Exercise Solutions Ch6-Ch9

Exercise Solutions Ch6-Ch9 - Microelectronic Circuit Design...

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1 Microelectronic Circuit Design Third Edition - Part II Solutions to Exercises CHAPTER 6 Page 280 NM L = 0.8 V 0.4 V = 0.4 V NM H = 3.6 V 2.0 V = 1.6 V Page 282 V 10% = V L + 0.1 Δ V ( ) = 2.6 V + 0.1 0.6 − − 2.6 ( ) [ ] = 2.4 V or V 10% = V H 0.9 Δ V ( ) = 0.6 V 0.9 0.6 − − 2.6 ( ) [ ] = 2.4 V V 90% = V L + 0.9 Δ V ( ) = 2.6 V + 0.9 0.6 − − 2.6 ( ) [ ] = 0.8 V or V 90% = V H 0.1 Δ V ( ) = 0.6 V 0.1 0.6 − − 2.6 ( ) [ ] = 0.8 V V 50% = V H + V L 2 = 0.6 2.6 2 = 1.6 V t r = t 4 t 3 = 3 ns t f = t 2 t 1 = 5 ns Page 283 At P =1 mW: PDP = 1 mW 1 ns ( ) = 1 pJ At P = 3 mW: PDP = 3 mW 1 ns ( ) = 3 pJ At P = 20 mW: PDP = 20 mW 2 ns ( ) = 40 pJ Page 285 Z = A + B ( ) B + C ( ) = AB + AC + BB + BC = AB + BB + AC + BB + BC Z = AB + B + AC + B + BC = B A + 1 ( ) + AC + B C + 1 ( ) = B + AC + B Z = B + B + AC = B + AC Page 288
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2 I DD = P V DD = 0.4 mW 2.5 V = 160 μ A R = V DD V L I DD = 2.5 V 0.2 V 160 A = 14.4 k Ω 1.6 x 10 4 A = 10 4 A V 2 W L S 2.5 0.6 0.2 2 0.2 V 2 W L S = 4.44 1 Page 289 I DD = V DD V L R = 3.3 V 0.1 V 102 k Ω = 31.4 A 31.4 x 10 6 A = 6 x 10 5 A V 2 W L S 3.3 0.75 0.1 2 0.1 V 2 W L S = 2.09 1 Page 290 0.15 V = R on R on + 28.8 k Ω 2.5 V R on = 1.84 k Ω W L S = 1 10 4 2.5 0.60 0.15 2 1.84k Ω ( ) W L S = 2.98 1 Page 291 i ( ) R on = 1 6 x 10 5 1.03 1 3.3 0.75 0.2 2 = 6.61 k Ω V L = 6.61 k Ω 6.61 k Ω + 102 k Ω 3.3 V = 0.201 V ii ( ) 1 K n R = V 2 A 1 Ω = V Page 293 K n R = 6 x 10 5 ( ) 1.03 1 1.02 x 10 5 ( ) = 6.30 V NM H = 3.3 0.75 + 1 2 6.30 ( ) 1.63 3.3 6.30 = 1.45 V NM L = 0.75 + 1 6.30 2 3.3 ( ) 3 6.30 ( ) = 0.318 V Page 297
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3 i ( ) V H = 2.5 0.6 + 0.75 H + 0.6 0.6 ( ) [ ] V H = 1.416 V ii ( ) 80 x 10 6 A = 100 x 10 6 A V 2 W L S 1.55 0.60 0.15 2 0.15 V 2 W L S = 6.10 1 V TNL = 0.6 + 0.5 .15 + 0.6 0.6 ( ) = 0.646 V 80 x 10 6 A = 100 x 10 6 2 A V 2 W L L 2.5 0.15 0.646 ( ) 2 V 2 W L L = 0.551 1 = 1 1.82 ( iii ) 80 x 10 6 A = 100 x 10 6 A V 2 W L S 1.55 0.60 0.1 2 0.1 V 2 W L S = 8.89 1 V TNL = 0.6 + 0.5 .1 + 0.6 0.6 ( ) = 0.631 V 80 x 10 6 A = 100 x 10 6 2 A V 2 W L L 2.5 0.1 0.631 ( ) 2 V 2 W L L = 0.511 1 = 1 1.96
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4 Page 300 The high logic level is unchanged: V H = 2.11 60 x 10 6 A = 50 x 10 6 A V 2 W L S 2.11 0.75 0.1 2 0.1 V 2 W L S = 9.16 1 V TNL = 0.75 + 0.5 .1 + 0.6 0.6 ( ) = 0.781 V 60 x 10 6 A = 50 x 10 6 2 A V 2 W L L 3.3 0.1 0.781 ( ) 2 V 2 W L L = 0.410 1 = 1 2.44 Page 302 γ = 0 V TN = 0.6 V and V H = 2.5-0.6 =1.9 V I DD = 0 for v O = V H 100 x 10 6 10 1 1.9 0.6 V L 2 V L = 100 x 10 6 2 2 1 2.5 V L 0.6 ( ) 2 6 V L 2 116.8 V L + 3.61 = 0 V L = 0.235 V I DD = 100 x 10 6 10 1 1.9 0.6 0.235 2 0.235 = 278 μ A Checking: I DD = 100 x 10 6 2 2 1 2.5 0.235 0.6 ( ) 2 = 277 A Page 307 V TNL = 1.5 + 0.5 0.2 + 0.6 0.6 ( ) = 1.44 V 60.6 x 10 6 = 100 x 10 6 2 W L
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This note was uploaded on 09/09/2009 for the course EECS 335 taught by Professor Islam during the Spring '09 term at University of Tennessee.

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Exercise Solutions Ch6-Ch9 - Microelectronic Circuit Design...

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