ch02_solutions

# ch02_solutions - CHAPTER 2 2.1 Based upon Table 2.1 a...

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15 CHAPTER 2 2.1 Based upon Table 2.1, a resistivity of 2.6 μΩ -cm < 1 m -cm, and aluminum is a conductor. 2.2 Based upon Table 2.1, a resistivity of 10 15 -cm > 10 5 -cm, and silicon dioxide is an insulator. 2.3 I max = 10 6 A cm 2 5 m m ( 29 1 m m ( 29 10 - 8 cm 2 m m 2 = 50 mA 2.4 n i = BT 3 exp - E G 8.62 x 10 - 5 T For silicon, B = 1.08 x 10 31 and E G = 1.12 eV: n i = 2.01 x10 -10 /cm 3 6.73 x10 9 /cm 3 8.36 x 10 13 /cm 3 . For germanium, B = 2.31 x 10 30 and E G = 0.66 eV: n i = 35.9/cm 3 2.27 x10 13 /cm 3 8.04 x 10 15 /cm 3 . 2.5 Define an M-File: function f=temp(T) ni=1E14; f=ni^2-1.08e31*T^3*exp(-1.12/(8.62e-5*T)); n i = 10 14 /cm 3 for T = 506 K n i = 10 16 /cm 3 for T = 739 K 2.6 n i = BT 3 exp - E G 8.62 x 10 - 5 T with B =1.27x10 29 K - 3 cm - 6 T = 300 K and E G = 1.42 eV: n i = 2.21 x10 6 /cm 3 T = 100 K: n i = 6.03 x 10 -19 /cm 3 T = 500 K: n i = 2.79 x10 11 /cm 3

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16 2.7 v n = - m n E = - 710 cm 2 V - s 3000 V cm  = - 2.13 x 10 6 cm s v p = + m p E = + 260 cm 2 V - s 3000 V cm = + 7.80 x 10 5 cm s j n = - qnv n = - 1.60 x 10 - 19 C ( 29 10 17 1 cm 3  - 2.13 x 10 6 cm s = 3.41 x 10 4 A cm 2 j p = qnv p = 1.60 x 10 - 19 C ( 29 10 3 1 cm 3 7.80 x 10 5 cm s = 1.25 x 10 - 10 A cm 2 2.8 n i 2 = BT 3 exp - E G kT B =1.08 x 10 31 10 10 ( 29 2 = 1.08 x 10 31 T 3 exp - 1.12 8.62 x 10 - 5 T Using a spreadsheet, solver, or MATLAB yields T = 305.22K Define an M-File: function f=temp(T) f=1e20-1.08e31*T^3*exp(-1.12/(8.62e-5*T)); Then: fzero('temp',300) | ans = 305.226 K 2.9 j = Qv = 0.5 C cm 3 10 7 cm sec  = 5 x 10 6 A cm 2 = 5 MA cm 2 2.10 v = j Q = - 1000 A / cm 2 0.02 C / cm 2 = - 5 x 10 4 cm s 2.11 j p = qp m p E = 1.602 x 10 - 19 C ( 29 10 16 cm 3 400 cm 2 v - s - 2000 V cm = - 1280 A cm 2 j n = qn m n E = 1.602 x 10 - 19 C ( 29 10 4 cm 3 1100 cm 2 v - s - 2000 V cm  = - 3.92 x 10 - 9 A cm 2 2.12 a ( 29 E = 5 V 10 x 10 - 4 cm = 5000 V cm b ( 29 V = 10 5 V cm 10 x 10 - 4 cm ( 29 = 100 V
17 2.13 j p = qpv p = 1.602 x 10 - 19 C ( 29 10 18 cm 3 10 7 cm s = 1.60 x 10 6 A cm 2 i p = j p A = 1.60 x 10 6 C A cm 2 2 x 10 - 4 cm ( 29 25 x 10 - 4 cm ( 29 = 0.80 A 2.14 For intrinsic silicon, s = q m n n i + m p n i ( 29 = qn i m n + m p ( 29 s 1000 Ω- cm ( 29 - 1 for a conductor n i s q m n + m p ( 29 = 1000 Ω- cm ( 29 - 1 1.602 x 10 - 19 C 100 + 50 ( 29 cm 2 v - sec = 4.16 x 10 19 cm 3 n i 2 = 1.73 x 10 39 cm 6 = BT 3 exp - E G kT with

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ch02_solutions - CHAPTER 2 2.1 Based upon Table 2.1 a...

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