1219_Physics ProblemsTechnical Physics

# 1219_Physics ProblemsTechnical Physics - 560 Molecules and...

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560 Molecules and Solids Section 43.7 Semiconductor Devices P43.42 IIe eVkT B =− 0 1 af ej . Thus, e I I B =+ 1 0 and V kT e I I B F H G I K J ln 1 0 . At T = 300 K, V I I I I = × × + F H G I K J F H G I K J 138 10 300 160 10 12 5 91 23 19 00 . . ln . ln JK K C mV . (a) If II = 900 0 ., V == 25 9 10 0 59 5 .l n . . mV mV a f . (b) If 0900 0 V 25 9 0 100 59 5 n . . mV mV a f a f . The basic idea behind a semiconductor device is that a large current or charge can be controlled by a small control voltage. P43.43 The voltage across the diode is about 0.6 V. The voltage drop across the resistor is 0 025 150 3 75 .. A V a f Ω= . Thus, ε −−= 06 38 0 V V and = 44 . V. P43.44 First, we evaluate I 0 in B 0 1 , given that I = 200 mA when V = 100 mV and T = 300 K. eV B = × × = 0100 386 19 23 . . C V K so I I e e B 0 1 200 1 428 = = = mA mA . . If V 100 mV, B . ; and the current will be e B = = 0 1 4 28 1 4 19 . mA mA . *P43.45 (a) The currents to be plotted are Ie D V 10 1 60 0 2 5 A V e j . , I V W = 242 745 . V The two graphs intersect at V =
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