CHAPTER
8
Additional Problems
Solved Problems
8.1
A current commutated chopper is fed from a d.c. source of 236 V. Its commutating components
are L = 20 mH and C = 50 mF. If the load current of 200 A is assumed constant during the
commutation process, compute the following:
(i) turn-off time of main thyristor.
(ii) total commutation interval.
(iii) turn-off time of auxiliary thyristor.
Sol. (i) Peak commutating current is given by
ICP = Edc c / L \ 230
\
\
x=
I cp
Io
=
50
= 363.66 A
20
363.66
= 1.82.
200
Turn-off time, tq = [p – 2sin–1(1/1.82)]
20 ¥ 50 ¥ 10-12
= 62.52 usec
(ii)
ÊI ˆ
Ê 200 ˆ
q1 = sin–1 Á o ˜ = sin–1 Á
Ë 363.66 ˜¯
I
Ë cp ¯
= 33.365°.
Total commutation interval is given by
Ê 5p
ˆ
ÁË 2 - q1 ˜¯
\
Lc + 2 C ES ◊
sin 2 q1/2
.
Io
Ê 5p 33.365 ¥ p ˆ
Ê 1 - cos 33.365 ˆ
1000 ¥ 10-6 + 50 ¥ 10-6 ¥ 230 Á
ÁË 2 ˜
˜¯
¯
Ë
180
200
= 229.95 ¥ 10–6 + 9.48 ¥ 10–6
= 239.43 usec.
(iii) Turn-off time of auxiliary thyristor is given by
(p – q1) Lc
Solution Manual 2
Ê p - 33.365 ˆ
–6
ÁË
˜¯ 1000 ¥ 10 = 80.93 usec.
180
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
= 6 ¥ 10–3 s.
R
R = 10 W.
\
L = 6 ¥ 10–3 ¥ 10 = 60 mH.
Chopping period,
T = 1/f =
1
¥ 1000 = 0.5 ms.
2000
Average or mean load voltage, E0 = a Es.
a=
\
50
= 0.5
1000
Ton = 0.5 ¥ 05 = 0.25 ms.
Toff = 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 Iomin to Iomax (maximum current), must be taken as linear.
\ During Ton period,
Es – Eo = L
\
\
(Iomax – Iomin) =
Iomax - Iomin
Ton
( Es - Eo ) Ton
L
=
˘
Eo È Es
- 1˙ Ton.
Í
L Î Eo
˚
=
Eo È 1
˘
- 1˙ Ton
Í
L Îa
˚
=
Eo
L
Iomax – Iomin =
ÈT
˘
- 1˙ Ton
Í
Î Ton
˚
Eo
Toff = DI, current ripple.
L
\ Magnitude of ripple current,
DI = Iomax – Iomin =
=
Mean load current,
Io =
Eo
. Toff.
L
50
¥ 0.25 ¥ 10–3 = 0.21 A.
60 ¥ 10-3
Iomax + Iomin
2
3
Power Electronics
=
Eo 50
=
= 5 A.
R 10
Maximum value of load current Iomax is given by
E
DI
= Io + o . Toff
2
2L
Iomax = Io +
50 ¥ 0.25 ¥ 10-3
2 ¥ 60 ¥ 10-3
= 5+
= 5.104 A.
Minimum value of load current
DI
= 5 – 0.104 = 4.896 A.
2
Iomin = Io –
8.3
For a type-A chopper feeding an RLE load, show that the maximum value of rms current rating
3/ 2
ÈE Ê
Eˆ ˘
s
Í
for free wheeling diode is given by 0.39
1 - ˜ ˙ . Consider load current to be ripple
Es ¯ ˙
Í R ÁË
Î
˚
free.
Sol. Average load current for chopper with RLE load is given by
Io =
Eo - E a ◊ Es - E
=
R
R
During period Toff, current flows through freewheeling diode. Therefore, rms value of
freewheeling diode current, when Io is ripple free is given by
1/ 2
Toff
È T - Ton ˘
◊ Io Í
˙
t
Î t ˚
Iof =
È a ◊ Es - E ˘
.Í
˙
R
Î
˚
=
1
(a ◊ Es . - E )
R
=
1 È
◊ a 1 - a Es - 1 - a ◊ E ˘˚
R Î
=
1 È 2
◊ a - a 3 ◊ Es - 1 - a ◊ E ˘˙
˚
R ÍÎ
(
1-a
)
(i)
This current will have maximum value when
(
da
(2a - 3a ) E =
2
or
2
a -a
(2a - 3a )
2
or
3
1-a
a 1-a
)
2
È
1 Í 1 2a - 3a Es 1 E
=
◊
+
2
3
R Í2
2 1-a
a
a
Î
d I Df
s
=
–
E
1 - a.
