperes Maximum switching frequency for typical thyristors is around 1kHz. But it can rise up to 20kHz for some special types.

It is inherently slow switching device compared compare d to BJTs BJTs or MOSFETs. MOSFETs. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

3

Structure 

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It is a four-layer semiconductor device of pnpn structure with three pn-junctions; e.g. J1, J2 and J3. Terminals; Anode (A), Cathode (K) and Gate (G) Forward blocking mode: When Anode voltage is made positive with respects to Cathode, the junctions J1 and J3 are forward biased, and J2 is reverse biased, and therefore a small leakage current can flow through the device. Reverse blocking mode: When Cathode has positive voltage than the Anode, J1 and J3 are reverse biased, and J2 is forward biased, and therefore a small leakage current can flow through the device. For both blocking modes the gate terminal is assumed as open circuit. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

4

Avalanche breakdown 

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Forward leakage current is very small. If the V AK is increased to a sufficiently Latching current large value, the leakage current increases and the junction J2 breaks down.

IT

Forward breakdown voltage

IL

This voltage is called as forward breakdown voltage (VBO). Since the other junctions J1 and J3 are already fwd biased, then the device will be in conducting state or on-state.

Forward leakage current

VBO

V AK

The anode current must be more than a value known as latching current to keep the thyristor is on-state. Otherwise the device will revert to the blocking condition as the V AK is reduced. Latching current is the minimum anode current to maintain the thyristor in the on-state immediately after it is turned on (no gate signal) EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

5

Holding current 

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Once the thyristor conducts, it behaves as a conducting diode and there is no control over the device. Therefore the device current must be limited by external components, such as resistors.

IT Forward voltage drop

Latching current

IL

Forward breakdown voltage

IH Holding current

The forward voltage drop is due to the ohmic drop in the four layers and it is small, typically 1V.

VBO

V AK

If the fwd anode current is reduced below a level known as holding current IH, a depletion region is developed around junction J2 due to the reduced number of carriers, and the thyristor will be in blocking state. The holding current is very small in the order of milliamperes, and it is less than the latching current.


6

Gate triggering 

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A thyristor can be turned on by increasing forward voltage beyond breakdown voltage V BO, but such a turn-on could be destructive. In practice, a positive gate-cathode voltage V GK, is applied to the thyristor. As the gate current is increased, the forward blocking voltage is decreased as shown in the figure Once the thyristor is turned on by gating signal and its forward current is greater than the holding current, the device continues to conduct. Thyristor is a latching device.


7

Two-transistor model of thyristor 

The latching action due to positive feedback can be demonstrated by using two-transistor model of thyristor.

I A  I C 1  I C 2   1 I A  I CBO1    2 I K  I CBO2  

ICBO1 and I CBO2 are leakage currents.

For a gating current I K  I A  I G and solving for I A

I A 

 2 I G  I CBO1  I CBO 2 1  ( 1   2 )


8

Two-transistor model of thyristor I A 

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 2 I G  I CBO1  I CBO 2

1  ( 1   2 ) The current 1 varies with the emitter current I A, and 2 varies with IK=I A+IG. When gate current I G is increased, I A increases too. An increase in IG also increases 1 and 2. The increase in 1 and 2 causes further increase Typical variation of current gain with emitter current in in I A, and go on… thyristor If (1 + 2) tends to be unity denominator approaches zero, resulting in large current of I A. Therefore can be turned on by a small gate current.


9

Thyristor turn-on A thyristor can be turned on by the following ways; Thermals: If the temperature is high, there will be an increase in the number of electron-hole pairs. This increase would cause ( 1 + 2) to increase which may turn on the thyristor. This type of turn-on may cause thermal runaway and is normally avoided. 

