CHAPTER
2
Additional Problems
Solved Problems
2.1 The latching current for a thyristor inserted between a dc source voltage of 100V and a load being
75mA. Calculate the minimum width of the gate-pulse required to turn-on the thruster when the
load is
(i) Purely inductive having an inductance of 100 mH and (ii) Consisting of resistance and
inductance of 10 ohm and 100 mH respectively.
Sol. (i)
When the load is purely inductive,
V =L
di
dt
i.e.,
di =
V
dt
L
\
i =
V
t.
L
\
t =
L · i 100 ¥ 10-3 ¥ 75 ¥ 10-3
=
V
100
= 75 ms.
(ii) When load is R-L types
V =R·i+L
i =
\
100 ¥ 10
–3
(
di
dt
R
- ·t
V
1- e L
R
(
)
10
t
-3
100
=
1 - e 100 ¥ 10
10
)
t = 100 msec
2.2
An SCR with
Úi
2
dt rating of 20 Amp2 sec is used to act as a rectifier of feed a load. If a earth
fault occurs at the output of the rectifier when the input ac voltage (100 sin w t) is at its positive
peak, find the fault current and the safe time that the SCR can withstand the fault without
damage. Assume the wire resistance and thyristor resistance to be 1 ohm.
Sol. Let if be the fault current,
\
if =
100
input peak voltage
= 100 A.
=
1
total circuit resistance
Solution Manual 2
i2 dt = 20
Since,
tf =
\
\
20
tf
Úo 100
2
· dt = 20
= 2 m sec.
1002
Thus, the safe time the fault of 100 A can be withstood without the damage of SCR is 2 m Sec.
2.3
The specification sheet for an SCR gives maximum rms on state current as 35 A. If this SCR is
used in a resistance circuit, computer average. On state current rating for half sine wave current
for conduction angles of (a) 180° (b) 90° (c) 30°
Sol. Half sine wave current waveform is shown in Fig. 3.1
i
Conduction
angle.
Im
π
Q1
O
π
π
Fig. 3.1
Iav =
Irms
I
1 p
I m sin q dq = m (1 + cos q1)
Ú
q
2p
2p 1
È1
= Í
ÍÎ 2p
p
Ú
I m2
q1
˘
sin q dq ˙
˙˚
12
2
È I 2 Ïq sin 2 q ¸p ˘
˝ ˙
= Í m Ì 4 ˛q1 ˙˚
ÍÎ 2p Ó 2
ÈI2
= Í m
ÍÎ 2p
12
{
}
È I 2 p - q1 1 2 ˘
= Í m
+ sin q1 ˙
2
4
Î 2p
˚
12
(a) For 180° conduction angle, q1 = 0°
\
and
Iav =
Irms
Im
I
[(1 + cos 0°)] = m
2p
p
{
}
ÈI2 p 1
˘
= Í m
- (0) ˙
Î 2p 2 4
˚
\
Form factor (FF) =
\
ITAV =
12
=
Im
2
I rms
I
p
p
= m·
=
2 Im
I av
2
I rms 35 ¥ 2
= 22.282 A.
=
FF
p
(b) For 90°, conduction angle, q1 = 90°.
p
Ï 1 cos 2 q ¸˘
Ú ÌÓ 2 - 2 ˝˛˙˙
q1
˚
12
3
Power Electronics
Iav =
\
and
Irms
Im
I
1 + cos q0 ] = m
[
2p
2p
{
}
ÈI2 p 1
˘
+ (0) ˙
= Í m
Î 2p 4 4
˚
\
Form factor =
\
ITAV =
Im
2 2
¥
12
=
Im
2 2
2p p
.
Im
2
35 ¥ 2
= 15.755 A.
p
(c) Fro 30°, conduction angle, q1 = 150°.
Iav =
\
Irms
{
}
ÈI2 p 1
˘
= Í m
+ ( -0.866) ˙
Î 2p 12 4
˚
\
Form factor =
\
ITav =
2.4
Im
[1 + ( -0.866)] = 0.021 Im.
2p
12
= 0.085 Im.
0.0849035 I m
= 3.98
0.021 I m
35
= 8.79 A
3.98
Repeat example (3) in case the current has rectangular waveshape.
Sol. Rectangular waveform is shown in Fig. 4.1
i
360°
I
t
T
nT
Fig. 4.1
Conduction angle =
\
Here
\
h =
Iav =
T
¥ 360°
hT
360°
conduction angle
I ¥T
I
=
hT
h
ÈI2 ¥T ˘
Irms = Í
˙
Î hT ˚
12
=
I
h
Solution Manual 4
(a) For 180° conduction angle, h =
\
Iav =
\
Form factor =
\
ITav =
I
I
.
and Irms =
2
2
I
2
=
2 I
·
35
Iav =
\
Form factor =
\
ITav =
360
=4
90
I
I
I
= .
and Irms =
4
2
4
I 4
· =2
2 I
35
= 17.5 A.
2
(c) For 30° conduction angle and h =
360
= 12
12
Iav =
I
and Irms =
12
Form factor =
I 12
= 12
12 I
ITAV =
2.5
2
= 24.75 A.
2
(b) For 90° conduction angle, h =
\
360
= 2.
