CHAPTER
8
Additional Problems
Solved Problems
8.1
A current commutated commutated chopper is fed from a d.c. source of 236 V. Its commutating commutating components components are L L = 20 m H and C = 50 m F . If the load current of 200 A is assumed constant during the commutation commutation process, compute the following: (i) turn-off time of main thyristor. (ii) total commutation commutation interval. interval. (iii) turn-off time of auxiliary thyristor.
Sol.
(i) Peak commutating commutating current is given by I CP CP = Edc c / L
\ \
x = x =
I cp I o
=
\
363.66 200
230
50 20
= 363.66 A
= 1.82.
–1
2
Turn-off time, t q = [p – – 2sin (1/1.8 )]
20 ¥ 50 ¥ 10-12
= 62.52 usec (ii)
–1
q 1 = sin
Ê I o ˆ –1 Ê 200 ˆ Á I ˜ = sin ÁË 363.66 ˜ ¯ Ë cp ¯
= 33.365°. Total commutation interval is given by
Ê 5p - q ˆ ÁË 2 1 ˜ ¯ \
Lc
+ 2 C E S ◊
Ê 5p - 33.365 ¥ p ˆ ÁË 2 ˜¯ 180
1000
–6
= 239.43 usec. (iii) Turn-off time of auxiliary thyristor is given by
I o
.
1 - cos 33.365 ˆ ¥ 10-6 + 50 ¥ 10-6 ¥ 230 ÊÁ ˜¯ Ë 200
–6
= 229.95 ¥ 10 + 9.48 ¥ 10
(p – – q 1) Lc
sin 2 q 1/ 2
Solution Manual 2
Ê p - 33.365 ˆ ÁË 180 ˜ ¯
1000
¥ 10 –6 = 80.93 usec.
8.2
A type a chopper operating at 2 kHz from a 100 V d.c. source as a load time constant of 6 ms and load resistance of 10 W. Find the mean load current and the magnitude of current ripple for a mean load voltage of 50 V, Also, compute the minimum and maximum values of load current.
Sol.
Load time constant,
L R
= 6 ¥ 10 –3 s.
R = 10 W. –3
\
L = 6 ¥ 10
Chopping period,
T = 1/ f =
¥ 10 = 60 mH.
1 2000
¥ 1000 = 0.5 ms.
Average or mean load voltage, E 0 = a E s.
\
a =
50 1000
= 0.5
T on = 0.5 ¥ 05 = 0.25 ms.
\
T off = 0.25 ms.
As chopping period T = 0.5 ms is much less than the load time constant = 6 ms, the current variation from minimum current I omin to Iomax (maximum current), must be taken as linear.
\ During T on period, E s – E o = L
\
(Iomax – Iomin) =
\
L Eo
=
E o
Iomax – Iomin =
T on
( E s - Eo ) T on
=
=
- I omin
Iomax
L
L E o L E o L
È E s ˘ Í E - 1˙T on. Î o ˚ È 1 - 1˘T ÍÎ a ˙˚ on È T ˘ Í T - 1˙T on Î on ˚ T off = D I , current ripple.
\ Magnitude of ripple current, D I = =
Mean load current,
Io =
Iomax – Iomin = 50 60 ¥ 10 Iomax
-3
L
. T off .
¥ 0.25 ¥ 10 –3 = 0.21 A.
+ Iomin 2
E o
3
Power Electronics
=
E o
50
=
R
10
= 5 A.
Maximum value of load current I omax is given by
D I
Iomax = I o +
2
= I o +
E o 2 L
. T off
50 ¥ 0.25 ¥ 10-3
= 5+
2 ¥ 60 ¥ 10-3
= 5.104 A. Minimum value of load current
D I
Iomin = I o – 8.3
2
= 5 – 0.104 = 4.896 A.
For a type-A chopper feeding an RLE load, show that the maximum value of rms current rating for free wheeling diode is given by 0.39
È E Ê E ˆ 3/ 2 ˘ Í s Á1 - ˜ ˙ . Consider load current to be ripple ÍÎ R Ë E s ¯ ˙˚
free. Sol.
Average load current for chopper with RLE load is given by I o =
Eo
-E
R
=
a ◊ Es
- E
R
During period T off , current flows through freewheeling diode. Therefore, rms value of freewheeling diode current, when I o is ripple free is given by T off
I of =
= = =
t 1 R 1 R 1 R
1/ 2
T - Ton ˘ ◊ I o ÈÍ Î t ˙˚
È a ◊ Es - E ˘ Î R ˙˚
.Í
(a ◊ E s . - E ) ( 1 - a ) ◊ ÈÎa ◊ ÈÍ Î
1 - a E s a2
-
- a 3 ◊ E s -
(i)
1 - a ◊ E ˘
˚
1 - a ◊ E ˘
˚˙
This current will have maximum value when d I Df d a
=
(2a - 3a ) E =
È 1 ( 2a - 3a 2 ) E s 1 ◊Í + 2 3 R Í 2 2 a a Î 1
2
or
a
2
(2a - 3a ) 2
or
- a
3
s
1 - a
a 1 - a
=
–
E 1 - a .
- E E s .
