Transcript Thyristor Converters or Controlled Converters The controlled rectifier circuit is divided into three main circuits:Power Circuit Control Circuit Triggering circuit The commutation of thyristor.

Thyristor Converters or Controlled Converters
The controlled rectifier circuit is divided into three main circuits:Power Circuit
Control Circuit
Triggering circuit
The commutation of thyristor is
Natural Commutation
Forced Commutation
Half Wave Single Phase Controlled Rectifier With Resistive Load

Vdc
Vm
Vm
1

Vm sin(t ) dt 
( cos  cos( )) 
(1  cos )
2
2
2


Vdm  Vm / 
Vn  Vdc / Vm  0.5 (1  cos )

Vrms
1
Vm 1 
sin(2  ) 
2
Vm sin( t ) d t 

   


2 
2 
2 
Example 1 In the rectifier shown in Fig.3.1 it has a load of R=15  and,
Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain an
average output voltage of 70% of the maximum possible output voltage,
calculate:- (a) The firing angle, (b) The efficiency, (c) Ripple factor (d)
Peak inverse voltage (PIV) of the thyristor
Vdc
Vn 
 0.5 (1  cos )  0.7
Vdm
Vdc 49.02
Vm
I dc 

 3.268 A
Vdc  0.7 * Vdm  0.7 *
 49.02 V
R
15

Vm 1 
sin( 2  ) 
o
=66.42
, Vrms=95.1217V
Vrms 
   

2 
2

, Irms=95.122/15=6.34145A
Pdc
Vdc * I dc
49.02 * 3.268



 26.56%
Pac Vrms * I rms 95.121 * 6.34145
Vrms 95.121

FF 


 1.94
Vdc
49.02 2 2
Vac
RF 
 FF 2  1  1.942  1  1.6624
Vdc
The PIV is Vm
Half Wave Single Phase Controlled Rectifier With RL Load
Fig.3.4 Half wave single phase controlled rectifier with RL load.
Single-Phase Center Tap Controlled Rectifier With Resistive Load
b
a
Fig.3.8 Center tap controlled rectifier with resistive load.
Vdc 
1

Vm sin( t ) d t




Vm

( cos   cos( )) 
Vm

(1  cos  )
Vdc
Vn 
 0.5 (1  cos )
Vdm
Vrms 

Vm sin( t )


1
2

Vm
d t 
2
sin( 2  )
  
2
Example 4 The rectifier shown in Fig.3.8 has load of
R=15  and, Vs=220 sin 314 t and unity transformer
ratio. If it is required to obtain an average output voltage
of 70 % of the maximum possible output voltage,
calculate:- (a) The delay angle, (b) The efficiency, (c)
The ripple factor (d) The peak inverse voltage (PIV) of
the thyristor.
Vdc
Vn 
 0.5 (1  cos )  0.7
Vdm
then, =66.42o
Vm  220 , then, Vdc  0.7 *Vdm  0.7 *
2 Vm

Vm
Vdc 98.04
I dc 

 6.536 A Vrms 
R
15
2
 98.04 V
sin( 2  )
  
2
at =66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A

Pdc
V *I
98.04 * 6.536
 dc dc 
 53.04%
Pac Vrms * I rms 134.638* 8.976
Vrms 134.638
FF 

 1.3733
Vdc
98.04
Vac
2
2
RF 
 FF  1  1.3733  1  0.9413
Vdc
The PIV is 2 Vm
Single-Phase Fully Controlled Rectifier Bridge With Resistive Load
Fig.3.11 Single-phase fully controlled rectifier bridge with resistive load.
Vdc 
1

Vm sin( t ) d t 



Vm

 cos   cos( ) 
Vm

(1  cos )
Vn  Vdc / Vdm  0.5 (1  cos )
Vrms 

Vm sin( t )


1
2

Vm
d t 
2
sin( 2  )
  
2
Example 5 The rectifier shown in Fig.3.11 has load of R=15 
and, Vs=220 sin 314 t and unity transformer ratio. If it is required
to obtain an average output voltage of 70% of the maximum
possible output voltage, calculate:- (a) The delay angle , (b) The
efficiency, (c) Ripple factor of output voltage(d) The peak inverse
voltage (PIV) of one thyristor.
Vdc
Vn 
 0.5 (1  cos )  0.7 , then, =66.42o
Vdm
Vm  220 , then, Vdc  0.7 *Vdm  0.7 *
2 Vm

 98.04 V
Vdc 98.04
I dc 

 6.536 A
R
15
Vm
sin( 2  )
Vrms 
  
2
2
=66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A
Pdc
Vdc * I dc
98.04 * 6.536


 
 53.04%
Pac Vrms * I rms
134.638* 8.976
Vrms 134.638
FF 

 1.3733
Vdc
98.04
Vac
2
2
RF 
 FF  1  1.3733  1  0.9413
Vdc
The PIV is Vm
Full Wave Fully Controlled Rectifier With RL Load In Continuous Conduction Mode
Fig.3.14 Full wave fully controlled rectifier with RL load.
Full Wave Fully Controlled Rectifier With pure DC Load
i(t ) 
4 Io
1
1
1
1
* (sin t  sin 3t  sin 5t  sin 7t  sin 9t  ..........)

