CHAPTER
9
Additional Problems
Solved Problems
9.1
A single-phase bridge inverter delivers power to a series connected RLC load with R = 2W and
wL = 10W. The periodic time T = 0.1 msec. What value of C should the load have in order to
obtain load commutation for the SCRs. The thyristor turn-off time is 10 msec. Take circuit turnoff time as 1.5 tq. Assume that load current contains only fundamental component.
Sol. Value of C should be such that RLC load is underdamped. Moreover, when load voltage passes
through zero, the load current must pass through zero before the voltage wave, i.e. the load
current must lead the load voltage by an angle q as shown in Fig. E 9.1
Vo
Fundamental
component of voltage
Fundamental component
of current
Fig. E 9.1
From Fig,
tanq =
Xc - X L
R
As the current is leading voltage, hence XC > XL.
Now q | w | must be at least equal to circuit turn-off time. i.e.,
1.5 ¥ 10 = 15 msec.
\
Now,
q /w = 15 ¥ 10–6 sec.
f = 103/0.1 = 104 Hz
q = 2p ¥ 104 ¥ 15 ¥ 10–6 = 0.943 rad = 54°
\
tan 54° =
=
9.2
X c - 10
2
1
2p ¥ 104 ¥ C
\ Xc = 12.753
\ C = 1.25 mF
In a self commutated SCR circuit, the load consists of R = 10W in series with commutating
components of L = 10 mH and C = 10 mF. Check whether the circuit will commutate by itself
Solution Manual 2
when triggered from zero voltage condition on the capacitor. What will be voltage across the
capacitor and inductor at the time of commutation? Also, determine di/dt at t = 0
R = 10 ohm \ R2 = 100W
Sol. Given
4L
4 ¥ 10 ¥ 10-3
=
= 4000.
C
10 ¥ 10-6
And
4L
, circuit is underdamped. Hence, the series circuit will commutate on its own when
C
triggered from zero voltage condition on the capacitor.
As R2 <
Also,
e=
R
10 ¥ 1000
=
= 500.
2L
2 ¥ 10
w0 =
1
= 107 = 3.16 ¥ 103 rad/sec
Lc
wr =
wo2 - Œ2 = [107 – 2.5 ¥ 105]1/2
= 3.123 ¥ 103 rad/sec
Ê
ˆ
Ê e ˆ
500
Y = tan–1 Á ˜ = tan–1 Á
3˜
Ë wr ¯
Ë 3.123 ¥ 10 ¯
= 9.097°.
The load current is zero, i.e. SCR will commutate when w.r.t = p.
t=
p
= 1 msec.
3.123 ¥ 1000
e. t = 500 ¥ 1 ¥ 10–3 = 0.503
Now, voltage across inductance L is
VL = Es .
= Es .
wo - e t
e cos (wr t + Y)
wr
3.1623 –0.503
e
cos(180° + 9.097°)
3.1225
= – 0.605 Es
Similarly, voltage across capacitor C is given by
Vc = Es[1 – e–0.503 ¥
3.1623
cos (180° – 9.097°)]
3.1225
= 1.605 Es
Now,
Es
di
e
e
[eРt wr . cos wr.t - Π. eРt sin w r t ]
=
dl
wr ◊ L.
Es
Es
Ê di ˆ
= 100 Es . A/s.
=
=
ÁË dt ˜¯
L 10 ¥ 10-3
t =0
9.3
A McMurray inverter uses a commutation circuit consisting of C = 25 mF and L = 25 mH. The
source voltage Edc = 230 V d.c. The load current varies from 50 to 150 A at the instant of
3
Power Electronics
commutation. Would this circuit provide reliable commutation if the thyristor turn-off time is
25 ms.
Sol. Given:
C = 25 mF, L = 25 mH,
ILmax = 150 A toff = 25 msec, Edc = 230 V,
From Eq. (9.165),
C=
toff =
\
0.893 t off ◊ IL max
Edc min
Edc min ¥ C
0.893 ¥ I L max.
Edc min = 230 – 10% (230)
= 207 V
toff =
\
207 ¥ 25 ¥ 10-6
= 38.64 ms.
0.893 ¥ 150
Circuit will provide reliable commutation if
toff calculate > 2 toff givens
> 50 ms.
Since calculated toff is less than 50 ms, hence circuit will not provide reliable commutation.
9.4
Determine the peak load current and PIV rating of the thyristors for the parallel capacitor
commutated inverter for C = 100 mF and R = 10W.
Sol. From Eq. (9.203),
È1 - e -T / 2 Rc ˘
IL = I . Í
-T / 2 Rc ˙
Î1 + e
˚
R.C. = 1000 ms.
È 1 - e -20/ 2 ¥ 1000 ˘
20 A
IL = 20 Í
-20 / 2 ¥ 1000 ˙
Î1 + e
˚
\
PIV = VL = 20 ¥ 10 = 200 Volts.
9.5
A single-phase half bridge inverter has a resis load of 2.4 W and the d.c. input voltage of 48 V.
Determine:(i) RMS output voltage at the fundamental frequency
(ii) Output power P0
(iii) Average and peak currents of each transistor
(iv) Peak blocking voltage of each transistor.
(v) Total harmonic distortion and distortion factor.
(vi) Harmonic factor and distortion factor at the lowest order harmonic.
Sol.
(i) RMS output voltage of fundamental frequency, E1 = 0.9 ¥ 48 = 43.2 V.
(ii) RMS output voltage, Eorms = E = 48 V.
Solution Manual 4
\
Output power = E 2/R = (48)2/2.4 = 960 W.
(iii) Peak transistor current = Ip = Ed/R =
48
= 20 A.
2.4
Average transistor current = Ip/2 = 10 A.
(iv) Peak reverse blocking voltage,
VBR = 48 V.
(v) RMS harmonic voltage
1/ 2
È 8
˘
En2 ˙
En = Í
ÍÎ n = 3,5,7 ˙˚
Â
= (E2orms – E12rms)1/2
= [(48)2 – (43.2)2] 1/2
= 20.92 V.
\
THD =
20.92
= 48.43%.
43.2
È •
Í
En / n 2
Í n = 3,5,7
D.F. = Î
0.9
Â(
(vi)
=
2 ˘1/ 2
) ˙˙
˚
0.03424
= 3.8%
0.9
(vii) Lowest order harmonic is the third harmonic. RMS value of third harmonic is
E3rms = E1rms/3
\
H.F3 = E3rms/E1rms = 33.33%
and
D.F.3 = (E3rms/32)/E1rms
= 1/27 = 3.704%.