ee3161 spring10 lecture note 6

Then v0 at x xp and vvbi at xxn dv 0 x xp qna

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Unformatted text preview: = − xp K ￿0 K ￿0 xn + xp = W NA xp = ND xn Emax EE3161 Semiconductor Devices Sang-Hyun Oh q NA ND =− W K ￿0 NA + ND 25 UNIVERSITY OF MINNESOTA Solution for V(x), p-side dV = −E (x) = dx V (x) qNA KS ￿0 (xp + x), −xp ≤ x ≤ 0 qND KS ￿0 (xn − x), 0 ≤ x ≤ xn We can set the arbitrary reference potential equal to zero at x = -xp. Then V=0 at x=-xp, and V=Vbi at x=xn. dV = ￿ 0 ￿ x −xp qNA (xp + x￿ )dx￿ KS ￿0 −xp ≤ x ≤ 0 qNA V (x) = (xp + x)2 2KS ￿0 EE3161 Semiconductor Devices Sang-Hyun Oh 26 UNIVERSITY OF MINNESOTA Solution for V(x), n-side ￿ Vbi dV = ￿ V (x) ￿ xn x qND (xn − x￿ )dx￿ KS ￿0 0 ≤ x ≤ xn qND V (x) = Vbi − (xn − x)2 2KS ￿0 qNA V (x) = (xp + x)2 2KS ￿0 Evaluate both equations (n-side and p-side) at x=0: qND V (x) = Vbi − (xn − x)2 2KS ￿0 qNA 2 qND 2 V (0) = xp = Vbi − xn 2KS ￿0 2KS ￿0 EE3161 Semiconductor Devices Sang-Hyun Oh 27 UNIVERSITY OF MINNESOTA Solution for xn and xp It’s important to calculate xn and xp as a function of doping and applied bias voltage. qNA 2 qND 2 xp + xn = Vbi 2KS ￿0 2KS ￿0 xn and xp are the only unknowns here. Also, they are related: NA xp = ND xn qNA 2KS ￿0 q 2KS ￿0 EE3161 Semiconductor Devices Sang-Hyun Oh ￿ ￿ ND xn...
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