i've been trying to get my head around roughly how nMOS's work, this diagram has really been confusing me:

enter image description here

Now from what i understand, at first electrons cannot flow from source to drain due to the depletion layer that is formed, effectively creating a diode that is reverse biased. When turning the gate on, it attracts the electrons near the gate's electric field, shrinking the depletion layer and creating a channel which electrons can flow. However, what does not make sense to me is that if a voltage is applied to the gate won't it be negatively charged? Therefore repelling electrons? I feel like i'm missing something obvious.

  • 3
    \$\begingroup\$ If you apply a positive voltage to the gate why should it be negatively charged? \$\endgroup\$
    – Finbarr
    Commented Apr 1, 2019 at 14:40

1 Answer 1


As part (b) shows, when you apply a positive voltage to the gate, then it will have a positive charge. This positive charge attracts electrons to the area so that they can make the "jump" from drain to source.

Note: The electrons would flow to the gate if it weren't for the "Metal Oxide" in the Semiconductor layer, which insulates the gate from the substrate and the source and drain.

This area in the substrate where the electrons are attracted is called the channel. The more positive voltage applied to the gate (higher voltage), the larger( stronger potential and larger in size ) the field in the area known as the channel. Thus allowing more electrons to flow from the drain through the channel and to the source.

PS. I know that electrons actually come from the source, but we like to think of things in positive terms for simplicity. The math is the same.

  • \$\begingroup\$ Ahh i see, my confusion i think was with thinking that where they said positive voltage, they meant basically 'having' more electrons than the ground, which i guess is opposite from the truth in this case. So the ground in this case is more negatively charged than Vdd? \$\endgroup\$
    – Dom
    Commented Apr 1, 2019 at 20:33
  • \$\begingroup\$ I would say using the word electron might make things a little more confusing. If you stick with the word "charge", then it can be positive or negative and makes the flipping of the sign a lot easier (mentally comprehending) when doing the math. As for having more charge in the channel as compared to the substrate, it's a gradient. Very dense close to the oxide layer, and then less dense as the field radiates towards the bottom of the substrate tub. \$\endgroup\$
    – Aaron
    Commented Apr 2, 2019 at 0:13

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