thyristor commutation techniques

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

Thyristor Commutation Techniques


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

+

v ov sR


t

t

t

t

S u p p ly v o lta g e v s S in u s o id a l

Vo l ta g e a c r o s s S C R

L o a d v o lta g e v o

Tu rn o ffo c c u r s h e re

0

0

2

2

3

3

tc

G a te P u ls e


• Natural Commutation of Thyristors takes place in

– AC voltage controllers.

– Phase controlled rectifiers.

– Cyclo converters.


Forced Commutation

• Applied to dc circuits

• Commutation achieved by reverse biasing the SCR or by reducing the SCR current below holding current value.

• Commutating elements such as inductance and capacitance are used for commutation purpose.


Methods of Forced Commutation

• Self commutation.

• Resonant pulse commutation.

• Complementary commutation.

• Impulse commutation.

• External pulse commutation.

• Load Commutation.

• Line Commutation.


Forced Commutationis applied to

• Choppers.

• Inverters.


Self Commutation Or

Load Commutation Or

Class A Commutation (Commutation By Resonating

The Load)


• Circuit is under damped by including suitable values of L & C in series with load.

• Oscillating current flows.

• SCR is turned off when current is zero.


V

R L V (0)c

C

T i

L oad

+ -


Expression for Current

Fig. shows a transformed network

VS

R SL1

C S

V C (0 )S

C

T I(S) + +- -


2

2

1

0

1

0

1

0C

C

CCS V V

SI SRC

V V

SI SR SL

CS

C

S S LC

V V

RLC S S

L LC


2

2 22

0

1

0

12 2

C

C

V V

LI SR

S SL LC

V V

LI SR R R

S SL LC L L


2 2 22 2

2

0

12 2

Where,

0 1, ,

2 2

C

C

V VALI S

SR RS

L LC L

V V R RA

L L LC L


2 2

is called the natural frequency

Taking inverse Laplace transforms

sint

AI S

S

Ai t e t


2

Expression for current

0sin

Peak value of current

0

RtC L

C

V Vi t e t

L

V V

L


Expression For Voltage Across Capacitor At The Time Of Turn Off

V

R L V (0)c

C

T i

L oad

+ -


Applying KVL to figure

Substituting for i,

sin sin

c R L

c

t tc

v t V v V

div t V iR L

dt

A d Av t V R e t L e t

dt


sin cos sin

sin cos sin

sin cos sin2

sin cos2

t t tc

tc

tc

tc

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


Substituting for A,

0sin cos

2

0sin cos

2

SCR turns off when current goes to zero.

i.e., at

C tc

C tc

V V Rv t V e t L t

L

V V Rv t V e t t

L

t


2

Therefore at turn off

00 cos

0

0

C

c

c C

R

Lc C

V Vv t V e

v t V V V e

v t V V V e


2

For effective commutation

the circuit should be under damped.

1That is

2

With R = 0, and the capacitor initially uncharged

that is 0 0

sin

Note:

C

R

L LC

V

V ti t

L LC


1But

sin sin

and capacitor voltage at turn off is equal to 2V

Fig. shows the waveforms for the above conditions.

Once the SCR turns off voltage across it is

negative voltage.

LC

V t C ti t LC V

L LLC LC


C u r r e n t i

C a p a c i to r v o l t a g e

G a te p u l s e

Vo l ta g e a c r o s s S C R

0 / 2t

t

t

t

V

V

2 V

Conduction time of SCR


Problem

• Calculate the conduction time of SCR and the peak SCR current that flows in the circuit employing series resonant commutation (self commutation or class A commutation), if the supply voltage is 300 V, C = 1F, L = 5 mH and RL = 100 . Assume that the circuit is initially relaxed.


V = 300V

R LL

1 F100 5 m H

CT


2

2

3 6 3

1

2

1 100

5 10 1 10 2 5 10

10,000 rad/sec

Since the circuit is initially relaxed, initial voltage

across capacitor is zero as also the initial current

through and the expre

LR

LC L

L

ssion for current isi


3

sin , where 2

peak value of current

3006

10000 5 10

Conducting time of SCR = 0.314msec10000

t

P

V Ri t e t

L LV

L

I A


Problem

• Figure shows a self commutating circuit. The inductance carries an initial current of 200 A and the initial voltage across the capacitor is V, the supply voltage. Determine the conduction time of the SCR and the capacitor voltage at turn off.


