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Here it says that:

When a transistor is OFF, there exists a potential barrier between the source and the drain..

I don't know exactly what a potential barrier is. If I had to make an educated guess, it would be a barrier that an electron has to overcome in order to reach the drain, from the source. If you don't want the electron to reach the drain (i.e. the transistor is "off"), you make the barrier harder to overcome. However, this guess raises questions (the foremost being whether or not the guess is actually correct):

  • What kind of barrier is this? Obviously, a 'potential' barrier, but how does this work?

  • Is this barrier removed if you want a transistor to be on?

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  • \$\begingroup\$ Phaptitude - please make clear if you speak about FETs or BJTs. You mention the terms "source" and drain - on the other hand ypou speak about a "potential barrier", which is a term used for bipolar transistors (BJTs) only. \$\endgroup\$
    – LvW
    Commented Aug 7, 2016 at 9:43

4 Answers 4

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For the first answer, this barrier is caused due to the different materials used to construct the transistor. Think of each connection in the transistor as a diode.

Now, the source may be of n type material(i.e. have extra free electrons) and the substrate might be p type material (i.e. have extra holes - positive charges). Now if you don't apply any external voltage, there will be an equilibrium state where the electrons at the border of the n-type material will cross over to the p-type substrate. This would then result in the formation of a charge neural depletion region, and any new electron which now needs to cross-over to the p-type substrate would require an additional external voltage.

This voltage is your barrier potential.

When you apply an external voltage, this barrier potential reduces as electrons are able to flow more freely, thus turning the diode on.

Why do I refer you to diodes, it's because a transistor can be considered as device with two back-to-back connected diodes. Once you understand how a diode works, transistors become easy!

[1] http://www.electronics-tutorials.ws/diode/diode_1.html

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    \$\begingroup\$ I believe the article is referring to MOSFETs, not BJTs. \$\endgroup\$ Commented Mar 1, 2014 at 20:35
  • \$\begingroup\$ Is the potential barrier equal to the threshold voltage? \$\endgroup\$
    – Phaptitude
    Commented Mar 3, 2014 at 1:52
  • \$\begingroup\$ Yes, the barrier potential is also referred to as threshold voltage. Note, the threshold voltage is not a constant, it depends on different parameters - such as temperature, bulk-source voltage, etc. Usually the values are of 0.45-0.7V for different types of Silicon semiconductors. \$\endgroup\$
    – Sameed
    Commented Mar 3, 2014 at 2:24
  • \$\begingroup\$ @Sameed Thanks. Can you cite a source where this is mentioned? Because I'll have to use it for a project and I need to be 100% sure. \$\endgroup\$
    – Phaptitude
    Commented Mar 4, 2014 at 11:56
  • \$\begingroup\$ Here you go, ecee.colorado.edu/~bart/book/book/chapter7/ch7_4.htm \$\endgroup\$
    – Sameed
    Commented Mar 4, 2014 at 21:35
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It is the minimum potential of a transistor of given material to overcome the minimum energy for the electron to release from the substance

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My knowledge of transistor physics is outdated but let me try. When physicists talk of "potential" they mean energy (actually potential energy per unit charge ).

Imagine ( like in the old vacuum tube days ) an electron moving between two charged plates. Basic electromagnetism says there is a constant electric field between the plates. To cross the plates an electric charge ( eg an electron ) has to have certain amount of kinetic energy. The electric field created by the charges on the plate, contributes to the potential energy of the system. The difference between potential energy per unit charge at each plate due to electric field is called the potential difference or voltage.

We see this primarily in two areas electrons being pushed around between plates in a vacuum tube, and particle accelerators. Particle physicists are lazy. Instead of using ergs or joules to describe these sort of energies, they use a new unit the electron-volt or eV for short ( as well as the KeV--Kilo-eV, the Mev Million-eV, GeV, Giga-eV, TeV Tera-eV etc. )

So how does this apply to transistors? Inside a doped semiconductor there is a soup opf atoms, ions, electrons. The electrons are not constrained to go anywhere. It's just like a conductor. But at the surface of the material, ions collect to form a net charge.

So now imagine two oppositely doped semiconductors sitting in a vacuum. The surface of the two form something similar to the charged plates above. Now bring the surfaces close together. The potential difference is still there. Electrons will need energy to go from one side to the other. Well ( like a diode) through a p-n barrier. A n-p barrier does not offer an obstacle. But after it goes through one barrier and then encounters another barrier of the opposite type. So current cannot flow.

Now we need a little trick from quantum mechanics. Imagine a marble on a smooth surface with a hill. That marble can move around a lot, but it can only go about halfway up the hill. It doesn't have enough energy to go higher.

That marble can't get to the other side ( well assume that it can't go around the hill ). If the marble could get to the other side, it would could exist perfectly fine, but the hill is in the way. Enter quantum mechanics. If that hill were small enough, and the marble were small enough there would be some probability that the marble appears on the other side of the hill. Actually there is but for the standard case the probability is something like one in one billion times one billion times one billion ,,, effectively zero. However for electrons trying to pass through a potential barrier the numbers work out much better.

So now you have the possibility that electrons simply pop over to the other side if the barrier, btw this effect is called quantum tunneling. For an unbiased transistor this is still incredibly small, but if you bias the transistor, you draw away charges from the surface, and you reduce the barrier the electron has to cross, you increase the probability that electrons cross.

So to sum up --- on the surfaces where two semiconductors meet is a distribution of charges due to the doping. Those charges form an electric barrier to electrons crossing across the barrier. Biasing the transistor causes ions to be pulled away from the surface, reduces the barrier, Not enough to let electrons cross mechanically but to allow electrons to tunnel quantum mechanically. The more bias, the smaller the barrier, the greater probabiliuty that electrons tunnel, the more electrons that tunnel.

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N-type semiconductors have more free electrons than intrinsic semiconductors and than p-type semiconductors. P-type semiconductors have more empty spaces than both of the other cases. Naturally, when there is a proximity between the two, thermal energy makes them move so as to fill those empty spaces (they are of lower energy). In summary, we have electrons on the n-side that could be in a lower energy state on the other side because of the empty spaces added by acceptor atoms. This electrons diffuse because of thermal agitation and eventually fill those spaces.

That diffusion leaves cations behind (without their electrons to neutralize globally) and forms anions. That means that there is a space region with net charge forming around the junction. This charge means an electric field from the cations to the anions and this electric field opposes further diffusion (eventually reaching an equilibrium if there is no biasing). This region of electric field keeps electrons from moving from n to p and is the potential barrier that exists across semiconductor devices.

In particular, in enhancement MOSFETs the potential barrier impedes diffusion both from source to the substrate and from drain to the substrate. If you eliminate it you still get no current because those charrier flows oppose eachother. You need to provide an outside electric field to push them from source to drain. (They could a priori move from drain to source but usually the substrate terminal is shorted to one of those to and that one we call source. We do this to prevent forward biasing the pn-junction formed between any of them and the substrate).

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