CHAPTER 2.1
2
When thyristor is turned-ON, it behaves like a diode. Therefore, with SCR-ON, the device acts like a closed switch. Apply KVL to the loop, we get L
di 1 + i dt = ES. dt c Ú
Solution of this equation is given by C sin w0 t L
i(t) = ES Here, W0 =
1 LC
is called as resonant frequency of the circuit. =
1 –6
10 ¥ 10 ¥ 10 ¥ 10
(i) conduction time of SCR = t1 =
–6
= 105 rad/s.
p p = 5 = 31.42 ms w0 10
(ii) capacitor voltage is given by VC = ES – (ES – VCO) cos w0 t + w0 L IO sin w0 t
p = 200 – (200 – 0) cos ÊÁ105 ¥ 5 ˆ˜ Ë 10 ¯ p + 105 ¥ 10 ¥ 10–6 ¥ 250 sin ÊÁ105 ¥ 5 ˆ˜ Ë 10 ¯ VC = 400 V 2.2
Charging current of a capacitor is given by, ic = C
dv dt
5 ¥ 10–3 = 25 ¥ 10–12 ¥
\ 2.3
and
dv dt
= 200 V/ms
Vg Ig
= 130
dv dt
We have
Pg = Vg ◊ Ig = 0.5
…(i) (ii)
Solution Manual 2
Vg = 130 Ig, substitute in Eq. (ii).
\
0.5 = 130 Ig2
\
Ig = 62 mA
and,
Vg = 8.06 V
\ Now,
Egs = Rg ◊ Ig + Vg 15 = Rg (62 ¥ 10-3) + 8.06 Rg = 112 W.
\ 2.4
Given: Egs = 15 V,
Igmin = 25 mA
(i) Slope of the load line gives the required gate source resistance Rs
\
Rs = 120 W
(ii) For the gase circuit, we have the relation, Egs = Rg ◊ Ig + Vg
(i)
But, we have Vg ◊ Ig = 0.4 0.4 Ig
\
Vg =
\ (i) fi,
15 = 120 ◊ Ig +
120 Ig2 – 15 Ig + 0.4 = 0
\ \ \
0.4 Ig
Igmax = 86.45 mA Vg =
0.4 86.45 ¥ 10-3
= 4.63 V
\ For minimum gate current of 25 mA, take Vg = 4.63 V and Ig = 86.45 mA. 2.5 ic = C
dv dt
ic = 30 ¥ 10-6 ¥ 150 = 4.5 mA 2.6
Current i(t) = whose
\
V t (1 – e–t/ ) R
t = i(t) =
L 0.2 = = 0.01 sec R 20 200 (1 = e–t/0.01) 20
\
100 ¥ 10–3 = 10 (1 = e–ton/0.01)
\
ton = 100.5 m sec.
3
Power Electronics
2.7
We have the relation, PIV =
2 ¥ Vin ¥ Vf
1500 =
\ 2.8
2 ¥ 415 ¥ Vf
Vf = 2.56
(a) From Eq. (2.11), gate pulse width tp = =
0.2
= 50 msec.
4 ¥ 103
1w =5w 0.2
2.8
(b) From Eq. (2.12), Pgmax ≥
2.9
During conduction, Vgs = Rg ◊ Ig + Vg but
\
d f
Vg = (1 + 10 Ig) Vgs = Rg ◊ Ig + 1 + 10 Ig 20 = (Rg + 10) Ig + 1
Now,
Pgav = Pgmax ◊ 0.5 = Vg ◊ Ig
\ \
180∞ 360∞ 1 2
0.5 = (1 + 10 Ig) Ig ◊
1 2
Ig = 0.27 A
Substitute in (i), 20 = (Rg + 10) (0.27) + 1
\
Rg = 60 W
Now Vg = 1 + 10 Ig = 1 + 10 (0.27) = 3.7 V 2.10 R1 = 5 W, Edc = 120 V, toff = 15 ms For minimum value of c, take toff = 10 msec toff R1
We have,
c = 1.44
\
10 c = 1.44 ÊÁ ¥ 10-6 ˆ˜ Ë5 ¯
@ 3 mF 15 ms = R2 ¥ 3 ¥ 10–6 ¥ 0.693
Now,
R2 @ 5 W
\
2.11 R1 = 10 ohms, Edc = 100 V, toff = 50 mF we have,
\
toff = 0.6931 R1C C = 7.15 mF
(i)
