Chapter02

Microelectronic Circuit Design

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CHAPTER 2 2.1 Based upon Table 2.1, a resistivity of 2.6 μΩ -cm < 1 m Ω -cm, and aluminum is a conductor. 2.2 Based upon Table 2.1, a resistivity of 10 15 Ω -cm > 10 5 Ω -cm, and silicon dioxide is an insulator. 2.3 I max = 10 7 A cm 2 5 μ m () 1 m 10 8 cm 2 m 2 ⎟ = 500 mA 2.4 = T x E BT n G i 5 3 10 62 . 8 exp 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 20
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2.7 v n =− μ n E 700 cm 2 V s 2500 V cm 1.75 x 10 6 cm s v p =+ p E 250 cm 2 V s 2500 V cm 6.25 x 10 5 cm s j n qnv n 1.60 x 10 19 C () 10 17 1 cm 3 1.75 x 10 6 cm s = 2.80 x 10 4 A cm 2 j p = qnv p = 1.60 x 10 19 C 10 3 1 cm 3 6.25 x 10 5 cm s = 1.00 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 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 s cm cm C cm A Q j v 5 2 2 10 / 01 . 0 / 1000 = = = 2.10 2 2 6 7 3 4 10 4 sec 10 4 . 0 cm MA cm A x cm cm C Qv j = = = = 21
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2.11 v n =− μ n E 1000 cm 2 V s 2000 V cm =+ 2.00 x 10 6 cm s v p p E 400 cm 2 V s 2000 V cm 8.00 x 10 5 cm s j n qnv n 1.60 x 10 19 C () 10 3 1 cm 3 + 2.00 x 10 6 cm s 3.20 x 10 10 A cm 2 j p = qnv p = 1.60 x 10 19 C 10 17 1 cm 3 8.00 x 10 5 cm s 1.28 x 10 4 A cm 2 2.12 V 100 10 10 10 b 5000 10 10 5 4 5 4 = = = = cm x cm V V cm V cm x V E a 2.13 j p = qpv p = 1.60 x 10 19 C 10 19 cm 3 10 7 cm s = 1.60 x 10 7 A cm 2 i p = j p A = 1.60 x 10 7 A cm 2 1 x 10 4 cm 25 x 10 4 cm = 4.00 A 2.14 For intrinsic silicon, σ = q n n i + p n i = qn i n + p ( ) 1000 Ω− cm 1 for a conductor n i q n + p = 1000 cm 1 1.602 x 10 19 C 100 + 50 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 B = 1.08 x 10 31 K 3 cm 6 , k = 8.62x10 -5 eV/K and E G = 1.12 eV This is a transcendental equation and must be solved numerically by iteration. Using the HP solver routine or a spread sheet yields T = 2701 K. Note that this temperature is far above the melting temperature of silicon.
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Chapter02 - CHAPTER 2 2.1 Based upon Table 2.1, a...

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