ch06_solutions - CHAPTER 6 6.1 (a) Pavg = 6.2 1W 10-5 W...

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159 CHAPTER 6 6.1 ( a ) P avg = 1 W 10 5 gates = 10 m W / gate ( b ) I = 10 -5 W / 5 V = 2 m A / 6.2 ( a ) P avg = 40 W 10 7 gates = 4 m W / ( b I = 4 x 10 -6 W / 3.3 V = 1.21 m A / ( c ) I total = 1.21 m A 10 7 gates ( 29 = 12.1 A 6.3 ( a ) V H = 5 V | V L = 0 V P V H = I 2 R = mW P V L = 5 - 0 10 5 2 10 5 = 0.25 ( b ) V H = 3.3 V V L = 0 V P V H = I 2 R = P V L = 3.3 - 0 10 5 2 10 5 = 0.109 6.4 v O v I V H (3.3 V) V L (0V) 1.1 V (V REF ) 3.3V V + 6.5 6.6 v O v I V H (3.3 V) V L (0V) 1.1 V ) 3.3V V + Z = A ( 29 = A v I A V V REF 6.7 V H = 3 V V L = V V IH = 2 V V IL = 1 V A V = dv O dv I = - 3 V 1 V = - 3
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160 6.8 6.9 v O v I V H (3 V) V L (0V) 1.33 V 3V V + 1.67 V 1.5 V Slope = +9 V OH = 5 V V IH = V REF = 2 V NM H = 5 - 2 = 3 V V OL = 0 V V IL = V REF = V L = 2 - 0 = 2 V 6.10 We would like to achieve the highest possible noise margins for both states and have them be symmetrical. Therefore V REF = 3.3/2=1.65 V. 6.11 V H = 3.3 V | V L = 0 V V IH = 1.8 V V OL 2245 0.25 V V IL = 1.5 V V IH 2245 3.0 V H = 3.0 - 1.8 = 1.2 V L = 1.5 - 0.25 = 1.25 V 6.12 V H = 5 V V L = 0.25 V 6.13 V H = - 0.80 V V L = - 1.35 V 6.14 V IH = V OH - H = - 0.8 - 0.5 = - 1.3 V V IL = L + V OL = 0.5 + - 2 ( 29 = - 1.5 V 6.15 τ P = PDP/P = 10 -13 J/10 -4 W = 10 -9 s = 1 ns 6.16 ( a ) P avg = 1 W 2.5 x 10 5 gates = 4 m W / gate ( b ) I = 4 x 10 -6 W / 5 V = 0.80 m A / ( c ) PDP = 2 ns m W ( 29 = 8 fJ 6.17 ( a ) P avg = 40 W 8 x 10 6 gates = m W / ( b ) I = 5 m W / 3.3 V = 1.5 m A / ( c ) PDP =1 ns 5 m W ( 29 =
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161 6.18 1 10 50 100 P 1 250 Slope = 1 Slope = 2 PDP 6.19 ( a ) v t ( 29 = i t ( 29 R + v C t ( 29 | i t ( 29 = C dv c t ( 29 dt v t ( 29 = RC dv c t ( 29 dt + v C t ( 29 v t ( 29 = 1 for t 0 v t ( 29 = 1 - exp - t 0.9 =1 - - t 90% → t 90% = - ln 0.1 ( 29 0.1 =1 - - t 10% t 10% = - ln 0.9 ( 29 t r = t 90% - t 10% = ln 9 ( 29 = 2.20 ( b ) v t ( 29 = 0 v C 0 ( 29 = v t ( 29 = - t 0.9 = - t 90% t 90% = - ( 29 0.1 = - t 10% t 10% = - ln 0.1 ( 29 t f = t 10% - t 90% = ( 29 = 2.20 6.20 ( a ) V H = 5 V V L = 0.25 V ( b V 10% = V L + 0.1 V = 0.25 + 0.475 = 0.725 V t 10% 2245 22 ns for v O V 90% = V L + 0.9 V = 0.25 + 4.275 = 4.525 V t 2245 33 ns for v O t r = 33 - 22 = 11 ns For fall time : t 2245 2.5 ns for v O t 90% 2245 0.8 ns for v O t f = 1.7 ns For v I , t 10% 2245 0 ns t 90% 2245 1 ns t r = 1 ns t f 2245 1 ns c ( 29 t PHL 2245 1.5 ns - 0.5 ns = 1 ns t PLH 2245 26 ns - 21 ns = 5 ns ( d t P = 1 + 5 2 ns = 3 ns
