20111ee121B_1_EE121B_HW3_soln

# 20111ee121B_1_EE121B_HW3_soln - 2011 Winter EE121B HW3...

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2011 Winter EE121B HW3 SOLUTIONS 1. a) It is on the same graph in c). b) Dp E = kT/q x 140 = 3.626 cm 2 /s Dn B =kT/q x 230 = 5.957 cm 2 /s Lp E = (Dp E x τ p )^0.5 = 8.52 x 10 -4 cm Ln B = (Dn B x τ n )^0.5 = 1.14 x 10 -2 cm 2 , ln( ) 0.895 dE aB BE i NN kT Vbi qn  2 , ln( ) dc aB BC i kT Vbi W BE : depletion region in Emitter-Base junction W BC : depletion region in Base-Collector junction 2 ( , 0.5) 11 ( ) 2 ( , 10) ( ) BE BE dE aB BC BC dC aB Vbi W q N N Vbi W q N N X 1 = W BE * N dE / (N dE +N aB ) = 7.5 x 10 -6 cm. X 2 =W BC * N dC / (N dC +N aB ) = 3.73 x 10 -6 cm. W b =W Base – X1-X2 = 2.89 x 10 -4 cm. i) Emitter injection efficiency 1 1 1 () (1 ) pE b aB En En En Ep nB pE dE Ep D W N I I I I D L N I    = 0.99588 ii) Base transport factor B=sech (Wb/LnB) = 0.999679 iii) The common base current gain α = B r = 0.99588 x 0.999679 = 0.99556 iv) The common emitter current gain β= α / (1- α) = 224.24 c) with τ n = 10 -9 s. Dp E , Dn B , Lp E are the same. But Ln B = (Dn B x τ n )^0.5 = 7.72 x 10 -5 cm V bi,BE , V bi,BC , W b are the same.

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## This note was uploaded on 03/28/2011 for the course EE 121b taught by Professor Bjt-gamma during the Winter '08 term at UCLA.

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20111ee121B_1_EE121B_HW3_soln - 2011 Winter EE121B HW3...

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