3
5
7
9
Vdc 
1

 

Vm sin(t ) dt 

2Vm

cos
Vn  Vdc / Vdm  cos
Vrms 
1

 
 Vm sin( t )
2

Vm
d t 
2
 
Vm
 (1  cos(2 t ) d t  2

Example 6 The rectifier shown in Fig.3.14 has pure DC load
current of 50 A and, Vs=220 sin 314 t and unity transformer ratio.
If it is required to obtain an average output voltage of 70% of the
maximum possible output voltage, calculate:- (a) The delay
angle , (b) The efficiency, (c) Ripple factor (d) The peak inverse
voltage (PIV) of the thyristor and (e) Input displacement factor.
Vdc
Vn 
 cos  0.7
Vdm
then, =45.5731o= 0.7954
Vm  220 , Vdc  0.7 *Vdm  0.7 * 2 Vm /   98.04 V , Vrms  Vm / 2
At =45.5731 Vrms=155.563 V. Then, Irms=50 A
o
Pdc
Vdc * I dc
98.04 * 50



 63.02%
Pac Vrms * I rms
155.563* 50
Vrms 155.563
The PIV is Vm
FF 

 1.587
Vdc
98.04
Vac
RF 
 FF 2  1  1.37332  1  1.23195
Vdc
Input displacement factor. cos  0.7
Single Phase Full Wave Fully Controlled Rectifier With Source Inductance

2Ls 
u  cos cos  
 
Vm 

1
Vrd
4 Ls I o

 4 fLs I o
2
Vdc actual  Vdc without sourceinductan ce  Vrd 
2Vm
8I o
u
 u 
I S1 
* sin
Is 

2
2 u
  2 3 
I s1
u

p. f 
cos   
Is
2

2
2I o

cos  4 fLs I o
Inverter Mode Of Operation
Fig.3.25 Single phase SCR inverter.
Fig.3.27 SCR inverter with a DC voltage source.
Ed  Vd  Vdo cos  
2

 Ls I d
Three Phase Half Wave Controlled Rectifier with Resistive Load
Fig.3.31 Three phase half wave controlled rectifier.
  30
  30
3
Vdc 
2
5 / 6
3 3 Vm
Vm sin t d t  2 cos  0.827Vm cos
 / 6
3

VLL cos  0.675VLL cos
2
I dc
3 3 Vm
0.827 * Vm

cos  
cos 
2 * *R
R
Vrms
I rms
3

2
3 Vm

R
5 / 6
 Vm sin  t 
2
d t  3 Vm
 / 6
1
3

cos 2
6 8
Ir  IS 
1
3

cos 2
6 8
I rms
Vm

R
3
1
3

cos 2
6 8
 > 30
 > 30

3
3 Vm 





Vdc 
V
sin

t
d

t

1

cos



0
.
4775
V
1

cos






m
m


2  / 6
2 
6

6


I dc
3 Vm 



1  cos    

2 R 
6


Vrms
3
5 
1
2
Vm sin  t  d t  3 Vm



sin( / 3  2 )

2  / 6
24 4 8 
3 Vm
I rms 
R
Ir  IS 
5 
1


sin( / 3  2 )
24 4 8 
I rms
Vm

R
3
5 
1


sin( / 3  2 )
24 4 8 
Example 7 Three-phase half-wave controlled rectfier is connected to 380 V
three phase supply via delta-way 380/460V transformer. The load of the
rectfier is pure resistance of 5  . The delay angle   25o . Calculate: The
rectfication effeciency (b) PIV of thyristors
3
3
VLL cos 
460cos 25  281.5V
2
2
Vdc 281.5
I dc 

 56.3 A
1
3
R
5
Vdc 
Vrms  2 VLL *

6 8
cos 2
1
3
 2 * 460 *

cos 2 * 25  298.8 V
6 8
I rms
Vrms 298.8


 59.76 A
R
5
Vdc I dc

*100  88.75%
Vrms I rms
Example 8 Solve the previous example (evample 7) if the firing angle   60o
 2
 * 460
3 

3
3 Vm 





   


Vdc 
1  cos     
1  cos     179.33 V


2 
2
6

 6 3 

I dc 
Vdc 179.33

 35.87 A
R
5
From (3.65) we can calculate Vrms as following:
Vrms  3 Vm
5

1


sin( / 3  2 )  230V
24 4 8 
Vrms 230
I rms 

 46 A
R
5
Vdc I dc

*100  60.79 %
Vrms I rms
PIV  2 VLL  2 * 460  650.54 V
Three Phase Half Wave Controlled Rectifier With DC Load Current
Fig.3.36 Three Phase Half Wave Controlled Rectifier With DC Load Current
t=0
bn 
2

2 / 3

0
I dc sin(n t ) d t 
2 I dc 
2n 
1

cos


n 
3 
2 I dc 3
Then, bn 
*
n 2
And
bn  0
for n=1,2,4,5,7,8,10,…..
For n=3,6,9,12
3I dc 
1
1
1
1

i p (t ) 
sin t  sin 2t  sin 4t  sin 5t  sin 7t  ......

