Solution to HW3 - Problem 1 a) Boundary conditions include...

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Unformatted text preview: Problem 1 a) Boundary conditions include continuity of potential and electrical field (derivative of potential) at x = 0, and continuity of electrical field at x = −xp and x = xn , namely 1. Vp (0) = Vn (0) 2. ∂ ∂x Vp (0) 3. ∂ ∂x Vp (−xp ) 4. ∂ ∂x Vn (xn ) = ∂ ∂x Vn (0) =0 =0 We also have the constraint that Vn (xn ) − Vp (−xp ) = Vbi . This gives rise to the dependence of parameters xn , xp and Vbi which is asked to be found in part d). b) The charge density within each region is constant. By solving the Poisson’s equation, we have that the potential within each region is a quadratic function with respect to x. eNA 2 x + Ax + B Vp (x) = 2Ks ε0 eND 2 Vn (x) = − x + Cx + D 2Ks ε0 By boundary condition 1, we have B = D. We can always assume that the potential at x = 0 is zero so that B = D = 0. By boundary condition 3&4, we have eNA xp A= Ks ε0 eND xn C= K s ε0 By boundary condition 2, we have A = C , so that ND x n = NA x p c) ∂ eNA Vp (x) = − ( x + xp ) ∂x Ks ε0 eND ∂ (x − xn ) En (x) = − Vn (x) = ∂x Ks ε0 Ep (x) = − d) e (ND x2 + NA x2 ) n p 2Ks ε0 Combining with ND xn = NA xp we can easily obtain Vbi = Vn (xn ) − Vp (−xp ) = xp + xn = 2Ks ε0 Vbi NA + ND e NA ND 1 1/2 ...
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This note was uploaded on 02/02/2012 for the course ELECTRICAL ENEE380 taught by Professor D.romeo during the Fall '11 term at Maryland.

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