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Light: If light is allowed to strike the junctions, the electron-hole pairs will increase; and the thyristor may be turned on. High voltage: If VAK is greater than the VBO, thyristor turns on. dv/dt: If the rate of rise of anode-cathode voltage is high, the charging current of the junction capacitance may trigger the thyristor and make it turned-on. The device must be protected against dv/dt. Gate current: If a thyristor forward biased, a positive gate-cathode voltage would turn on the thyristor. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

10

Thyristor types 

Phase-control thyristors (SCRs)

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Fast-switching thyristors (SCRs)

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Gate-turn-of thyristors (GTOs)

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Bidirectional triode thyristors (TRIACs)

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Reverse-conducting thyristors (RCTs)

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Static Induction thyristors (SITHs)

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Light-activated silicon-controlled thyristor (LASCRs)

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FET-controlled thyristor (FET-CTHs)

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MOS-controlled thyristor (MCTs)

Triac

MCT


SIT

GTO

LASCR

11

Gate drive 

The following points should be considered in designing the gate control circuit; – The gate signal should be removed after the thyristor is turned on. A continues gate signal increases the power loss in the gate junction. – While thyristor is reversed biased, there should be not gate signal; otherwise, the thyristor may fail due to increased leakage current. – The width of gate pulse t G must be longer than the time required for the anode current reach to the holding current.

Opto isolated gate drive EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

Pulse transformer isolated gate drive 12

Unijunction transistor 

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Since the device has one pn-junction and three leads, it is commonly called a unijunction transistor. With only one pn-junction, the device is really a form of diode. The emitter is heavily doped having many holes. The n region, however, is lightly doped. For this reason, the resistance between the base terminals is very high ( 5 to 10 kΩ) when emitter lead is open.


13

Unijunction transistor 

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If voltage VBB is applied between B2 and B1 with emitter open, a voltage gradient is established along the n-type bar. Since the emitter is located nearer to B2, more than half of VBB appears between the emitter and B1. The voltage V1 between emitter and B1 establishes a reverse bias on the pn junction and the emitter current is cut off. Only leakage current flows. If a positive voltage is applied at the emitter the pn junction will remain reverse biased so long as the input voltage is less than V1 If the input voltage to the emitter exceeds V1, the pn junction becomes forward biased, and then the device is now in the ON state. If a negative pulse is applied to the emitter, the pn junction is reverse biased and the emitter current is cut off. The device is then in the OFF state.


14

Equivalent circuit of UJT 

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The resistance of the silicon bar is called the interbase resistance RBB The inter-base resistance is represented by two resistors in series; RB1 and RB2.

RB2 is constant. RB1 is variable and depends on the bias voltage across the pn junction

The pn junction is represented in the emitter by a diode D. The ratio V1/VBB is called intrinsic stand-off ratio () which is between 0.51 to 0.82.


15

Characteristics of UJT 

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Initially, in the cut-off region, as VE increases from zero, slight leakage current flows from terminal B2 to the emitter. This current is due to the minority carriers in the reverse biased diode. Above a certain value of VE, forward IE begins to flow, increasing until the peak voltage VP and current I P are reached at point P. After the peak point P, an attempt to increase V E is followed by a sudden increase in emitter current I E with a corresponding decrease in VE. This is a negative resistance portion. The negative portion of the curve lasts until the valley point Vv is reached. After the valley point, the device is driven to saturation.


16

Design UJT circuit 

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The load line formed by R and Vs should intersect the device characteristics to the right of peak point but to the left of valley point, otherwise UJT will not turn on.

I p

V s  V v I v

Therefore min&max values for R V s  V v I v

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V s  V p

 R 

V s  V p I p

where Vp can be calculated as

V p  V 1  V D   V BB  V D 

assume VS  VBB

V p   V S  V D

V D: One diode fwd voltage drop,  0.5V


17

Design UJT circuit 

The pulse width of triggering pulse t g is

t g  R B1C 

Oscillation frequency is T 

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1 f

 RC ln

1 1  

A resistor RB2 is generally connected in series with B2 to compensate for the decrease in Vp due to temperature rise and to protect the UJT from thermal runaway.

R B 2 

104  V S


18

Exercise 1 

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Design the UJT triggering circuit. The parameters of UJT are V s=30V, =0.51, Ip=10 A, Vv=3.5V, and Iv=10mA. The frequency of oscillation is f=60Hz, and width of pulse is t g=50s.

