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Consider a simple CMOS inverter. To calculate the switching power dissipation, we consider the case when capacitor is charged till by pmos as the current is drained from Vdd. But when it is being discharged by nmos the power supply is cut off from circuit the current simply goes to the ground. Since Vdd is not involved, so is nmos even dissipating power?

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  • \$\begingroup\$ Why does the CMOS inverter dissipate power, as the Cload is being discharged? \$\endgroup\$ Commented Jun 25, 2019 at 10:11

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Yes, the NMOS transistor is consuming power, and converting electrical energy to heat. An MOS transistor has an effective resistance from source to drain (\$R_{DS}\$) when it is conducting, and current flowing through this resistance consumes power.

In the ideal case, when the load capacitance is charged you find that exactly half of the energy drawn from the power supply will end up in the capacitor...the rest is lost as power consumed by the PMOS transistor. When the capacitor is discharged through the NMOS transistor then the remaining half of the total energy is consumed by the NMOS transistor.

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It doesn't consume power from the power source when discharging the effective load capacitance but it does dissipate power based on the stored energy in that capacitor. Then the cycle starts again - energy is taken from the power supply to charge the effective load capacitor and that energy is turned to heat in the 2nd part of the switching cycle.

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