Thyristor Commutations

THYRISTOR COMMUTATION TECHNIQUES INTRODUCTION In practice it becomes necessary to turn off a conducting thyristor. (Often thyristors are used as switches to turn on and off power to the load). The process of turning off a conducting thyristor is called commutation. The principle involved is that either the anode should be made negative with respect to cathode (voltage commutation) or the anode current should be reduced below the holding current value (current commutation). The reverse voltage must be maintained for a time at least equal to the turn-off time of SCR otherwise a reapplication of a positive voltage will cause the thyristor to conduct even without a gate signal. On similar lines the anode current should be held at a value less than the holding current at least for a time equal to turn-off time otherwise the SCR will start conducting if the current in the circuit increases beyond the holding current level even without a gate signal. Commutation circuits have been developed to hasten the turn-off process of Thyristors. The study of commutation techniques helps in understanding the transient phenomena under switching conditions. The reverse voltage or the small anode current condition must be maintained for a time at least equal to the TURN OFF time of SCR; Otherwise the SCR may again start conducting. The techniques to turn off a SCR can be broadly classified as • Natural Commutation • Forced Commutation. NATURAL COMMUTATION (CLASS F) This type of commutation takes place when supply voltage is AC, because a negative voltage will appear across the SCR in the negative half cycle of the supply voltage and the SCR turns off by itself. Hence no special circuits are required to turn off the SCR. That is the reason that this type of commutation is called Natural or Line Commutation. Figure 1.1 shows the circuit where natural commutation takes place and figure 1.2 shows the related waveforms. tc is the time offered by the circuit within which the SCR should turn off completely. Thus tc should be greater than tq , the turn off time of the SCR. Otherwise, the SCR will become forward biased before it has turned off completely and will start conducting even without a gate signal.

T +

vs

~

↑

R

↑ vo −

Fig. 1.1: Circuit for Natural Commutation

Supply voltage vs

Sinusoidal

π

3π

0

2π

ωt

ωt

α Load voltage vo Turn off occurs here

ωt

π

3π

0

2π

ωt

Voltage across SCR tc Fig. 1.2: Natural Commutation – Waveforms of Supply and Load Voltages (Resistive Load) This type of commutation is applied in ac voltage controllers, phase controlled rectifiers and cyclo converters. FORCED COMMUTATION When supply is DC, natural commutation is not possible because the polarity of the supply remains unchanged. Hence special methods must be used to reduce the SCR current below the holding value or to apply a negative voltage across the SCR for a time interval greater than the turn off time of the SCR. This technique is called FORCED COMMUTATION and is applied in all circuits where the supply voltage is DC - namely, Choppers (fixed DC to variable DC), inverters (DC to AC). Forced commutation techniques are as follows: • Self Commutation • Resonant Pulse Commutation • Complementary Commutation • Impulse Commutation • External Pulse Commutation. • Line Side Commutation.

SELF COMMUTATION OR LOAD COMMUTATION OR CLASS A COMMUTATION: (COMMUTATION BY RESONATING THE LOAD) In this type of commutation the current through the SCR is reduced below the holding current value by resonating the load. i.e., the load circuit is so designed that even though the supply voltage is positive, an oscillating current tends to flow and when the current through the SCR reaches zero, the device turns off. This is done by including an inductance and a capacitor in series with the load and keeping the circuit under-damped. Figure 1.3 shows the circuit. This type of commutation is used in Series Inverter Circuit.

T

i

L

R

Vc(0) + -

Load

C

V

Fig. 1.3: Circuit for Self Commutation

EXPRESSION FOR CURRENT At t = 0 , when the SCR turns ON on the application of gate pulse assume the current in the circuit is zero and the capacitor voltage is VC ( 0 ) . Writing the Laplace Transformation circuit of figure 1.3 the following circuit is obtained when the SCR is conducting.

T

R I(S)

sL

1 VC(0) S CS + - + C

V S

Fig.: 1.4.

I (S )

V − VC ( 0 )  S = 1 R + sL + CS

CS V − VC ( 0 )  S = RCs + s 2 LC + 1 =

C V − VC ( 0 )  R 1   LC  s 2 + s + L LC  

V − VC ( 0 ) L = R 1 2 s +s + L LC

(V − V ( 0 ) ) C

=

L 2 2 R 1  R   R  2 s +s + +  −  L LC  2 L   2 L 

(V − V ( 0 ) ) C

=

L 2 R   1  R     s + + −      2 L   LC  2 L      2

=

Where

A 2 (s +δ ) +ω2

,

(V − V ( 0 ) ) , A= C

L

R δ= , 2L

ω is called the natural frequency I (S ) =

A

2

ω

ω ( s + δ )2 + ω 2

ω=

1  R  −  LC  2 L 

2

Taking inverse Laplace transforms i (t ) =

A

ω

e −δ t sin ω t

Therefore expression for current V − VC ( 0 ) −2 RL t i (t ) = e sin ω t ωL Peak value of current =