Solution: V p  0.51 30  0.5  15.8V Let assume C=0.5 F 1 1 1   R  0.5 F ln T  RC ln 1   60 1  0.51

V s  V p V s  V v  R  I v I p

R  46.7k 

30  3.5

t g

10mA

R B1 

R B 2 

C



50 s 0.5 F

104 0.51 30

 100

 654

 R 

30  15.8 10  A

2.65k   R  1.42 M  Min&max limits for R


19

Principle of phase control 

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Diode rectifiers produce a fixed output voltage. In order to control the output voltage of a rectifier, phase control thyristors must be used instead of diodes. The output voltages of these rectifiers are varied by varying the delay or firing angle of the thyristor. Phase controlled thyristors are turned on by the application of a short pulse to the gate and they are turned off by the process of natural or line commutation.

V dc 


V m 2

1  cos 

20

Principle of phase control 

There are two types of phase control converters: – Single phase converters. – Three-phase converters

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Each type of converter can be divided into three categories: – S emic onverter : one quadrant converter, having one polarity of voltage and current

– F ull converter : two quadrant device whose output voltage polarity can be either positive or negative but whose output current has one polarity.

– Dual conver ter : four quadrant device whose output voltage and current can be of either positive of negative polarity.


21

Single Phase half-wave SCR rectifier with R Load

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For the positive half cycle of Vs, T1 is forward biased and when the thyristor is fired at wt=  it conducts and input voltage appears across the load. When the input voltage goes negative at wt=  thyristor is reverse biased and it is turned off. The firing angle (or delay angle) is defined as the time after the input voltage starts to go positive until the thyristor is fired at wt= . For this converter: 0 180° EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

22


d 3 3 e e / r u o f e d r / f f a t s / c e l e / s t p e d / t t . i w u . g n e . w w w / / : p t t h EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

23


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Average output voltage V 1 V dc  V m sin  t d  t  m  cos t  2 2 









 

V m 2

1  cos 

Rms output voltage V rms 

1



V  2

2 m

sin

2

 t d  t





V m 2

1 

      



V m2



1  cos 2 t  d  t  4 

sin 2  2

 


24

Exercise 2 

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For single phase half wave thyristor converter with R load, delay angle is =90°, determine, a) rectification efficiency, b) FF, c) RF, d) TUF, e) PIV Solution: V dc 

V m 1  cos 90  0.1592V m 2

V m 2

1 



    90 180  

sin2  90  2

 

 0.3536V m

V 0.1592V m I dc  dc  R R P dc  V dc I dc 

V rms 

I rms 

0.1592V m 2 R

V rms R



0.3536V m R

P ac  V rms I rms 

a) Rectification efficiency

0.3536V m 2 R

0.1592V m  P dc   20.27%   2 P ac 0.3536V m  2


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Exercise 2 (cont.) b) Form factor FF 

0.3536V m V ac   2.221or 222.1% 0.1592V m V dc

c) Ripple factor RF  FF 2  1 

2.2212  1  1.983 or 198.3%

d) VA power at transformer secondary is I s  I rms 

0.3536V m

VA  V s I s 

V s 

R

V m 2

e) PIV=Vm

 0.707V m

 0.3536V m   0.70 7V m    R 2  

V m

0.1592V 2 R   0.1014or (10.14%) TUF  0.707V 0.3536V  R VA P dc

m

m

m


26

Single phase semiconverters

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Load current is assumed continues and ripple free. During the positive half cycle, T1 & D2 are fwd biased, and when the T1 is fired at wt= , load is connected to input supply. During   wt  +, the input voltage is negative and freewheeling diode Dm conducts. At wt= +, T2 is fired and Dm is reverse biased. Then the load is connected to the supply through T2 and D1. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

27


d 3 3 e e / r u o f e d r / f f a t s / c e l e / s t p e d / t t . i w u . g n e . w w w / / : p t t h EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

