330hw3-solutions(1).pdf

# B using the depletion approximation sketch the charge

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(b) Using the depletion approximation, sketch the charge density as a function of x for this diode. (c) Derive an analytical solution for the electric field E ( x ) in this diode. (d) Derive an analytical solution for the voltage V ( x ) in this diode. x N D - N A - N A N D / 5 N D x 0 330hw3-solutions.tex Spring 2018

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P. R. Nelson 3 (a) Because x n > x 0 only the dopings at the edges of the deple- tion region are needed to determine the built-in voltage. It is determined by N A and N D . (b) x ρ - x p - qN A qN D / 5 N D x 0 x n (c) Solve Poisson’s equation in one dimension, d E dx = ρ K S ϵ 0 ρ = - qN A - x p x < 0 qN D / 5 0 x < x 0 qN D x 0 x < x n 0 elsewhere Let E = 0 for x < - x p and x > x n . Then p quasi-neutral region: ρ = 0 so E = 0 . p-side depletion region: E = - qN A K S ϵ 0 ( x + x p ) n quasi-neutral region: ρ = 0 so E = 0 . n-side depletion region: This region is more complicated be- cause there are two sections with different dopings. E = qN D 5 K S ϵ 0 ( x + A ) 0 x < x 0 QN D K S ϵ 0 ( x - x n ) x 0 x x n E is continuous at x 0 , so A = 4 x 0 - 5 x n 330hw3-solutions.tex Spring 2018
P. R. Nelson 4 E = - qN A K S ϵ 0 ( x + x p ) - x p x < 0 qN D 5 K S ϵ 0 ( x + 4 x 0 - 5 x n ) 0 x < x 0 QN D K S ϵ 0 ( x - x n ) x 0 x < x n 0 elsewhere (d) Let V = 0 for large negative x . Also, V ( x n ) = V bi . p quasi-neutral region: E = 0 so V = 0 p-side depletion region: V = - E dx = qN A 2 K S ϵ 0 ( x + x p ) 2 n-side depletion region: V = - E dx = - qN D 10 K S ϵ 0 ( x + 4 x 0 - 5 x n ) 2 + C 1

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• Spring '18
• Shaolei
• P-n junction, Xn, Vbi, P. R. Nelson

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