36407294 Thyristor Commutation Techniques 100403040622 Phpapp02

8/3/2019 36407294 Thyristor Commutation Techniques 100403040622 Phpapp02

http://slidepdf.com/reader/full/36407294-thyristor-commutation-techniques-100403040622-phpapp02 1/137

Power Electronics by Prof. M. Madhusudhan Rao 11Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Thyristor Commutation

Techniques



Power Electronics by Prof. M. Madhusudhan Rao 22 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Introduction

Commutation

± Process of turning off a conducting thyristor.

Current Commutation

Voltage Commutation




Methods of Commutation

Natural Commutation

Forced Commutation




Natural Commutation

Occurs in AC circuits

~

T+

vvs R o o




[t

[t

[t

[t

Supply oltage sSinusoidal

oltage across S

Load oltage o

urn ooccurs here

0

0

T

T

2T

2T

3T

3T

E

tc

Gate ulse

TE

E TE




Natural Commutation of hyristors takes place

in ± AC oltage controllers.

± Phase controlled rectifiers.

± Cyclo con erters.




Forced Commutation

Applied to dc circuits

Commutation achie ed by re erse biasing the

C or by reducing the C current belowholding current alue.

Commutating elements such as inductance and

capacitance are used for commutation purpose.




Methods of Forced

Commutation

elf commutation.

esonant pulse commutation. Complementary commutation.

Impulse commutation.

External pulse commutation.

oad Commutation.

ine Commutation.




Forced Commutationis applied to

Choppers.

In erters.




Self Commutation

Or

Load CommutationOr

Class A Commutation

(Commutation By ResonatingThe Load)




Circuit is under damped by including suitable

alues of & C in series with load. Oscillating current flows.

C is turned off when current is zero.




V

R V (0)c

C

T i

Load

+ -



Power Electronics by Prof. M. Madhusudhan Rao 1313

Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Expression for Current

Fig. shows a transformed network

1 ( )

( )- -





22

0

0

0C

C

C CS V V

S I S RC

V V

S I S

R SLCS

C

S S LC

V V

R LC S S

L LC

« »- ½

!

«« »- ½

!

»- ½

! « » ¬-

¼½





2

2 2

2

0

0

2 2

C

C

V V

L I S

RS S

L LC

V V

L I S R R R

S S L LC L L

!

! ¨ ¸ ¨ ¸ © ¹ © ¹

ª º ª º





2 2 22 2

2

0

2 2

Where,

0, ,

2 2

C

C

V V

A L I S

S

S L LC L

V V R R A

L L LC L

H [

H [

! !

« »¨ ¸ ¨ ̧¬ ¼ © ¹ © ¹ª º ª º¬ ¼- ½

¨ ¸! ! ! © ¹

ª º





2 2

is called the natural frequency

aking in erse aplace transforms

sint

A

I S S

Ai t e t H

[

[

[ H [

[[

!

!



Power Electronics by Prof. M. Madhusudhan Rao1818


2

Expression for current

0sin

Peak alue of current

0

Rt

C L

C

V V

i t e t L

V V

L

[[

[

@

!

!





Expression For Voltage AcrossCapacitor At The Time Of Turn Off

V

R V (0)c

C

T i

Load

+ -



Power Electronics by Prof. M. Madhusudhan Rao202 0 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

A pplying K L to igure

Substituting or i,

sin sin

c R L

c

t t

c

v t V v V

d iv t V i R Ld t

A d Av t V R e t L e t

d t

H H [ [

[ [

!

!

¨ ¸! © ¹

ª º



Power Electronics by Prof. M. Madhusudhan Rao212 1Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

? A

sin cos sin

sin cos sin

sin cos sin2

sin cos2

t t t

c

t c

t

c

t

c

A Av t V R e t L e t e t

Av t V e R t L t L t

A Rv t V e R t L t L t

L

A Rv t V e t L t

H H H

H

H

H

[ [ [ H [[ [

[ [ [ H [[

[ [ [ [[

[ [ [[

!

