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a) Design an NMOS inverter with resistive load for: VDD = 2.5V, VIN = 0.6V, Kn = 50X10-6 A/V², VoL=0.1V. Assume power dissip

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answer) a) Given VDD = 2,5V, VTW = 0.6V, kn - unCox= 50 MALV², VOL= 0.1V. Given Power dissipation P = 0.1 mW SRL • Vout Vin FNOH= NDD = 2.5v. from power formular 0.imw= 0] 0-Imw= 125 124] RL= (1-25)(2-4)x102 IDN- Uncox -Vo] Vou that can 0.1 RL=30kn P: Design of nmos inverter is VDP 2.5v & R=30kn e out 648 Vin | [w = 8.648 b) Tol an NNOS, RON=VDS - Vout - vost MnCox W voltDesigning an resistive load nmos inverter means we have to find out the load resitance RL value and W/L ratio of the transistor.So,I found out them.The on resitance of nmos is nothing but the voltage from the drain to source to the drain current.

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