(V − V ( 0 ) ) C

ωL

Expression for voltage across capacitor at the time of turn off Applying KVL to figure 1.3

vc = V − vR − VL vc = V − iR − L

di dt

Substituting for i, vc = V − R

vc = V − R vc = V −

A

ω

A

ω A

ω

e −δ t sin ω t − L

d  A −δ t   e sin ω t  dt  ω 

e −δ t sin ω t − L

(e ω A

−δ t

ω cos ω t − δ e −δ t sin ω t )

e −δ t [ R sin ω t + ω L cos ω t − Lδ sin ω t ]

vc = V −

R   e −δ t  R sin ω t + ω L cos ω t − L sin ω t  ω 2L  

vc = V −

R  e −δ t  sin ω t + ω L cos ω t  ω 2 

A

A

Substituting for A,

vc ( t ) = V −

(V − V ( 0 ) ) e δ

vc ( t ) = V −

(V − V ( 0 ) ) e δ

C

ωL C

ω

− t

− t

R   2 sin ω t + ω L cos ω t  R   2 L sin ω t + ω cos ω t 

SCR turns off when current goes to zero. i.e., at ω t = π . Therefore at turn off

vc = V −

(V − V ( 0 ) ) e ωδπ −

C

ω

vc = V + V − VC ( 0 )  e vc = V + V − VC ( 0 )  e

Therefore

( 0 + ω cos π )

−δπ

ω

− Rπ 2 Lω

Note: For effective commutation the circuit should be under damped. 2

1  R    < LC  2L 

That is •

With R = 0, and the capacitor initially uncharged that is VC ( 0 ) = 0

i=

But

ω=

Therefore

V t sin ωL LC

1 LC

i=

V t C t LC sin =V sin L L LC LC

and capacitor voltage at turn off is equal to 2V. • •

Figure 1.5 shows the waveforms for the above conditions. Once the SCR turns off voltage across it is negative voltage. π Conduction time of SCR = . ω

V

C L

Current i

π/2

0

ωt

π

2V Capacitor voltage

V

ωt

Gate pulse ωt

ωt −V Voltage across SCR Fig. 1.5: Self Commutation – Wave forms of Current and Capacitors Voltage Problem 1.1 : 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 = 1µF, L = 5 mH and RL = 100 Ω. Assume that the circuit is initially relaxed.

T

RL

L

C + −

100 Ω

5 mH

1 µF

V =300V

Fig. 1.6.

Solution:

ω=

1  RL  −  LC  2 L 

2

1  100  ω= − −3 −6 −3  5 ×10 × 1× 10  2 × 5 × 10 

2

ω = 10, 000 rad/sec Since the circuit is initially relaxed, initial voltage across capacitor is zero as also the initial current through L and the expression for current i is

Therefore peak value of

Conducting time of SCR

i=

V −δ t R , e sin ω t , where δ = ωL 2L

i=

V ωL

i=

300 = 6A 10000 × 5 ×10−3

=

π π = = 0.314msec ω 10000

Problem 1.2 : Figure 1.7 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.

L

T

i(t)

→ IO 10 µH C 50 µF

V =100V

Fig. 1.7.

+ − VC(0)=V

Solution : The transformed circuit of figure 1.7 is shown in figure 1.8.

sL

−

IOL +

I(S) +

+ −

V S −

VC(0) =V S 1 CS

Fig.1.8: Transformed Circuit of Fig. 1.7 The governing equation is V ( 0) V 1 = I ( S ) sL − I O L + C + I (S ) s s Cs

Therefore

V VC ( 0 ) − + IO L s s I (S ) = 1 sL + Cs

V VC ( 0 )   −  Cs s s  + I O LCs I (S ) =  2 s LC + 1 s 2 LC + 1 V − VC ( 0 )  C I O LCs I (S ) =  + 1  1    LC  s 2 + LC  s 2 +  LC  LC   

I (S ) =

V − VC ( 0 ) L  s + ω  2

2

+

sI O s +ω2 2

V − VC ( 0 )  ω sI I (S ) =  + 2 O 2 where ω = 2 2 ω L  s + ω  s + ω Taking inverse LT

i ( t ) = V − VC ( 0 ) 

