MD Singh Power Electronics Solution Manual to Chapter 02

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)