# Pass Transistor: discharging of the capacitance

let's consider this pass transistor logic:

Let's assume that the MOSFET is always active (CK = high for a N-MOSFET). Now, immagine that at the beginning Vin = 10V and Vx = 0: the capacitance Cx will be charged and at the end we will get Vx = 10V.

Now suppose that Vin becomes 1V: the capacitance will discharge itself until it reaches the value 1V. But, which is the resistance to which it discharges itself? If I look to the previous circuit I see, since the transistor is ON, simply a voltage source equal to 1V in parallel with a capacitor which is precharged to 10V.

Is it the internal resistance of the voltage source connected to Vin, the internal resistance of the transistor etc?

When Vin goes to 1V, the left end of the MOSFET becomes the source and the $$\V_{GS}\$$ is a high voltage so the MOSFET becomes conducting. Current flows from the capacitor, through the MOSFET, and into the voltage source. In any real circuit, all of the wires will have some resistance. The MOSFET has a drain-to-source voltage across it when it is conducting, so it also dissipates power. The capacitor has an ESR (effective series resistance) and the voltage source has some effective resistance.