HW5_soln

HW5_soln - T B D =75(40+40+10)+75(40+10)=0.0105ns If you...

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ECE124A VLSI Principles Solutions to Homework # 5 Problem 1:
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Problem 3:
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Problem 4:
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Problem 5:
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Problem 6:
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Problem 7:
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Problem 8:
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Problem 9: T1 A Æ B =150 (70+70+40+10+120+120+20) + 150 (70+40+10+120+120+20)=0.125ns T B Æ C =260(120+120+20)+260(120+20)=0.104ns
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Unformatted text preview: T B D =75(40+40+10)+75(40+10)=0.0105ns If you add a inverter after B C branch in node B there is input capacitance of inverter: Do the same procedure as previous part, and put h T =0, You find h=30,...
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HW5_soln - T B D =75(40+40+10)+75(40+10)=0.0105ns If you...

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