thyristor commutation techniques
DESCRIPTION
Thyristor Commutation Techniques with ckts and discription.TRANSCRIPT
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 22Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Introduction
• Commutation – Process of turning off a conducting thyristor.
• Current Commutation
• Voltage Commutation
Power Electronics by Prof. M. Madhusudhan Rao 33Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Methods of Commutation
• Natural Commutation
• Forced Commutation
Power Electronics by Prof. M. Madhusudhan Rao 44Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Natural Commutation
• Occurs in AC circuits
~
T
+
v ov sR
Power Electronics by Prof. M. Madhusudhan Rao 55Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 66Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• Natural Commutation of Thyristors takes place in
– AC voltage controllers.
– Phase controlled rectifiers.
– Cyclo converters.
Power Electronics by Prof. M. Madhusudhan Rao 77Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 88Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Methods of Forced Commutation
• Self commutation.
• Resonant pulse commutation.
• Complementary commutation.
• Impulse commutation.
• External pulse commutation.
• Load Commutation.
• Line Commutation.
Power Electronics by Prof. M. Madhusudhan Rao 99Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Forced Commutationis applied to
• Choppers.
• Inverters.
Power Electronics by Prof. M. Madhusudhan Rao 1010Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Self Commutation Or
Load Commutation Or
Class A Commutation (Commutation By Resonating
The Load)
Power Electronics by Prof. M. Madhusudhan Rao 1111Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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.
Power Electronics by Prof. M. Madhusudhan Rao 1212Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
V
R L V (0)c
C
T i
L oad
+ -
Power Electronics by Prof. M. Madhusudhan Rao 1313Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Expression for Current
Fig. shows a transformed network
VS
R SL1
C S
V C (0 )S
C
T I(S) + +- -
Power Electronics by Prof. M. Madhusudhan Rao 1414Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 1515Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 1616Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 1717Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
2 2
is called the natural frequency
Taking inverse Laplace transforms
sint
AI S
S
Ai t e t
Power Electronics by Prof. M. Madhusudhan Rao 1818Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
2
Expression for current
0sin
Peak value of current
0
RtC L
C
V Vi t e t
L
V V
L
Power Electronics by Prof. M. Madhusudhan Rao 1919Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Expression For Voltage Across Capacitor At The Time Of Turn Off
V
R L V (0)c
C
T i
L oad
+ -
Power Electronics by Prof. M. Madhusudhan Rao 2020Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 2121Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 2222Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 2323Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 2424Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 2525Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 2626Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 2727Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 2828Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
V = 300V
R LL
1 F100 5 m H
CT
Power Electronics by Prof. M. Madhusudhan Rao 2929Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 3030Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 3131Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 3232Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
V= 100V
L
50 F
10 HC
T
+
i(t)
I O
V C (0 )= V
Power Electronics by Prof. M. Madhusudhan Rao 3333Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
The transformed circuit of the previous figure is shown in figure below
SL
VS
V (0)S
C
I LO
+
+
+
-
-
-
1C S
I(S)
Power Electronics by Prof. M. Madhusudhan Rao 3434Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 3535Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 3636Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 3737Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 3838Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 3939Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 4040Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 4141Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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
Power Electronics by Prof. M. Madhusudhan Rao 4242Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
V
L
i(t) I0
C
T 1
Power Electronics by Prof. M. Madhusudhan Rao 4343Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 4444Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 4545Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 4646Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 4747Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 4848Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
1
1
conduction time of SCR
tan
1
Substituting the values tan
O
O
LI
CV
LC
LI
CV
Power Electronics by Prof. M. Madhusudhan Rao 4949Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 5050Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Resonant Pulse Commutation (Class B Commutation)
L
C
VLoad
FW D
ab
iT
IL
Power Electronics by Prof. M. Madhusudhan Rao 5151Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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
Power Electronics by Prof. M. Madhusudhan Rao 5252Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 5353Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 5454Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 5555Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 5656Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
+
-
Power Electronics by Prof. M. Madhusudhan Rao 5757Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
2
22
1
111
1
VSI S
SLCs
VCI S
LC SL
VCS
sS LC
V
LLC
SC
Power Electronics by Prof. M. Madhusudhan Rao 5858Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 5959Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
1Where
sin sin
amps
p
p
LCV
or i t t I tL
CI V
L
Power Electronics by Prof. M. Madhusudhan Rao 6060Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Expression For Conduction Time Of SCR
1
Conduction time of SCR
sin L
p
t
II
Power Electronics by Prof. M. Madhusudhan Rao 6161Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 6262Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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
Power Electronics by Prof. M. Madhusudhan Rao 6363Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 6464Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
1
1
Instant at
which the thyristor turns off is at
& sin
0
L p
p C
t
tI I
LC
CI V
L
Power Electronics by Prof. M. Madhusudhan Rao 6565Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 6666Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 6767Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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)
Power Electronics by Prof. M. Madhusudhan Rao 6868Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Resonant Pulse CommutationWith Accelerating Diode
Power Electronics by Prof. M. Madhusudhan Rao 6969Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
VLOADFW D
LCT 1
IL
T 3
T 2iC (t)
V C (0 )+-
D 2 iC (t)
Power Electronics by Prof. M. Madhusudhan Rao 7070Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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.
