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 CommutationHalf Wave Single Phase Controlled Rectifier With Resistive Load

Fig.3.1 Half wave single phase controlled rectifier.

)cos1(2

))cos(cos(2

)sin(2

1

mmmdc

VVtdtVV

/mdm VV )cos1(5.0/ mdcn VVV

2

)2sin(1

2)sin(

2

1 2

mmrms

VtdtVV

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

7.0)cos1(5.0 dm

dcn V

VV

VV

VV mdmdc 02.49*7.0*7.0

A

R

VI dc

dc 268.315

02.49

2

)2sin(1

2

mrms

VV =66.42o, Vrms=95.1217V

, Irms=95.122/15=6.34145A

%56.2634145.6*121.95

268.3*02.49

*

*

rmsrms

dcdc

ac

dc

IV

IV

P

P

94.12202.49

121.95

dc

rms

V

VFF

6624.1194.11 22 FFV

VRF

dc

ac

The PIV is mV

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

Fig.3.8 Center tap controlled rectifier with resistive load.

a b

)cos1())cos(cos()sin(1

mmmdc

VVtdtVV

)cos1(5.0 dm

dcn V

VV

2

)2sin(

2)sin(

1 2

mmrms

VtdtVV

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.

7.0)cos1(5.0 dm

dcn V

VV then, =66.42o

220mV , then, VV

VV mdmdc 04.98

2*7.0*7.0

AR

VI dc

dc 536.615

04.98

2

)2sin(

2

mrms

VV

at =66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A

%04.53976.8*638.134

536.6*04.98

*

*

rmsrms

dcdc

ac

dc

IV

IV

P

P

3733.104.98

638.134

dc

rms

V

VFF

9413.013733.11 22 FFV

VRF

dc

ac

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.

)cos1(5.0/ dmdcn VVV

)cos1()cos(cos)sin(1

mmmdc

VVtdtVV

2

)2sin(

2)sin(

1 2

mmrms

VtdtVV

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.

7.0)cos1(5.0 dm

dcn V

VV , then, =66.42o

220mV , then, VV

VV mdmdc 04.98

2*7.0*7.0

AR

VI dc

dc 536.615

04.98

2

)2sin(

2

mrms

VV

=66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A

%04.53976.8*638.134

536.6*04.98

*

*

rmsrms

dcdc

ac

dc

IV

IV

P

P

3733.104.98

638.134

dc

rms

V

VFF

9413.013733.11 22 FFV

VRF

dc

ac

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

)..........9sin9

17sin

7

15sin

5

13sin

3

1(sin*

4)( ttttt

Iti o

cos2

)sin(1 m

mdcV

tdtVV

cos/ dmdcn VVV

2

)2cos(1(2

)sin(1 2 mm

mrmsV

tdtV

tdtVV

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.

7.0cos dm

dcn V

VV then, =45.5731o= 0.7954

220mV , VVVV mdmdc 04.98/2*7.0*7.0 , 2/mrms VV

At =45.5731o Vrms=155.563 V. Then, Irms=50 A

%02.6350*563.155

50*04.98

*

*

rmsrms

dcdc

ac

dc

IV

IV

P

P

587.104.98

563.155

dc

rms

V

VFF

23195.113733.11 22 FFV

VRF

dc

ac

The PIV is Vm

Input displacement factor. 7.0cos

Single Phase Full Wave Fully Controlled Rectifier With Source Inductance

m

s

V

Lu

2coscos 1

osos

rd IfLIL

V 42

4

osm

rdceinducsourcewithoutdcactualdc IfLV

VVV 4cos2

tan

32

2 2 uII o

s

2sin*

2

81

u

u

II o

S

2cos. 1 u

I

Ifp

s

s

Inverter Mode Of Operation

Fig.3.25 Single phase SCR inverter.

Fig.3.27 SCR inverter with a DC voltage source.

dsdodd ILVVE

2cos

Three Phase Half Wave Controlled Rectifier with Resistive Load

Fig.3.31 Three phase half wave controlled rectifier.

