homework3_sol

# homework3_sol - x = 6C(5/3/f = 3.16C 4 Problem 4.11 from...

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Computer-aided IC Design/VLSI I (Prof. David Pan) Homework #3 Solution 1. Problem 4.3 from the Exercises for Chapter 4 The rising delay is (R/2)*(8C)+R*(4C+2C)=10RC and the falling delay is (R/2)*C+R*(2C+4C)=6.5RC. (Note: these are only the parasitic delays, and the additional effort delay of the gate is not included.) *the answer is based on the transistor schematic and the stick diagram you draw. 2. Problem 4.4 from the Exercises for Chapter 4 The output node has capacitance of 3nC, and each internal node has 2nC. The resistance through each pMOS is R/n. The propogation delay is: 3. Problem 4.10 from the Exercises for Chapter 4 (a) should be faster than (b) because the NAND has the same parasitic delay but lower logical effort than the NOR. In each design, H = 6, B = 1, P = 1 + 2 = 3. For (a), G = (4/3) * 1 = 4/3. F = GBH = 8. f = 8 1/2 = 2.8. D = 2f + P = 8.6 τ . x = 6C * 1/f = 2.14C. For (b), G = 1*(5/3). F = GBH = 10. f = 10 1/2 = 3.2. D = 2f + P = 9.3 τ

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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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homework3_sol - x = 6C(5/3/f = 3.16C 4 Problem 4.11 from...

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