28


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Average output voltage 2V m 2  cos t  V dc  V m sin  t d  t  2 2 











V m

1  cos 

 


2



V  2

2 m

sin

2

 t d  t





V m

1 

     2  



V m2



1  cos 2 t  d  t  2 

sin 2  2

 


29

Exercise 3 

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Single phase semi converter is connected to a 120V 60Hz supply. Load current Ia can be assumed continues and its ripple content is negligible. Turns ratio is unity. Delay angle is =/2. a) express the input line current in a Fourier series; b) determine Vdc, Vrms, Harmonic factor, displacement factor, input power factor

Solution a) I dc 



1

 i (t ) d  t  0

2 

s

1 

 an    I a cos n t d  t   I a cos n t d  t         2





2 I a

n

   I a sin n t d  t   I a sin n t d  t         2



bn 

sin n for n  1,3,5...

0

1 



2 I a

n 0

for n  2,4,6...

1  cos n 

for n  1,3,5...

for n  2,4,6...



 a

i s (t )  I dc 

n

cos n t  bn sin n t 

n 1, 2...



i s (t ) 

 n 1,3,5...

2 I sn sin n t   n 

 n  tan 

1

an bn



n 2

I sn  an  bn 


2

2

2 2 I a n

cos

n 2 30

Exercise 3 (cont.) b)

V dc 

V m

V rms 

V m

I s1 

1  cos   54.02V



1 

      

2

sin 2 

  84.57V 2 

2 2 I a  cos  0.6366I a  4

2

I rms 



2 

 1  

PF 



I a2 d  t  I a 1 

2

 

 0.7071I a

 I  HF   s   1  0.4835 or 48.35%  I s1 

    DF  cos    0.7071 4  4 



I s1 I s

cos

 2

 0.6366(lagging )


31

Single phase semiconverter with RL load 

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The load current of a semiconverter is dependent on the load resistance inductance and the battery voltage E in series with the load. The converter operation can be divided into two modes: Mode 1 (0 wt  ): Freewheeling diode is forward biased and load current is provided by the energy stored in the magnetic field.

Mode 2 (  wt 

): Load current is provided by the supply.

The current avg and rms values can be calculated by solving first the differential equations above. (See the book for solution.) EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

32

Single phase full converter

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Load current is assumed continues and ripple free. During the positive half cycle, T1 & T2 are fwd biased, and when the SCRs are fired at wt= , load is connected to input supply. T1&T2 will continue to conduct beyond wt= due to inductive load. At wt=+, T3&T4 are fired that makes T1&T2 are reverse biased. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

33






For the period of   wt   the input voltage and current are positive and power flows from supply to load and the converter is said to be in rectification mode. For the period   wt  +, the input voltage is negative, input current is positive and power flows from the load to the supply. The converter is said to be operated in the inversion mode. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

34




Average output voltage 2 V dc  2



2V m 

 



V m sin  t d  t 



2V m  cos t    2

cos Note that: For purely resistive loads the output voltage will be similar to semiconverter


35

Single phase full converters



Rms output voltage

V rms 

2 2

 





V m2 sin 2  t d  t 

V m2 2

 

 1  cos 2 t  d  t 




V m 2

 V s

36

Exercise 4 



Single phase full converter is connected to a 120V 60Hz supply. Load current Ia can be assumed continues and its ripple content is negligible. Turns ratio is unity. Delay angle is =/3. a) express the input line current in a Fourier series; b) determine Vdc, Vrms, Harmonic factor, displacement factor, input power factor

Solution a) I dc 

 

1 2

 i (t ) d  t  0 s



 

2 

1 

 an    I a cos n t d  t   I a cos n t d  t         

4 I a

n

sin n for n  1,3,5...

0

for n  2,4,6...

 

bn 



2 

1 

   I a sin n t d  t   I a sin n t d  t        

4 I a

n 0

cos n for n  1,3,5...

for n  2,4,6...



 a

i s (t )  I dc 

n

cos n t  bn sin n t 

n 1, 2...



i s (t ) 

 n 1,3,5...

