# hw4 - 5.58 (a) Substituting iC = 0 in Eq. 5.30 gives 1 1...

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5.58 (a) Substituting i C = 0 in Eq. 5.30 gives V CESAT = V T ln 1 α R = 0.025 V ( 29 ln 1 0.5 = 0.0173 V = 17.3 mV (b) By symmetry V ECSAT = V T ln 1 F or by using i E = 0 and i C = -i B , V CESAT = V T ln 1 R 1 - 1 β R + 1 1 + 1 F = V T ln 1 R R R + 1 F + 1 F = V T ln 1 R R 1 F V CESAT = V T ln F ( 29 and V ECSAT = V T ln 1 F V ECSAT = V T ln 1 F = 0.025 V ( 29 ln 1 0.99 = 0.000251 V = 0.251 mV 5.93 Writing a loop equation starting at the 9 V supply gives: 9 = 1500 I C + I B ( 29 + 10000 I B + V BE Assuming forward-active region operation, V BE = 0.7 V and I C = β F I B . 9 = 1500 F I B + I B ( 29 + 10000 I B + 0.7 I B = 9 - 0.7 1500 F + 1 ( 29 + 1000 and I C = F I B = F 9 - 0.7 ( 29 1500 F + 1 ( 29 + 1000

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## This note was uploaded on 01/12/2012 for the course ECE 255 taught by Professor Staff during the Fall '08 term at Purdue.

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hw4 - 5.58 (a) Substituting iC = 0 in Eq. 5.30 gives 1 1...

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