!

« »! ¬ ¼- ½

« »! ¬ ¼- ½





Substituting or A,

0sin cos

2

0sin cos

2

S turns o hen current goes to zero.

i.e., at

C t

c

C t

c

V V Rv t V e t L t

L

V V Rv t V e t t

L

t

H

H

[ [ [[

[ [ [[

[ T

« »! ¬ ¼- ½

« »! ¬ ¼- ½

!





2

here ore at turn o

00 cos

0

0

C

c

cC

R

L

c C

V V

v t V e

v t V V V e

v t V V V e

H T

[

H T

[

T

[

[ T[

!

« »! - ½

« »! - ½





2

or e ecti e commutationthe circuit should be under damped.

hat is2

With = 0, and the capacitor initially uncharged

that is 0 0

sin

ote:

C

R

L LC

V

V t i t

L LC [

¨ ¸

© ¹ª º

!

!





1But

sin sin

and capacitor oltage at turn o is equal to 2

ig. sho s the a e orms or the abo e conditions.

nce the S turns o oltage across it is

negati e oltage.

LC

V t C t i t LC V

L L LC LC

[ !

@ ! !





Current i

Capacitor oltage

Gate pulse

oltage across SCR

0 TT/2[t

[t

[t

[t

2

Conduction time of SCR T

[!





Problem

Cal culate the conduction time of SCR and the

peak

SCR

curr ent that f

l ows in the circuit

empl oying ser ie s r e sonant commut ation ( sel f

commut ation or cla ss A commut ation), i f the

su ppl y vol t a g e i s 300 V , C = 1 Q F , L = 5 mH

and R L = 100 ; . A ssume that the circuit i s

initiall y r elaxed .





V=300V

R L L

1 Q100 ; 5 H





2

2

3 6 3

12

1 100

10 1 10 2 10

10,000 rad/sec

Since the circuit is initially relaxed, initial oltageacross capacitor is zero as also the initial current

through and the expre

L R LC L

L

[

[

[

¨ ¸! © ¹ª º

¨ ¸! © ¹v v v v vª º

!

ssion or current isi




Problem

F i gur e shows a sel f commut ating circuit. The

induct a

nce ca

rr ie s a

n initial

curr ent of 200 Aand the initial vol t a g e across the ca pacit or i s

V , the su ppl y vol t a g e. Deter mine the

conduction time of the SCR and the ca pacit or

vol t a g e at t urn off .





V100V

0 FQ

10 Q

C

i(t)

p IO

VC(0)=V





The transformed circuit of the pre ious figure is shown in figure below

L

V

S

V (0)

SC

I LO

+

+

+

-

-

-

1CS

I(S)





2 2

The go erning equation is

0 1

0

1

0

1 1

C

O

C

O

C

O

V V I S S L I L I S

S S C S

V V I L

S S I S

S LC S

V V C S S S I LC S

I S S LC S LC

!

@ !

« »¬ ¼- ½

!





2 2

2 22 2

2 22 2

0

1 1

0

0 1; where

C O

C O

C O

V V C I LCS I S

LC S LC S

LC LC

V V SI I S

S L S

V V SI I S

S L S LC

[[

[[

[[ [

« »- ½!

« » « » ¬ ¼ ¬ ¼

- ½ - ½

! « » - ½

« »- ½! !

« » - ½





0

0

Taking in erse T

0 sin cos

The capacitor oltage is gi en by

10

10 sin cos 0

C O

t

c C

t

c C O C

C i t V V t I t

L

v t i t d t V

C

C v t V V t I t d t V

C L

[ [

[ [

! « »- ½

!

® ¾± ±! « »¯ ¿- ½

± ±° À

´

´





01cos sin 0

011 cos sin 0

1sin 0 1 cos 0

sin cos 0 0 cos 0

sin 0 cos

C O

c C

C O

c C

O

c C C

c O C C C

c O C

V V t t I C v t t t V

o oC L

V V I C v t t t V

C L

I C v t LC t V V LC t V

C C L

Lv t I t V V t V V t V C

Lv t I t V V t V

C

[ [[ [

[ [[ [

[ [

[ [ [

[ [

« »! ¬ ¼¬ ¼- ½

« »! ¬ ¼

¬ ¼- ½

!

