In a MOSFET, the current saturates at VGS-VT = VDS. Why does more current pass at higher values of VGS? Is it due to an increase in channel depth or more electrons in the channel? Is it due to higher VDS required for saturation, resulting in higher electric field pushing electrons through the point of pinch off?

As far as I am aware, a battery results in constant electric field, which means that no matter which value of voltage VGS-VT = VDS occurs at, the effect on channel effective length is the same, so it should have no effect on the current.

  • \$\begingroup\$ To avoid ambiguity, show your circuit. \$\endgroup\$
    – Andy aka
    Nov 25, 2022 at 20:27
  • \$\begingroup\$ I don't have a particular circuit that I am studying. I was look the ID-VDS graph as it is shown in Sedra Smith. The question is in regards to the theory of operation. At saturation, Id is proportional to the square of VGS. But what happens physically in the channel? \$\endgroup\$ Nov 25, 2022 at 21:35

1 Answer 1


Assuming that \$V_{DS}\$ is supplied from an ideal dc voltage source and \$V_{DS}=V_{GS}-V_{T}\$, then the channel is at pinch-off and length of the channel is maximum.

Increasing \$V_{GS}\$ will not change the channel length, but will attract more electrons into the channel, increasing the current \$I_{D}\$.

The channel will move into the resistive region, away from pinch-off. Increasing \$V_{DS}\$ will bring the channel back to pinch-off at the higher current.


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