HW2Solutions

# HW2Solutions - EE105 HW 2 Solutions 1 A C#t 25"10#15 \$...

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EE105 HW 2 Solutions 1. a. C = " # A t \$ A = C # t " , so for SiO 2 , A = 25 " 10 # 15 \$ 25 " 10 # 7 3.45 " 10 # 13 = 1.8 " 10 # 7 cm 2 . For Si 3 N 4 , A = 25 " 10 # 15 \$ 25 " 10 # 7 8.63 " 10 # 13 = 750 " 10 # 10 cm 2 . b. Q = CV = 25 " 10 # 15 \$ 2.5 = 62.5 " 10 # 15 Coulombs 2. a. μ p = 198 cm 2 V " s , since N d =5x10 17 . D p = μ p V th = 198 " 25.85 m = 5.11 cm 2 s . dp ( x ) dx = " 10 18 cm " 4 , so J p diff = " qD p dp dx = ( " 1.602 # 10 " 19 )(5.11)( " 10 18 ) = 818.6 mA cm 2 . b. " ( x ) = # V th \$ ln p ( x ) n i % & ( ) * Note: This problem has a bug, in that the range for plotting Phi included the point where p(x) reaches zero, which is not physically possible and results in the graph “blowing up” as can be seen. Because of this bug, we have decided not to deduct points from people who were unable to solve this problem.

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Unformatted text preview: bi = V th ln N a N d n i 2 # \$ % & ’ ( = 25.85 m ln 10 15 ) 5 * 10 16 10 20 # \$ % & ’ ( = 696.3 mV b. c. x p = 2 s # bi qN a \$ % & ’ ( ) N d N a + N d \$ % & ’ ( ) = 2 * 1.035 + 10 , 12 * .6963 1.602 + 10 , 19 * 10 15 \$ % & ’ ( ) 5 + 10 16 10 15 + 5 + 10 16 \$ % & ’ ( ) = 93.9 + 10 , 6 cm x n = 2 s bi qN d \$ % & ’ ( ) N a N a + N d \$ % & ’ ( ) = 2 * 1.035 + 10 , 12 * .6963 1.602 + 10 , 19 * 5 + 10 16 \$ % & ’ ( ) 10 15 10 15 + 5 + 10 16 \$ % & ’ ( ) = 1.88 + 10 , 6 cm...
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