Unformatted text preview: . Solution: Begin with: ∂ Δn
d 2 Δn Δn
= Dn
−
+ GL ∂t
dx 2
τn Simplify for steady state: 0 = Dn d 2 Δn Δn
−
+ GL dx 2
τn No generation: GL = 0 ; ECE 606 17 Spring 2013 Mark Lundstrom the simplified MDE equation is: d 2 Δn Δn
d 2 Δn Δn
Dn
−
= 0 2 −
= 0 dx 2
τn
dx
Dnτ n 2/24/2013 d 2 Δn Δn
− 2 = 0 dx 2
Ln Ln = Dnτ n d 2 Δn Δn
− 2 = 0 where Ln = Dnτ n is the minority carrier diffusion length. dx 2
Ln 10b) Specify the initial and boundary conditions, as appropriate for this problem. Solution: Since this is a steady state problem, there is no initial condition. As x → ∞ , we expect all of the minority carriers to have recombined, so: Δn ( x → ∞ ) = 0 At the surface, the excess electron concentration is held constant, so Δn ( x = 0 ) = 1012 cm 3 10c) Solve the problem. Solution: d 2 Δn Δn
− 2 = 0 solutions is Δn ( x ) = Ae− x / Ln + Be+ x / Ln dx 2
Ln To satisfy the first boundary condition in 10b): B = 0. Now consider the second: Δn ( 0 ) = 1012 cm 3 () Δn ( x ) = Δn ( 0 ) e− x / Ln = 1012 e− x / Ln 10d) Provide a sketch of the solution, and explain it in words. ECE 606 18 Spring 2013 Mark Lundstrom 2/24/2013 Solution: Looks just like the solutions for prob. 8). Only difference is that instead of creating Δn ( 0 ) by generation at the surface, we just specify Δn ( 0 ) directly. 11) The sample is in the dark, and 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. Assume that the semiconductor is only 5 μm long. You may also assume that there is an “ideal ohmic contact” at x = L = 5 μm, which enforces equilibrium conditions at all times. Make reasonable approximations, and approach the problem as follows. 11a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ∂ Δn
d 2 Δn Δn
Begin with: = Dn
−
+ GL ∂t
dx 2
τn d 2 Δn Δn
Simplify for steady state: 0 = Dn
−
+ GL dx 2
τn Generation is zero for this problem: GL = 0 ; the simplified MDE equation...
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 Fall '08
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