Unformatted text preview: his problem. Since this is a steady state problem, there is no initial condition. At x = L , we expect all of the minority carriers to have recombined, so: Δn ( x = L ) = 0 At the surface, the total number of e h pairs generation per cm2 per second is GS = GL Δx = 102410−6 = 1018 cm 2s1 . In steady state, these must diffuse away at the same rate that they are generated, so − Dn d Δn
= GS dx x =0 9c) Solve the problem. Solution: d 2 Δn
= 0 solutions is Δn ( x ) = Ax + B dx 2 To satisfy the first boundary condition in 9b): Δn ( L ) = AL + B = 0 . B = − AL Now consider the second: D
GS
d Δn
1018
− Dn
→ − n A = GS → A = −
=−
= −6.4 × 1013 cm 3 −4
dx x=0
L
( Dn L) 7.8 5 × 10 ECE 606 16 Spring 2013 Mark Lundstrom Δn ( x ) = ( 2/24/2013 ) GS
( L − x ) = 6.4 × 1013 ( L − x ) ( Dn L) 9d) Provide a sketch of the solution, and explain it in words. Concentration increases towards surface, because generation occurs the. Is zero at x = L because of the boundary condition there. Variation is linear with position because there is no recombination. Electron QFL(x) follows from n ( x ) ≈ Δn ( x ) = ni e − ( Fn ( x )− Ei ) / k BT . 10) The sample is in the dark, but the excess carrier concentration at x = 0 is held constant at Δn ( 0 ) = 1012 cm 3. Find the steady state excess minority carrier concentration and QFL’s vs. position. You may assume that the sample extends to x = +∞ . Make reasonable approximations, and approach the problem as follows. 10a) Simplify the Minority Carrier Diffusion Equation for this problem...
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 Fall '08
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