121B_1_bjt

# Jason woo 8 ee121b spring 2010 to solve the equations

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Unformatted text preview: ed the boundary conditions: E B NDE -xE-WE -xE 0 NAB WB C NDC xC WB is neutral base width, WE is neutral emitter width Recall that when a bias VJ is applied to a PN junction, the minority carrier concentrations at the depletion edges are: ni2 qVJ n ( xP ) = exp NA kT Jason Woo ni2 qVJ p ( xn ) = exp ND kT 9 EE121B Spring 2010 Therefore, in the BJT case: In the quasi-neutral base: ni2 qVBE n ( 0 ) = exp kT N AB - 1 ni2 qVBC n ( W B ) exp N AB kT -1 ( 0) In the quasi-neutral emitter: ni2 p ( - x E ) = N DE qVBE exp kT - 1 p ( -W E - x E ) 0 Also in the quasi-neutral collector: ni2 p ( xC ) = N DC Jason Woo qVBC exp kT 10 - 1 EE121B Spring 2010 The general solution for the electron equation in the base: 2 n n = 2 2 LB x is given by: x n ( x ) = A exp LB -x + B exp LB With the B.C. at x=0 and x=W, we have: n ( x ) = and, n ( 0 ) sinh ( (WB - x ) / LB ) sinh (WB / LB ) n (WB ) sinh ( x / LB ) + sinh (WB / LB ) qDn n ( 0 ) cosh ( (WB - x ) / LB ) n (WB ) cosh ( x / LB ) - Jn ( x ) = + LB sinh (WB / LB ) sinh (WB / LB ) 11 EE121B Spring 2010 Jason Woo n n(0) Decreasing WB/Lp n(W) 0 0.2 0.4 0.6 0.8 1 x/WB We see that, if WB LB I Cn I En Jason Woo 12 EE121B Spring 2010 Similarly, in the quasi-neutral emitter, 2 p p = 2 2 x LE with B.C. p ( x ) = p ( i.e., sinh x...
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## This note was uploaded on 06/01/2010 for the course EE EE161 taught by Professor Wang during the Spring '09 term at UCLA.

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