20085eeM216A_1_Hw3-F08-Sol

# 20085eeM216A_1_Hw3-F08-Sol - Electrical Engineering...

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Electrical Engineering Department Fall 2008 1 EEM216A – Design of VLSI Circuits and Systems Homework #3 Solution D. Markovic Problem 1: Delay and Energy Capacitance In this problem we have to determine linear equivalent of the inverter input capacitance using Spectre. Use simulation setup similar to the one discussed in class. All inverters are INVX1. To develop models for supply voltage optimization, do the following for V DD = 0.4V, 0.7V, and 1.0V. a) Determine capacitance C D from the simulation setup to match propagation delays of the inverter loaded with another identical inverter (include extra stage to suppress Miller capacitance). b) Determine capacitance C E in such a way as to match energy of the two inverters from the simulation setup. Calculate energy by integrating current through V DD . c) Based on signal slopes from (1a) and transistor thresholds from homework 1, estimate the capacitance C SC that corresponds to short-circuit energy dissipation. Discuss C SC / C E (V DD ). d) Are C D and C E different? How does V DD affect results? Comment your results. SOLUTION Capacitance values are obtained by simulation as shown in lecture 3 ( slide 40 ). Figure 1 illustrates the estimation method for a fixed V DD (1V in this case). 1.5 1.7 1.9 2.1 2.3 2.5 10 12 14 16 18 20 C (fF) Delay (ps) C D = 1.95 fF 1.5 2 2.5 3 3.5 4 4.5 2 3 4 5 C (fF) Energy (fJ) C E = 2.45 fF C Ec = 3.95 fF Figure 1. Simulated delay (left) and energy (right) versus capacitance. Plot on the right estimates two values: C E represents energy on the gate capacitance, C Ec represents energy stored on gate and parasitic capacitances. In both cases, we are matching energy dissipated from V DD of the test gate. Discussion: The difference between C E and C Ec is that C E considers gate capacitance only, while C Ec lumps both C gate and C parasitic . In fact, we can use these values to estimate γ as = (C Ec – C E )/C E = 0.61. As you may expect, this matches our result from homework 2 , problem 1c .

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Electrical Engineering Department Fall 2008 2 Cap / V DD 0.4 V 0.7 V 1.0 V C D (fF) 1.5 1.7 1.9 C E (fF) 2.1 2.2 2.4 (1c) Short circuit power exists when both NMOS and PMOS are “on” simultaneously during a single switching transition. For the output rising transition, we can assume that PMOS is “on” and that NMOS is linear (V out close to 0). For the output falling transition, we can assume that NMOS is “on” and that PMOS is linear (V out close to V DD ). Strategy: Approximate short circuit current with a triangular current waveform. Given the slope values from previous homeworks, we can approximate t SC as the time spent between V tn and V DD + V tp . (factor 0.8 accounts for 10
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## This note was uploaded on 02/06/2011 for the course EE M216A taught by Professor Marković during the Spring '08 term at UCLA.

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20085eeM216A_1_Hw3-F08-Sol - Electrical Engineering...

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