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Considering an NMOS transistor with separate body and source terminals, with the following voltages: Vd - 10V; Vg = Vs = 5V and Vb = 0V, why wouldn't it conduct?

Isn't the channel of electrons formed mainly because of the potential difference between the gate and the body?

Is the source voltage disturbing the channel?

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2 Answers 2

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The electrons come from the source, which is an N-doped region. This source quotes Chenming Hu's book "Modern Semiconductor Devices for Integrated Circuits":

...there are few electrons in the P-type body, and it can take minutes for thermal generation to generate the necessary electrons to form the inversion layer... The inversion electrons are supplied by the N+ junctions...

I also ran across a (probably illegal) PDF copy of Ali Niknejad's book "Electromagnetics for High-Speed Analog and Digital Communication Circuits". I'm not going to link to it, but here's the relevant quotation from section 2.3 regarding MOS capacitors:

In the above discussion you may have wondered, “Where do the electrons come from to form the inversion layer?” In the body of the MOS-C structure, electrons are minority carriers and few and far between. So when inversion occurs, where do we find all the electrons necessary to invert the surface? Well, there was a subtle assumption that if we apply a change in gate voltage, we wait long enough for thermal generation to create a sufficient number of electrons to form the surface layer. We may have to wait a very long time! In other words, if we apply a fast enough signal to the gate, there isn’t enough time for the minority carriers to be generated and thus the capacitance remains at the low value given by depletion.

While the depletion region can respond very quickly to our gate voltage since it is formed by majority carriers, the minority carrier generation is slow. There is a simple way to solve this problem, as shown in Fig. 2.25, where a n+ grounded contact is placed adjacent to the gate. Normally electrons are prevented from entering the body, like any good pn-junction. But as we raise the surface potential, electrons can easily diffuse into the surface of the structure. Since the energy distribution of electrons in thermal equilibrium is exponential, changing the potential barrier linearly results in an exponential increase in the number of electrons that can cross the n+-surface junction and likewise an exponential increase in surface conductors.

I redrew Figure 2.25 to illustrate what he's talking about. You can see how similar this is to a MOSFET:

enter image description here

In an NMOSFET, when \$V_S > V_B\$, you get a secondary effect called the body effect or the substrate bias effect. This effect acts to increase the threshold voltage. According to Wikipedia, the body is acting more like a second gate.

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  • \$\begingroup\$ Thank you for your response. Considering what you wrote, what happens exactly when we "negatively" bias the substrate? Holes are attracted by the terminal which leaves more electrons in the rest of the substrate. Then why is a higher voltage necessary to form the channel? And another question which is not really related to the subject, for which I apologize: why if the drain voltage is increased, the channel gets narrower near it? \$\endgroup\$
    – user42768
    Commented Dec 27, 2014 at 17:50
  • \$\begingroup\$ I'm not an expert in semiconductor physics, but my understanding is that there just aren't that many conducting electrons in the p-substrate to begin with. If you put a large enough negative voltage on the substrate, I suppose the source-body and drain-body diodes would avalanche. Regarding pinch-off: As you move along the channel, the voltage increases from Vs to Vd. If the channel voltage exceeds Vg - Vt, the capacitive action no longer works. It's like putting the same voltage on both sides of a capacitor -- there's no excess charge. \$\endgroup\$
    – Adam Haun
    Commented Dec 27, 2014 at 18:45
  • \$\begingroup\$ But what stops electrons from the getting closer to the drain terminal if it is at a very high potential? Because the gate and the drain electric fields are not really cancelling each other out. \$\endgroup\$
    – user42768
    Commented Dec 27, 2014 at 19:01
  • \$\begingroup\$ Nothing stops them. Conduction continues after channel pinch-off. (I think it's because of the depletion region; I don't fully understand the physics.) But the electrons can't accumulate near the drain because the drain voltage is sucking them out of the MOSFET. \$\endgroup\$
    – Adam Haun
    Commented Dec 27, 2014 at 19:08
  • \$\begingroup\$ Ok. Just to see if I understood the answer to the first question. The voltage difference between the gate and the source is so important because the channel is formed by electrons which were in the source? \$\endgroup\$
    – user42768
    Commented Dec 27, 2014 at 19:17
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One possible explanation, MOSFET are usually constructed symmetrically with the rvoltage between gate and base controlling the resistance between the two other terminals. Then we somehow got into a habit of shorting the base to one of the other terminals and start calling that the source terminal and the third drain.

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