!

!





we get

cos

sin

C

O

c O

V O V

i t I t

Lv t I t V

C

[

[

!

@

!

!

Q

I0

i(t)

T/

T/

[t

[t

c(t)





6 6

6

Turn o occurs at a time ' ' so that

20.

0.

0. 10 10 0 10

0. 10 00 3 .1 seconds

O

O

O

O

O

t

t

t LC

t

t

T[

TT

[

T

T Q

!

@ ! !

! v v v v

! v v !




60

6

and the capacitor oltage at turn off is gi en by

sin

10 10200 sin 90 100

0 10

89.4 100 189.4

c O O O

c O

c O

L

v t I

t V

C

v t

v t V

[

!

!

! !




I n the circuit shown in f i gur e V = 600 vol t s, initial ca pacit or vol t a g e i s zero, L = 20 Q H , C =50 Q F and the curr ent throug h theinduct ance at the time of SCR t r i gg er ing i s I

O= 350 A. Deter mine

± Peak val ue s of ca pacit or vol t a g e and curr ent ± C onduction time of T 1.

Problem





62 2

6

Substituting the alues, the peak capacitor current

0 103 0 600 1011.19

20 10The expression or capacitor oltage is

sin 0 cos

with 0 0, sin cos

c O C

C c O

A

Lv t I t V V t V

C

LV v t I t V t V

C

[ [

[ [

v! v !

v

!

! !





2 2 1

2 2

62 2

6

This can be rewritten as

sin ; Where tan

The peak alue of capacitor oltage is

ubstituting the alues, the peak alue of capacitor oltage

20 10600 3 0

0 10

c O

O

O

C V

L Lv t V I t V

C I

LV I V

C

[ F F

! !

!

! 600 639. 600 1239. V ! !





To calculate conduction time of T1

The wa eformof capacitor

current is shownin figure.

When thecapacitor

current becomeszero the Cturns off.

[t

Capacitor current

TE

E





61

6

0

6 6

3 0 20 10tan

600 0 10

20.2 i.e., 0.3 34 radians

1 131622.8 rad/sec

20 10 0 10

conduction time o SCR

0.3 3488.17 sec

31622.8

LC

E

E

[

T Q

v!v

!

! ! !v v v

@

! !



Power Electronics by Prof. M. Madhusudhan Rao 505 0 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Resonant Pulse Commutation

(Class B Commutation)

L

C

VLoad

FWD

a b

iT

IL



Power Electronics by Prof. M. Madhusudhan Rao 515 1Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Series LC circuit connected across thyristor µT¶.

Initially µC¶ is charged to µV¶ olts with plateµa¶ as positi e.

Current in LC oscillates when SC is ON.

µ T¶ turns off when capacitor discharges throughthyristor in a direction opposite to I

L





GSCR

Capacitor oltagevab

Tt1t

t

t

I p i

IL

tC

(t

T[





t

t

Volta e acr ossSCR

ISCR Thyristor rre t





Expression For tC

,

the Circuit Turn Off Time

Assume that at the time o turn o o the SCR

the capacitor oltage and a constant

load current lows.

is the time taken or the capacitor oltage to reach 0 olts rom olts and

it i

ab

L

c

v V

I

t V

}

s deri ed as ollows





0

1

seconds

For proper commutation should be greater than .

Magnitude of the peak alue of should be

greater than the load current .Expression for is deri ed as foll

ct

L c

L

c

L

c q

p

L

I t V I

d t C C

VC t

I

t t

I i

I

i

! !

!

´

ows





2

22

1

1

11

1

V

S I S

SLC s

VC

I S

LC S L

V CS

s

S LC

V

L

LC S

C

! !

