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I understand that in an n channel MOSFET as Vgs increases, electrons are attracted towards the gate forming an n channel. These electrons are responsible for the conduction of the MOSFET. As Vds crosses saturation voltage a pinch-off occurs. This pinch-off results in channel tapering and the n channel effectively reduces. My question is if tapering occurs, shouldn't the available charge carriers in the channel reduces thus reducing current flow?

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As the channel current increases, so does the drop across it (the channel is like a resistance), and as a result the drain voltage rises above source voltage. This does start to constrict the channel but as the channel gets pinched the drop gets limited as well and so a form of equilibrium is reached which limits the maximum current the channel can carry.

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No, current flow does not drop in saturation. It increases very slightly as VDS increases. I think you already know this and are just wondering where the carriers go since the channel is pinched off.

Basically, they spread out in the two dimensions at right angles to the channel so that conduction is no longer just taking place through a narrow channel.

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  • \$\begingroup\$ If they spread out at right angles and if I imagined it right, then the carriers are still in the tapered regions. Please correct me if wrong. \$\endgroup\$ – jrvinayak Oct 8 '19 at 16:27
  • \$\begingroup\$ Yes, that's correct. You might be confused about how the current can flow, given that that there is a region with no mobile carriers. The answer is that the electric field across the depletion region can still move current across the region. This will explain much better than I possibly can: electronics.stackexchange.com/questions/77198/… \$\endgroup\$ – Annie Oct 8 '19 at 19:17

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