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Semiconductor Physics and Devices: Basic Principles, 4 th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ EO B EO B N N K N N K A B 1 2 1 EO B EO B N N K N N K 1 2 1 EO B EO B N N K N N K 2 1 EO B N N K 2 1 or E B B E EO B x x D D N N A B 2 1 (ii) Now E B E B x x K x x K 1 1 1 Then EO B EO B x x K x x K A C 1 2 1 EO B EO B x x K x x K 1 2 1 EO B EO B x x K x x K 21 EO B x x K 1 or finally EO B B E E B x x D D N N A C 1 (b) We have 2 2 1 1 B B T L x (i) 1 A B T T (ii) 1 A C T T (c) Neglect any change in space charge width. t BE sO rO V V J J 2 exp1 1 sO sO J K J K 1 1 1 (i) sOAsOB sOA sOB J K J K J K J K A B 11 1 1 sOAsOB J K J K 1 Now EE sO xN J 1 so EOEO NKNK A B 21 or EONK A B 1 Recombination factor decreases (ii) We have EO EO xK x K A C 2 1 or EOxK A C 2 1 1 Recombination factor increases _______________________________________ 12.26 (b) 3 17 2102 1025.2 10 105.1 B i BO N n n cm 3 Then t BC BOB V V nn exp0 0259.0 6.0 exp1025.2 3 131059.2 cm 3 Now