Power Electronics by Prof. M. Madhusudhan Rao 7171Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
I L
0
V C
0
V 1V (O )C
iC
t
tt1 t2
tC
Power Electronics by Prof. M. Madhusudhan Rao 7272Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 7373Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
VLOADFW D
LCT 1
IL
T 3
T 2iC (t)
V C(0 )+
Power Electronics by Prof. M. Madhusudhan Rao 7474Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 7575Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 7676Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 7777Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
2
1
Repeat above problem
For 200 if an antiparallel diode is
connected across thyristor as shown in figureLI A D
T
Problem
Power Electronics by Prof. M. Madhusudhan Rao 7878Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
VLOADFW D
LC
T 1IL
T 3
T 2iC (t)
V C (0 )+-
D 2 iC (t)
Power Electronics by Prof. M. Madhusudhan Rao 7979Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 8080Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 8181Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 8282Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 8383Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 8484Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 8585Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 8686Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
1 1
Conduction time of SCR
1sin sin2
16666 166660.523
radians 0.00022 seconds16666
0.22 msec
L
p
II
Power Electronics by Prof. M. Madhusudhan Rao 8787Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 8888Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
V= 100V L i IL
IL
C
Power Electronics by Prof. M. Madhusudhan Rao 8989Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 9090Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 9191Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 9292Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 9393Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 9494Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Complementary Commutation (Class C Commutation,
Parallel Capacitor Commutation)
Power Electronics by Prof. M. Madhusudhan Rao 9595Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
V
R 1 R 2
T 1 T 2
IL
iC
C
a b
Power Electronics by Prof. M. Madhusudhan Rao 9696Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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
Power Electronics by Prof. M. Madhusudhan Rao 9797Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
G a te p u ls eo f T 1
G a te p u ls eo f T 2
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
Power Electronics by Prof. M. Madhusudhan Rao 9898Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 9999Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 100100Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 101101Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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
Power Electronics by Prof. M. Madhusudhan Rao 102102Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 103103Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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
Power Electronics by Prof. M. Madhusudhan Rao 104104Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
V= 100V
R 1 R L
M ainSC R
A uxiliarySC R
iC
C
ILi1
Power Electronics by Prof. M. Madhusudhan Rao 105105Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 106106Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 107107Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 108108Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Impulse Commutation(Class D Commutation)
VLOADFW D
C
T 1
T 3
IL
T 2
V C (O )+
L
Power Electronics by Prof. M. Madhusudhan Rao 109109Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• ‘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.
Power Electronics by Prof. M. Madhusudhan Rao 110110Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
G a te p u ls eo f T 2
G a te p u ls eo f T 3
Vo lta g e a c r o s s T 1
C a p a c ito rv o lta g e
G a te p u ls eo f T 1
V S
t
t
t
tC
V C
-V C
Power Electronics by Prof. M. Madhusudhan Rao 111111Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 112112Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 113113Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
An Alternative Circuit For Impulse Commutation
V
C
D
T 1
IT 1
IL
i
T 2
L
R L
V C (O )+
_
Power Electronics by Prof. M. Madhusudhan Rao 114114Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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.
Power Electronics by Prof. M. Madhusudhan Rao 115115Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 116116Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
L o a d c u r re n t
Vo lta g e a c r o s s T 1
t
t
tC
IL
V
2 VR L
Power Electronics by Prof. M. Madhusudhan Rao 117117Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 118118Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
C
T 1
T 2
V (0)C
V +
+
-
-
R
Power Electronics by Prof. M. Madhusudhan Rao 119119Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
When T2 is triggered the circuit is as shown in figure
C
i(t)
T 2V
++ -
-
R
V (O )C
Power Electronics by Prof. M. Madhusudhan Rao 120120Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Writing the transformcircuit, we obtain
Vs
V (0)s
C
+
+
I(s)
1C s
R
Power Electronics by Prof. M. Madhusudhan Rao 121121Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 122122Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 123123Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 124124Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
The waveform of vC(t) is shown in figure
t
tC
V
V (0)C
v (t)C
Power Electronics by Prof. M. Madhusudhan Rao 125125Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 126126Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 127127Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 128128Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
C
T 1
T 2 I0
I0
V (0)= VC+
V
Power Electronics by Prof. M. Madhusudhan Rao 129129Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 130130Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
Power Electronics by Prof. M. Madhusudhan Rao 131131Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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
+
Power Electronics by Prof. M. Madhusudhan Rao 132132Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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.
Power Electronics by Prof. M. Madhusudhan Rao 133133Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
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.
Power Electronics by Prof. M. Madhusudhan Rao 134134Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
Line Side Commutation
• In this type of commutation the discharging and recharging of capacitor takes place through the supply.
Power Electronics by Prof. M. Madhusudhan Rao 135135Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
LOAD
FW D
L r
CT 3
IL
L
T 2
V S
T 1
+
_
+
_
Power Electronics by Prof. M. Madhusudhan Rao 136136Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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.
Power Electronics by Prof. M. Madhusudhan Rao 137137Prof. T.K. Anantha Kumar, E&E Dept., MSRIT
• 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.