2 I sn sin n t   n 

 n  tan 1

an bn

 n

I sn  an2  bn2 


2 2 I a n 37

Exercise 4 (cont.) b)

V dc 

2V m

V rms 

V m

I s1 

cos  54.02V



2

 V s  120V

2 2 I a  0.90032I a 

2

 I  HF   s   1  0.4834 or 48.34%  I s1 

I rms  I a

     0.5  3 

 1    DF  cos 

PF 

I s1 I s

cos    0.45(lagging )


38

Single phase fullconverter with RL load 



The load current of a fullconverter is dependent on the load resistance inductance and the battery voltage E in series with the load. The converter operation can be divided into two identical modes.



Mode 1 (



Mode 2 ( +



For both mode the load current is provided by supply, then



 wt 

+ ): T1 & T2 conducts

 wt  2 +

): T3 & T4 conducts

The current avg and rms values can be calculated by solving first the differential equations above. (See the book for solution.)


39

Single phase fullconverter 





Harmonic content of output voltage vs. delay angle for single phase full converter. Load is continues The rms load current can be calculated approximately using voltage harmonics instead of analytical solution.


40

Single phase dual converter









When two full converter are connected back to back, and the system will provide four-quadrant operation and is called as dual converter . If 1 and 2 delay angles of converter 1 and 2 respectively, the corresponding average output voltages are Vdc1 and Vdc2. The delay angles are controlled such that one converter operates as a rectifier and the other converter operates as an inverter, but both converter produce the same average voltage. Since the instantaneous voltage difference will result in circulating current between converters, two circulating current reactor Lr/2 are used . EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

41

Single phase dual converter

V dc1 

2V m 

cos 1

V dc 2 

2V m 

cos 2

One converter is rectifying, the other is inverting V dc1

 V dc 2

therefore

cos 2   cos 1  cos   1   2

    1 EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

42

Circulating current Assume circulating current İr is discontinues (İr(0)=0 )  t  t 1 1 ir  vr d  t  vo1  vo 2 d  t  Lr 2   Lr 2  1 1



V  m  Lr



   2

 t



  1



  sin  t d  t   sin  t d  t   2   t

  1

2V m cos t  cos 1   Lr

The circulating current depends on 1 The maximum value can be

ir ,max 

4V m  Lr

If the peak load current is Ip, then the converters should carry a peak current of

I p  ir , max  I p 

4V m  Lr


43

Circulating current 



The dual converter can be operated with or without a circulating current. If only one converter operates at a time and carries load current, t he other converter is completely blocked by inhibiting gate pulses. However, operating with a circulating current has the following advantages; – Both converters are in continues conduction, independent of the load – Power flow in either direction at any time is possible. – Time response for changing from one quadrant to another is faster

Exercise 5: Dual converter supplied from 120V, 60Hz source, resistive load R=10 . Lr=40mH, 1=60°, 2=120°.

Peak circulating current ir

Peak load current i p 



2V m  Lr

1  cos 1  



2

 1  cos 60  11.25 A

2 120

3770.04

2 120  16.97 A 10

Peak current of converter1 is 16.97+11.25=28.22A EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

44

Three phase half-wave converter









Three phase converters provide higher average output voltage Output voltage ripple frequency is higher than single phase converters. Therefore filtering requirement is simpler. For this reason 3-phase converters are used extensively in highpower applications.

