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 inductance of 100 mH and (ii) Consisting of resistance and and inductance of 10 ohm and 100 mH respectively. Sol. (i)
When the load is purely inductive, di
V = L
\
i =
\
t =
i.e., i.e.,
dt
V L
di = di =
V L
dt
t .
L · i V
=
100 ¥ 10-3
¥ 75 ¥ 10-3
100
= 75 m s. s. (ii) When load load is R-L types types di
V = R · i + L + L
\
i =
100 ¥ 10
–3
=
V R
(1 - e- )
100 10
dt R
· t
L
(1 - e-
10
100 ¥ 10-
3
t
)
t = 100 m sec sec 2.2
An SCR with
dt rating of 20 Amp sec is used to act as a rectifier of feed a load. If a earth Ú i dt rating 2
2
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 =
input input peak peak volta voltage ge total total circuit circuit resistan resistance ce
=
100 1
= 100 A.
Solution Manual 2
i dt = 20
\
t f =
t f
Ú 100 · dt = 20
\
2
Since,
20 1002
2
o
= 2 m sec.
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.
I m
Q1
O
π
π
π
Fig. 3.1
I av =
I rms =
1
p
I 2p Ú
m
q 1
sin q d q =
I m 2p
(1 + cos q 1)
È 1 p 2 2 ˘1 2 Í Ú I m sin q d q ˙ = ÍÎ 2p q ˙˚
È I m2 p Ï 1 cos 2 q ¸˘ 1 2 ˝˙ Í Ú Ì 2 ˛˙ ÍÎ 2p q Ó 2 ˚
1
=
1
È I 2 Ïq sin 2 q ¸p ˘ 1 2 Í mÌ ˝ ˙ 4 ˛q ˙˚ ÍÎ 2p Ó 2 1
=
È I m2 Í 2p Î
{
p - q 1 2
+
1 4
2
sin q 1
}
˘ ˙ ˚
12
(a) For 180° conduction angle, q 1 = 0°
\ and
I av =
I rms =
\
Form factor (FF) =
\
I TAV =
I m 2p
[(1 + cos 0°)] =
È I m2 Í 2p Î I rms I av I rms FF
{
}
p 1 - (0) 2 4
=
=
I m
p
˘ 1 2 I m ˙ = 2 ˚
p p = 2 I m 2
I m
·
35 ¥ 2
p
(b) For 90°, conduction angle, q 1 = 90°.
= 22.282 A.
3
Power Electronics
\
I av =
and
I rms =
\
Form factor =
\
I TAV =
I m
I m
[1 + cos q 0 ] = 2p
È I m2 Í 2p Î I m
{
}
p 1 + (0) 4 4
¥
2 2
2p I m
35 ¥ 2
p
-
p 2
2p
˘1 2 ˙ = ˚
I m 2 2
.
= 15.755 A.
(c) Fro 30°, conduction angle, q 1 = 150°.
\
I av =
I rms =
\
Form factor =
\
I Tav =
2.4
I m 2p
[1 + ( -0.866)] = 0.021 I m.
È I m2 Í 2p Î
{
0.0849035 I m 0.021 I m 35 3.98
˘1 2 ˙ = 0.085 I m. ˚
}
p 1 + ( -0.866) 12 4 = 3.98
= 8.79 A
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 = I av =
I rms =
T
hT
¥ 360° 360°
conduction angle I
¥ T = I
hT
h
È I 2 ¥ T ˘ 1 2 Í hT ˙ = Î ˚
I
h
Solution Manual 4
(a) For 180° conduction angle, h = I
\
I av =
\
Form factor =
\
I Tav =
360 180
I
·
2
2 I 35
\
Form factor =
\
I Tav =
I 4
=
360
2
I
and I rms =
4
=
I 2
.
I 4 · = 2 2 I
I av =
35 2
I 12
= 17.5 A.
12 I 35 12
360 12
= 12
and I rms =
I 12
Form factor =
2.5
.
= 4
90
(c) For 30° conduction angle and h =
I TAV =
2
= 24.75 A.
