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MIT 6.002 Circuits and Electronics, Spring 2007
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Midterm Q2: VS=5.0V, VOH=4.5V, VIH=4.0V, VIL=1.5V, VOL=1.0V VT=2.0V and RON=6000.0Ω.
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low noise margin=VIL-VOL=0.5 HIGH NOISE MARGIN=VOH-VIH=0.5 forbidden region=VIH-VIL=2.5 RpuI=4*RON=24000 RpuA=8*RON=48000 RpuO=4*RON=24000 power for inverter=0.000833 power for NAND=0.000416 power for NOR=0.000925
Danke!!!
if RON=16000.0Ω. power for inverter=? power for NAND=? power for NOR=?
?
if ron=16000, for inverter==>p=19.992/Rpul for nand,==>p=19.968/Rpua for nor,==>p=22.2/rpuo bcz,P=(v^2)/R i gave you v^2 from above answer in reverse way
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@apurvdixit713
@apurvdixit713 power for inverter=(Vs^2)/(RpuI+RON) power for NAND=(Vs^2)/(RpuA+2RON) power for NOR=(Vs^2)/(RpuO+(RON//RON))
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