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Unformatted text preview: . x = 6C * (5/3)/f = 3.16C. 4. Problem 4.11 from the Exercises for Chapter 4 D = N(GH) 1/N + P. 5. Problem 4.13 from the Exercises for Chapter 4 Assumptions: XOR gate has g=p=4, and neglecting the branch on A which could be buffered. G = 4*(9/3)*(6/3)*(5/3) = 40 B=16 driving the final ANDs. H=10/10=1 Path Effort F=GBH=640. N=4, Best stage effort f=(640) 1/4 = 5.03. P=4+4+4+2=14 D=Nf+P = 34.12 = 6.8 FO4 delays. Input capacitances, z = 10*(5/3)/5.03 = 3.18; y = 16*z*(6/3)/5.03 = 20.27; x = y*(9/3)/5.03 = 12.09. 6. 7. (a) Assume that all load and internal capacitances have been charged to V DD . Which input input vector will result in the longest t pdf ? ABCDE = 10111 This presents the maximum capacitance to the output node. (b) Assuming all the capacitors are initially charged to V DD , use the Elmore delay approximation to find the value of t pdf for the input vector ABCDE = 10111. t pHL = 20 * 5 + 15 * 10 + 20 * 15 + 50 * 15 = 1.3nS...
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 Spring '10
 PAN

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