 
2
4
5
7

3I dc
I 2p  I 2p1
2
I p1 
THD 
I

*
I
p
dc
2
I 2p1
3
THD 
2 2
9 2
I dc 
I dc
2
3
2
 68 %
9 2
I
2 dc
2
2
2
2
2
2
2
2
2
2
 1  1 1  1 1  1   1   1   1 
THD                             ....  68 %
 2   4   5   7   8   10   11   13   14 
Example 9 Three phase half wave controlled rectfier is connected to
380 V three phase supply via delta-way 380/460V transformer. The load
of the rectfier draws 100 A pure DC current. The delay angle,   30 o .
Calculate: (a) THD of primary current. (b) Input power factor.
460
the peak value of primary current is 100 *
 121.05 A .
380
2
I P, rms  121.05 *
 98.84 A
3
3I dc 3 *121.05
I P1 

 81.74 A
2
2
THD I P
2
2
 I P, rms 
98
.
84

  1 *100  
 
  1 *100  67.98 %
 81.74 
 I P1 
  81.74

  
P. f 
* cos    
* cos    0.414 Lagging
I P, rms
6  98.84

6 6
I P1
Three Phase Full Wave Fully Controlled Rectifier With Resistive Load
  60
o
Vdc 
3
 / 2 

  / 6 
Vdm 
3 Vm sin( t  ) d t 
6
3 3 Vm


3 3 Vm

cos 
 / 2 
2
 

Vrms 
3 Vm sin( t  )  d t  3 Vm

6 

 / 6 
3

1 3 3

 

cos
2

2

4


  60o
Vdc 
3

5 / 6

 / 6 

3 Vm sin( t  ) d t 
6
3 3 Vm

1  cos   / 3   
Vdm 
Vrms
3 3 Vm

5 / 6
Vdc
Vn 
 1  cos   / 3   
Vdm
 
3 
 



3
V
sin(

t

)
d

t

3
V
1

2


cos
2



 m



m

  / 6 
6 
4 
6 

3
2
Example 10 Three-phase full-wave controlled rectifier is connected to
380 V, 50 Hz supply to feed a load of 10  pure resistance. If it is
required to get 400 V DC output voltage, calculate the following: (a) The
firing angle,  (b) The rectfication effeciency (c) PIV of the thyristors.
3 3 Vm
3 3
2
Vdc 
cos  
*
* 380 cos   400V .

R
3
Vdc 400
o
Then   38.79 , I dc 

 40 A
R
10
From (3.84) the rms value of the output voltage is:
Vrms  3 Vm
1 3 3

 
cos 2   412.412 V
4
2

Vrms 412.412
Then, Vrms  412.412 V Then, I rms 

 41.24 A
R
10
Vdc * I dc
400 * 40
Then,  
*100 
*100  94.07%
Vrms * I rms
412.4 * 41.24
The PIV= 3 Vm=537.4V
Example 11 Solve the previous example if the required dc voltage is 150V.
Solution: From (3.81) the average voltage is :
3 3*
Vdc 

2
* 380
3
1  cos   / 3     150V
  73o
t is not acceptable result because the above equation valid only for
  60
. Then we have to use the (3.85) to get
3 3*
Vdc 
3 3 Vm
I dc 
Vdc 150

 15 A
R
10

cos  

2
* 380
3
cos   150V
  75.05o
From (3.88) the rms value of the output voltage is:
Vrms  3Vm 1 
I rms 
3 
 
2

* 380 *
 2  cos 2     3 *
4 
6 
3

Vrms 198.075

 19.8075 A
R
10
The PIV=
3
Vm=537.4V


3 


1 
 cos 2* 75.05  30 
 2 * 75.05 *
180

 4 
Vdc * I dc
150*15
*100 
*100  57.35 %
Vrms * I rms
198.075*19.81
Vrms  198.075 V
Three Phase Full Wave Fully Controlled Rectifier With pure DC Load Current
 > 60o
1 
2 LS I o 
u  cos cos  
 
VLL 

Vdc actual  Vdc without sourceinductan ce  Vrd 
2 I o2  u 
IS 
 

  3 6
I S1
u

pf 
cos     
IS
2

I S1
2 6 Io  u 
sin 
u
2
3 2

VLL cos  6 fLs I o
2 6 Io
u

sin  
u
2
u
2 3 * sin 
u
u
2 


cos     
cos   
2
2
2


 u 
2I o  u 
u   

 3 6
  3 6 
Inverter Mode of Operation