When we draw the electric field in the oxide of an MOS capacitor in the depletion regime, we take the charge in the depletion layer (Qd) and calculate the oxide electric field (Eox) as Qd/epsilon of Ox. (Ideal case with no charges in the oxide). My doubt is, in this approach do we consider the charges in the oxide as that on an infinite two dimensional sheet? Can anybody help ? Thanks in advance.
Correction: please read..." do we consider the charges in the depletion layer as that on an infinite sheet ?"
The depletion charge has a thickness (Xd, the depletion layer width) so I don't believe assuming it is a sheet is reasonable. The inversion charge and the Si-SiO2 fixed interface charge (Qf) are often approximated as a sheet, because their thickness is usually very small compared to other magnitudes, in many devices.
Dear Rajeev, you are right. Considering a finite sheet, the electric field vector changes considerably (as regards both direction and magnitude) as one approaches the boundary of the flat metallic sheet; it becomes a mathematically difficult problem to handle (although now-a-days one can solve the problem relatively easily with the aid of the available finite-element methods). The reason the approximation works as well as it does is that the ratio of the circumference of the plate to its area goes to zero for the size of this area going to infinity; physically, the surface effects are more important than edge effects for sufficiently large surfaces (similarly as regards the significance of volume effects to surface effects for sufficiently large volumes).
Dear Rajeev,
The depletion region has a thickness xd as you said. It is considered as sheet with a thickness xd and an electrostatic potential difference epsi s. Which is termed the surface potential and or the band bending. The metal counter electrode on the other side of the oxide sees this charge on the semiconductor electrode as if it is coming from a surface located at the semiconductor surface. This is because of the nature of the electrostatic field. The electrostatic field lines remain unchanged in the oxide because it is free of charges.Therefore for sake of calculating the potential difference on the oxide one can consider that as if the depletion charges are located at the surface of semiconductor.
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