Ch01s - Chapter 1 Problem Solutions 1.1 E 2 kT ni = BT 3 2...

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Chapter 1 Problem Solutions 1.1 / 2 3/ 2 g E kT i n BT e = (a) Silicon (i) ( ) ( ) ( ) ( ) [ ] 3/ 2 15 6 19 8 3 1.1 5.23 10 250 exp 2 86 10 250 2.067 10 exp 25.58 1.61 10 cm i i n n = × × = × = × (ii) ( ) ( ) ( ) ( ) [ ] 3/ 2 15 6 19 11 3 1.1 5.23 10 350 exp 2 86 10 350 3.425 10 exp 18.27 3.97 10 cm i i n n = × × = × = × (b) GaAs (i) ( ) ( ) ( ) ( ) ( ) [ ] 3/ 2 14 6 17 3 3 1.4 2.10 10 250 exp 2 86 10 250 8.301 10 exp 32.56 6.02 10 cm i i n n = × × = × = × (ii) ( ) ( ) ( ) ( ) ( ) [ ] 3/ 2 14 6 18 8 3 1.4 2.10 10 350 exp 2 86 10 350 1.375 10 exp 23.26 1.09 10 cm i i n n = × × = × = × 1.2 a. 3/ 2 exp 2 i Eg n BT kT = 12 15 3/ 2 6 1.1 10 5.23 10 exp 2(86 10 )( ) T T = × × 3 4 3/ 2 6.40 10 1.91 10 exp T T × × = By trial and error, 368 K T b. 9 3 10 cm i n = ( ) ( ) 9 15 3/ 2 6 1.1 10 5.23 10 exp 2 86 10 T T = × × 3 7 3/ 2 6.40 10 1.91 10 exp T T × × = By trial and error, 268 K T ° 1.3 Silicon
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