C sin ω t + I O cos ω t L

1 LC

The capacitor voltage is given by t

1 vc ( t ) = ∫ i ( t ) dt + VC ( 0 ) C0 t  1  C vc ( t ) = ∫  V − VC ( 0 )  sin ω t + I O cos ω t  dt + VC ( 0 ) C 0  L 

vc ( t ) =

 t I t 1  (V − VC ( 0 ) ) C ( − cos ω t ) + O ( sin ω t ) + VC ( 0 )   o ω o C  L ω 

vc ( t ) =

 I 1  (V − VC ( 0 ) ) C (1 − cos ω t ) + O ( sin ω t ) + VC ( 0 )  C  L ω ω 

vc ( t ) =

IO 1 C × LC sin ω t + (V − VC ( 0 ) ) LC (1 − cos ω t ) + VC ( 0 ) C C L

vc ( t ) = I O

L sin ω t + V − V cos ω t − VC ( 0 ) + VC ( 0 ) cos ω t + VC ( 0 ) C

vc ( t ) = I O

L sin ω t − (V − VC ( 0 ) ) cos ω t + V C

In this problem VC ( 0 ) = V Therefore we get,

i ( t ) = I O cos ω t and

vc ( t ) = I O

L sin ω t + V C

The waveforms are as shown in figure 1.9

I0

i(t)

π/2

ωt

vc(t)

V

π/2 Fig.: 1.9 Turn off occurs at a time to so that ω tO =

Therefore

tO =

0.5π

ω

π 2

= 0.5π LC

tO = 0.5 × π 10 ×10−6 × 50 ×10−6 tO = 0.5 × π × 10−6 500 = 35.1µ seconds and the capacitor voltage at turn off is given by

vc ( tO ) = I O vc ( tO ) = 200

L sin ω tO + V C 10 × 10−6 sin 900 + 100 −6 50 × 10

 35.12  vc ( tO ) = 200 × 0.447 × sin   + 100  22.36  vc ( tO ) = 89.4 + 100 = 189.4 V

ωt

Problem 1.3: In the circuit shown in figure 1.10. V = 600 volts, initial capacitor voltage is zero, L = 20 µH, C = 50µF and the current through the inductance at the time of SCR triggering is Io = 350 A. Determine (a) the peak values of capacitor voltage and current (b) the conduction time of T1. T1 L

→ I0 V

i(t) C

Fig. 1.10 Solution: (Refer to problem 1.2). The expression for i ( t ) is given by

i ( t ) = V − VC ( 0 ) 

C sin ω t + I O cos ω t L

It is given that the initial voltage across the capacitor, VC ( O ) is zero. Therefore

i (t ) = V

C sin ω t + I O cos ω t L

i ( t ) can be written as

i ( t ) = I O2 + V 2

where

α = tan −1

and

ω=

C sin (ω t + α ) L

L C

IO V

1 LC

The peak capacitor current is

I O2 + V 2

C L

Substituting the values, the peak capacitor current

50 × 10−6 = 1011.19 A 20 ×10−6

= 3502 + 6002 ×

The expression for capacitor voltage is

vc ( t ) = I O

with

L sin ω t − (V − VC ( 0 ) ) cos ω t + V C

VC ( 0 ) = 0, vc ( t ) = I O

L sin ω t − V cos ω t + V C

This can be rewritten as

vc ( t ) = V 2 + I O2

Where

β = tan −1

V

L sin (ω t − β ) + V C

C L IO

The peak value of capacitor voltage is

= V 2 + I O2

L +V C

Substituting the values, the peak value of capacitor voltage 20 × 10 −6 = 600 + 350 × + 600 50 × 10−6 2

2

= 639.5 + 600 = 1239.5V

To calculate conduction time of T1 The waveform of capacitor current is shown in figure 1.11. When the capacitor current becomes zero the SCR turns off.

Capacitor current

α

ωt

0 π−α

Fig. 1.11. π −α Therefore conduction time of SCR = ω  L  IO C π − tan −1   V   = 1 LC

     

Substituting the values

 L  IO C α = tan −1   V  

     

350 20 × 10−6 α = tan 600 50 ×10−6 −1

α = 20.250 i.e., 0.3534 radians ω=

1 1 = = 31622.8 rad/sec LC 20 ×10 −6 × 50 × 10−6

Therefore conduction time of SCR

=

π − 0.3534 31622.8

= 88.17 µ sec

RESONANT PULSE COMMUTATION (CLASS B COMMUTATION) The circuit for resonant pulse commutation is shown in figure 1.12.

