Chapter05

Microelectronic Circuit Design

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CHAPTER 5 5.1 Base Contact = B Collector Contact = A Emitter Contact = C n-type Emitter = D n-type Collector = F Active Region = E 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 = β F I B or F = I C I B = 275 μ A 4 A = 68.8 R = α R 1 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 A exp 0.64 0.025 = 2.10 fA 5.3 + + - v BC i E i C i B C E B - + - V v BE For V BC > 0 and V BE = I E =− R I B or R I E I B 275 A 125 A = 2.20 F = F 1 F = 0.975 1 0.975 = 39 I E I S exp V BC V T or I S = I C exp V BE V T = 275 A exp 0.63 0.025 = 3.13 fA 123
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5.4 Using β = α 1 and = + 1 : Table 5.P1 & & 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 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 () exp 5 .025 I S R exp 5 .025 1 I = I C = I S 1 + 1 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 exp V BE V T exp V BC V T ⎥ + I S F exp V BE V T 1 ⎥ = 0 which gives exp V BE V T = 1 1 + F + F 1 + F exp V BC V T . Substituting this result into I C : I CBO = I S 1 + F 1 exp V BC V T ⎥ − I S R exp V BC V T 1 . For V BC =-5 V , I CBO = I S 1 1 + F + 1 R = 10 15 A 1 101 + 1 1 = 1.01 fA and V BE = V T ln 1 1 + F = 0.025 V ln 1 101 =− 0.115 V 0! 5.6 124
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(a) - (c) 150 μ A C B E + - V BE + - V BC I E I C I B (b) npn transistor (d) V BE = V BC e () I C =− I S β R exp V BE V T 1 I E =+ I S F exp V BE V T 1 I B = I S 1 F + 1 R exp V BE V T 1 I E I B = 1 1 + F R and I E I C R F f Using I C F R I E 400 I E and I B = I E I C = 401 I E For the circuit I B = 150 μ A Therefore I E = 150 A 401 = 0.374 A , and I C 149.6 A . V BC = V BE = V T ln I B I S 1 F + 1 R = 0.025 V ln 150 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 175 μ A npn transistor For V BC = 0, I E = I S 1 + 1 F exp V BE V T 1 | I B = I E F + 1 I C = F I B I E = 175 A I B = 175 A 101 = 1.73 A I C = 100 101 175 A = 173 A V BE = V T ln F F + 1 I E I S + 1 = 0.025 V ln 100 101 175 A 2 fA + 1 = 0.630 V 125
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5.8 C B E + - + - V BE V BC I E I B I C 175 μ A npn transistor For V BE = 0, I C =− I S 1 + 1 β R
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This note was uploaded on 03/05/2008 for the course EE 307 taught by Professor Braun during the Spring '08 term at Cal Poly.

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Chapter05 - CHAPTER 5 5.1 Base Contact = B n-type Emitter =...

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