180
35
12
I
12
= 10.10 A
In a class-C commutations circuit, determine the value of R, RL, and C for commutating the main
SCR when it is conducting a full current of 15 Amp. The minimum time for which this SCR is to
be reverse biased for proper commutation is 30 msec. It is given that the complementary SCR
will undergo natural commutation when its forward current falls below the holding current of
3mA. Assume supply voltage to be 100 volts.
Edc 100
= 6.66 W.
=
15
IL
Sol.
RL =
Now,
t ˆ
Ê
RC
˜
Vc = Edc . Á1 - e
ÁË
˜¯
t = 30 ms, Edc = 200 V, Vc = 100 V.
\
t ˆ
Ê
100 = 200 Á1 - e RL ◊ C ˜
ÁË
˜¯
5
Power Electronics
– -
t
RC ◊ C
= loge 0.5 = – 0.693
t
30 ¥ 10-6
=
0.693 RL 0.693 ¥ 6.66
\
C=
\
C = 6.49 mF
Select R such that
R>
Edc
3 mA
\
R>
100 V
3 ¥ 10 - 3
2.6
\ R > 33.33 k W
Determine the values of R and C in Fig. E 2.6. The load current through the main SCR 1 is 25 A.
The minimum time for which this SCR is to be reverse biased is 30 ms. The auxiliary SCR 2
should undergo natural commutation when its forward current goes below its holding current of
2.5 mA.
+
R1
R2
C
100 V
SCR 1
SCR 2
–
Fig. E 2.6
Sol. In order to ensure that SCR2 will undergo natural commutation
R2 ≥
\
100
2.5 ¥ 10-3
R2 ≥ 40 K W
When SCR1 is turned-off, the charging of C takes place through R1
Where
R1 =
100
=4W
25
The charging equation for C is
tq ˆ
Ê
R1 C
˜
Vc = V Á1 - e
ÁË
˜¯
\
tq ˆ
Ê
50 = 100 Á1 - e 4C ˜
Ë
¯
\
tq = 0.693 ¥ 4C = 2.77 C.
\
C=
30 ¥ 10-6
= 10.82 mF.
2.77
The commutation capacitor C ≥ 10.82 mF
Solution Manual 6
E 2.7 For the class-C commutation, R1 = R2 = 5 W, C = 10 mF, supply voltage Edc = 100 V. Determine
the turns-off time.
Sol. Capacitor charging equation is
tc ˆ
Ê
Vc = Edc Á1 - e R1 C ˜
ÁË
˜¯
\
tc ˆ
Ê
R1 C
˜
100 = 200 Á1 - e
ÁË
˜¯
tc = 0.693 R1C = 0.693 ¥ 5 ¥ 10 ¥ 10–6 = 34.65 msec.
\
E 2.8 Determine the commutating components for an auxiliary commutation method if IL = 10A, Edc
= 100 V and tq = 50 ms.
Sol. We have the equation
C=
=
Also,
Lmin =
I L max ◊ tq
Edc
10 ¥ 50 ¥ 10-6
= 5 mF
100
Edc 2
¥C
I L2
2
È100 ˘
= Í
¥ 5 ¥ 10–6 = 0.5 mH
˙
10
Î
˚
Lmax =
0.01T 2
p2 ◊C
Assume operating frequency of the circuit to be 400 Hz
\
T = 2.5 ms
\
Lmax = 1.26 mH.
E 2.9 In class-D commutation circuit, Edc = 200 V, L = 10 mH, C = 50 mF. Determine the minimum
on time of SCR 1 and peak-value of capacitor current.
Sol. Minimum on time of SCR 1 is given by
ton(min) = p LC = 10 = 10-6 ¥ 50 ¥ 10-6
= 69 msec.
Peak SCR current is given by,
Ip = Edc
= 200
C
L
50 ¥ 10-6 /10 ¥ 10-6
= 447.21 A.
7
Power Electronics
E 2.10 Thyristor in Fig. 2.10 has a latching current level of 50 mA and is fired by a pulse of width 50
ms. Show that without R, SCR will fail to remain ON when the firing pulse ends. Find the
maximum value of R to ensure firing.
Neglect the SCR volt-drop and assume that the initial value of rate of rise of current remains
constant over the entire pulse width.
+
i
20 W
100 V
R.
0.5 H
–
Fig. E 2.10
Sol. For the given load,
t = L/R =
0.5
= 0.025 sec.
20
100
= 5 A.
20
Without R, SCR current i will grow exponentially as,
t
Ê
- ˆ
i = I0. Á1 - e t ˜
Ë
¯
Max. value of steady state current =
at
Ê t ˆ˘
È
-Á
˜
= 5 Í1 - e Ë 0.025 ¯ ˙
Í
˙
Î
˚
ton = t = 50 msec,
50 ¥ 10-6 ˘
È
i = 5 Í1 - e 0.025 ˙ = 9.99 ¥ 10–3
Í
˙
Î
˚
= 10 mA
OR
Since given that initially
\
\
di
is constant over entire time ion,
dt
ton =
I0
5
=
= 200 A/sec.
t 0.025
i
= 50 ¥ 10–6 = 200 ¥ 50 ¥ 10–6 = 10 mA
t
\ Hence if trigger pulse with is 50 msec, SCR will fail to reach its latching current level of 50 mA.
SCR current is less by (50 – 10) = 40 mA. from its latching current value.
This current must be supported by additional R.
\
\
40 ¥ 10–3 R = 100 V
R=
100
= 2.5 kW.
40 ¥ 10-3