Consider the nMOS for simplicity, even in saturation, Ids grows with Vds because the depletion region around the drain grows too with Vds and then the effective length of the channel is shorter.

What i don't understand is that, why we consider the depletion region around the drain a good conductor ?

For example, usually we want to work with a Vgs higher than Vt so that the channel is 'filled of' electrons (0 < Vgs < Vt we have that the channel is made by a depletion region and we just have some leakage of current...! but we don't use the MOSFET in this region...)

  • \$\begingroup\$ yet a full-length channel still exists, because the gate voltage is higher than the bulk voltage by at least Vthreshold. \$\endgroup\$ Jun 6, 2019 at 16:20

1 Answer 1


The channel isn't completely cut off due to the strength of the electric field between the drain and the channel. We use this region because the current is no longer dependent on Vds and we can amplify waveforms without distortion as long as we make care to stay in saturation.

I mean strictly speaking you're right the depletion region around the drain in saturation is smaller than it is in the triode region, so it technically isn't as good, but as for the reason stated above it doesn't matter, current is still flowing, but now largely without regard to Vds.


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