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Unformatted text preview: ECE 132 Quiz 5 Name __________________________ November 12, 2010 In a semiconductor the diffusion of excess holes is described in general by the diffusion equation, given below. The semiconductor below has reached steady‐state and has no recombination in the region 0 to x1. ∂δp ∂ 2δp δp − = Dp ∂t ∂x 2 τ p
1) The only solution for δp(x) in a region which has no recombination and has reached steady‐state is a straight line. Given this and the boundary conditions δp(x=x1) and δp(x=0) drawn below: a. Sketch the solution for δp(x) in the region 0 to x1. b. Write the expression for δp(x) without doing any calculus. δp(x) δp(x=x1) δp(x=0) 0 x1 x 2) Explain how the diffusion current density Jp(x) is related to δp(x). Give an equation if you are able. ‐More problems on reverse side‐ 3) In a region with no recombination, what do you expect to be the shape of the hole diffusion current density Jp(x)? (Circle one) a. Straight line: J p ( x) = c1 ⋅ x + c 2 b. Exponential: J p ( x) = c1 + c 2 ⋅ e ± x / c3 c. Constant: J p ( x) = c 4) Returning to the full diffusion equation, what value must τ p (the hole recombination time) have if there is no recombination in a region? (Circle one) a. 0 b. ∞ c. finite and non‐zero 5) In a system which has reached steady state, what will the value of (Circle one) a. 0 b. ∞ c. finite and non‐zero 6) Given your responses to (4) and (5), simplify the full diffusion equation as much as you can. ∂δp ( x) be? ∂t ...
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