! !¨

¨ ¸© ¹ª º

©ª

¸¹ º





2

2

1

11 1

1

1

Taking in erse LT sin

V LC I S L

S LC LC

C LC I S V

LS

LC

C i t V t

L[

¨ ¸© ¹ª º!

¨ ¸ © ¹

ª º

¨ ¸© ¹ª º

!

!



Power Electronics by Prof. M. Madhusudhan Rao 606 0 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Expression For

Conduction Time Of SCR

1

Conduction time o SCR

sin

L

p

t

I

I

T[

T

[ [

! (

¨ ¸

© ¹© ¹ª º!



Power Electronics by Prof. M. Madhusudhan Rao 616 1Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Alternate Circuit

For Resonant Pulse Commutation

V

LOAD

FWD

L

T1 IL

T3

T2

iC(t)

iC(t)

VC( )

a b

+

C





sin

Where 0 and the capacitor oltage

1.

1

0 sin .

0 cos

c p

p C

c C

c C

c C

i t I t

C I V

L

v t i t d t C

C

v t V t d t C L

v t V t

[

[

[

!

!

!

!

!

´

´





1

1

Instant at

which the thyristor turns o is at

sin

0

L p

p C

t

t I I

LC

C I V

L

!

!





1

1

1 1 1

or sin 0

the corresponding capacitor oltage is

0 cos

L

C

c C

I L

t LC V C

v t V V t [

¨ ¸

! © ¹© ¹ª º

! !





Expression For tC

1

1

Assuming a constant load

current which charges the capacitor

seconds

Normally 0

For reliable commutationdepends on & becomes smaller for

higher alues of

L

c

L

C

c q

C L

L

I

CV

t I

V V

t t t I

I

!

}

"





t

t

t

t1

V1

V

V ( )

it r lt

urr nt i (t)





Resonant Pulse CommutationWith Accelerating Diode





V

LOADFWD

L

1IL

3

2iC(t

VC(0)-

D2 iC(t)



Power Electronics by Prof. M. Madhusudhan Rao 707 0


Diode D2 connected as shown to acceleratedischarge.

T 2 turned on to turn off T 1.

Once T 1 is off at t 1. iC (t ) flows through D2 untilcurrent reduces to I L

at time t 2.

From t = t 2 , µ C ¶ charges through load, T 2 self

commutates.

But thyristor reco ery process low hence

longer re erse bias time.





IL

0

VC

0

V1

V (O)C

iC

t

tt

1t2

tC





V

LOADF D

LC1

IL

3

2i (t)

VC(0)





2

1

1

11

66 6 1

1 6

When 200 ; Let be triggered at 0

Capacitor current reaches a alue at

when turns off

sin0

200 3 103 10 30 10 sin 3.0 sec

200 30 10

1 1

3 10

L

c L

L

C

I A T t

i t I t t

T

I Lt LC

V C

t

LC

Q

[

! !

!

¨ ¸! © ¹© ¹

ª º

¨ ¸! !© ¹

© ¹ª º

! ! 6

6 60.10 10 / sec

30 10r ad

!





1

1 1

6 6

1

1

1

1

6

At the magnitude o capacitor oltage is,

0 cos

i.e., 200cos 0.10 10 3.0 10

200 0.9487189.75 Volts

30 10 189.7528.46 sec

200

C

c

L

c

t t

V V t

V

V V

CV t

I

t

[

Q

!

!

! v v v

! v!

!

v v! !





66 6 1

1 6

6 6

1

6

1

When 50

50 3 103 10 30 10 sin 0.749 sec

200 30 10

200cos0.105 10 0.749 10 200 1 200 Volts

30 10 200120 sec

50

It is obser ed that as load increases the alue of re

L

c

L

c

I A

t

V

CV t

I

t

Q

Q

!

¨ ¸! !© ¹

© ¹ª º

! ! !

! ! !

duces.





2

1

R epeat abo e problem

For 200 if an antiparallel diode is

connected across thyristor as shown in figure

L I A D

T

!