˘ ˙ = 0. 1 - a ˙ ˚ E
Solution Manual 4
or
3a – 2 =
Ê Ë
E
or a = 1/3 Á 2 +
E s
E ˆ
E s ˜ ¯
Substituting this value of a in Eq. (i), we get maximum value of rms current rating I Dfm of freewheeling diode as, I f Dm =
=
1 È1 Ê
Í
R ÍÎ 3 ÁË
1/ 2
=
È 2 E s - 2 E ˘ È Es - E ˘ ˙˚ Í 3E ˙ R ÍÎ 3 Î s ˚
1
1/ 2
1
1 2 E s . E s
◊
R 3 2
3 E s .
◊
E s
3 3 R
= 0.39
Sol.
˘ È 2 Es + E ˘ + Eˆ E E ◊ ˙ Í1 ˙ s E s ˜ 3E s ˚ ¯ ˙˚ Î
=
=
1/ 2
1 È 1 Ê 2 E s
È 2 E s + E - 3E ˘ È 3Es - 2 Es - E ˘ ˙ ˙˚ Í R ÍÎ 3 3E s Î ˚
=
8.4
1/ 2
˘È 1Ê ˆ E ˆ ˘ Í Á 2 + ˜ ◊ E s - E ˙ Í1 - Á 2 + ˜ ˙ R ÎÍ 3 Ë E s ¯ ˚˙ ÎÍ 3 Ë E s ¯ ˚˙ E
E s R
=
1 3
◊ ÈÎ E s - E ˘˚3/ 2 R 2
1 3 E s
3/ 2
È E ˘ Í1 - E ˙ s ˚ Î 3/ 2
È E ˘ Í1 - E ˙ s ˚ Î
3/ 2
È E ˘ ◊ Í1 - ˙ Î E s ˚
A current commutated chopper is operating on a 150 V d .c. supply. The maximum load current is 50 A. The turn-off time of main SCR is 15 m sec. If the value of the commutating L is 10 m H and C = 4.4 m F ., find out if it is possible commutate the main SCR. If not, calculate the values of L and C to commutate the main SCR successfully. For current commutated chopper, L =
)
(
3 Edc Loff + Dt 4p t om
\ Circuit turn-off time t q = (t off + Dt ) L ¥ 4p I om =
=
3 E dc 10 ¥ 10-6
¥ 4p ¥ 50
3 ¥ 150
= 13.96 m sec.
\ Circuit turn-off time t q < t off . \ This commutation circuit cannot commutate the main SCR successfully. In order to have successful commutation, t q = 2 t off = 2 ¥ 15 = 30 m s.
5
Power Electronics
\
L =
and,
C =
3 ¥ 150 ¥ 30 ¥ 10 -6 4 ¥ p ¥ 50
¥ t q
3 Iom
p ◊ E dc
=
= 21.48 m H.
3 ¥ 50 ¥ 30 ¥ 10-6 p ¥150
C = 9.54 m F For successful commutating, L = 21.48 m H and C = 9.54 m F. 8.5
A voltage commutated chopper is operating with an input d.c. voltage of 100 V. Commutating components are L = 15 m H and C = 4.7 m F . Determine:(i) Minimum duty cycle of the chopper. (ii) Peak capacitor current (iii) Peak current through the device if the load current is 28 A and operating frequency is 400 Hz
Sol.
p Lc
(i) Minimum duty cycle, Dmin =
T
(
= p Lc ◊ ¥ f = p ¥ 15 ¥ 10-6
1/ 2
¥ 4.7 ¥ 10-6 )
¥ 400
= 1%. (ii) Peak capacitor current Icpeak = E dc
c/L
= 100
4.7 /15 = 55.97 A.
(iii) Peak thyristor current = Iom + Icpeak = 28 + 55.97 = 83.97 A 8.6
A parallel capacitor type A chopper operates at 600 Hz from a 72 V battery and feeds to a load of 10 A. If the turn-off time of main SCR is 45 m s and that of auxiliary SCR is 60 m s. Determine the values of commutating components. Assume a 100% safety margin for turn-off times of main and auxiliary SCRs.
Sol.
(i) Capacitor
C =
=
(ii)
(iii)
E dc 10 ¥ 90 ¥ 10-6 72
L £ C
£ L
≥
)
(
I om ◊ toff + Dt
E dc
I om
= 12.5 m F .
£ 12.5 ¥ 10-6
72 10
0.65 mH. 0.01 T 2 E dc 2
p
◊ C .
T = 1/600 = 1.66 ms.
Solution Manual 6
\
L
≥
\
L
≥
(
0.01 ¥ 1.66 ¥ 10-3 2
p
)
2
¥ 72
¥ 12.5 ¥ 10-6
16.21 mH.
(iv) The time for which SCR 2 remains reverse biased is approximately half of the time (t 1 – t 0). But
(t 1 – t 0) = p Lmin ◊ C .
\
(t 1 – t 0) = p 0.65 ¥ 10-3
¥ 12.5 ¥ 10-6
= 282.74 m s.
\ Time for which SCR 2 is reverse-biased =
282.74 2
= 141.37 m s. This is larger than 2 ¥ 60 = 120 m s.
\ The value of commutating components calculated in (i), (ii), and (iii) can ensure successful commutation of the SCRs.