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let's consider this pass transistor logic:

enter image description here

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?

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The notion of "resistance to which it discharges itself" is not very meaningful. It's better to think about the path of the current.

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.

The energy lost from the capacitor as it discharges is converted to heat by all of those elements, in proportion to their effective resistance.

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  • \$\begingroup\$ So it is in practice a parasitic resistance, not a designed resistance. I was thinking that therefore it is very low. But this means that if there is a big variation of Vin, there may be a very high current. Is this a problem? \$\endgroup\$ – Kinka-Byo Aug 16 at 15:12
  • \$\begingroup\$ Whether or not the current is "a problem" depends on whether your wires and transistor can safely pass that amount of current. In pass-transistor logic circuits the capacitance is usually pretty small and the resistance relatively high, so the current is not large enough to cause any damage. \$\endgroup\$ – Elliot Alderson Aug 16 at 15:21
  • \$\begingroup\$ Won't the capacitor discharge via the body-diode primarily when Vin goes to 1V? \$\endgroup\$ – Captainj2001 Aug 16 at 15:53
  • \$\begingroup\$ Not if the body (p-type) is connected to ground. In CMOS integrated circuits the NMOS bodies are usually tied to ground and the PMOS bodies are tied to Vdd. There need not be, and often is not, an ohmic connection from the source to the body. In the pass transistor example, the notion of which terminal is actually the source will change depending on the relative voltages...the source for an NMOS is the n-type node with the lower voltage. \$\endgroup\$ – Elliot Alderson Aug 16 at 16:14

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