-E
Es .
˘
˙ = 0.
˙
˚
Solution Manual 4
or
3a – 2 =
Ê
Eˆ
E
or a = 1/3 Á 2 + ˜
Es ¯
Es
Ë
Substituting this value of a in Eq. (i), we get maximum value of rms current rating IDfm of
freewheeling diode as,
1
If Dm =
R
=
1
R
1/ 2
È1 Ê
˘È 1Ê
Eˆ
E ˆ˘
Í Á 2 + ˜ ◊ Es - E ˙ Í1 - Á 2 + ˜ ˙
Es ¯
Es ¯ ˙˚
ÍÎ 3 Ë
˙˚ ÍÎ 3 Ë
1/ 2
È 1 Ê 2 Es + E ˆ
˘ È 2E + E ˘
◊ Es - E ˙ Í1 - s
Í Á
˙
˜
3 Es ˚
˙˚ Î
ÎÍ 3 Ë Es ¯
1/ 2
1 È 2 Es + E - 3 E ˘ È 3 Es - 2 Es - E ˘
=
˙
˙Í
R ÍÎ
3
3 Es
˚Î
˚
1/ 2
1 È 2 Es - 2 E ˘ È Es - E ˘
=
˙ Í 3E ˙
R ÍÎ
3
˚Î
s ˚
=
1 2 Es . Es
◊
R 3 3 Es .
=
Es
3 3 R
2
◊
È
E˘
Í1 - ˙
Î Es ˚
È
E˘
Í1 - ˙
E
s˚
Î
E È
E˘
= 0.39 s ◊ Í1 - ˙
R Î Es ˚
8.4
=
1 3
3/ 2
◊ ÈÎ Es - E ˘˚
R 2
1
3 Es
3/ 2
3/ 2
3/ 2
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 msec. If the value of the commutating L is 10 mH
and C = 4.4 mF., 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.
Sol. For current commutated chopper,
L=
(
3Edc Loff + Dt
)
4p tom
\ Circuit turn-off time tq = (toff + Dt)
=
L ¥ 4p I om
3 Edc
=
10 ¥ 10-6 ¥ 4p ¥ 50
= 13.96 msec.
3 ¥ 150
\ Circuit turn-off time tq < toff.
\ This commutation circuit cannot commutate the main SCR successfully. In order to have
successful commutation,
tq = 2 toff = 2 ¥ 15 = 30 ms.
5
Power Electronics
\
L=
and,
C=
3 ¥ 150 ¥ 30 ¥ 10-6
= 21.48 mH.
4 ¥ p ¥ 50
3 Iom ¥ tq
p ◊ Edc
=
3 ¥ 50 ¥ 30 ¥ 10-6
p ¥150
C = 9.54 mF
For successful commutating, L = 21.48 mH and
C = 9.54 mF.
8.5
A voltage commutated chopper is operating with an input d.c. voltage of 100 V. Commutating
components are L = 15 mH and C = 4.7 mF. 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. (i) Minimum duty cycle, Dmin =
= p
p Lc
T
(
Lc ◊ ¥ f = p ¥ 15 ¥ 10-6 ¥ 4.7 ¥ 10-6
1/ 2
)
¥ 400
= 1%.
(ii) Peak capacitor current
Icpeak = Edc
c/L
4.7 /15 = 55.97 A.
= 100
(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 ms and that of auxiliary SCR is 60 ms. 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)
(
I om ◊ toff + Dt
)
Edc
10 ¥ 90 ¥ 10-6
= 12.5 mF.
72
L£ C
Edc
I om
£ 12.5 ¥ 10-6
72
10
£ 0.65 mH.
(iii)
L ≥
0.01 T 2 Edc
T = 1/600 = 1.66 ms.
p 2 ◊ C.
Solution Manual 6
2
(
)
2
-6
0.01 ¥ 1.66 ¥ 10-3
\
L ≥
\
L ≥ 16.21 mH.
p ¥ 12.5 ¥ 10
¥ 72
(iv) The time for which SCR2 remains reverse biased is approximately half of the time (t1 – t0).
But
(t1 – t0) = p Lmin ◊ C.
\
(t1 – t0) = p 0.65 ¥ 10-3 ¥ 12.5 ¥ 10-6
= 282.74 ms.
\ Time for which SCR2 is reverse-biased =
282.74
2
= 141.37 ms.
This is larger than 2 ¥ 60 = 120 ms.
\ The value of commutating components calculated in (i), (ii), and (iii) can ensure successful
commutation of the SCRs.