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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)
∂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.
x + Ax + B
Vp (x) =
Vn (x) = −
x + Cx + D
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
K s ε0
By boundary condition 2, we have A = C , so that
ND x n = NA x p
Vp (x) = −
( x + xp )
(x − xn )
En (x) = − Vn (x) =
Ks ε0 Ep (x) = − d) e
(ND x2 + NA x2 )
Combining with ND xn = NA xp we can easily obtain
Vbi = Vn (xn ) − Vp (−xp ) = xp + xn = 2Ks ε0 Vbi NA + ND
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.
- Fall '11