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I have very little experience with electrical engineering, so please bear with me. I was hoping to get a qualitative explanation as well as pointers to the maths/formulas which could formally explain the principles behind the following question:

In a MOSFET, I understand the concept of why an inversion layer can be formed by the gate-body voltage, but why is it that the required gate-body voltage to create this inversion layer is so much lower than the voltage required to overcome the reverse bias - i.e. It takes only 6V to create an inversion layer via the gate, but it would take 100V to create an "inversion layer" between the source and drain (overcoming the reverse bias junction). Is it the positioning of the gate perpendicularly to the npn junctions which somehow affects the ease with which the electrical field can make an inversion channel across the device? Or perhaps that the gate-created "inverted" channel across the middle-p-section is a smaller area than would be inverted if the source-drain voltage were high enough to create a channel? Or is it the proximity of the gate (and thus its field) to the area on the semiconductor it is trying to invert?

If so, what are the physics formulas which explain this? (I have a hunch that the electrical field of the gate is like chipping a small channel across the top of a brick wall rather than trying to push it down from the front)

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  • \$\begingroup\$ Have a look at this. \$\endgroup\$ Jul 9, 2016 at 21:40

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why is it that the required gate-body voltage to create this inversion layer is so much lower than the voltage required to overcome the reverse bias?

Or is it the proximity of the gate (and thus its field) to the area on the semiconductor it is trying to invert?

That's it. The gate is separated by mere 10's of nanometers from the channel by the gate oxide, whereas the channel length might be separated by µm's or more (depending on VDSmax), so the electric field from the gate is hundreds of times stronger.

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