Delay angle  is defined starting from natural commutation points where phase voltages are equal. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

45

Three phase half-wave converter 











Load current is assumed constant. When T1 is fired at wt= /6+, the load voltage is Vo=Van, until T2 is fired at wt=5/6+. When thyristor T2 is fired, thyristor T1 is reverse biased, because line-to-line voltage Vab is negative, and T1 is turned off. For resistive loads and when /6, the load current will be discontinues. The frequency of output ripple is 3f s This converter is not used frequently in practice due to high ripple content. EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

46


For continues load current:



Output dc voltage 3 V dc  2

5  6 

 V

m



sin  t d  t

6 

3 3V m  cos 2 


 3V m

3 2 1 6

5 6 



V m2 sin 2  t d  t

 6 



3 8

co s 2


47

3-phase half wave converter with RL load 

Van

V bn

Vcn

 V0 0

 =30

=300

0

0

30

0

60

0

90

0

120



0

150

0

180

0

210

Van

0

240

0

270

0

300

0

330

0

360

V bn

0

390

t

0

420

Vcn

 V0

=60

0

=600 0

0

30

0

60

0

90

0

120

0

150

0

180

0

210

0

240

0

270

0

300

0

330

0

360


0

390

0

t

420

48

3-phase half wave converter with RL load

=900


49








For resistive loads: For </6, the Vdc and Vrms are same with constant load current case.

Vbn

Vcn

Van

200

100

0

For /6: Output dc voltage

-100



3 V dc  V m sin  t d  t 2  6 



 

3V m         1 cos   2  6   

Rms output voltage

V rms 

3

-200

Vload 200

150

100





V m2 sin 2  t d  t

50

2  6

0

 3V m

     sin  2  24 4 8  3  5



1

0.3


0.31 Time (s)

0.32

0.33

50

3-phase half wave converter with R load V bn

Van

Vcn

=0 Vs

=00

0

0

30

0

60

0

90

0

120



0

150

0

180

0

210

0

240

0

270

0

300

0

330

0

360

V bn

Van

0

390

0

420

t

Vcn

=150 =150

V0

0

0

30

0

60

0

90

0

120

0

150

0

180

0

210

0

240

0

270

0

300


0

330

0

360

0

390

0

420

t

51

3-phase half wave converter with R load 

V bn

Van

Vcn

=300 =300

V0

0 0

30

0

60

0

90

120

0

150

0

180

0

0

210

0

240

300

0

330

0

0

360

V bn

Van



0

270

0

390

0

420

t

Vcn

=600 =600

V0

0

0

30

0

60

0

90

120

0

150

0

180

0

0

210

0

240

0

270

300

0

330


0

0

360

0

390

0

420

t

52

Exercise 6 



3-phase half wave converter is operated from Y-connected 208V 60Hz supply. Load resistance R=10 . Half of the maximum possible average voltage is requested at the output. Calculate a)delay angle, b) rms and average output current, c)average and rms thyristor currents, d) rectification efficiency, e)Transformer utilization factor, f) input power factor Solution:

The maximum output voltage is V dc , max 

3 3V m 3 3 169.83   140.45V 2 2

Desired output voltage, then 0.5*140.45=70.23V a) For resistive load and </6,

3 3 169.83  cos    60  then it is not valid ! 2 6 Therefore, the equation for >/6 should be used 70.23 

70.23 

3 3 169.83     1  cos        67.7  2  6   EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

53

Exercise 6 b) rms and average output current V 70.23 I dc  dc   7.02 A 10

R

V rms  I rms 

5

3 16 9.83 V rms R



24

94.74 10



67.7



     180  1 cos   2 67.7    94.74 V 4 8 18 0     3

 9.47 A d) rectification efficiency

c) average and rms thyristor currents I A 

I dc

I R 

3



I rms 3

7.02 3



 2.34 A

9.47 3

 

P dc P ac



70.23  7.02 94.74  9.47

 54.95%

 5.47 A


54

Exercise 6 e)Transformer utilization factor The rms line current is the same as the thyristor rms current

I s  I R  5.47 A

V s  120.1V

VA  3V s I s  3  120.1 5.47  1970.84W

TUF 

P dc 70.23  7.02   0.25 or ( 25%) VA 1970.84

f)

2 2   9 . 47 10  896.81W P I R Output power is o rms

Then input power factor is

PF 

896.81 1970.84

 0.455 (lagging )

Note that due to the delay angle, , the fundamental component of input line current is also delayed with respect to the input phase voltage