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162 6.21 ( a ) V H = - 0.78 V | V L = - 1.36 V ( b ) V 10% = V L + 0.1 V = - 1.36 + 0.1 0.58 ( 29 = - 1.30 V t 10% 2245 32.5 ns for v O V 90% = V L + 0.9 V = - 1.36 + 0.9 0.58 ( 29 = - 0.84 V t 90% 2245 42 ns for v O t r = 42 - 32.5 = 9.5 ns For fall time : t 10% 2245 11.5 ns for v O t 90% 2245 2 ns for v O t f = 9.5 ns For v I , t 10% 2245 0 ns t 90% 2245 1 ns t r = ns t f 2245 ns c ( 29 V 50% = - 0.78 - 1.36 2 = - 1.07 V t PHL 2245 4 ns t PLH 2245 ns ( d t P = 4 + 4 2 ns = ns 6.22 6.23 6.24 A + B ( 29 A + C ( 29 AA + AC + BA + BC A + AC + + A 1+C ( 29 + AB + A + AB + A 1 + B ( 29 + A + Z = ABC + ABC + A Z = ABC + + + + A Z = AB C + C ( 29 + A + A ( 29 Z = AB 1 ( 29 + 1 (29 Z = AB + A B C Z 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 1 1 1 1 1 Z = AB+BC
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163 6.25 6.26 6.27 Z = A B C + ABC + A BC + AB C Z = C A B + AB + A B + AB ( 29 Z = C A B + A B + AB + ( 29 Z = C A B + B ( 29 + A B + B ( 29 ( 29 Z = C A 1 (29 + A 1 ( 29 ( 29 Z = C A + A ( 29 Z = C 1 ( 29 Z = C A B C Z 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 Z = C A B C D Z 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 Z = AB + CD Z = AB ( 29 ( 29 Z = ABCD 6.28 6.29 (a) Fanout = 2 (b) Fanout = 1 A B C Z 1 Z 2 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 1 0 1 1 0 0 0 0 1 0 1 0 1 1 1 0 1 1 1 1 1 1 1 Z 1 = AB = AB Z 2 = AB + C 6.30 (a) Diode AND; Z = ABC (b) Diode OR; Z = A+B+C 6.31 (a) Diode AND; Z = AB (b) Diode OR; Z = A+B
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164 6.32 ( a ) I B = 3.3 - 0.6 - 0.7 10000 = 200 m A | I C = 3.3 - 0.05 3000 = 1.083 mA The transistor is saturated since I B I C b F = 54.2 m A (b) I = 3.3 - 0.6 10000 = 270 m A (c) P = VI = 3.3 V 0.200mA +1.083mA ( 29 = 4.23 mW (d) P = VI = 3.3 V 0.270mA ( 29 = 0.891 6.33 I B = 3.3 - 0.6 - 0.7 30000 = 66.67 m A I C = 3.3 - 0.05 3000 = 1.083 mA The transistor is saturated for b F I C I B = 1.083 0.06667 =16.2 6.34 6.35 Z = AB ( 29 C = ABC A B Z D 4 D 2 D 1 +3.3 V R B R C
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This note was uploaded on 01/17/2011 for the course ECE 3040 taught by Professor Hamblen during the Fall '07 term at Georgia Institute of Technology.

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ch06_solutions - CHAPTER 6 6.1 (a) Pavg = 6.2 1W 10-5 W...

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