L T

i a b

C

IL

V

Load FWD Fig. 1.12: Circuit for Resonant Pulse Commutation This is a type of commutation in which a LC series circuit is connected across the SCR. Since the commutation circuit has negligible resistance it is always under-damped i.e., the current in LC circuit tends to oscillate whenever the SCR is on. Initially the SCR is off and the capacitor is charged to V volts with plate ‘a’ being positive. Referring to figure 1.13 at t = t1 the SCR is turned ON by giving a gate pulse. A current I L flows through the load and this is assumed to be constant. At the same time SCR short circuits the LC combination which starts oscillating. A current ‘i’ starts flowing in the direction shown in figure. As ‘i’ reaches its maximum value, the capacitor voltage reduces to zero and then the polarity of the capacitor voltage reverses ‘b’ becomes positive). When ‘i’ falls to zero this reverse voltage becomes maximum, and then direction of ‘i’ reverses i.e., through SCR the load current I L and ‘i’ flow in opposite direction. When the instantaneous value of ‘i’ becomes equal to I L , the SCR current becomes zero and the SCR turns off. Now the capacitor starts charging and its voltage reaches the supply voltage with plate a being positive. The related waveforms are shown in figure 1.13.

Gate pulse of SCR t

π

t1 V

Capacitor voltage vab t tC

Ip

i

IL

π ω

t ∆t

ISCR

t Voltage across SCR t

Fig. 1.13: Resonant Pulse Commutation – Various Waveforms

EXPRESSION FOR tc , THE CIRCUIT TURN OFF TIME Assume that at the time of turn off of the SCR the capacitor voltage vab ≈ −V and load current I L is constant. tc is the time taken for the capacitor voltage to reach 0 volts from – V volts and is derived as follows. t

V=

1 c I L dt C ∫0

V=

I L tc C

tc =

VC seconds IL

For proper commutation tc should be greater than tq , the turn off time of T. Also, the magnitude of I p , the peak value of i should be greater than the load current I L and the expression for i is derived as follows The LC circuit during the commutation period is shown in figure 1.14.

L T

i C

+ VC(0) − =V

Fig. 1.14. The transformed circuit is shown in figure 1.15.

I(S) sL T

1 Cs + V − s Fig. 1.15.

I (S ) =

V s sL +

1 Cs

V    Cs s I ( S ) = 2  s LC + 1 I (S ) =

VC 1   LC  s 2 +  LC  

I (S ) =

V 1 × L s2 + 1 LC

 1  V  LC  1 I (S ) = × × L s2 + 1  1  LC  LC 

 1  C  LC  I (S ) = V × L s2 + 1 LC Taking inverse LT

C sin ω t L

i (t ) = V

Where

ω=

1 LC

Or

i (t ) =

V sin ω t = I p sin ω t ωL

Therefore

Ip =V

C amps . L

EXPRESSION FOR CONDUCTION TIME OF SCR For figure 1.13 (waveform of i), the conduction time of SCR =

π + ∆t ω

I  sin −1  L   Ip  π   = +

ω

ω

ALTERNATE CIRCUIT FOR RESONANT PULSE COMMUTATION The working of the circuit can be explained as follows. The capacitor C is assumed to be charged to VC ( 0 ) with polarity as shown, T1 is conducting and the load current I L is a constant. To turn off T1 , T2 is triggered. L, C, T1 and T2 forms a resonant circuit. A resonant

current ic ( t ) flows in the direction shown, i.e., in a direction opposite to that of load current

IL . ic ( t ) = I p sin ω t (refer to the previous circuit description). Where I p = VC ( 0 ) and the capacitor voltage is given by

vc ( t ) =

1 iC ( t ).dt C∫

vc ( t ) =

1 C VC ( 0 ) sin ω t.dt . ∫ C L

vc ( t ) = −VC ( 0 ) cos ω t

C

ab

T1

iC(t)

L

iC(t)

IL T2

− + VC(0) V

T3 FWD

Fig. 1.16: Resonant Pulse Commutation – An Alternate Circuit

L O A D

C L

&

When ic ( t ) becomes equal to I L (the load current), the current through T1 becomes zero and T1 turns off. This happens at time t1 such that I L = I p sin

t1 LC

I p = VC ( 0 )

C L

 I L t1 = LC sin −1  L   VC ( 0 ) C  and the corresponding capacitor voltage is

vc ( t1 ) = −V1 = −VC ( 0 ) cos ω t1 Once the thyristor T1 turns off, the capacitor starts charging towards the supply voltage through T2 and load. As the capacitor charges through the load capacitor current is same as load current I L , which is constant. When the capacitor voltage reaches V, the supply voltage, the FWD starts conducting and the energy stored in L charges C to a still higher voltage. The triggering of T3 reverses the polarity of the capacitor voltage and the circuit is ready for another triggering of T1 . The waveforms are shown in figure 1.17.

