Hi, probably a dumb question but I can't figure it out. For an n-channel enhancement MOSFET, why does the saturation current occur after Vds = Vgs - Vt. From what I understand is that this causes the channel to pinch-off at the gate/drain edge and inverts back into the p-type region. In this case, how is there constant current flow between the drain and source if the channel becomes disconnected.

Thank you,
JP

 

Okay, I got a rough picture of what is going on now, so if someone could clarify this further that would be great.

When Vds = Vgs - Vthres, the n-channel pinches off at the drain. With further increase in Vds, (Vds > Vgs - Vthres), the pinch-off point moves farther from the drain, shrinking the channel, merging the depletion regions of the drain and channel together. The pinch-off point has a voltage of Vgs - Vthres and still attracts electrons from the source. When the electrons reach the pinch-off point, they are injected into the depletion region between the drain. Because the pinch-off voltage remains more or less constant and Vds is subject to increase, the field between them becomes stronger for increasing Vds. The electrons that are injected are accelerated through the depletion region to the drain. Thus, (conventional+) current actually flows from drain to source.

Is that what happens?

 

Ya you got it.
And remember any depletion region has an electric field. So if a charge is injected into the depletion region, due to the E field it is swept out of the region.

This is the reason whenever pinchoff occurs there is no channel but still current flows due to this sweeping effect of E-field.
Many people initially feel that since there is no channel no current will flow. But that's wrong perception