solution3 - ECE 442 Spring 2009 HW#3 Solutions 1 Note VCC...

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ECE 442 Spring 2009 HW#3 Solutions 1. Note: V CC denotes the top (positive) power supply, V EE denotes the negative one. a) V E = - 0 . 7 V I E = V E - V EE R E = - 0 . 7 - ( - 10 . 7) 10 k = 1 mA β ± 1 I 1 = I C = I E = 1 mA b) I C = V C - V EE R E = - 4 - ( - 10) 2 . 4 k = 2 . 5 mA β ± 1 I E = I C = 2 . 5 mA V 2 = V CC - I E R E = 12 - 2 . 5 m · 5 . 6 k = - 2 V c) I C = V C - V EE R C = 0 - ( - 10) 10 k = 1 mA β ± 1 I 3 = I C = I E = 1 mA Also, β ± 1 I B = 0 no current through R B V 4 = V CC = 1 V d) β ± 1 I B = 0 no current through 10 k Ω resistor V 6 = V C = V E + V BE = V E + 0 . 7 V I 5 = I E = V E - V EE R E = ( V C - 0 . 7) - ( - 10) 5 k I C = V CC - V C R C = 10 - V C 15 k β ± 1 I E = I C V C +9 . 3 5 k = 10 - V C 15 k V C = V 6 = - 4 . 475 V Now I 5 = - 4 . 475 - 0 . 7+10 5 k = 0 . 965 mA 2. I C = 5 mA, V C = 3 V R C = V CC - V C I C = 15 - 3 5 m = 2 . 4 k Ω I C = I S e v BE V T v BE = ln ( I C I S ) · V T Since v BE = 0 . 7 V at 1 mA , the value at desired
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This note was uploaded on 09/19/2009 for the course ECE 442 taught by Professor Schutt-aine during the Spring '08 term at University of Illinois at Urbana–Champaign.

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solution3 - ECE 442 Spring 2009 HW#3 Solutions 1 Note VCC...

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