I have a question about the saturation mode. The cross-section of an NMOS transistor in saturation mode is usually drawn like this:

enter image description here

But it seems to me that there can be no current from source to drain in this case (because there is no contact between the channel and the drain). From the diagram that plots \$V_{\text{DS}}\$ against \$I_{\text{D}}\$, it can be seen that in saturation mode, there is a current (it just doesn't increase when we increase \$V_{\text{DS}}\$).

So what is going on in saturation mode?


1 Answer 1


At onset of saturation, inversion charge near drain decreases until it becomes zero, close to drain region. VD at which this happens is called pinchoff and drain current starts to saturate. At saturation regime (VD > VD-sat), pinchoff moves towards source leaving behind a depleted region; inversion charge drifts down conducting channel and is then injected into depletion region. ID will essentially remain constant because inside channel potential is fixed at VD-sat.

Please note that this is kind of "first-order" behavior of the MOS transistor, in reality the drain current in saturation regime is not constant and is affected by many other phenomena e.g. channel length modulation, velocity saturation, etc.

  • 1
    \$\begingroup\$ Good answer. Just to add for the original poster, just because there is no channel doesn't mean there aren't any carriers. As you mentioned, there is an incredibly thin sheet of carriers (electrons in N-type) present (but not drawn) but these are only swept into the channel as they travel from source to drain, they aren't free carriers since it's a depletion region. \$\endgroup\$ Apr 7, 2017 at 23:34

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