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

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Chapter 1 Problem Solutions 1.1 /2 3/2 g E kT i nB Te = (a) Silicon (i) () [] 15 6 19 83 1.1 5.23 10 250 exp 286 10 250 2.067 10 exp 25.58 1.61 10 cm i i n n ⎡⎤ ⎢⎥ × ⎣⎦ (ii) 15 6 19 11 3 1.1 5.23 10 350 exp 350 3.425 10 exp 18.27 3.97 10 cm i i n n × (b) GaAs (i) 14 6 17 33 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) 14 6 18 1.4 2.10 10 350 exp 1.375 10 exp 23.26 1.09 10 cm i i n n × 1.2 a. exp 2 i Eg T kT ⎛⎞ = ⎜⎟ ⎝⎠ 12 15 3/ 2 6 1.1 10 5.23 10 exp 2(86 10 )( ) T T × 3 43 / 2 6.40 10 1.91 10 exp T T × ×= By trial and error, 368 K T b. 93 10 cm i n = 91 5 3 / 2 6 1.1 10 5.23 10 exp T T × 3 73 / 2 6.40 10 1.91 10 exp T T × By trial and error, 268 K T ≈° 1.3 Silicon
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(a) () [] 3/2 15 6 18 10 3 1.1 5.23 10 100 exp 286 10 100 5.23 10 exp 63.95 8.79 10 cm i i n n −− ⎡⎤ ⎢⎥ × ⎣⎦ (b) 15 6 19 10 3 1.1 5.23 10 300 exp 300 2.718 10 exp 21.32 1.5 10 cm i i n n × (c) 15 6 19 14 3 1.1 5.23 10 500 exp 500 5.847 10 exp 12.79 1.63 10 cm i i n n × Germanium. (a) 15 18 6 3 0.66 1.66 10 100 exp 1.66 10 exp 38.37 2 86 10 100 35.9 cm i i n n × = (b) 15 18 6 13 3 0.66 1.66 10 300 exp 8.626 10 exp 12.79 2.40 10 cm i i n n = × × (c) 15 19 6 15 3 0.66 1.66 10 500 exp 1.856 10 exp 7.674 8.62 10 cm i i n n = × × 1.4 a. 15 3 51 0 cm n t y p e d N ⇒ − 15 3 0 d nN == × 2 10 2 43 00 15 0 1.5 10 4.5 10 cm 0 i n pp n × ⇒=× × b. 15 3 t y p e d Nn ⇒− 15 3 od × 14 6 14 12 63 1.4 2.10 10 300 exp 2(86 10 )(300) 2.10 10 300 1.65 10 1.80 10 cm i n ⎛⎞ ⎜⎟ × ⎝⎠ × 2 6 2 15 0 1.8 10 6.48 10 cm 0 i n n × ⇒= × × 1.5
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(a) n -type (b) () 16 3 2 10 2 33 16 51 0 cm 1.5 10 4.5 10 cm 0 od i o o nN n p n == × × × (c) 16 3 × From Problem 1.1(a)(ii) 11 3 3.97 10 cm i n 2 11 63 16 3.97 10 3.15 10 cm 0 o p × × × 1.6 a. 16 3 10 cm type a Np =⇒ 16 3 0 a pN 2 10 2 43 00 16 0 1.5 10 2.25 10 cm 10 i n nn p × = × b. Germanium 16 3 type a 16 3 0 a 3/2 15 6 15 6 13 3 0.66 1.66 10 300 exp 286 10 300 1.66 10 300 2.79 10 2.4 10 cm i n ⎛⎞ ⎜⎟ × ⎝⎠ × 2 13 2 10 3 16 0 2.4 10 5.76 10 cm 10 i n p × = × 1.7 (a) p -type (b) 17 3 2 10 2 17 21 1.5 10 1.125 10 cm 0 oa i o o n n p × × = × × (c) 17 3 o p From Problem 1.1(a)(i) 83 1.61 10 cm i n 2 8 3 17 1.61 10 0.130 cm 0 o n × × 1.8 (a) 15 3 2 10 2 15 1.5 10 4.5 10 0 o i oo o nc m n pp n c m × ⇒=× × (b) n-type np ± (c) 15 3 c m ≅= × 1.9 a. Add Donors 15 3 71 d N
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b. Want 63 2 10 cm / oi d p nN == So () ( ) 26 1 5 23 10 7 10 7 10 exp i n Eg BT kT = × ⎛⎞ = ⎜⎟ ⎝⎠ 2 1 2 21 15 3 6 1.1 71 0 5 . 2 31 0 e x p 86 10 T T ×= × × By trial and error, 324 K T ≈° 1.10 ( ) 4 2.2 15 10 3.3 IJA E A II σ =⋅= =⇒ = m A 1.11 1 85 12 7.08 (ohm cm) J JE E σσ = = =− 1.12 19 11 1 1.6 10 480 0.80 a pa p gN eN eg μμ ≈⇒ = = × 16 3 1.63 10 cm a N 1.13 19 15 3 0.5 1.6 10 1350 2.31 10 cm nd d n d N e N μ = × 1.14 (a) For n-type, ( ) 19 1.6 10 8500 d N σμ ≅= × For 1 15 19 3 4 10 10 1.36 1.36 10 d Nc m c m ≤≤ × Ω (b) 32 0.1 0.136 1.36 10 / J EJ ≤× A c m 1.15 15 2 19 4 2 10 10 1.6 10 180 0.5 10 576 A/cm nn n n dn n Je D e D dx x J Δ Δ × = 1.16
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() 15 19 15 4 / 1 10 exp 1.6 10 15 10 exp 10 10 2.4 p pp p p p xL p dp Je D dx x eD LL x J L =− ⎛⎞ ⎛⎞ −− ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ × ⎛⎞ =
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This homework help was uploaded on 02/10/2008 for the course ECE 138 taught by Professor N/a during the Spring '08 term at CSU Fresno.

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

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