MCD3EExerciseSolnsPart1

MCD3EExerciseSolnsPart1 - Microelectronic Circuit Design...

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1 Microelectronic Circuit Design Third Edition - Part I Solutions to Exercises CHAPTER 1 Page 11 V LSB = 5.12 V 2 10 bits = 5.12 V 1024 bits = 5.00 mV V MSB = 5.12 V 2 = 2.560 V 1100010001 2 = 2 9 + 2 8 + 2 4 + 2 0 = 785 10 V O = 786 5.00 mV ( ) = 3.925 V or V O = 2 1 + 2 2 + 2 6 + 2 10 ( ) 5.12 V = 3.912 V Page 12 The dc component is V A = 4V. The signal consists of the remaining portion of v A : v a = (5 sin 2000 π t + 3 cos 1000 π t) Volts. Page 19 i SC = i 1 + β i 1 = v s R 1 + v s R 1 = + 1 ( ) v s R 1 v OC = + 1 ( ) R S + 1 ( ) R S + R 1 v S R th = v OC i SC = + 1 ( ) R S + 1 ( ) R S + R 1 R 1 + 1 ( ) = 1 + 1 ( ) R 1 + 1 R S R th = R S R 1 + 1 ( ) Page 23 v o = 5cos 2000 π t + 25 o ( ) = − − 5sin 2000 t + 25 o 90 o ( ) [ ] = 5sin 2000 t 65 o ( ) V o = 5 ∠− 65 o V s = 0.001 0 o A v = 5 65 o 0.001 0 o = 5000 65 o
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2 Page 26 v s = 0.5sin 2000 π t ( ) + sin 4000 t ( ) + 1.5sin 6000 t ( ) [ ] The three spectral components are f 1 = 1000 Hz f 2 = 2000 Hz f 3 = 3000 Hz a ( ) The gain of the band- pass filter is zero at both f 1 and f 3 . At f 2 , V o = 10 1 V ( ) = 10 V , and v O = 10.0sin 4000 t ( ) volts. b ( ) The gain of the low- pass filter is zero at both f 2 and f 3 . At f 2 V o = 6 0.5 V ( ) = 3 V , and v O = 3.00sin 2000 t ( ) volts. Page 27 39k Ω 1 0.1 ( ) R 39k Ω 1 + 0.1 ( ) or 35.1 k Ω≤ R 42.9 k Ω 3.6k Ω 1 0.01 ( ) R 3.6k Ω 1 + 0.01 ( ) or 3.56 k R 3.64 k Ω Page 29 P = V S 2 R 1 + R 2 P nom = 15 2 54 k Ω = 4.17 mW P max = 1.1 x 15 ( ) 2 0.95 x 54 k Ω = 5.31 mW P min = 0.9 x 15 ( ) 2 1.05 x 54 k Ω = 3.21 mW Page 33 R = 10 k Ω 1 + 10 3 o C 55 25 ( ) o C = 9.20 k Ω R = 10 k Ω 1 + 10 3 o C 85 25 ( ) o C = 10.6 k Ω
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3 CHAPTER 2 Page 47 n i = 2.31 x 10 30 K 3 cm 6 ( ) 300 K ( ) 3 exp 0.66 eV 8.62 x 10 5 eV / K ( ) 300 K ( ) = 2.27 x 10 13 cm 3 n i = 1.08 x 10 31 K 3 cm 6 ( ) 50 K ( ) 3 exp 1.12 eV 8.62 x 10 5 eV / K ( ) 50 K ( ) = 4.34 x 10 39 cm 3 n i = 1.08 x 10 31 K 3 cm 6 ( ) 325 K ( ) 3 exp 1.12 eV 8.62 x 10 5 eV / K ( ) 325 K ( ) = 4.01 x 10 10 cm 3 L 3 = cm 3 4.34 x 10 39 10 2 m cm 3 L = 6.13 x 10 10 m Page 48 v p = μ p E = 500 cm 2 V s 10 V cm = 5.00 x 10 3 cm s v n = n E = 1350 cm 2 V s 1000 V cm = 1.35 x 10 6 cm s E = V L = 1 2 x 10 4 V cm = 5.00 x 10 3 V cm Page 48 n = v n E = 4.3 x 10 5 cm / s 100 V / cm = 4300 cm 2 s p = v p E = 2.1 x 10 5 cm / s 100 V / cm = 2100 cm 2 s n = v n E = 8.5 x 10 5 cm / s 100 V / cm = 8500 cm 2 s
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4 Page 50 n i 2 = 1.08 x 10 31 400 ( ) 3 exp 1.12 8.62 x 10 5 400 ( ) = 5.40 x 10 24 / cm 6 ρ = 1 σ = 1 1.60 x 10 19 2.32 x 10 12 ( ) 1350 ( ) + 2.32 x 10 12 ( ) 500 ( ) [ ] = 1450 Ω− cm n i 2 = 1.08 x 10 31 50 ( ) 3 exp 1.12 8.62 x 10 5 50 ( ) = 1.88 x 10 77 / cm 6 = 1 = 1 1.60 x 10 19 4.34 x 10 39 ( ) 6500 ( ) + 4.34 x 10 39 ( ) 2000 ( ) [ ] = 1.69 x 10 53 cm Page 54 n i 2 = 1.08 x 10 31 400 ( ) 3 exp 1.12 8.62 x 10 5 400 ( ) = 5.40 x 10 24 / cm 6 p = N A N D = 10 16 2 x 10 15 = 8 x 10 15 holes cm 3 n = n i 2 p = 5.40 x 10 24 8 x 10 15 = 6.75 x 10 8 electrons cm 3 n = N D = 2 x 10 16 electrons cm 3 n = n i 2 p = 10 20 2 x 10 16 = 5.00 x 10 3 holes cm 3 n > p n - type silicon Page 55 Reading from the graph for N T = 10 16 /cm 3 , 1250 cm 2 /V-s and 400 cm 2 /V-s.
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MCD3EExerciseSolnsPart1 - Microelectronic Circuit Design...

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