V= 100V

L

50 F

10 HC

T

+

i(t)

I O

V C (0 )= V


The transformed circuit of the previous figure is shown in figure below

SL

VS

V (0)S

C

I LO

+

+

+

-

-

-

1C S

I(S)


2 2

The governing equation is

0 1

0

1

0

1 1

CO

CO

C

O

VVI S SL I L I S

S S CSVV

I LS SI S

SLCS

VVCS

S S I LCSI S

S LC S LC


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 LCSI S

LC S LC SLC LC

V V SII S

SL S

V V SII S

SL S LC


0

0

Taking inverse LT

0 sin cos

The capacitor voltage is given by

10

10 sin cos 0

C O

t

c C

t

c C O C

Ci t V V t I t

L

v t i t dt VC

Cv t V V t I t dt V

C L


01cos sin 0

011 cos sin 0

1sin 0 1 cos 0

sin cos 0 0 cos 0

sin 0 cos

C Oc C

C Oc C

Oc C C

c O C C C

c O C

V V t tICv t t t V

o oC L

V V ICv t t t V

C L

I Cv 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


we get

cos &

sin

C

O

c O

V O V

i t I t

Lv t I t V

C

I0i(t)

/2

/2

t

t

v c(t)

V


6 6

6

Turn off occurs at a time ' ' so that

20.5

0.5

0.5 10 10 50 10

0.5 10 500 35.1 seconds

O

O

O

O

O

t

t

t LC

t

t


60

6

and the capacitor voltage at turn off is given by

sin

10 10200 sin 90 100

50 10

89.4 100 189.4

c O O O

c O

c O

Lv t I t V

C

v t

v t V


• In the circuit shown in figure V = 600 volts, initial capacitor voltage is zero, L = 20 H, C=50F and the current through the inductance at the time of SCR triggering is IO = 350 A. Determine – Peak values of capacitor voltage and current– Conduction time of T1.

Problem


V

L

i(t) I0

C

T 1


The expression for is given by

sin cos

It is given that the initial voltage across

the capacitor O is zero.

sin cos

C O

C

O

i t

Ci t V V O t I t

L

V

Ci t V t I t

L


2 2

1

2 2

can be written as

sin

1where tan and

The peak capacitor current is

O

O

O

i t

Ci t I V t

L

LI

CV LC

CI V

L


62 2

6

Substituting the values, the peak capacitor current

50 10350 600 1011.19

20 10The expression for capacitor voltage 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


2 2 1

2 2

62 2

6

This can be rewritten as

sin ; Where tan

The peak value of capacitor voltage is

Substituting the values, the peak value of capacitor voltage

20 10600 350

50 10

c OO

O

CV

L Lv t V I t VC I

LV I V

C

600 639.5 600 1239.5V


To calculate conduction time of T1

• The waveform of capacitor current is shown in figure.When the capacitor current becomes zero the SCR turns off.

t

C apacito rcurrent

0


1

1

conduction time of SCR

tan

1

Substituting the values tan

O

O

LI

CV

LC

LI

CV


61

6

0

6 6

350 20 10tan

600 50 10

20.25 i.e., 0.3534 radians

1 131622.8 rad/sec

20 10 50 10 conduction time of SCR

0.353488.17 sec

31622.8

LC


Resonant Pulse Commutation (Class B Commutation)

L

C

VLoad

FW D

ab

iT

IL


• Series LC circuit connected across thyristor ‘T’.

• Initially ‘C’ is charged to ‘V’ volts with plate ‘a’ as positive.

• Current in LC oscillates when SCR is ON.• ‘T’ turns off when capacitor discharges through

thyristor in a direction opposite to IL


G a te p u ls eo f S C R

C a p a c ito r v o lta g ev ab

t1

V

t

t

t

I p i

IL

tC

t


t

t

Vo lta g e a c r o s sS C R

ISC RT h y r isto r C u rre n t


Expression For tC , the Circuit Turn Off Time

Assume that at the time of turn off of the SCR

the capacitor voltage and a constant

load current flows.

is the time taken for the capacitor

voltage to reach 0 volts from volts and

it i

ab

L

c

v V

I

t

V

s derived as follows


0

1

seconds

For proper commutation should be greater than .

Magnitude of the peak value of should be

greater than the load current .

