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enter image description here All of the textbook that I have read mention that the total capacitance is just the oxide capacitance (Cox) for a MOS capacitor in an inversion and accumulation mode. It makes sense that is true for the accumulation case, since there is no depletion region formed in the semiconductor. It also makes sense that the total capacitance for the depletion mode is the oxide capacitance and the depletion capacitance in series. But why is it that for an inversion mode, the total capacitance is just Cox, i.e. why do we ignore capacitance due to the depletion region formed in the semiconductor?

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It is important to remember that the capacitance of a MOS capacitor is the differential capacitance. In a "normal" capacitor you can calculate capacitance by dividing charge by voltage. But here we mean the derivative of charge with respect to voltage. The differential capacitance tells us how much the voltage will change if we add or subtract a small amount of charge.

Depletion mode is the odd one out. As you stated, you take the oxide and depletion capacitances in series. You do this because adding or removing charges happens at the edge of the depletion region. The thickness of the capacitor is the oxide plus the depletion region.

But in both accumulation and inversion modes the additional charges are added or removed right under the oxide, at the semiconductor surface, not at the edge of the depletion region. So the thickness of the depletion region has no effect.

In inversion you have a depletion region but you get to ignore it when calculating the capacitance because all those charges in the depletion region remain unchanged when you vary the voltage within inversion mode. They dont affect the differential capacitance since they are constants.

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  • \$\begingroup\$ That makes sense! I wish the textbooks explained it this way. Thank you. \$\endgroup\$
    – user207787
    Commented Nov 13, 2021 at 16:49
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Main concept: Electric Field Lines can start at a +ve charge and end at a -ve charge.

With that in mind:

  • Accumulation: Capacitance is Cox as the diffusion region charges don't come into play. The negative voltage pushes the free electrons in n-type material leaving behind a stationary donor impurity with net +ve charge. The field lines go from this positive ion terminate on the other side of the gate that is negative. The plate thickness of the capacitor is the gate thickness.

  • Depletion: Electrons pushed away previously are still far away from the gate but as the voltage becomes more positive, some make their way to the ions neutralizing this charge (deposited in close proximity under the gate). When gate voltage becomes positive, field lines start on the top side and are looking to terminate somewhere in the bulk, but those terminations are found deeper inside the bulk on average. It is as if plates are pushed further and further apart, so in this region capacitance keep decreasing.

  • Inversion at threshold: The channel inverts meaning that now there are plenty of carriers (electrons) available and the field lines can terminate at those points, essentailly right under the gate. Thus plate thickness is reduced to oxide thickness and diffusion region stops playing a part. The capacitance bounces back.

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  • \$\begingroup\$ Your description of accumulation is wrong. You dont deplete the semiconductor in accumulation mode. You instead accumulate more carriers of the same type as the semiconductor. I cant tell if depletion is right, its a confusing description, but you dont neutralize the ions in depletion, so I tend to think this is wrong too unless Im misunderstanding what you mean. \$\endgroup\$
    – Matt
    Commented Nov 13, 2021 at 12:46
  • \$\begingroup\$ @Matt, you can write an answer. \$\endgroup\$
    – Syed
    Commented Nov 14, 2021 at 12:19
  • \$\begingroup\$ I did, before you. Which I assumed you would have read before writing your own incorrect answer. \$\endgroup\$
    – Matt
    Commented Nov 14, 2021 at 13:12
  • \$\begingroup\$ I write my own answers and not comment on other people's work. Thanks for your interest. \$\endgroup\$
    – Syed
    Commented Nov 14, 2021 at 13:39

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