2
(b) For 90° conduction angle, h = I av =
I
and I rms =
2
\
= 2.
=
I 12
12
= 10.10 A
In a class-C commutations circuit, determine the value of R, R L, 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 m sec. 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.
Sol.
Now,
R L =
E dc I L
V c = E dc .
=
100 15
= 6.66 W.
Ê - t ˆ Á1 - e R ˜ ÁË ˜ ¯ C
t = 30 m s, E dc = 200 V, V c = 100 V.
\
100 =
Ê - t ◊ ˆ 200 Á1 - e R C ˜ ÁË ˜ ¯ L
5
Power Electronics
–
-
t RC ◊ C
= loge 0.5 = – 0.693 t
\
C =
\
C = 6.49 m F R >
\
R >
2.6
=
0.693 R L
Select R such that
30 ¥ 10-6 0.693 ¥ 6.66
E dc 3 mA 100 V 3 ¥ 10 - 3
\ 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 m s. The auxiliary SCR 2 should undergo natural commutation when its forward current goes below its holding current of 2.5 mA. +
R 1
R 2
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 25
= 4 W
The charging equation for C is V c =
Ê - tq ˆ V Á1 - e R C ˜ ÁË ˜ ¯
\
50 =
Ê - tq ˆ 100 Á1 - e 4C ˜ Ë ¯
\
tq = 0.693 ¥ 4C = 2.77 C.
\
C =
1
30 ¥ 10-6 2.77
The commutation capacitor C ≥ 10.82 m F
= 10.82 m F .
Solution Manual 6
E 2.7 For the class-C commutation, R1 = R2 = 5 W, C = 10 m F , supply voltage E dc = 100 V. Determine the turns-off time. Sol. Capacitor charging equation is
Vc = E dc
\
100 =
\
Ê - tc ˆ Á1 - e R C ˜ ÁË ˜ ¯ 1
Ê - tc ˆ 200 Á1 - e R C ˜ ÁË ˜ ¯ 1
tc = 0.693 R1C = 0.693 ¥ 5 ¥ 10 ¥ 10 = 34.65 m sec. –6
E 2.8 Determine the commutating components for an auxiliary commutation method if I L = 10A, E dc = 100 V and tq = 50 m s. Sol. We have the equation
C =
=
Also,
Lmin =
=
Lmax =
I L max ◊ tq E dc 10 ¥ 50 ¥ 10 -6 100 E dc 2 I L2
= 5 m F
¥ C
È100 ˘2 ¥ 5 ¥ 10 –6 = 0.5 mH ÍÎ 10 ˙˚ 0.01T 2
p 2 ◊ C
Assume operating frequency of the circuit to be 400 Hz
\ \
T = 2.5 m s Lmax = 1.26 mH.
E 2.9 In class-D commutation circuit, E dc = 200 V, L = 10 m H, C = 50 m F . 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 m sec. Peak SCR current is given by, Ip = E dc = 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 m s. 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
R .
100 V 0.5 H
–
Fig. E 2.10 Sol. For the given load,
t = L/ R =
0.5 20
= 0.025 sec.
100
Max. value of steady state current =
= 5 A. 20 Without R, SCR current i will grow exponentially as, i =
= at
Ê - t ˆ I 0. Á1 - e t ˜ Ë ¯ È - ÊÁË t ˆ ˜ ¯ ˘ 5 Í1 - e 0.025 ˙ Í ˙ Î ˚
t on = t = 50 m sec, È - 50 ¥ 10-6 ˘ –3 i = 5 Í1 - e 0.025 ˙ = 9.99 ¥ 10
ÍÎ
˙˚
= 10 mA OR
Since given that initially
\
di dt
t on = i
\
t
is constant over entire time ion, I 0
t
=
5 0.025
= 200 A/sec.
= 50 ¥ 10 = 200 ¥ 50 ¥ 10 = 10 mA –6
–6
\ Hence if trigger pulse with is 50 m sec, 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 R = 100 V
\
R =
–3
100 40 ¥ 10-3
= 2.5 k W.