Expression for is derived as foll

ct

L cL

cL

c q

p

L

I tV I dt

C C

VCt

I

t t

I i

I

i

ows


The LC Circuit During The Commutation Period

LC Circuit Transformed Circuit

i

L

C

T

+

V C(0 )=V

I(S )

SL

T 1C S

VS

+

-


2

22

1

111

1

VSI S

SLCs

VCI S

LC SL

VCS

sS LC

V

LLC

SC


2

2

11

1 1

1

1

Taking inverse LT sin

V LCI SL S

LC LC

C LCI S VL S

LC

Ci t V t

L


1Where

sin sin

amps

p

p

LCV

or i t t I tL

CI V

L


Expression For Conduction Time Of SCR

1


sin L

p

t

II


Alternate CircuitFor Resonant Pulse Commutation

VLOADFW D

L

T 1 IL

T 3

T 2

iC (t)

iC (t)

V C (0 )

a b

+

C


• Initially C charged with polarity as shown in figure.

• T1 is conducting & IL is constant.

• To turn off T1, T2 is fired.

• iC(t) flows opposite to IL & T1 turns off at iC(t) = IL


sin

Where 0 & and the capacitor voltage

1.

10 sin .

0 cos

c p

p C

c C

c C

c C

i t I t

CI V

L

v t i t dtC

Cv t V t dt

C L

v t V t


1

1

Instant at

which the thyristor turns off is at

& sin

0

L p

p C

t

tI I

LC

CI V

L


11

1 1 1

or sin0

& the corresponding capacitor voltage is

0 cos

L

C

c C

I Lt LC

V C

v t V V t


Expression For tC

1

1

Assuming a constant load

current which charges the capacitor

seconds

Normally 0

For reliable commutation

depends on & becomes smaller for

higher values of

L

cL

C

c q

C L

L

I

CVt

I

V V

t t

t I

I


t

t

tC

t1

V 1

V

V C (0 )

C a p a c ito rv o lta g e v ab

C u r re n t iC (t)


Resonant Pulse CommutationWith Accelerating Diode


VLOADFW D

LCT 1

IL

T 3

T 2iC (t)

V C (0 )+-

D 2 iC (t)


• Diode D2 connected as shown to accelerate discharge.

• T2 turned on to turn off T1.

• Once T1 is off at t1. iC(t) flows through D2 until current reduces to IL at time t2.

• From t = t2 , ‘C’ charges through load, T2 self commutates.

• But thyristor recovery process low hence longer reverse bias time.


I L

0

V C

0

V 1V (O )C

iC

t

tt1 t2

tC


Problem

• The circuit in figure shows a resonant pulse commutation circuit. The initial capacitor voltage VC(O)=200V, C = 30F and L = 3H. Determine the circuit turn off time tC, if the load current IL is 200 A and 50 A.


VLOADFW D

LCT 1

IL

T 3

T 2iC (t)

V C(0 )+


2

1

1

11

66 6 1

1 6

When 200 ; Let be triggered at 0

Capacitor current reaches a value at

when turns off

sin0

200 3 103 10 30 10 sin 3.05 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

6

6 60.105 10 / sec

30 10rad


1

1 1

6 61

1

1

1

6

At the magnitude of capacitor voltage is,

0 cos

i.e., 200cos 0.105 10 3.05 10

200 0.9487

189.75 Volts

&

30 10 189.7528.46 sec

200

C

cL

c

t t

V V t

V

V

V

CVt

I

t


66 6 1

1 6

6 61

61

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 observed that as load increases the value of re

L

cL

c

I A

t

V

CVt

I

t

duces.


2

1

Repeat above problem

For 200 if an antiparallel diode is

connected across thyristor as shown in figureLI A D

T

Problem


VLOADFW D

LC

T 1IL

T 3

T 2iC (t)

V C (0 )+-

D 2 iC (t)


2

1

1

11

66 6 1

1 6

1

200

Let be triggered at 0

Capacitor current reaches the value at ,

when turns off

sin

200 3 103 10 30 10 sin

200 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


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


2 1

2

2 1

6 6 62

2

6 6

6

flows through diode after turns off.

current falls 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 D T

i t I t

t LC t

t

t

LC


2

6 62 2

2 2

6 62 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


Problem• For the circuit shown in figure calculate the

value of L for proper commutation of SCR. Also find the conduction time of SCR.

V= 30V L

i

4 F

R L

IL30


The load current

301 Amp

30

For proper SCR commutation ,

the peak value of resonant current i, should be

greater than

Let 2 , 2 Amps

LL

p

L

p L p

VI

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 CI V

L LLLC

L mHL

LC


1 1


1sin sin2

16666 166660.523

radians 0.00022 seconds16666

0.22 msec

L

p

II


Problem

• For the circuit shown in figure given that the load current to be commutated is 10 A, turn off time required is 40sec and the supply voltage is 100 V. Obtain the proper values of commutating elements.


