In the case of a mosfet, it is more like a capacitor: the voltage applied on the gate opens / closes the channel of conduction between the drain and source.
Yes. I'm looking specifically at the dielectric barrier between this capacitor and what I forget the name for, let's call the channel. This meeting point
is one of the diode boundaries I'm discussing.* And I'm trying to understand why the minimum voltage of this junction is 0.7V (or maybe it's not!), by necessity, before significant conduction can be achieved.
So the thickness of this barrier determines how far apart are the, say the positively charged plate of that gate capacitor and the semiconductor channel. And this is (I think) at least a very significant reason for the
minimum voltage. (Although full saturation can be adjusted to however high you want via size/shape/distance/doping). This is (I think) possibly related to the VFD of an LED*. This is (I think) the reason for the
minimum threshold for a FET. If you could metaphysically superimpose the two plates, you could reduce distance to nothing, and force of attraction between positive and negative charges would be infinite. But there is a discrete distance of the barrier, itself, and of the materials carrying those charges.
Stepping back one step.... the attraction is necessary to hold the electrons in the channel... so farther apart your positive gate cap plate from the channel the higher the charge before you can start packing the channel with conductors. And this attraction being dependent on inverse of distance squared, the closer you can put these charges and still keep them separate, the lower the threshold voltage where channel saturation may begin. Furthermore, the construction of these semiconductors and thickness of the dielectrics that can be "grown" on these dies are dependent on many things including the materials used. Which is why I brought up germanium and silicon diode junctions in my earlier post analogy. So apologies if I skipped a few steps.
Looking at the explanations for how a FET works, at least the basic searches I did, does not include this information how if/why there might be a minimum threshold. Yet article such as I linked suggests 0.7V min. Experience with FETs says it's same.
I suspect if I were to research how a diode works, and why it has a certain VFD, perhaps there are more clues.
*I know the diode boundary in a regular diode is different than a gate. Maybe I should leave that out. And perhaps I'm misusing the terminology of what a diode junction technically means (my understanding of the Art of Electronics is certainly not complete). I mean the boundary between two semiconductors, in this case between the gate capacitor and the semiconducting channel. I keep connecting the two because of the 0.7V thing, but perhaps a diode has 0.7V drop for a completely different reason and it's coincidence that this number is the same.
Maybe I'm just stating common knowledge. Or academic minutia which no one should bother to remember, or maybe even to learn in the first place.
Or maybe I'm inventing new physics, as another member so kindly suggests. If someone can help me to understand why (or even if ) FET needs about 0.7V min to switch? And if this is the case, is it a coincidence that this is the VFD of a silicon diode? I already asked with no one responding. So I have tried to figure this out myself.
Maybe this is a question for Jim Williams.