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HW5_solutions

# HW5_solutions - Homework 5 1 Problem A silicon p-n junction...

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Homework 5 1. Problem: A silicon p-n junction is formed between n-type Si doped with N D = 10 17 cm - 3 and p-type Si doped with N A = 10 16 cm - 3 . (a) Sketch the energy band diagram. Label all axes and all important energy levels. (b) Find n n 0 , n p 0 , p p 0 , and p n 0 . Sketch the carrier concentration (of both electrons and holes) as a function of position. (c) Calculate the built-in potential V bi in eV. Solution: (a) The energy band diagram with labeled important energy levels and axes is (b) Given n i = 1 . 5 × 10 10 cm - 3 , in the quasi-neutral p-region, ECE344 Fall 2009 1

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p p 0 = N A = 10 16 cm - 3 . (1) In the quasi-neutral n-region, n n 0 = N D = 10 17 cm - 3 . (2) In the depletion region, n p 0 = n 2 i N A = 2 . 25 × 10 4 cm - 3 (3) for p-side and p n 0 = n 2 i N D = 2 . 25 × 10 3 cm - 3 (4) for n-side. The diagram for carrier concentration is: ECE344 Fall 2009 2
(c) Using Eq. (123) or (124) on lecture notes Part 2, the build-in potential is V bi = k B T q ln ( N A N D n 2 i ) = k B T q ln ( p p p n ) = 0 . 7543 eV. (5) 2. Problem: A Si p-n junction has dopant concentrations N D = 2 × 10 15 cm - 3 and N A = 2 × 10 16 cm - 3 . Calculate the built-in potential V bi in eV and the total width of the depletion region W = x n 0 + x p 0 at zero bias (that is, V a = 0 ) and under a reverse bias V a = - 8 V. Solution: (a) Given n i = 1 . 5 × 10 10 cm - 3 , the build-in potential is: V bi = k B T q ln ( N A N D n 2 i ) = k B T q ln ( p p p n ) = 0 . 6709 eV. (6) (b) According to Eq. (140) &

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HW5_solutions - Homework 5 1 Problem A silicon p-n junction...

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