MD Singh Power Electronics Solution Manual to Chapter 13

CHAPTER

13

13.1 Fusing time,

tc =

\

£

Edc Io

=

3 I 2t I p2 3 ¥ 150 ¥ 103 4

\

Ip = 335.41 A.

13.2 Tj = 130°C,

TA = 50°C, P = 30 W.

(a)

Q =

Thermal resistance,

130 - 50 30

= 2.7∞ C/W. (b)

Thermal resistance associated with given time Qth = =

Temperature difference (rise) Power - loss in device over defined time 100 3

330 ¥ 10

= 0.303 ¥ 10-3 C/W.

Now, power supplied/dissipated =

150 - 50 0.303 ¥ 10-3

= 330 kW

Hence circuit can supply 330 kW load for 5 minutes. 13.3

P =

\

Q =

Now,

T1 - T2 Q 30 - 20 3

100 ¥ 10

= 0.1 ¥ 10-3 ∞C/W .

temp. rise = 30∞ + ( Power loss × total thermal resistance) = 30∞ + (100 ¥ 103 ¥ 0.1 ¥ 10-3 ) = 40°C.

13.4 A non polarised form of selenium limiter is to be used in this case. Vrms Number of plates in each direction = 30

68

Power Electronics

=

230 20

= 7.67 @ 8.

\

= 2 ¥ 8 = 16

Total no. of plates

the following arrangement can be used.

1

2

16

Fig. 13.1 13.5 we have power-loss P =

T1 T2 Q =

125 - 40 = 22.4 W 2 t 1.8

13.6 The steady junction temperature at 300 W loss = 30 + (300 ¥ 0.2) = 90°C. Hence, during 100 ms overload, the junction temperature can rise another. 125 – 90 = 35°C.

\

Additional power

\

Total overload

loss =

35 = 700 W. 0.05

power = 300 + 700 = 1000 W.

13.7 On state power loss is given by P =

120 - 45 = 4160.67 W 0.1 + 0.08