55

Three phase semiconverter





Three phase semiconverters are used in industrial applications up to 120kW level, where one quadrant operation is required. The power factor of this converter decreases as the delay angle increases, but it is better than 3-phase half wave converters.    v an  V m sin  t l vac  van  vcn  3V m sin  t   a 6   r t 2    u s e s  5   e e vbn  V m sin   t  n e i 3  3 V sin  t  n g l g vab  van  vbn     m 6 a a   e t e t l l 2    n n i o v i o  V m sin   t      L v cn L v 3  vcb  vcn  vbn  3V m sin  t    2   EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

56

3-phase Semiconverter for  > 600









Frequency of output voltage is 3fs The delay angle can be varied from 0 to  At wt=7/6, Dm conducts due to Vac starts to be negative If there were no Dm, the freewheeling action would be accomplished by T1 and D2.

=90°


57

3-phase Semiconverter for   600 

Output dc voltage 7  6

3 V dc  3V m sin  t d  t 2  6 







3 3V m 1  cos  2

Output rms voltage

V rms 

 3V m

3

7  6



3V m2 sin 2  t d  t

2  6 

3  1       sin 2  4  2 

=90° EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

58

3-phase Semiconverter for   600





Each thyristor conducts for 120° and freewheeling diode Dm doesnt conduct. Average output voltage

3   vab d  t  V dc  2   6  

2





3 3V m 1  cos  2

 2vac d  t  

5 6 



=30°


59




Rms output voltage

V rms 

 3V m

3  



2



2   6

 v d  t   v d  t   2  2 ab

5 6  2 ac 

3  2   3 cos2    4  3  =30° EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

60


Current waveforms for thyristors and diodes


=30°

61

Exercise 7 



3-phase semiconverter is operated from Y-connected 208V 60Hz supply. Load resistance R=10 . Half of the maximum possible average voltage is requested at the output. Calculate a)delay angle, b) rms and average output current, c)average and rms thyristor currents, d) rectification efficiency, e)Transformer utilization factor, f) input power factor Solution:


3 3V m



3 3 169.83



 280.9V



Desired output voltage, then 0.5*280.9=140.45V a) The output voltage for semiconverter is

V dc 

3 3V m 1  cos  2

3 3 169.83 140.45  1  cos     90 2 EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

62

Exercise 7 b) rms and average output current I dc 

V dc



R

14 0.45 10

   3  1     sin 2   18 0.13V 4  2 2 2 

V rms  3 16 9.83

I rms 

V rms



R

 14.05 A

18 0.13 10

 18.01 A


I R 

I dc 3



I rms 3

14.05



3

 4.68 A

18.01 3

d) rectification efficiency  

P dc P ac



140.45  14.05 18 0.13  18.01

 60.8%

 10.4 A


63

Exercise 7 e)Transformer utilization factor Since a thyristor conducts 2 /3, the rms input line current is I s  I rms

2 3

 18.01

2 3

 14.71 A

V s  12 0.1V

VA  3V s I s  3  120 .1 14.71  5300VA

TUF 

P dc 140.45  14.05   0.372 or (37.2%) VA 5300

f)

2 2   18 . 01 10  3243.6W P I R Output power is o rms


PF 

3243.6 5300

 0.61 2 (lagging )

Note that the power factor is better than that of 3-phase half-wave converter


64

Three phase full converters 





Known as a 6-pulse converter, or 3 phase bridge converter Used in industrial applications up to 120kW output power. Two quadrant operation is possible.



Output ripple frequency is 6fs



Thyristor is fired at interval of /3



Filtering requirement is less than semi- and half converters



Firing sequence is 12,23,34,45,56,61,12… l a r t u s e e n - g a e t l n i o L v

v an  V m sin si n  t 2    3   2   vcn  V m sin   t   3  

vbn  V m sin   t 

e s n e i l g a e t l n i o L v

    6      vbc  vbn  vcn  3V m sin  t   2   5   vca  vcn  van  3V m sin  t   6   vab  van  vbn  3V m sin  