Problem





V

L

O

A

DFWD

LC

T IL

T3

T2iC(t)

VC(0)+

-

D2 iC(t)





2

1

1

1

1

6

6 6 11 6

1

200

Let be triggered at 0

Capacitor current reaches the alue at ,

when turns o

sin

200 3 103 10 30 10 sin200 30 10

3.05 sec

L

C L

L

C

I A

T t

i t I t t

T

I Lt LC

V O C

t

t Q

!

!

!

« »@ ! ¬ ¼

- ½

¨ ¸v! v v v © ¹© ¹vª º

!





6 6

6

1

1 1 1

6 61

1

1 1

3 10 30 10

0.105 10 radians/sec

cos

200cos 0.105 10 3.05 10

189.75

C C

C

C

LC

At t t

V t V V O t

V t

V t V

[

[

[

! !v v v

! v

!

! !

! v v v

!





2 1

2

2 1

6 6 62

2

6 6

6

lows through diode a ter turns o .

current alls back to at .

3 10 30 10 3.05 10

26.75 sec

1 1

3 10 30 10

0.105 10 rad/sec.

C

C L

i t T

i t I t

t LC t

t

t

LC

T

T

Q

[

[

!

! v v v v

!

! !v v v

! v





2

6 6

2 2

2 2

6 6

2 1

200cos0.105 10 26.75 10

189.02

26.75 10 3.05 10

23.7 secs

C

C

C

C

At t t V t V

V t V V

t t t

t Q

!

! !

! !

@ ! !

!





Problem

F or the circuit shown in f i gur e cal culate theval ue of L for pro per commut ation of SCR. Al so f ind the conduction time of SCR.

V

=30V L

i

4 FQ

R L

IL30 ;





The load current30

1 Amp30

For proper SCR commutation ,

the peak alue of resonant current i, should be

greater than

Let 2 , 2 Amps

L

L

p

L

p L p

V I

R

I

I

I I I

! ! !

! @ !





6

3 6

1

0.9

Also

4 102 30

1 116666 rad/sec

0.9 10 4 10

p

V V C I V

L L L

LC

L mH

L

LC [

[

! ! !

!

@ !

! ! !





11

Conduction time of SCR

1sinsin 2

16666 16666

0.523radians 0.00022 seconds16666

0.22 msec

L

p

I

I T T

[ [

T

¨ ¸

¨ ¸© ¹© ¹ © ¹ª º ª º! !

! !

!





Problem

F or the circuit shown in f i gur e g iven that the

l oad curr ent t o be commut ated i s 10 A , t urn off

time r equir ed i s 40 Q sec and the su ppl y vol t a g e

i s 100 V . Obt ain the pro per val ue s of

commut ating el ement s.





V

=100Vi I

I





peak alue o this should be greater than

Let 1.5

1.5 10 100 ...

Also, assuming that at the time o turn o the capacitor

oltage is approximately equal to V, and the load curren

p L

p L

C I i V I

L

I I

C a

L

!

!

@ v !

t

linearly charges the capacitor

secondsc

L

CV t

I !





6

6

4 6

2 2

4

& this is gi en to be 40 sec.

10040 10

10

4

Substituting this in equation (a)

4 101.5 10 100

10 4 101.5 10

1.777 10 0.177

ct

C

C F

L

L

L H mH

Q

Q

@ v ! v

@ !

vv !

v vv !

@ ! v !





Problem

I n a r e sonant commut ation circuit su ppl y

vol t a g e i s 200 V . Load curr ent i s 10 A and the

d evice t urn off time i s 20 Q s. The r atio of peak

r e sonant curr ent t o l oad curr ent i s 1.5.

Deter mine the val ue of L and C of the

commut ation circuit.





Gi en 1.5

1.5 1.5 10 15

. ., 15 ...

It is gi en that the de ice turn off time is 20 sec.

Therefore the circuit turn off time should be

greater than this,

Let 30 sec &

p

L

p L

p

c

c c

L

I

I I I A

C i e I V A a

L

t

CV t t

I

Q

Q

!

