1
\$\begingroup\$

If a positive tension V is applied in a MOSCap, the depletion zone width will increase until the threshold voltage is reached. If the applied tension is above the threshold voltage the depletion layer will not increase in width. What are the reasons for this behavior? Why over the threshold voltage the bands will only bend slightly, leaving the depletion width unchanged?

EDIT: What I don't understand is that the threshold is defined as the condition in which the density of electrons near the oxide is equal the density of donors. In other words I've applied a tension which have caused the "conversion" of a p-type semiconductor in a n-type one. If I further increase the tension more electron will be collected under the oxide but the depletion zone will not increase in width. Why?

\$\endgroup\$
  • \$\begingroup\$ You should be aware that threshold voltage is defined differently by different groups. I define the threshold voltage as where the channel is 1/2 drift and 1/2 diffusion. The physicists like the version in your edit, but for undoped channels, we tend to use my first definition. \$\endgroup\$ – b degnan Jul 13 '17 at 17:06
1
\$\begingroup\$

Let's consider an N-channel MOSFET as shown below. We dope the substrate with the p-type material, which causes the silicon to have somewhat excess of what called holes. Now we are going to dope right beside the ends of dielectric layer with an n-type material, as you see below:

enter image description here

N-type dopants have excess of electrons. We dope highly these regions with n-type materials in such a way that they become highly conductive. Now we apply a positive voltage to the gate of the FET with respect to the p-type substrate. Now what this does is attract electrons from N-channel region and repel holes outward to the bulk silicon. If you keep increasing the gate voltage you'd attract more and more electrons until you have collected up all electrons. From now on, increasing gate voltage would have no effect on the width of the channel.

EDIT:

The region below the oxide layer is populated with a p-type material. Placing a positive charge on the gate would drive the holes away from the region below the oxide layer, leaving a depleted region that is insulating because no mobile holes remain. At the same time it attracts electrons from N+ regions, making a an inversion layer containing of electrons. Increasing the voltage on the gate leads to a higher electron density in the inversion layer.

The electrons come from the N+ regions. The n-type doping is made in such a way that after a certain voltage on the gate is reached you would get no more electrons. Therefore further increasing the gate voltage would have no effect whatsoever on the inversion layer width. In other words, you don't have any more free electrons to bring into the inversion layer.

\$\endgroup\$
  • \$\begingroup\$ The answer explain how DEMOSFET works. I was asking about a simple MOS capacitor composed only of Metal, Oxide and a p-Si bulk. \$\endgroup\$ – skdys Jul 13 '17 at 12:47
  • \$\begingroup\$ Sorry, I changed the picture. A DE-NMOS is the same as an E-NMOS except that it has a separate N-type channel formed within the substrate. \$\endgroup\$ – dirac16 Jul 13 '17 at 13:03
  • \$\begingroup\$ What I was saying is that you said that the threshold is reached when "you have collected up all electrons". But all the electron of what? All the electrons in conduction band of the bulk maybe? \$\endgroup\$ – skdys Jul 13 '17 at 13:06
  • \$\begingroup\$ I edited the question. Don't know if you're still stuck on that \$\endgroup\$ – dirac16 Jul 13 '17 at 13:51
  • \$\begingroup\$ The device I referring (I call it MOS capacitor, I don't know if there is a better name for that) has not the N+ regions. It is a simple p-Si bulk with a layer of oxide and a metal gate. If I apply a reverse tension, the holes are attracted toward the bulk terminal creating a depletion zone. If I further increase the voltage at a certain point holes are not more neutralized and electrons from the terminal flown under the oxide and form the inversion layer. Why this "certain point" exists and why it is reached at the threshold voltage? \$\endgroup\$ – skdys Jul 13 '17 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.