V= 100V L i IL

IL

C


peak value of & this should be greater than

Let 1.5

1.5 10 100 ...

Also, assuming that at the time of turn off the capacitor

voltage is approximately equal to V, and the load curren

p L

p L

CI i V I

LI I

Ca

L

t

linearly charges the capacitor

secondscL

CVt

I


6

6

4 62 2

4

& this is given to be 40 sec.

10040 10

104

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


Problem

• In a resonant commutation circuit supply voltage is 200 V. Load current is 10 A and the device turn off time is 20s. The ratio of peak resonant current to load current is 1.5. Determine the value of L and C of the commutation circuit.


Given 1.5

1.5 1.5 10 15

. ., 15 ...

It is given that the device 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 cL

I

I

I I A

Ci e I V A a

L

t

CVt t

I


6

6

62 2

20030 10

101.5

Substituting in (a)

1.5 1015 200

1.5 1015 200

0.2666 mH

C

C F

L

LL


Complementary Commutation (Class C Commutation,

Parallel Capacitor Commutation)


V

R 1 R 2

T 1 T 2

IL

iC

C

a b


• Two SCRs are used, turning ON one SCR turns off the other.

• T1 is fired, IL flows through R1.

• At same time ‘C’ charges towards ‘V’ through R2 with plate ‘b’ positive.

• To turn off T1, T2 is fired resulting in capacitor voltage reverse biasing T1 and turning it off.

• When T2 is fired current through load shoots up as voltage across load is V+VC


G a te p u ls eo f T 1


C u rr e n t th ro u g h R 1

I LV

t

t

t

t

C u rr e n t th ro u g h T 1

C u rr e n t th ro u g h T 2

2

2

VR

2

1

V

R

V

R 1

VR 2

V

R1

2

1

V

R


t

t

t

Vo lta g e a c r o s sc a p a c ito r v a b

Vo lta g e a c r o s s T 1

C u r re n t th r o u g h R 2

tC tC

tC

V

- V


Expression For Circuit Turn Off Time tC

1 1

1

Where is the final voltage, is the initial

voltage and is the time constant.

At , 0, , C, &

0 2c c

tc 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 Ve


1 1

1

1

1

2 2

2 ; 0.5

Taking natural logarithms on both sides

ln 0.5

0.693

This time should be greater than the turn off time of

Similarly when is commutated 0.693

And this

c ct t

R C R C

c

c

q

c

V Ve e

t

R C

t R C

t T

T t R C

2

1 2

time should be greater than of

Usually

qt T

R R R


• In the circuit shown in figure, the load resistances R1 = R2 = R = 5 & the capacitance C = 7.5 F, V = 100 volts. Determine the circuit turn off time tC

Problem


V

R 1 R 2

T 1 T 2

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


• Calculate the values of RL and C to be used for commutating the main SCR in the circuit shown in figure. When it is conducting a full load current of 25 A flows. The minimum time for which the SCR has to be reverse biased for proper commutation is 40sec. Also find R1, given that the auxiliary SCR will undergo natural commutation when its forward current falls below the holding current value of 2 mA.

Problem


V= 100V

R 1 R L

M ainSC R

A uxiliarySC R

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 the auxiliary SCR is the sum of the capacitor charging current iC and the current i1 through R1, iC reduces to zero after a time tC and hence the auxiliary SCR turns off automatically after a time tC, i1 should be less than the holding current.


6

6

Given 25

100. ., 25

4

40 sec 0.693

. ., 40 10 0.693 4

40 10

4 0.69314.43

L

L L

L

c L

I A

Vi e A

R R

R

t R C

i e C

C

C F


11

1

1 3

1

should be less

than the holding current of auxiliary SCR

100 should be 2

100

2 10. ., 50

Vi

R

mAR

R

i e R K


Impulse Commutation(Class D Commutation)

VLOADFW D

C

T 1

T 3

IL

T 2

V C (O )+

L


• ‘C’ charged to a voltage VC(O) with polarity as shown.

• T1 is conducting and carries load current IL.• To turn off T1, T2 is fired.• Capacitor voltage reverse biases T1 and turns it

off.• ‘C’ Charges through load.• T2 self commutates.• To reverse capacitor voltage T3 is turned ON.





