homework 5 solutions - despite perhaps lowered mobility and...

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1 Homework #5, Due 10/30/07 I. (25 points) Problem 5.9 of text. Assume complete ionization. (a) eV 407 . 0 cm / 10 5 . 1 cm / 10 ln eV 0259 . 0 ln 3 10 3 17 , , = × = = i p o B F p i n p T k E E eV 347 . 0 cm / 10 5 . 1 cm / 10 ln eV 0259 . 0 ln 3 10 3 16 , , = × = = i n o B n i F n n T k E E 0.754eV eV 347 . 0 eV 407 . 0 ) ( ) ( , , = + = + = n i F F p i o E E E E qV eV 754 . 0 ) cm / 10 5 . 1 ( ) cm / 10 )( cm / 10 ( ln eV 0259 . 0 ln ln 2 3 10 3 16 3 17 2 2 , , = × = = i d a B i n o p o B o n N N T k n n p T k qV II. (25 points) Problem 5.21 of text. (a) increases (W decreases, so capacitance increases) (b) increases (Fermi level closer to conduction band on n-side so contact potential is larger, as per problem 5.9 above) (c) decreases (W decreases, so peak field magnitude must increase to produce the same electrostatic potential drop across the depletion region. Or the potential drop and thus the reverse bias voltage can be less to produce the same fields, allowing either Zener tunneling or avalanche breakdown whichever comes first to occur at lower reverse biases). (d) decreases (More doping leads to more conductivity
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Unformatted text preview: despite perhaps lowered mobility and thus less resistive or ohmic loss.) 0.467eV 0.347eV 0.754eV 2 III. (25 points) Problem 5.27 of text. Again assume complete ionization on both sides. ( or more roughly note that L D is around an order of magnitude smaller than W.) 3 IV. (25 points) Problems 5.35 of text. Note that x x x x x x x x e e e e x x e e x e e x + = = + = = ) sinh( ) cosh( ctnh(x) & 2 ) cosh( , 2 ) sinh( Also be careful of notation: Here x n is just x on the n-side and x n = 0 is the n-side edge of the depletion region (not x n = x n,o as in figures in the text). And p n is the equilibrium value of the minority hole concentration (what I have labeled p n-side,o in my notes)...
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homework 5 solutions - despite perhaps lowered mobility and...

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