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Week7HW Solutions

# Solution n d 2 n n begin with dn gl t dx 2 n

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Unformatted text preview: ncentration is virtually unchanged. But the electron QFL is much closer to the conduction band because there are orders of magnitude more electrons. The sample has been uniformly illuminated with light for a long time. The optical generation rate is GL = 1020 cm- 3 sec- 1. At t = 0, the light is switched off. Find the excess minority carrier concentration and the QFL’s vs. time. Assume spatially uniform conditions, and approach the problem as follows. 6a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ∂ Δn d 2 Δn Δn Begin with: = Dn − + GL ∂t dx 2 τn ECE- 606 7 Spring 2013 Mark Lundstrom 2/24/2013 Simplify for spatially uniform conditions with no generation: d Δn Δn =− dt τn 6b) Specify the initial and boundary conditions, as appropriate for this problem. Solution: Because there is no spatial dependence, there is no need to specify boundary condition. The initial condition is (from prob. 5): Δn ( t = 0 ) = 1014 cm -3 6c) Solve the problem. Solution: d Δn Δn The solution is: Δn ( t ) = Ae− t /τ n =− dt τn Now use the initial contition: Δn ( 0 ) = 1014 = A Δn ( t ) = 1014 e− t /τ n 6d) Provide a sketch of the solution, and explain it in words. Solution: ⎛ Δn ( t ) + n0 ⎞ For the electron QFL: Fn ( t ) = Ei + kBT ln ⎜ ⎟ ni ⎝ ⎠ ECE- 606 8 Spring 2013 Mark Lundstrom 7) 2/24/2013 Initially, Δn ( t ) >> n0 and Δn ( t ) = Δn ( 0 ) e− t /τ n , so Fn(t) initially drops linearly with time towards EF. The sample is uniformly illuminated with light, resulting in an optical generation rate GL = 1020 cm- 3 sec- 1. The minority carrier lifetime is 1 μsec, except for a thin layer (10...
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