HW5_Solutions

# HW5_Solutions - Homework Solutions HW 5 5.38 Base-Emitter...

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Homework Solutions ECE302 Spring 2007 HW 5 1 5.38 Base-Emitter Voltage Base-Collector Voltage 0.7 V -5.0 V -5.0 V Reverse Active Cutoff 0.7 V Saturation Forward Active 5.46 a ( 29 pnp transistor with V EB =- 3 V and V CB =- 3 V Cutoff | Using Eq. (5.17): I C =+ I S b R = 10 - 15 A 2 = 0.5x10 -15 = 0.5 fA | I E =- I S b F = 10 - 15 A 75 = 13.3x10 -18 = 13.3 aA I B =- I S 1 b F + 1 b R = 10 - 15 A 1 75 + 1 4 = 0.263 x 10 - 15 = 0.263 fA b (29 npn transistor with V BE =- 5 V and V BC =- 5 V Cutoff | The currents are the same as in part (a). 5.48 An npn transistor with V BE = 0.7 V and V BC =- 0.7 V Forward- active region Using Eq. (5.45): I E = b F + 1 ( 29 I B | b F = I E I B - 1 = 10 mA 0.15 mA - 1 = 65.7 I E = I S 1 + 1 b F exp V BE V T | I S = 0.01 A 1 + 1 65.7 exp 0.7 0.025 = 6.81 x 10 - 15 A = 6.81 fA

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Homework Solutions ECE302 Spring 2007
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HW5_Solutions - Homework Solutions HW 5 5.38 Base-Emitter...

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