hw12sol - EE105 Problem Set#12 Solutions Spring 2006 1...

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EE105 Problem Set#12 Solutions Spring 2006 1. SOLUTIONS: α F = β F /(1+ β F ) = 100/(100+1) = 0.99 α R = β R /(1+ β R ) = 0.1/(0.1+1) = 0.091 I ES = I S / α F = 1E-16/0.99=1.01E-16A I CS = I S / α R = 1E-16/0.091=1.1E-15A (a) V BE = 0.7V, V BC = V BE - V CE = 0.7-5 = -4.3V The transistor is in forward active region, I C = α F I ES [exp(V BE /V th )-1]-I CS [exp(V BC /V th )-1] = 0.99*1.01E-16*[exp(0.7/0.026)-1]-1.1E-15*[exp(-4.3/0.026)-1] = 4.927E-5A I E = -I ES [exp(V BE /V th )-1]+ α R I CS [exp(V BC /V th )-1] = -1.01E-16*[exp(0.7/0.026)-1]+0.091*1.1E-15*[exp(-4.3/0.026)-1] = -4.976E-5A I B = -I B E -I C = 4.9E-7A Or I B = I B C / β F = 4.926E-5/100=4.926E-7A (b) V BE = 0.7V, V BC = V BE - V CE = 0.7-0.2 = 0.5V The transistor is in saturation region, I C = α F I ES [exp(V BE /V th )-1]-I CS [exp(V BC /V th )-1] = 0.99*1.01E-16*[exp(0.7/0.026)-1]-1.1E-15*[exp(0.5/0.026)-1] = 4.902E-5A I E = -I ES [exp(V BE /V th )-1]+ α R I CS [exp(V BC /V th )-1] = -1.01E-16*[exp(0.7/0.026)-1]+0.091*1.1E-15*[exp(0.5/0.026)-1] = -4.974E-5A I B = -I B E -I C = 7.2E-7A NOTE: in saturation we cannot use the following equation,
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This homework help was uploaded on 04/01/2008 for the course EECS 105 taught by Professor Mingwu during the Spring '06 term at Berkeley.

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hw12sol - EE105 Problem Set#12 Solutions Spring 2006 1...

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