C a p a c ito rv o lta g e


V S

t

t

t

tC

V C

-V C


Expression For Circuit Turn Off Time tC

0

1

depends on & is given by the expression

1

(assuming the load current to be constant)

seconds

For proper commutation , turn off time of

c

c L

t

C L

L c CC c

L

c q

t I

V I dtC

I t V CV t

C I

t t T


T1 is turned off by applying a negative voltage across its terminals. Hence this is voltage commutation.

tC depends on load current. For higher load currents tC is small. This is a disadvantage of this circuit.

When T2 is fired, voltage 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

C

D

T 1

IT 1

IL

i

T 2

L

R L

V C (O )+

_


• Initially ‘C’ is charged to VC with top plate positive.

• T1 is fired, load current IL flows.

• ‘C’ discharges at the same time & reverses its polarity.

• ‘D’ ensures bottom plate remains positive.

• To turn off T1, T2 is fired.


G a te p u l s eo f T 1

G a te p u l s eo f T 2

C u r r e n t t h r o u g h S C R

T h is i s d u e to i

C a p a c i to rv o l ta g e

t

t

t

t C

V C

I L

IT 1

- V

VR L


L o a d c u r re n t


t

t

tC

IL

V

2 VR L


Problem

• An impulse commutated thyristor circuit is shown in figure. Determine the available turn off time of the circuit if V = 100 V, R = 10 and C = 10 F. Voltage across capacitor before T2 is fired is V volts with polarity as shown.


C

T 1

T 2

V (0)C

V +

+

-

-

R


When T2 is triggered the circuit is as shown in figure

C

i(t)

T 2V

++ -

-

R

V (O )C


Writing the transformcircuit, we obtain

Vs

V (0)s

C

+

+

I(s)

1C s

R


Expression for capacitor voltage

is obtained as follows,

10

1

0

1

0

1

C

C

C

V VSI S

RCS

C V VI S

RCS

V VI S

R SRC


Voltage across capacitor

01

0 011

0 0 0

1

CC

C CC

C C CC

VV s I S

CS SV V V

V sRCS SS

RC

V V V V VV S

S SSRC


01 1

1 0

In the given problem 0

1 2

CC

t tRC RC

c C

C

tRC

c

VV VV S

S S SRC RC

v t V e V e

V V

v t V e


The waveform of vC(t) is shown in figure

t

tC

V

V (0)C

v (t)C


At , 0

0 1 2

1 2

1

2

c

c

c

c c

tRC

tRC

tRC

t t v t

V e

e

e


6

Taking natural logarithms

1log

2

ln 2

10 10 10 ln 2

69.3 sec

ce

c

c

c

t

RC

t RC

t

t


Problem

• In the commutation circuit shown in figure C = 20 F, the input voltage V varies between 180 and 220 V and the load current varies between 50 and 200 A. Determine the minimum and maximum values of available turn off time tC.


C

T 1

T 2 I0

I0

V (0)= VC+

V


It is given that varies

between 180 and 220 V &

varies between 50 and 200 A

The expression for available

turn off time is given by

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

O

c

cO

c O

V

I

t

CVt

I

t I


maxmax

min

6max

minmin

max

6min

22020 10 88 sec

50

18020 10 18 sec

200

cO

c

cO

c

CVt

I

t

CVAnd t

I

t


External Pulse Commutation (Class E Commutation)

V S V A U X

L

C

T 1 T 3T 2

R L 2V A U X

+


• T1 is conducting & RL is connected across supply.• T3 is fired & ‘C’ is charged to 2VAUX with upper

plate positive.• T3 is self commutated.• To turn off T1, T2 is fired.• T2 ON results in a reverse voltage VS – 2VAUX

appearing across T1.


Load Side Commutation

• In load side commutation the discharging and recharging of capacitor takes place through the load. Hence to test the commutation circuit the load has to be connected. Examples of load side commutation are Resonant Pulse Commutation and Impulse Commutation.


Line Side Commutation

• In this type of commutation the discharging and recharging of capacitor takes place through the supply.


LOAD

FW D

L r

CT 3

IL

L

T 2

V S

T 1

+

_

+

_


• Thyristor T2 is fired to charge the capacitor ‘C’. When ‘C’ charges to a voltage of 2V, T2 is self commutated. To reverse the voltage of capacitor to -2V, thyristor T3 is fired and T3 commutates by itself. Assuming that T1 is conducting and carries a load current IL thyristor T2 is fired to turn off T1.


• The turning ON of T2 will result in forward biasing the diode (FWD) and applying a reverse voltage 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.

thyristor commutation techniques

Education

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commutation purpose

complementary commutation

impulse commutation

line commutation

commutation period lc

external pulse commutation

dc circuits commutation