EXPRESSION FOR tc Assuming a constant load current I L which charges the capacitor tc =

CV1 seconds IL

Normally V1 ≈ VC ( 0 ) For reliable commutation tc should be greater than tq , the turn off time of SCR T1 . It is to be noted that tc depends upon I L and becomes smaller for higher values of load current.

Current iC(t)

t

V

Capacitor voltage vab t

t1

V1 tC VC(0) Fig. 1.17: Resonant Pulse Commutation – Alternate Circuit – Various Waveforms RESONANT PULSE COMMUTATION WITH ACCELERATING DIODE

D2

C

T1 L

iC(t)

IL iC(t)

T2

+ VC(0) V

T3 FWD

Fig. 1.17(a)

L O A D

iC IL 0

t

VC 0

t1

V1 VC(O)

t2

t

tC Fig. 1.17(b)

A diode D2 is connected as shown in the figure 1.17(a) to accelerate the discharging

of the capacitor ‘C’. When thyristor T2 is fired a resonant current iC ( t ) flows through the

capacitor and thyristor T1 . At time t = t1 , the capacitor current iC ( t ) equals the load current

I L and hence current through T1 is reduced to zero resulting in turning off of T1 . Now the

capacitor current iC ( t ) continues to flow through the diode D2 until it reduces to load current

level I L at time t2 . Thus the presence of D2 has accelerated the discharge of capacitor ‘C’. Now the capacitor gets charged through the load and the charging current is constant. Once capacitor is fully charged T2 turns off by itself. But once current of thyristor T1 reduces to zero the reverse voltage appearing across T1 is the forward voltage drop of D2 which is very small. This makes the thyristor recovery process very slow and it becomes necessary to provide longer reverse bias time. From figure 1.17(b) t2 = π LC − t1 VC ( t2 ) = −VC ( O ) cos ω t2 Circuit turn-off time tC = t2 − t1

Problem 1.4 : The circuit in figure 1.18 shows a resonant pulse commutation circuit. The initial capacitor voltage VC ( O ) = 200V , C = 30µF and L = 3µH. Determine the circuit turn off time tc , if the load current I L is (a) 200 A and (b) 50 A.

T1 L

C

IL iC(t)

T2

− + VC(0)

L O A D

T3

V

FWD

Fig. 1.18. Solution (a) When I L = 200 A Let T2 be triggered at t = 0 .

The capacitor current ic ( t ) reaches a value I L at t = t1 , when T1 turns off  I L t1 = LC sin −1  L   VC ( 0 ) C 

 200 3 ×10−6 t1 = 3 ×10−6 × 30 ×10−6 sin −1   200 30 ×10−6 

   

t1 = 3.05µ sec .

ω=

1 1 = −6 LC 3 × 10 × 30 × 10−6

ω = 0.105 ×106 rad / sec . At t = t1 , the magnitude of capacitor voltage is V1 = VC ( 0 ) cos ω t1 That is

V1 = 200 cos 0.105 ×106 × 3.05 ×10−6

V1 = 200 × 0.9487 V1 = 189.75 Volts and

tc =

CV1 IL

tc = (b)

30 × 10−6 × 189.75 = 28.46 µ sec . 200

When I L = 50 A

 50 3 ×10−6 t1 = 3 ×10−6 × 30 ×10−6 sin −1   200 30 ×10−6  t1 = 0.749µ sec .

   

V1 = 200 cos 0.105 ×106 × 0.749 ×10−6

V1 = 200 ×1 = 200 Volts . tc =

CV1 IL

tc =

30 × 10−6 × 200 = 120 µ sec . 50

It is observed that as load current increases the value of tc reduces.

Problem 1.4a : Repeat the above problem for I L = 200 A , if an antiparallel diode D2 is connected across thyristor T1 as shown in figure 1.18a.

D2

C

T1 L

iC(t)

IL iC(t)

T2

- + VC(0) V

T3 FWD

Fig. 1.18(a)

L O A D

Solution

I L = 200 A Let

T2 be triggered at t = 0 .

Capacitor current iC ( t ) reaches the value I L at t = t1 , when T1 turns off Therefore

 I L t1 = LC sin −1  L  VC ( O ) C 

 200 3 ×10−6 t1 = 3 ×10−6 × 30 ×10−6 sin −1   200 30 ×10−6  `

t1 = 3.05µ sec .

