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I am having lots of trouble trying to understand how the mosfet is triggered. The text I read assumes the source of the NMOS connect to ground, while a positive voltage is applied at the gate.

enter image description here

Because the source is grounded, VGS (voltage between gate and the source) = VG (voltage of the gate). And then, it claims that because a higher VG accumulates positive charge at the surface of the gate, which repel the holes beneath the oxide layer, leaving the negative ions. So the charge density beneath the oxide layer is: enter image description here

where Qd is the charge density at the depletion region created by applying VG.

Now this is my question: WHY does the term VGS instead of VG? If, say the source is not grounded, but connect to something like 0.01v, then VGS is not equal to VG, and to overcome VTH (threshold voltage), I didn't see why we need to consider the source voltage To me, this doesn't make sense because repelling holes and creating the depletion region beneath the oxide layer has nothing to do with the voltage level at the source; only the gate voltage matters. Why VGS then?

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  • \$\begingroup\$ Because capacitor. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 11 '14 at 5:34
  • \$\begingroup\$ @IgnacioVazquez-Abrams, what do you mean by capacitor \$\endgroup\$ – kuku Sep 12 '14 at 18:38
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The transistor only knows about the voltage between its terminals. It doesn't know or care what the voltage is between any of its terminals and what you consider "zero volts" or circuit ground.

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Building on @Peter Bennett's answer, the voltage level at the source matters very much. The transistor, along with many other electrical components, could care less about the voltage level of an individual terminal. What matters is the relationship between the different terminals.

For example, if the source was at 10 volts, the gate at 15 volts, and the drain was at 20 volts, the transistor would react exactly the same as if the source, gate, and drain were at 0, 5, and 10 volts, respectively.

Similarly, if your body was at 1000V and you touched something else that was at 1000V volts, you wouldn't get shocked. Further still, any multimeter I've used has two test leads to measure voltage. If absolute voltage mattered, you would only need one of the leads to measure it. However, what really matters is the difference between voltages. The zero volt reference (Not ground. Reference or return are much better descriptions.) can be placed wherever you want.

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I would understand this by recognizing that VGS is what allows charge carriers to enter the semiconductor channel. The gate charge is able to attract or repel charge carriers, but the source charge determines the amount of potential that's required to do so. If the source voltage is very low then there are more carriers available and it requires a lower gate charge to repel them and keep them out.

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The substrate (body) is (almost always) connected to the source on discrete MOSFETs. See this diagram.

http://www.doitpoms.ac.uk/tlplib/semiconductors/images/mosfet.jpg

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  • \$\begingroup\$ So, are you saying that, since B is almost always connect to S, then VGS is just VGB, which is as if you are applying potential to a capacitor with the poly is one plate, and the bottom of the body as the second plate? \$\endgroup\$ – kuku Sep 12 '14 at 3:57

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