Based on: Basic MOS Structure by Bill Wilson

This module is adapted from the Connexions module entitled *Basic MOS Structure* by Bill Wilson.

Figure 1 shows the basic steps necessary to make the MOS structure. It will help us in our understanding if we now rotate our picture so that it is pointing sideways in our next few drawings. Figure 2 shows the rotated structure. Note that in the p-silicon we have positively charged mobile holes, and negatively charged, fixed acceptors. Because we will need it later, we have also shown the band diagram for the semiconductor below the sketch of the device. Note that since the substrate is p-type, the Fermi level is located down close to the valance band.

Let us now place a potential between the gate and the silicon substrate. Suppose we make the gate negative with respect to the substrate. Since the substrate is p-type, it has a lot of mobile, positively charged holes in it. Some of them will be attracted to the negative charge on the gate, and move over to the surface of the substrate. This is also reflected in the band diagram shown in Figure 3. Remember that the density of holes is exponentially proportional to how close the Fermi level is to the valence band edge. We see that the band diagram has been bent up slightly near the surface to reflect the extra holes which have accumulated there.

An electric field will develop between the positive holes and
the negative gate charge. Note that the gate and the substrate
form a kind of parallel plate capacitor, with the oxide acting
as the insulating layer in-between them. The oxide is quite thin
compared to the area of the device, and so it is quite
appropriate to assume that the electric field inside the oxide
is a uniform one. (We will ignore fringing at the edges.) The
integral of the electric field is just the applied gate voltage *V _{g}*. If the oxide has a thickness

If we focus in on a small part of the gate, we can make a little
"pill" box which extends from somewhere in the oxide, across the
oxide/gate interface and ends up inside the gate material
someplace. The pill-box will have an area Δ*s*. Now we will invoke Gauss' law which we reviewed
earlier. Gauss' law simply says that the surface integral over a
closed surface of the displacement vector *D* (which is, of course, ε x *E*) is equal to the total charge
enclosed by that surface. We will assume that there is a surface
charge density *-Q _{g}* Coulombs/cm

Note that we have used ε_{ox}*E* in place of *D*. In this
particular set-up the integral is easy to perform, since the
electric field is uniform, and only pointing in through one
surface - it terminates on the negative surface charge inside
the pill-box. The charge enclosed in the pill box is just -(*Qg*Δ*s*), and so we have
(keeping in mind that the surface integral of a vector
pointing into the surface is negative), Equation 3, or Equation 4.

Now, we can use Equation 1 to get Equation 5 or Equation 6.

The quantity *c*_{ox} is called the oxide
capacitance. It has units of Farads/cm^{2}, so it is really a capacitance per unit
area of the oxide. The dielectric constant of silicon
dioxide, ε_{ox}, is about 3.3 x 10^{-13} F/cm. A typical oxide thickness might be 250 Å (or 2.5 x 10^{-6} cm). In this case, *c*_{ox} would be about 1.30 x 10^{-7} F/cm^{2}. The units we are using here, while they might seem a
little arbitrary and confusing, are the ones most commonly used
in the semiconductor business.

The most useful form of Equation 6 is when it is turned around, Equation 7, as it gives us a way to find the charge on the gate in terms of the gate potential. We will use this equation later in our development of how the MOS transistor really works.

It turns out we have not done anything very useful by apply a negative voltage to the gate. We have drawn more holes there in what is called an accumulation layer, but that is not helping us in our effort to create a layer of electrons in the MOSFET which could electrically connect the two n-regions together.

Let's turn the battery around and apply a
positive voltage to the gate (Figure 5). Actually,
let's take the battery out for now, and just let *V _{g}* be a positive value, relative to the
substrate which will tie to ground. Making

The electric field now extends into the
semiconductor. We know from our experience with the p-n junction
that when there is an electric field, there is a shift in
potential, which is represented in the band diagram by bending
the bands. Bending the bands down (as we should moving towards
positive charge) causes the valence band to pull away from the
Fermi level near the surface of the semiconductor. If you
remember the expression we had for the density of holes in terms
of *E _{v}* and

The electric field extends further into the semiconductor, as more negative charge is uncovered and the bands bend further down. But now we have to recall the electron density equation, which tells us how many electrons we have:

A glance at Figure 6 reveals that with this
much band bending, *E _{c}* the conduction band edge, and

Now, let's increase *V _{g}* above

Where did these electrons come from? We do not have any donors in this material, so they can not come from there. The only place from which electrons could be found would be through thermal generation. Remember, in a semiconductor, there are always a few electron hole pairs being generated by thermal excitation at any given time. Electrons that get created in the depletion region are caught by the electric field and are swept over to the edge by the gate. I have tried to suggest this with the electron generation event shown in the band diagram in the figure. In a real MOS device, we have the two n-regions, and it is easy for electrons from one or both to "fall" into the potential well under the gate, and create the inversion layer of electrons.