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As the title says, for a MOSCap with, say, a p-doped semiconductor substrate, when a gate voltage \$V_{g}\$ is applied across the gate and the grounded substrate, why doesn't \$V_{g}\$ drop entirely across the oxide or why does there have to be a non-zero potential at the oxide-semiconductor interface?

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  • \$\begingroup\$ Is 'contact potential' the answer you looking for? \$\endgroup\$ – nidhin Oct 18 '15 at 14:03
  • \$\begingroup\$ No. I want to why there has to be a voltage drop in the semiconductor also. Contact potential doesn't feature here. \$\endgroup\$ – transistor Oct 18 '15 at 14:38
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The differential form of Gauss's Law is

$$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}$$

This means that the electric field is continuous across space. Sometimes when there is a highly concentrated charge (such as a surface charge on a conductor), we might obtain an approximate discontinuity in the electric field. However, if the MOSFET is designed well, no such surface charge exists at the interface between the oxide and the semiconductor, so the field cannot be discontinuous.

Therefore if the field in the oxide is non-zero, the field in the semiconductor must be nonzero as well.

Then, we also define the electrostatic potential by

$$V=-\int_C\vec{E}\cdot\mathrm{d}\vec{l}$$

So that if there is a field in the semiconductor, there must also be a variation of the potential.

Aside: In fact, it is possible for surface states to occur at this interface, which can trap electrons at the interface, resulting in a reduced field in the semiconductor. This generally leads to poorer MOSFET performance, so MOSFET makers try to avoid conditions that lead to surface states.

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