Unformatted text preview: Spring 2013 Exam 2: ECE 606 Spring 2013 3c) Solve the MDE for this problem. Solution is: !n ( x ) = Ax + B !n ( 0 ) = B d !n ( x ) dx = A ! Dn Solve for A: A = ! d "n
= ! Dn A = S B "n ( L ) = S B AL + "n ( 0 ) dx ( "n ( 0 ) L + Dn S B ) # ( L " x ) + Dn S B &
!n ( x ) = !n ( 0 ) %
( %
(
$ L + Dn S B '
Check limits: When SB = 0, !n ( x ) = !n ( 0 ) , as expected. When S B ! " , !n ( x ) = !n ( 0 ) (1 " x L ) as expected. 4) This problem concerns a junction with a heavily doped N
type region, a thin intrinsic layer, and a moderately doped P
type region as sketched below. Assume the depletion approximation and assume that the width of the depletion region on the P
side is gre...
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 Fall '08
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