@ ! ! !

! !

! !





Complementary Commutation

(Class C Commutation,

Parallel Capacitor Commutation)





Two SCR s are used, turning ON one SCR turnsoff the other.

T 1 is fired, I L

flows through R1.

At same time µ C ¶ charges towards µ V ¶ through R2 with plate µb¶ positi e.

To turn off T 1, T 2 is fired resulting in capacitor

oltage re erse biasing T 1

and turning it off.

When T 2 is fired current through load shoots up

as oltage across load is V+V C





Gate p lseof T1

Gate p lseof T2

C rre t thro gh R 1

T

IL

V

t

t

t

t

C rre t thro gh T1

C rre t thro gh T2

2

2

V

R

21

V

R

V

R1

V

R2

V

R1

2

1

V

R





t

t

t

Voltage acr osscapacitor vab

Voltage acr oss T1

Currentthr ough R 2

tC tC

tC

V

- V





Expression For

Circuit Turn Off Time tC

1 1

1

Where is the f inal oltage, is the initial

oltage and is the time constant.

At , 0, , C, &

0 2c c

t

c f i f

f i

c c f i

t t

R C R C

v t V V V e

V V

t t v t V V R V V

V V V e V V e

X

X

X

!

! ! ! ! !

@ ! !





V

R 1 R 2

T1 T2

C

6

The circuit turn-off time

0.693 RC seconds

0.693 5 7.5 10

26 sec

c

c

c

t

t

t Q

!

!

!





Calculate the alues of R L and C to be used for

commutating the main SCR in the circuitshown in figure. When it is conducting a full

load current of 25 A flows. The minimum timefor which the SCR has to be re erse biased for proper commutation is 40Qsec. Also find R 1,gi en that the auxiliary SCR will undergo

natural commutation when its forward currentfalls below the holding current alue of 2 mA.

Problem





V=100V

R 1 R L

MainSCR AuxiliarySCR

iC

C

ILi1





In this circuit only the main SCR carries the load and

the auxiliary SCR is used to turn off the main SCR .

Once the main SCR turns off the current through theauxiliary SCR is the sum of the capacitor charging

current iC

and the current i1 through R1, iC

reduces to

zero after a time t C

and hence the auxiliary SCR turns

off automatically after a time t C , i1 should be less thanthe holding current.





6

6

Gi en 25

100. ., 25

4

40 sec 0.693

. ., 40 10 0.693 4

40 104 0.693

14.43

L

L L

L

c L

I A

V i e A

R R

R

t R C

i e C

C

C F

Q

Q

!

! !

@ ! ;

! !

!

@ !

!





11

1

1 3

1

should be less

than the holding current of auxiliary SCR

100 should be 2

100

2 10. ., 50

V

i R

mA R

R

i e R K

!

@

@ "

" ;





Impulse Commutation

(Class D Commutation)

V

LOADFWD

C

T1

T3

IL

T2

VC(O)+

L





µ C ¶ charged to a oltage V C ( O ) with polarity asshown.

T 1 is conducting and carries load current I L.

To turn off T 1 , T 2 is fired. Capacitor oltage re erse biases T 1 and turns it

off.

µ C ¶ Charges through load.

T 2 self commutates.

To re erse capacitor oltage T 3 is turned ON.





Gate pulseof T2

Gate pulseof T3

Voltage across T1

Capacitor oltage

Gate pulseof T1

VS

t

t

t

tC

VC

-VC





y T1 is turned off by applying a negati e oltage across

its terminals. ence this is oltage commutation.

y tC depends on load current. For higher load currents tC

is small. This is a disad antage of this circuit.

y When T2 is fired, oltage across the load is V VC;

hence the current through load shoots up and then

decays as the capacitor starts charging.





An Alternative Circuit

For Impulse Commutation

V

D

T1

IT1

I

i

T2

R L

V (O)+

_





Initially µ C ¶ is charged to V C

with top plate

positi e.

T 1 is fired, load current I L

flows.