65

Three phase full converters 









The thyristors are triggered at an interval of  / / 3. T1 is triggered at t = ( /6 + ), T6 is already conducting when T1 is turned ON. During the interval ( /6 + ) to (/2 + ), T1 and T6 conduct together & the output load voltage is equal to v ab = (v an – v bn )

T2 is triggered at t = (/2 + ), T6 turns off naturally as it is reverse biased as soon as T 2 is triggered. During the interval ( /2 + ) to (5/6 + ), T1 and T2 conduct together & the output load voltage v O = v ac = (v an – v cn ) EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

66

Three phase full converters

Current waveforms for thyristors and diodes

Line period 2  EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

67

Three phase full converters

Output rms voltage V rms 

 3V m

3

 2 



  6  1 2



 

3V m2 sin 2   t 

3 3 4

co s 2

Output dc voltage  

 d  t 6 

V dc 



3 





3 3V m

2 



 

3V m sin  t 

6 

 

 d  t

6 

cos




68

Harmonic content of output voltage



Three phase full converters for continues load current (highly inductive) without Dm


69

Exercise 8


70

Exercise 8 (cont.)


71

Output voltage for various  values

Maximum delay angle max = 180 °


72

Three phase full converters for or Dm is used 



/3 for R load

For   /3, the instantaneous output voltage vo will have a negative part. Since the current through thyristors can not be negative, the load current will always be positive. Thus for resistive loads, or when freewheeling diode is used, the instantaneous voltage can not be negative, and the full converter behave as a semiconverter.



Output dc voltage

V dc 

3 



2





3V m sin  t d  t

V rms 

6 

   3 3V m    1  cos      3    

Output rms voltage



 3V m

3

 2 



3V m2 sin 2  t d  t

  6 

3  1       sin 2  2   


73

Three phase full converters for or Dm is used

/3 for R load

Maximum delay angle max = 120 °


74

Exercise 9 



3-phase fullconverter is operated from Y-connected 208V 60Hz supply. Load resistance R=10 . Half of the maximum possible average voltage is requested at the output. Calculate a)delay angle, b) rms and average output current, c)average and rms thyristor currents, d) rectification efficiency, e)Transformer utilization factor, f) input power factor Solution:


3 3V m



3 3 169.83



 280.9V



Desired output voltage, then 0.5*280.9=140.45V a) For resistive load and /6,

V dc 

3 3V m

cos



140.45 

3 3 169.83

cos    60

  /6 OK!




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Exercise 9 b) rms and average output current I dc 

V dc R



14 0.45 10

V rms  3 169.83 I rms 

V rms R



 14.05 A

1 2

15 9.29 10



3 3 4

cos 2  60  159.29V

 15.93 A


I dc 3

I R  I rms



14.05 3

2 6

 4.68 A

 15.93

2 6

d) rectification efficiency  

P dc P ac



140.45  14.05 159.29  15.93

 77.8%

 9.2 A

Thyristor conducts 1/3 of the line period EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

76

Exercise 9 e)Transformer utilization factor Since a thyristor conducts during 4/6 of line period, then I s  I rms

4 6

 15.93

4 6

 13 A

V s  12 0.1V

VA  3V s I s  3  120.1  13  4683.9 W

TUF 

P dc 140.45  14.05   0.421or ( 42.1%) VA 4683.9

f)

2 2   10  2537.6W P I R 15 . 93 Output power is o rms


PF 

2537.6 4683.9

 0.542 (lagging )

Note that the power factor is less than 3-phase semiconverter, but better than that of 3-phase half-wave converter


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Three phase dual converters



 





For four quadrant operation in many industrial variable speed dc drives , 3 phase dual converters are used. Used for applications up to 2 mega watt output power level. Dual converter consists of two 3 phase full converters which are connected in parallel & in opposite directions across a common load. The delay angles are controlled such that one converter operates as a rectifier and the other converter operates as an inverter, but both converter produce the same average voltage. Lr stands for the circulating currents EE328 Power Electronics, Dr. Mutlu Boztepe, Ege University 2014

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tiristores_caracteristicas

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