ω=

1 1 = −6 LC 3 × 10 × 30 × 10−6

ω = 0.105 ×106 radians/sec At t = t1

VC ( t1 ) = V1 = −VC ( O ) cos ω t1

VC ( t1 ) = −200 cos ( 0.105 ×106 × 3.05 ×10−6 ) VC ( t1 ) = −189.75V iC ( t ) flows through diode D2 after T1 turns off. iC ( t ) current falls back to I L at t2 t2 = π LC − t1

t2 = π 3 ×10−6 × 30 ×10−6 − 3.05 ×10−6

t2 = 26.75µ sec .

ω=

1 1 = LC 3 × 10−6 × 30 × 10−6

ω = 0.105 ×106 rad/sec.

   

At t = t2

VC ( t2 ) = V2 = −200 cos 0.105 × 10+6 × 26.75 × 10−6 VC ( t2 ) = V2 = 189.02 V

Therefore

tC = t2 − t1 = 26.75 ×10−6 − 3.05 ×10−6

tC = 23.7 µ secs Problem 1.5: For the circuit shown in figure 1.19 calculate the value of L for proper commutation of SCR. Also find the conduction time of SCR.

4 µF V =30V

L RL

i

30 Ω

IL Fig. 1.19.

Solution: V 30 = = 1 Amp RL 30 For proper SCR commutation I p , the peak value of resonant current i, should be

The load current I L =

greater than I L , Let

I p = 2I L ,

Also

Ip =

Therefore

2 = 30 ×

Therefore

L = 0.9mH .

ω=

Therefore

V = ωL

I p = 2 Amps .

V C =V 1 L ×L LC

4 × 10−6 L

1 1 = = 16666 rad/sec −3 LC 0.9 × 10 × 4 × 10−6

I  sin −1  L   Ip  π   Conduction time of SCR = +

ω

ω

1 sin −1   π 2 = + 16666 16666 =

π + 0.523

radians

16666

= 0.00022 seconds = 0.22 msec

Problem 1.6: For the circuit shown in figure 1.20 given that the load current to be commutated is 10 A, turn off time required is 40µsec and the supply voltage is 100 V. Obtain the proper values of commutating elements.

C V =100V

L

i

IL IL

Fig. 1.20. Solution

I p peak value of i = V

C and this should be greater than I L . Let I p = 1.5I L . L

Therefore 1.5 ×10 = 100

C L

... ( a )

Also, assuming that at the time of turn off the capacitor voltage is approximately equal to V, (and referring to waveform of capacitor voltage in figure 1.13) and the load current linearly charges the capacitor tc =

CV seconds IL

and this tc is given to be 40 µsec. Therefore

40 × 10−6 = C ×

100 10

Therefore

C = 4µ F

Substituting this in equation (a) 4 × 10−6 1.5 × 10 = 100 L 10 4 × 4 × 10−6 1.5 × 10 = L 2

Therefore

2

L = 1.777 ×10−4 H L = 0.177 mH .

Problem 1.7 : In a resonant commutation circuit supply voltage is 200 V. Load current is 10 A and the device turn off time is 20µs. The ratio of peak resonant current to load current is 1.5. Determine the value of L and C of the commutation circuit. Solution Given

Ip IL

= 1.5

Therefore

I p = 1.5I L = 1.5 ×10 = 15 A .

That is

Ip =V

C = 15 A L

... ( a )

It is given that the device turn off time is 20 µsec. Therefore tc , the circuit turn off time should be greater than this, Let

tc = 30µ sec .

And

tc =

Therefore

30 × 10−6 =

Therefore

C = 1.5µ F .

CV IL 200 × C 10

Substituting in (a) 15 = 200

1.5 × 10−6 L

152 = 2002 × Therefore

1.5 × 10−6 L

L = 0.2666 mH

COMPLEMENTARY COMMUTATION (CLASS C COMMUTATION, PARALLEL CAPACITOR COMMUTATION) In complementary commutation the current can be transferred between two loads. Two SCRs are used and firing of one SCR turns off the other. The circuit is shown in figure 1.21.

IL R1

R2 ab

iC

V C T1

T2

Fig. 1.21: Complementary Commutation The working of the circuit can be explained as follows. Initially both T1 and T2 are off; Now, T1 is fired. Load current I L flows through R1 . At the same time, the capacitor C gets charged to V volts through R2 and T1 (‘b’ becomes positive with respect to ‘a’). When the capacitor gets fully charged, the capacitor current ic becomes zero. To turn off T1 , T2 is fired; the voltage across C comes across T1 and reverse biases it, hence T1 turns off. At the same time, the load current flows through R2 and T2 . The capacitor ‘C’ charges towards V through R1 and T2 and is finally charged to V volts with ‘a’ plate positive. When the capacitor is fully charged, the capacitor current becomes zero. To turn off T2 , T1 is triggered, the capacitor voltage (with ‘a’ positive) comes across T2 and T2 turns off. The related waveforms are shown in figure 1.22.

