ch05_solutions - CHAPTER 5 5.1 Base contact = G collector...

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121 CHAPTER 5 5.1 Base contact = G collector contact = F emitter contact = D n-type emitter = E n-type collector = B active region = A 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 5 m A = 55. 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 V BE V T = 275 m A 0.63 0.025 = 3.13 fA 5.3 + + - v BC i E i C i B C E B - + - V v BE For V BC and V BE = 0, I E = - b R I B or b R = - I E I B = - - 275 m A 125 m A = 2.20 b F = a F 1 - a F = 0.015 1 - 0.015 = 0.015 I E = - I S V V T or I S = I C V BE V T = 275 m A 0.63 0.025 = 3.13 5.4 Using b = a 1 - a and a = b b + 1 : Table 5.P1 α β 0.167 0.200 0.400 0.667 0.750 3.00 0.909 10.0 0.980 49.0 0.995 200 0.999 1000 0.9998 5000
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122 5.5 (a) For this circuit, V BE = 0 V, V BC = -5 V and I = I C . Substituting these values into the collector current expression in Eq. (5.13): I C = I S exp 0 ( 29 - exp - 5 .025 - I S b R - 5 .025 - 1 I = I C = I S 1 + 1 b R = 10 - 15 A 1 + 1 1 = 2 fA . (b) For this circuit, the constraints are V BC = -5 V and I E = 0. Substituting these values into the emitter current expression in Eq. (5.13): I E = I S V BE V T - V BC V T + I S b F V BE V T - 1 = 0 which gives V BE V T = 1 1 + b F + b F 1 + b F V V T . Substituting this result into I C : I CBO = I S 1 + b F 1 - V V T - I S b R V V T - 1 . For V BC = -5V, I CBO = I S 1 1 + b F + 1 b R = 10 - A 1 101 + 1 1 = 1.01 , and V BE = V T ln 1 1 + b F = 0.025 V 1 101  = - 0.115 V 0! 5.6 (a) - (c) 150 μ A C E B i B i E i C + - + - v BC v BE (b) npn transistor (d) V BE = V BC e ( 29 I C = - I S b R V BE V T - 1 I E = + I S b F V BE V T - 1 I B = I S 1 b F + 1 b R V BE V T - 1 I E I B = 1 1 + b F b R and I E I C = - b R b F
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123 f ( 29 Using I C = - b F b R I E = - 400 I E and I B = I E - I C = 401 I E For the circuit I B = 150 m A Therefore I E = 150 m A 401 = 0.374 m A , and I C = - 149.6 m A . V BC = V BE = V T ln I B I S 1 b F + 1 b R = 0.025 V ( 29 150 m A 2 fA 1 100 + 1 0.25 = 0.591 V 5.7 C B E + - + - V BC V BE I C I B I E 150 μ A npn transistor For V = 0, I E = I S 1 + 1 b F exp V BE V T - 1 | I B = I E b F + 1 I C = b F I B I E = 150 m A I B = 150 m A 101 = 1.49 m A I C = 100 101 150 m A = 149 m A V BE = V T b F b F + 1 I E I S + 1 = 0.025 V 100 101 150 m A 2 + 1 = 0.626 V 5.8 C B E + - + - V BE V BC I E I B I C 150 μ A For V BE = 0, I C = - I S 1 + 1 b R V V T - 1 I B = - I C b R + 1 I E = - b R I B I C = - 150 m A I B = - - 150
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This note was uploaded on 01/17/2011 for the course ECE 3040 taught by Professor Hamblen during the Fall '07 term at Georgia Institute of Technology.

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ch05_solutions - CHAPTER 5 5.1 Base contact = G collector...

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