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In an NMOS we have a p-substrate, and we use a positive voltage to attract negative charge "to the top". But could we have used negative charge to attract holes instead and gotten a "hole-current"? I am confused by this because the p-substrate have more holes than negative minority carriers so why not attract the holes to the top instead?

Is there a simple explanation for this?

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  • \$\begingroup\$ You mean negative charge to attract holes? \$\endgroup\$ Commented May 2, 2022 at 13:23
  • \$\begingroup\$ @user253751 Yes sorry. \$\endgroup\$
    – user394334
    Commented May 3, 2022 at 2:48

2 Answers 2

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In a N-channel MOSFET with Vgs = 0, there is a depletion region between the source and drain N+ regions and the P-type substrate that is a barrier to current flow. In normal operation with Vgs > 0, minority carriers are attracted to the oxide interface in sufficient quantities to create an inverted channel that acts as a N-type "bridge" between the source and drain (ignoring channel length modulation etc). If Vgs < 0, you are attracting more substrate majority carriers to the oxide interface but the depletion region still exists so there is no change in current between source and drain.

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In an NMOS, the source and drain are N diffusions. When a channel (electrons) is formed, you can imagine a continuous conductive sheet is formed between S and D at the top surface of the channel.

If you bias the gate negatively, more holes will accumulate under the gate, but the electrons in the S can't enter this P-type region (because of the P-N potential barrier), so no current can flow.

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