EXPRESSION FOR CIRCUIT TURN OFF TIME tc From the waveforms of the voltages across T1 and capacitor, it is obvious that tc is the time taken by the capacitor voltage to reach 0 volts from – V volts, the time constant being RC and the final voltage reached by the capacitor being V volts. The equation for capacitor voltage vc ( t ) can be written as

vc ( t ) = V f + (Vi − V f ) e −t τ Where V f is the final voltage, Vi is the initial voltage and τ is the time constant.

t = tc , vc ( t ) = 0 ,

At

τ = R1C , V f = V , Vi = −V , Therefore 0 = V + ( −V − V ) e

0 = V − 2Ve Therefore

V = 2Ve 0.5 = e

− tc R1C

− tc R1C

− tc R1C

− tc R1C

Taking natural logarithms on both sides −t ln 0.5 = c R1C

tc = 0.693R1C This time should be greater than the turn off time tq of T1 . Similarly when T2 is commutated

tc = 0.693R2C And this time should be greater than tq of T2 . Usually

R1 = R2 = R

Gate pulse of T2

Gate pulse of T1 p

t

V

IL

Current through R1

V R1

2V R1

t 2V R2

Current through T1 V R1

t Current through T2

2V R1

V R2

t

V Voltage across capacitor vab t

-V

tC

tC Voltage across T1

tC

Fig. 1.22

t

Problem 1.8 : In the circuit shown in figure 1.23 the load resistances R1 = R2 = R = 5Ω and the capacitance C = 7.5 µF, V = 100 volts. Determine the circuit turn off time tc .

R1

R2

V C T1

T2

Fig. 1.23. Solution The circuit turn-off time

tc = 0.693 RC seconds tc = 0.693 × 5 × 7.5 ×10−6

tc = 26µ sec . Problem 1.9: Calculate the values of RL and C to be used for commutating the main SCR in the circuit shown in figure 1.24. 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 40µsec. 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.

i1

IL

R1

RL iC

V =100V

C Auxiliary SCR

Main SCR

Fig. 1.24. Solution 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.

Given

I L = 25 A

That is

25 A =

Therefore

RL = 4Ω

V 100 = RL RL

tc = 40µ sec = 0.693RLC That is

40 ×10−6 = 0.693 × 4 × C

Therefore

C=

40 × 10 −6 4 × 0.693

C = 14.43µ F V should be less than the holding current of auxiliary SCR. R1 100 Therefore should be < 2mA. R1 i1 =

100 2 ×10−3

Therefore

R1 >

That is

R1 > 50 K Ω

IMPULSE COMMUTATION (CLASS D COMMUTATION) The circuit for impulse commutation is as shown in figure 1.25.

IL

T1 T3 V

VC(O) L

− +

C T2 FWD

Fig. 1.25: Circuit for Impulse Commutation

L O A D

The working of the circuit can be explained as follows. It is assumed that initially the capacitor C is charged to a voltage VC ( O ) with polarity as shown. Let the thyristor T1 be conducting and carry a load current I L . If the thyristor T1 is to be turned off, T2 is fired. The capacitor voltage comes across T1 , T1 is reverse biased and it turns off. Now the capacitor starts charging through T2 and the load. The capacitor voltage reaches V with top plate being positive. By this time the capacitor charging current (current through T2 ) would have reduced to zero and T2 automatically turns off. Now T1 and T2 are both off. Before firing T1 again, the capacitor voltage should be reversed. This is done by turning on T3 , C discharges through

T3 and L and the capacitor voltage reverses. The waveforms are shown in figure 1.26.

Gate pulse of T3

Gate pulse of T2

Gate pulse of T1 t

VS Capacitor voltage t

VC tC

Voltage across T1 t

VC

Fig. 1.26: Impulse Commutation – Waveforms of Capacitor Voltage, Voltage across T1 .

EXPRESSION FOR CIRCUIT TURN OFF TIME (AVAILABLE TURN OFF TIME) tc

tc depends on the load current I L and is given by the expression t

1 c VC = ∫ I L dt C0 (assuming the load current to be constant) VC =

tc =

I L tc C

VC C seconds IL

For proper commutation tc should be > tq , turn off time of T1 .

