hw11_solutions

hw11_solutions - ECE 440 HW XI Solutions 16 Spring 2006 3...

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ECE 440 HW XI Solutions Spring 2006 1. For a p-n silicon junction, Na=5x10 16 /cm 3 in the p-side and Nd=10 15 /cm 3 in the n- side. Determine the depletion capacitance per unit area of cm 2 at –4 V. The junction capacitance is (as calculated in homework #10) W A C j ε = so the capacitance per unit error is W C j = . First find the depletion width. V n N N q kT V i d a 677 . 0 ) 10 5 . 1 ( ) 10 ( ) 10 5 ( ln 0259 . ln 2 10 15 16 2 0 = × × = = cm N N q V V W a d 4 15 16 19 14 0 10 50 . 2 10 1 10 5 1 10 6 . 1 ) 4 677 . 0 ( 10 85 . 8 8 . 11 2 1 1 ) ( 2 × = + × + = + = So then 2 9 4 14 10 18 . 4 10 50 . 2 10 85 . 8 8 . 11 cm F C j × = × × =

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2. A Schottky barrier is formed between a metal having a work function of 4.7 eV and n-type Si (electron affinity = 4 eV). The donor doping in the Si is 5x10 17 /cm 3 . (a) Draw the equilibrium band diagram showing numerical values for qVo and schottky barrier, b q Φ . The semiconductor work function is the energy difference between the vacuum level
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hw11_solutions - ECE 440 HW XI Solutions 16 Spring 2006 3...

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