15
$\begingroup$

I have studied about the two types of doping which result in p and n type semiconductors. I also came to know that they are neutral. But, how can it be?

Is it that the positive charge(holes) in p-type and negative charge in n-type are negligibly small to affect the overall neutrality of the substance? But, in that case with very few holes or negative charges how can semiconductors work the way it actually does?

So, I think there is another possible explanation and I would like to know the correct one.

$\endgroup$
13
$\begingroup$

The terms n- and p-type doped do only refer to the majority charge carriers. Each positive or negative charge carrier belongs to a fixed negative or positive charged dopant.

p and n type materials are NOT positively and negatively charged.

An n-type material by itself has mainly negative charge carriers (electrons) which are able to move freely, but it is still neutral because the fixed donor atoms, having donated electrons, are positive.

Similarly p-type material by itself has mainly positive charge carrier (holes) which are able to move relatively freely, but it is still neutral because the fixed acceptor atoms, having accepted electrons, are negative.

http://www.halbleiter.org/en/fundamentals/doping/

http://uk.answers.yahoo.com/question/index?qid=20110422034224AAINiGl

Google is your friend, too.

$\endgroup$
5
$\begingroup$

Well there is no reason for the doped semi-conductor to be charged; either positive, or negative. Every single atom in the semiconductor came with its full complement of electrons, so the whole thing has no charge surplus or deficiency.

For example, if the semiconductor is say Silicon, every Si atom has 4 valence electrons. If you dope it P type with boron, which is a group III atom, each B atom has just 3 valence electrons. If you dope it N type with phosphorus, that is a group V element so each atom comes with 5 valence electrons. But either type of dopant atom is still charge neutral.

If a B atom replaces a Si atom in the crystal, one of the Si valence electrons has no mate from the Boron, and if a P atom replaces a Si atom, then the P atom has one electron with no mate from a Si atom; but it is always electrically neutral (the whole crystal)

$\endgroup$
3
$\begingroup$

Simply- intrinsic semi conductor is neutral to that if we add a neutral dopant then it is obvious that the overall charge will be neutral :)

$\endgroup$
1
  • $\begingroup$ this doesn't explain why the dopant is neutral which is inherently part of the question. $\endgroup$ Aug 25 '17 at 18:16
1
$\begingroup$

If semiconductor is doped with both n- and p- dopants equally - electrons and holes "annihilate", and you are getting almost "neutral" material.

Of course high level of such "neutral" doping will degrade semiconductor specs (like electron/hole mobility).

If there will be more n-type dopants than p-type ones - you will get n-type semiconductor which will work almost like there is no p-type impurities and vice-versa.

$\endgroup$
1
$\begingroup$

An atom contains not only the electrons but also the nucleus which consists of an equal number of protons. Hence an atom is neutral.

The reason why your doped semiconductor carries a neutral charge is because it has equal number of electrons as there are protons, be it boron doped or phosphorous doped.

While the whole crystal remains neutral, by doping you are vastly increasing the conductivity of the semiconductor. And hence, the specialty of semiconductors and doping.

$\endgroup$
1
$\begingroup$

Overall, the p-n crystal is neutral. If it wasn't we would immediately feel it. In fact, it is a fun exercise to calculate how many electrons would be needed to generate a substantial force equivalent to, say, the weight of a car. The answer is surprisingly not very many. This is because the electrostatic force is relatively strong. I am mentioning this to show that the substance would probably be not stable if it had even a small imbalance in the number of positive and negative carriers.

Now, when people talk about P or N doping they mean the following. Each atom of P or N type material is also neutral. However, when placed near each other they cause interesting things to happen. Namely, one of the outer electrons of the N atom "feels" that P atom has a "hole" available (more precise terminology would be a "quantum state"). Due to quantum mechanical effects, there is a non-zero possibility that the electron will transition to that hole. Once it transitions, the N now has one fewer electron and P has one extra electron. Now we have a bit of imbalance: N - slightly positive and P - slightly negative. This is precisely how the PN junction is formed.

when we go about biasing the junction, I.e, applying voltage or difference of potentials to it, we simply encourage certain quantum transitions. Forward bias means that we encourage the electrons to travel from N region towards P region. Of course, on the way between the two, they have to overcome the "natural" barrier that formed as described above. This barrier is the typical 0.1-0.8V potential drop of a forward-biased semiconducting device.

$\endgroup$

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