Note: • 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 Is shown in figure 1.27.

i +

T1 VC(O)

IT1 T2 V

_

C

D L

IL RL

Fig. 1.27: Impulse Commutation – An Alternate Circuit

The working of the circuit can be explained as follows: Initially let the voltage across the capacitor be VC ( O ) with the top plate positive. Now T1 is triggered. Load current flows through T1 and load. At the same time, C discharges through T1 , L and D (the current is ‘i’) and the voltage across C reverses i.e., the bottom plate becomes positive. The diode D ensures that the bottom plate of the capacitor remains positive. To turn off T1 , T2 is triggered; the voltage across the capacitor comes across T1 . T1 is reverse biased and it turns off (voltage commutation). The capacitor now starts charging through T2 and load. When it charges to V volts (with the top plate positive), the current through T2 becomes zero and T2 automatically turns off. The related waveforms are shown in figure 1.28. Gate pulse of T2

Gate pulse of T1

t

VC Capacitor voltage t

−V tC This is due to i IT 1

IL Current through SCR

V RL t

2V RL IL Load current t

V

Voltage across T1 t

tC

Fig. 1.28: Impulse Commutation – (Alternate Circuit) – Various Waveforms

Problem 1.10: An impulse commutated thyristor circuit is shown in figure 1.29. Determine the available turn off time of the circuit if V = 100 V, R = 10 Ω and C = 10 µF. Voltage across capacitor before T2 is fired is V volts with polarity as shown.

+ C V

T1

-

VC(0) T2

+

R

Fig. 1.29. Solution When T2 is triggered the circuit is as shown in figure 1.30.

VC(O) - +

+

i(t)

C T2

V

R

Fig. 1.30. Writing the transform circuit, we obtain

1 Cs

VC(0) s − +

I(s)

+ R

V s −

Fig. 1.31.

We have to obtain an expression for capacitor voltage. It is done as follows: 1 (V + VC ( 0 ) ) s I (S ) = 1 R+ Cs

I (S ) = I (S ) =

Voltage across capacitor

C (V + VC ( 0 ) ) 1 + RCs

(V + V ( 0 ) ) C

1   Rs +  RC  

VC ( s ) = I ( s )

VC ( s ) =

VC ( s ) =

VC ( s ) =

1 VC ( 0 ) − Cs s

1 V + VC ( 0 ) VC ( 0 ) − 1  RCs  s s+  RC   V + VC ( 0 ) V + VC ( 0 ) VC ( 0 ) − − 1  s s  s +  RC   V ( 0) V V − − C 1 s s+ 1 s+ RC RC

(

vc ( t ) = V 1 − e

−t

RC

In the given problem VC ( 0 ) = V Therefore

(

vc ( t ) = V 1 − 2e

−t

RC

)

The waveform of vc ( t ) is shown in figure 1.32.

) −V (0) e C

−t

RC

V vC(t) t

VC(0) tC

Fig. 1.32. At t = tc , vc ( t ) = 0 Therefore

− tc   0 = V 1 − 2e RC   

1 = 2e

− tc

RC

− tc 1 = e RC 2

Taking natural logarithms  1  −t log e   = c  2  RC tc = RC ln ( 2 ) tc = 10 ×10 × 10−6 ln ( 2 )

tc = 69.3µ sec . Problem 1.11 : In the commutation circuit shown in figure 1.33. C = 20 µF, 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 . T1 I0

C V

− +

VC(0)=V

T2 Fig. 1.33.

I0

Solution It is given that V varies between 180 and 220 V and I O varies between 50 and 200 A. The expression for available turn off time tc is given by tc =

CV IO

tc is maximum when V is maximum and IO is minimum. Therefore

tc max =

CVmax I O min

tc max = 20 ×10−6 ×

and

tc min =

220 = 88µ sec 50

CVmin I O max

tc min = 20 ×10 −6 ×

180 = 18µ sec 200

EXTERNAL PULSE COMMUTATION (CLASS E COMMUTATION)

T1

VS

T2

RL

L

2VAUX

+ −

C

T3

VAUX

Fig. 1.34: External Pulse Commutation In this type of commutation an additional source is required to turn-off the conducting thyristor. Figure 1.34 shows a circuit for external pulse commutation. VS is the main voltage source and VAUX is the auxiliary supply. Assume thyristor T1 is conducting and load RL is connected across supply VS . When thyristor T3 is turned ON at t = 0 , VAUX , T3 , L and C from an oscillatory circuit. Assuming capacitor is initially uncharged, capacitor C is now charged to a voltage 2VAUX with upper plate positive at t = π LC . When current through T3 falls to zero, T3 gets commutated. To turn-off the main thyristor T1 , thyristor T2 is turned ON. Then

T1 is subjected to a reverse voltage equal to VS − 2VAUX . This results in thyristor T1 being turned-off. Once T1 is off capacitor ‘C’ discharges through the load RL