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I think n-type semiconductors are easier to understand. You have an atom with 5 valence electrons, 4 of them are used up to connect it to the lattice and the remaining 1 electron needs only a little energy to enter the conduction band. So we have a positive impurity ion that's shackled in place with 4 bonds and a free electron that can roam around. So there are lots of electrons to carry current. That was simple to understand.

In a p-type semiconductor holes carry most of the current and electrons are somehow excluded from current conduction. I fail to understand how this happens. The acceptor has 3 valence electrons so it will use up all to bind in the lattice. And then now what? This where I'm lost. Many tutorial videos then just go on and assume there is automatically a hole there without a corresponding electron that then start migrating... Why is that hole next to the acceptor? And when that hole is filled why don't the acceptor let that electron go?

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    $\begingroup$ Because the neighboring silicon is extending an electron to make a bond with whatever is on the adjacent lattice site. And, the energy level of putting an electron there is right down by the valence band, so it is quite energetically favorable. And your last bit about a "real" hole does not make sense in terms of band structure. Real holes are no different than real electrons, both being states in extended bands, not tied to specific atoms. $\endgroup$
    – Jon Custer
    Jan 14, 2019 at 22:53
  • $\begingroup$ @JonCuster Edited the question a bit. Actually a good answer would explain two things: 1. why is it energetically favorable to force an extra electron on the acceptor. 2. Why don't the acceptor let that electron go. $\endgroup$
    – Calmarius
    Jan 14, 2019 at 23:16
  • $\begingroup$ Well, looking at the defect levels (say in silicon) for different impurities, it isn't always favorable. One finds potential donor and acceptor (and amphipolar) levels clear across the gap. But, if an acceptor state is below the Fermi energy, well, it is energetically favorable. And, that works out for a fairly small number of dopants (at least for those levels to be within a few kT of the valence band). $\endgroup$
    – Jon Custer
    Jan 14, 2019 at 23:26

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Okay, let's use the bond-model, as you suggest. We've got a huge lattice of Si or Ge (ideally an infinite lattice, so we don't have distortion due to boundaries).

This lattice of atoms usually have 4 charges in the outer band, like silicon or germanium, so they are like the left hand side of the picture. Now, we add one impurity per millions of silicon. That's the right-hand side of the picture:

P-type semiconductor atom structure

IF we want a P-type semconductor, we choose the impurity to have $3$ outer charges. That's because we want less charges than silicon, but still close to it in the periodic table.

That atom, which can be boron, for example, tries to imitate the solid structure, so it tries to make bonds with the surrounding atoms. But it fails in the last one, because boron has only 3 outer electrons, so it cannot make the last covalent bond.

As a result, there is a place which is not occupied and it is capable to hold an electron. That's a hole.

What's mroe, if an electron from other bond moves to that place, it leaves a hole in the other place. That can be regarded as a movement of a hole, altough it is "reverse movement of electrons" actually.

So now let's try to answer your questions.

Why is that hole next to the acceptor?

We should rather say "the acceptor level is close to the valence band". It is clear that this acceptor level:

  1. Is a level, and not a band, because it is an isolated atom. It's a perturbation of the periodic potential, but the strange atom is too far from other impurities, so there are many impurities that can have the same quantum numbers (and same energy) because they're too far from each other. Let's call $E_A$ to the energy of such aceptor levels.

  2. $E_A$ is located in the forbidden band. That acceptor level is "empty by default", so if an electron went there, the electrno would need some energy. An electron needs energy to escape from its own initial place, and then go and occupy that boron hole. So if electrons need some energy to reach the level, it must have more energy than $E_V$. So $E_A>E_V$

However, the electron that went there had less energy than a conducting electron. A conducting electron would have enough energy to get to the conduction band, where they can move. An electron that has gone to $E_A$ occupies a covalent bond, which are quite stable, so it can move less than a conducting electron. Consequently, $E_A<E_C$.

  1. And why is it close to the valence band (close to $E_V$? Well, because we just want to. We choose specific impurities so that $E_A$ is next to $E_V$. There are many impurities that lift that level up, but we don't want that. We want it to be close.

That's because we want that the $E_A$ levels get easily filled when heating the sample. We want electrons to have enough energy to reach $E_A$ at usual temperatures ($20ºC$), but not as much as $E_C$, so that it is still non-conducting but now there are holes in the VB. I mean, at usual temperatures, electrons have enough energy to get to $E_A$ but not to $E_C$. That makes electrons go to $E_A$, but that leaves holes in the VB, without putting electnos in the CB!

Excess of holes without associated electrons in the CB... that's like a possitive-charges semiconductor. a P-type semiconductor.

And when that hole is filled why don't the acceptor let that electron go?

When that "boron-hole" is filled, that electron can actually go. Why not? It can leave that place again. In fact, at high temperatures, it does (easily).

But what I said is that it is not much likely to do that. A covalent bond is strong enough to be stable. You'd need a lot of energy to liberate the electron. Such energy is $E_C-E_A$.

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  • $\begingroup$ It takes much longer to write it than reading it. When I wrote it, I thought it was clear, but let me know if you don't understand anything. $\endgroup$
    – FGSUZ
    Jan 17, 2019 at 12:35
  • $\begingroup$ Is a level, and not a band, because it is an isolated atom. and You'd need a lot of energy to liberate the electron. Such energy is $E_C−E_A$. This made it clear that electrons in those levels cannot conduct because acceptor atoms are too far away from each other. And also explains why they are locked in place. Thanks! $\endgroup$
    – Calmarius
    Jan 17, 2019 at 12:43
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A boron acceptor in silicon replaces a silicon atom and makes 4 bonds. However there is one electron short per boron. That hole is weakly bound to the boron at zero kelvin. the acceptor. At room temperature electrons can move between the B-Si bonds and si-si bonds. Effectively the hole that was weakly bound to the boron now moves around through the lattice. So you have effectively hole conduction but of course it is really the electrons that conduct. However there are no or better very few electrons in the conduction band.

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  • $\begingroup$ So basically you are saying that there is no 4th bond that is forced to the boron, but instead the constant reconnection of those 3 bonds that create these holes, when electron to reconnect the bond comes from the silicon? $\endgroup$
    – Calmarius
    Jan 15, 2019 at 8:01
  • $\begingroup$ At low temperature a hole is weakly localised at B. At higher T the B holes become mobile. I will extend my answer. $\endgroup$
    – my2cts
    Jan 15, 2019 at 18:17

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