HW4_Solutions

HW4_Solutions - Homework Solutions ECE302 Spring 2007 HW 4...

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Unformatted text preview: Homework Solutions ECE302 Spring 2007 HW 4 1 5.2 V C B E +- +- v BE- + v BC i C i E i B For V BE 0 and V BC = 0, I C = b F I B or b F = I C I B = 275 m A 4 m A = 68.8 b R = a R 1- a R = 0.5 1- 0.5 = 1 I C = I S exp V BE V T or I S = I C exp V BE V T = 275 m A exp 0.64 0.025 = 2.10 fA 5.8 C B E +- +- V BE V BC I E I B I C 175 A npn transistor For V BE = 0, I C = - I S 1 + 1 b R exp V BC V T - 1 | I B = - I C b R + 1 | I E = - b R I B I C = - 175 m A | I B = -- 175 m A 1.25 = 140 m A | I E = - 0.25 1.25 175 m A = - 35 m A V BC = V T ln- b R b R + 1 I C I S + 1 = 0.025 V ln 0.25 1.25 175 m A 2 fA + 1 = 0.590 V Homework Solutions ECE302 Spring 2007 HW 4 2 5.15 (a) pnp (b)-(c) + +- V EB I C I E I B C E B- +- V V CB (d) Using Eq. (5.17) with v CB = 0 and droping the "-1" terms:...
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This note was uploaded on 03/30/2008 for the course ECE 302 taught by Professor Barlage during the Spring '07 term at N.C. State.

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HW4_Solutions - Homework Solutions ECE302 Spring 2007 HW 4...

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