µ C ¶ discharges at the same time & re erses its

polarity.

µ D¶ ensures bottom plate remains positi e.

To turn off T 1 , T 2 is fired.





Gate pulseof T1

Gate pulseof T2

Current through SCR

This is due to i

Capacitor oltage

t

t

t

tC

VC

ILIT1

-V

V

R L





Load c rre t

Voltage across T1

t

t

tC

IL

V

2VR L





Problem

An impul se commut ated th yr i st or circuit i s

shown in f i gur e. Deter mine the availabl e t urn

off time of the circuit i f V = 100 V , R = 10 ;

and C = 10 Q F . V ol t a g e across ca pacit or

be for e T2 i s f ir ed i s V vol t s with polar it y a s

shown.





C

T1

T2

V ( )C

V +

+

-

-

R





When T2 is triggered

the circuit is as shown in figure

C

i(t)

T2V

++-

-

R

V (O)C





Writing the transform

circuit, we obtain

Vs

V (0)

s

+

+

I(s)

1s

Expression for capacitor oltage





Expression for capacitor oltage

is obtained as follows,

10

1

0

1

01

C

C

C

V V S

I S

R

CS

C V V I S

RCS

V V I S

R S RC

!

!

!¨ ¸

© ¹ª º





Voltage across capacitor

01

0 01

1

0 0 0

1

C

C

C C

C

C C C C

V V s I S

C S S

V V V

V s RC S S S

RC

V V V V V

V S S S

S RC

!

! ¨ ¸© ¹

ª º

!

¨ ¸© ¹

ª º





The waveform of v C (t)

is shown in figure

t

t(0)

v (t)





6

Taking natural logarithms

1log

2

ln 2

10 10 10 ln 2

69.3 sec

c

e

c

c

c

t

RC

t RC

t

t Q

¨ ¸!

© ¹ª º!

!!





Problem

I n the commut ation circuit shown in f i gur e

C = 20 Q F , the in put vol t a g e V var ie s bet ween

180 and 220 V and the l oad curr ent var ie s

bet ween 50 and 200 A. Deter mine the minimum

and maximum val ue s of availabl e t urn off time

t C .





C

T1

T2 I0

I0

V (0)=VC+

V

It i i th t iV





It is gi en that aries

between 180 and 220 V &

aries between 50 and 200 A

The expression f or a ailableturn off time is gi en by

is max. when V is max. & is min.

O

c

cO

c O

V

I

t

CV

t I

t I

!





maxmax

min

6

max

minmin

max

6

min

22020 10 88 sec

50

18020 10 18 sec

200

c

O

c

c

O

c

CV

t I

t

CV And t

I

t

Q

Q

@ !

! v v !

!

! v v !





External Pulse

Commutation (Class E Commutation)

VS VAUX

L1 32

R L 2VAUX

+




Load Side Commutation In load side commutation the discharging and

recharging of capacitor takes place through the

load. ence to test the commutation circuit theload has to be connected. Examples of load

side commutation are R esonant Pulse

Commutation and Impulse Commutation.




Line Side Commutation In this type of commutation the discharging

and recharging of capacitor takes place through

the supply.




L

OAD

FWD

Lr

C

T3

IL

L

T2

VS

T1

+

_

+

_




Thyristor T2 is fired to charge the capacitor µC¶.

When µC¶ charges to a oltage of 2V, T2 is self

commutated. To re erse the oltage of capacitor to -2V, thyristor T3 is fired and T3

commutates by itself. Assuming that T1 is

conducting and carries a load current ILthyristor T2 is fired to turn off T1.



The turning ON of T2 will result in forward

biasing the diode (FWD) and applying a

re erse oltage of 2V acrossT1. This turns off T1, thus the discharging and recharging of

capacitor is done through the supply and the

commutation circuit can be tested without load.

36407294 Thyristor Commutation Techniques 100403040622 Phpapp02

Documents

Transcript of 36407294 Thyristor Commutation Techniques 100403040622 Phpapp02