EE331HW9solutions - EE-331 Devices and Circuits 1 Prof. R....

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+3.3V 0V C 40pF V in V out I C 0 0 +3.3V V out (t) 1ns 0 0 0.132A I C (t) 1ns The current waveform just looks like the square pulse shown, essentially the derivative of the voltage waveform: 64 I C 8.448 amp = If all 64 drivers switch at the same time, the total peak current will be: I C 0.132 amp = I C C V t := t 1.0 ns := V 3.3 volt := C4 0 p F := (Use this value of 1.0 ns shown in Fig. P6.31, not the 0.8 ns in the problem text.) The average current to discharge each of the capacitive loads is just the dV/dt of the voltage waveform multiplied by the capacitance of the load: Problem 1: Problem 6.31 of Jaeger and Blalock. µ W1 0 6 watt := mW 10 3 watt := ns 10 9 sec := M 10 6 ohm := EE-331 Devices and Circuits 1 Prof. R. B. Darling Homework # 9 Solutions First, define some useful units and constants: µ m1 0 6 m := eV 1.602 10 19 joule := ms 10 3 sec := µ s1 0 6 sec := q 1.602 10 19 coul := k B 8.62 10 5 eV K 1 := nm 10 9 m := ε rSiO2 3.9 := ε o 8.854 10 14 farad cm 1 := n i 10 10 cm 3 := V T 0.025 volt := ε rSi 11.7 := ohm := k 10 3 ohm := EE-331 HW#9 Solutions Page 1 R. B. Darling
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I D V DD V OL R := I D 6.2 µ A = P D V DD I D := P D 15.59 µ W = Problem 3: Problem 6.53 (a) of Jaeger and Blalock. Find the values of R, W, and L for the resistively loaded nMOS inverter: V DD 3.0 volt := V OL 0.30 volt := I DD 33 µ A := R V DD V OL I DD := R 81.8 k = k n 60 µ A volt 2 := V Tn 0.75 volt := WLratio I DD k n V DD V Tn () V OL 0.5 V OL 2 := WLratio 0.873 = Choose L 2.0 µ m := W WLratio L := W 1.746 µ m = Problem 2: Problem 6.39 (a,b) of Jaeger and Blalock. (a) V OH is just equal to the upper power supply rail, +2.5 Volts. V OL is determined by setting the current through the resistor equal to the nonsaturated current of the MOSFET: k n 100 µ A volt 2 := V Tn 0.60 volt := V DD 2.5 volt := R 200 k := V OL 0.25 volt := (initial guess at root) V OL root V DD V OL R k n 3 1 V DD V Tn V OL 1 2 V OL
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EE331HW9solutions - EE-331 Devices and Circuits 1 Prof. R....

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