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I have difficulties understanding what happens in MOS-FET-pinch-off: Take an N-MOSFET:

Near the source the gate-bulk voltage is high enough to form an inversion layer. So we have electrons as majorities in the inversion layer. As the channel potential increases towards the drain, there is less voltage difference to form an inversion layer. Ultimately, close to the drain the difference is insufficient to form an inversion-layer.

So we have a thin depleted p-layer or i-layer. I don't understand why this is still conducting.

To me this looks like an N(source), N(channel), P(pinchoff-area), N(drain) structure. I would expect that to isolate or at least require an Uf between drain and source to conduct, which is not the case.

What am I getting wrong ?

I would be interested to see a band-model-diagram from source to drain for a MOSFET in pinch-off, but I have never found any, neither in a book nor on the web.

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The channel can pinch off, leaving a depletion region. As you said, the channel potential increases towards the drain. This means there is an electric field, and said field can pull the electrons in the channel through the depletion region.

I tried to explain it as best as I could, there is a more comprehensive answer below.

How does the drain current flow in a MOSFET when the channel is pinched off?

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  • \$\begingroup\$ Meanwhile I came up with the analogy of an NPN-structure in punch-through: One N is the drain end of the inverted channel. The other N is the reverse-biased N-Diffusion at the drain. The p is the p-type pinch-off area. Now the NP-Junction at the drain is reverse-biased. The corresponsing space-charge-region extends up to the channel-end. This leaves an electric-field at the "junction" channel-to-pinch. This field forward-biases this junction and e- get injected as minorities into the space-charge region were they are swept out to the drain. Isn't this just like in a npn punch-through? \$\endgroup\$
    – wolfgang6444
    Commented Aug 22, 2019 at 16:09

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