30

cos675.0cos2

3

cos827.0cos2

33sin

2

36/5

6/

LLLL

mm

mdc

VV

VV

tdtVV

30

cos*827.0

cos**2

33

R

V

R

VI mm

dc

2cos8

3

6

13sin

2

36/5

6/

2

mmrms VtdtVV

2cos8

3

6

13

R

VI m

rms

2cos8

3

6

1

3

R

VIII mrms

Sr

> 30

6cos14775.0

6cos1

2

3sin

2

3

6/

mm

mdc VV

tdtVV

> 30

6cos1

2

3

R

VI m

dc

)23/sin(8

1

424

53sin

2

3

6/

2

mmrms VtdtVV

)23/sin(8

1

424

53

R

VI m

rms

)23/sin(8

1

424

5

3

R

VIII mrms

Sr

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 o25 . Calculate: The rectfication effeciency (b) PIV of thyristors

VVV LLdc 5.28125cos4602

3cos

2

3

AR

VI dc

dc 3.565

5.281

V

VV LLrms

8.29825*2cos8

3

6

1*460*2

2cos8

3

6

1*2

AR

VI rms

rms 76.595

8.298

%75.88100* rmsrms

dcdc

IV

IV

Example 8 Solve the previous example (evample 7) if the firing angle o60

VV

V mdc 33.179

36cos1

2

460*3

23

6cos1

2

3

AR

VI dc

dc 87.355

33.179

From (3.65) we can calculate rmsV as following:

VVV mrms 230)23/sin(8

1

424

53

AR

VI rms

rms 465

230

%79.60100* rmsrms

dcdc

IV

IV

VVPIV LL 54.650460*22

Fig.3.36 Three Phase Half Wave Controlled Rectifier With DC Load Current

Three Phase Half Wave Controlled Rectifier With DC Load Current

t=0

3

2cos1

2)sin(

23/2

0

n

n

ItdtnIb dc

dcn

Then, 2

3*

2

n

Ib dc

n for n=1,2,4,5,7,8,10,…..

And 0nb For n=3,6,9,12

......7sin

7

15sin

5

14sin

4

12sin

2

1sin

3)( ttttt

Iti dc

p

21

21

2

p

pp

I

IITHD

dcp II *

3

2 2

31

dcp

II

%68

2

92

9

3

2

22

22

2

dc

dcdc

I

II

THD

%68....14

1

13

1

11

1

10

1

8

1

7

1

5

1

4

1

2

1222222222

THD

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, o30 . Calculate: (a) THD of primary current. (b) Input power factor.

the peak value of primary current is A05.121380

460*100 .

AI

I dcP 74.81

2

05.121*3

2

31

%98.67100*174.81

84.98100*1

22

1

,

P

rmsPI I

ITHD

P

AI rmsP 84.983

2*05.121,

LaggingI

IfP

rmsP

P 414.066

cos*84.98

74.81

6cos*.

,

1

Three Phase Full Wave Fully Controlled Rectifier With Resistive Load

o60

cos33

)6

sin(33

2/

6/

mmdc

VtdtVV

m

dmV

V33

2cos4

33

2

13)

6sin(3

32/

6/

2

mmrms VtdtVV

o60

3/cos133

)6

sin(33

6/5

6/

mmdc

VtdtVV

m

dmV

V33

3/cos1dm

dcn V

VV

6

2cos24

313)

6sin(3

36/5

6/

2

mmrms VtdtVV

VR

VV m

dc 400cos380*3

2*

33cos

33

.

Then o79.38 , AR

VI dc

dc 4010

400

From (3.84) the rms value of the output voltage is:

VVV mrms 412.4122cos4

33

2

13

Then, VVrms 412.412 Then, AR

VI rms

rms 24.4110

412.412

Then, %07.94100*24.41*4.412

40*400100*

*

*

rmsrms

dcdc

IV

IV

The PIV= 3 Vm=537.4V

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.

VV

V mdc 150cos

380*3

2*33

cos33

o73

60

VVdc 1503/cos1380*

3

2*33

o05.75

AR

VI dc

dc 1510

150

3005.75*2cos

180*05.75*2

4

31*380*

3

2*3

62cos2

4

313

mrms VV VVrms 075.198

AR

VI rms

rms 8075.1910

075.198 %35.57100*

81.19*075.198

15*150100*

*

*

rmsrms

dcdc

IV

IV

3

Example 11 Solve the previous example if the required dc voltage is 150V.Solution: From (3.81) the average voltage is :

It is not acceptable result because the above equation valid only for

.Then we have to use the (3.85) to get

From (3.88) the rms value of the output voltage is:

The PIV= Vm=537.4V

Three Phase Full Wave Fully Controlled Rectifier With pure DC Load Current

> 60o

LL

oS

V

ILu

2coscos 1

osLLrdceinducsourcewithoutdcactualdc IfLVVVV 6cos23

tan

63

2 2 uII o

S

2sin

621

u

u

II o

S

2cos

63

2sin*32

2cos

63

2

2sin

62

2cos

21 u

uu

uu

uI

u

u

Iu

I

Ipf

o

o

S

S

Inverter Mode of Operation