N quasi neutral region v e dx qn d 2 k s ϵ x x n 2 c

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n quasi-neutral region: V = - E dx = - qN D 2 K S ϵ 0 ( x - x n ) 2 + C 2 V ( x n ) = V bi C 2 = V bi From the continuity of V at x = x 0 (and after a bit of algebra) C 1 = V bi + 4( x 0 - x n ) 2 Finally, V = qN A 2 K S ϵ 0 ( x + x p ) 2 - x p x < 0 - qN D 10 K S ϵ 0 ( x + 4 x 0 - 5 x n ) 2 + V bi + 4( x 0 - x n ) 2 0 leqx < x n - qN D 2 K S ϵ 0 ( x - x n ) 2 + V bi x 0 x < x n 0 elsewhere 4. Problem 6.10 in the text (a) Reverse biased. The excess carrier concentrations, which are proportional to (exp( qV A /kT ) - 1) , are negative. Thus V A is negative. are negative, 330hw3-solutions.tex Spring 2018
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P. R. Nelson 5 (b) Yes, low-level injection applies. The excess minority carrier can- centrations are negligiblecompared to the majority carrier con- centrations everywhere. (c) N A = 10 14 cm - 3 = p p N D = 10 15 cm - 3 = n n (d) Note that the semiconductor is Si because np = n 2 i far from the depletion region. In the depletion region, np = n 2 i e q V A /kT V A = kT q ln ( np n 2 i ) Using the values in the graph for - x p or + x n for np , V A = 0 . 0259 V ln ( 10 17 10 20 ) = - 0 . 18 V 5. For the p-n junction in problem 1 above, the minority carrier lifetimes τ n and τ p are both 1 μ s . Calculate the following values (a) The minority carrier diffusion lengths L N and L P (b) the built-in voltage V bi (c) The excess minority carrier concentrations n p ( - x p ) and p n (+ x n ) for an applied voltage V A = 0 . 5 V (d) J N ( - x p ) , J P (+ x n ) , and J for V A = 0 . 5 V (e) The cross-sectional area of a device for which the total current at V A = 0 . 5 is 10 μ A .
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