inverter

1. 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 turn-
off 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
Fundamental
component of voltage
Fundamental component
of current
Vo
Fig. E 9.1
From Fig, tanq =
-c LX X
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.
q /w = 15 ¥ 10–6
sec.
Now, f = 103
/0.1 = 104
Hz
q = 2p ¥ 104
¥ 15 ¥ 10–6
= 0.943 rad = 54°
tan 54° =
10
2
-cX
Xc = 12.753
= 4
1
2 10 Cp ¥ ¥
C = 1.25 mF
9.2 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

2. 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
Sol. Given R = 10 ohm R2
= 100W
And
4L
C
=
3
6
4 10 10
10 10
-
-
¥ ¥
¥
= 4000.
As R2
<
4L
C
, circuit is underdamped. Hence, the series circuit will commutate on its own when
triggered from zero voltage condition on the capacitor.
Also, e =
2
R
L
=
10 1000
2 10
¥
¥
= 500.
w0 =
1
cL
= 7
10 = 3.16 ¥ 103
rad/sec
wr = 2 2
- Œow = [107
– 2.5 ¥ 105
]1/2
= 3.123 ¥ 103
rad/sec
Y = tan–1
rw
eÊ ˆ
Á ˜Ë ¯
= tan–1
3
500
3.123 10
Ê ˆ
Á ˜¥Ë ¯
= 9.097°.
The load current is zero, i.e. SCR will commutate when w.r.t = p.
t =
3.123 1000
p
¥
= 1 msec.
e. t = 500 ¥ 1 ¥ 10–3
= 0.503
Now, voltage across inductance L is
VL = Es . o
r
w
w
e- e t
cos (wr t + Y)
= Es .
3.1623
3.1225
e–0.503
cos(180° + 9.097°)
= – 0.605 Es
Similarly, voltage across capacitor C is given by
Vc = Es[1 – e–0.503
¥
3.1623
3.1225
cos (180° – 9.097°)]
= 1.605 Es
Now,
di
dl
=
.
s
r
E
w L◊
[e– e t
wr . cos wr.t - Π. eРe t
sin w r t ]
0=
Ê ˆ
Á ˜Ë ¯t
di
dt
= 3
10 10-
=
¥
s sE E
L
= 100 Es . A/s.
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. 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 = off Lmax
min
0.893 t I◊
dcE
toff = min
Lmax.0.893 I
¥
¥
dcE C
Edc min = 230 – 10% (230)
= 207 V
toff =
6
207 25 10
0.893 150
-
¥ ¥
¥
= 38.64 ms.
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),
IL = I .
/2
/2
1
1
T Rc
T Rc
e
e
-
-
È ˘-
Í ˙
+Î ˚
R.C. = 1000 ms.
IL = 20
20/2 1000
20/2 1000
1
20
1
e
A
e
- ¥
- ¥
È ˘-
Í ˙
+Î ˚
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.

4. Solution Manual 4
Output power = E 2
/R = (48)2
/2.4 = 960 W.
(iii) Peak transistor current = Ip = Ed/R =
48
2.4
= 20 A.
Average transistor current = Ip/2 = 10 A.
(iv) Peak reverse blocking voltage,
VBR = 48 V.
(v) RMS harmonic voltage
En =
1/2
8
2
3,5,7
n
n
E
=
È ˘
Í ˙
Í ˙Î ˚
Â
= (E2
orms – E1
2
rms)1/2
= [(48)2
– (43.2)2
] 1/2
= 20.92 V.
THD =
20.92
43.2
= 48.43%.
(vi) D.F. =
( )
1/22
2
3,5,7
/
0.9
n
n
E n
•
=
È ˘
Í ˙
Í ˙
Î ˚
Â
=
0.03424
0.9
= 3.8%
(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%.