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I have a question regarding the Fermi-Dirac function. I would like to confirm that my basic understanding is correct.

Is any space under the curve in the CB and VB describe the probability of finding an electron? And everything above the curve describes the probability of finding a hole? For example, for the n-type semiconductor, there is a slight probability of finding electrons in the conduction band. But you will still find more electrons in valence band compared to the conduction band? Again, for the p-type, the conduction band is filled with holes, but there is still a slight probability of finding holes in the valence band? Are these statements correct?

Regards

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    $\begingroup$ That "space under the curve" refers to figures that are rotated by 90 degrees, so probability $f(E)$ as a function of energy. $\endgroup$ – Pieter Apr 15 '18 at 9:00
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I'd say that your interpretation is basically right. Just keep in mind that a hole is simply an absence of an electron, so you can view any problem from the electron or hole point of view.

So, in the n-type example, one way of looking at it is that there are no holes in the VB because the VB is full of electrons. I.e. there is no absence of electrons, so there are no holes. In the CB, there are some electrons.

In the p-type example, one way of looking at it is that there are holes in the VB because there is an absence of electrons. You could either say that there are no electrons in the CB (which is what most people would say) or that the CB is full of holes (which is still a correct thing to say.)


EDIT: To clarify the point that we are free to make the choice about wheter to use electrons or holes, I quote Ashcroft and Mermin ch 12, p 226:

"Thus the current produced by occupying with electrons a specified set of levels is precisely the same as the current that would be produced if (a) the specified levels were unoccupied and (b) all other levels in the band were occupied but with particles of charge +e (opposite to the electronic charge.)

"Thus, even thought the only charge carriers are electrons, we may, whenever it is convenient, consider the current to be carried entirely by fictitious particles of positive charge that fill all those levels in the band that are unoccupied by electrons. The fictitious particles are called holes."


EDIT 2: With regards to charge neutrality. If you commit to the full electron/hole picture, then holes are the positive charges of the ions, so charge neutrality would mean an equal number of holes and electrons. It is easy to accidentally double count positive charges by naively counting ions and holes separately.

To be fair, you can count holes and ions carefully and adopt a hybrid electron/hole/ion point of view that holes only exist in the VB and the levels in the CB are not occupied by anything (holes or electrons), but in some ways that is more complicated since you now have to keep track of three particles (ions, holes, and electrons) rather than just two (electrons and holes).

In any case, the hybrid electron/hole/ion point of view is no more or less valid than the full electron/hole point of view, where all unoccupied electron levels are filled with holes.

Electrons and holes

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  • $\begingroup$ Okay thank you for the response. That helps a lot. $\endgroup$ – user367640 Apr 15 '18 at 15:59
  • $\begingroup$ In my opinion, your statement "You could either say that there are no electrons in the CB (which is what most people would say) or that the CB is full of holes (which is still a correct thing to say." is wrong! You cannot consider the missing electrons in the conduction band as positively charged holes as in the valence band. The missing electrons in the valence band are equivalent to positively charged (quasi-)particles whose positive charge is ultimately due to the positive lattice atom ions that are not neutralized by a negative electron. $\endgroup$ – freecharly Apr 15 '18 at 18:10
  • $\begingroup$ You cannot consider the missing electrons in the conduction band as holes with positive charges due to unneutralized lattice ions. If all missing electrons to fill the conduction band were positively charged holes (like the missing electrons in the valence band) this would constitute an enormous positive charge causing a massive violation of the charge neutrality of the semiconductor. $\endgroup$ – freecharly Apr 15 '18 at 18:11
  • $\begingroup$ My statement is correct. I have added an excerpt from Ashcroft and Mermin Ch 12, which emphasizes that we can always view unoccupied electrons states as being filled with holes. It is just as valid to say that the VB in the n-type example is filled with electrons as it is to say that the CB in the p-type example is filled with holes. The quasiparticle you describe in your first comment is called a hole. There is no need to invent a new quasiparticle. $\endgroup$ – lnmaurer Apr 16 '18 at 15:38
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    $\begingroup$ The (absolutely necessary) concept of detailed balance for electrons and holes in semiconductors is hard to reconcile with your opening paragraph. $\endgroup$ – Jon Custer Apr 17 '18 at 0:11
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Electronic states in an almost full valence band that are not occupied by an electron can be considered to be holes. This means, that they can be treated as particles with a positive elementary charge and a positive effective hole mass. It can be shown theoretically that this is a consequence of the collective behavior of all remaining electrons in the (almost full) valence band. Electronic states of the conduction band (which is almost empty) that are not occupied by an electron cannot be considered as holes.

Added explanation: Missing electrons in the conduction band cannot be fully considered positively charged holes like the missing electrons in the valence band. The missing electrons (holes) in the valence band are equivalent to positively charged (quasi-) particles whose positive charge is ultimately due to the positive lattice atom ions that are not neutralized by the missing negative electrons. In contrast, the missing electrons in the conduction band cannot be considered as holes with positive charges due to unneutralized lattice ions. If all electrons that are missing to fill the conduction band completely were positively charged holes (like the missing electrons in the valence band) this would constitute an enormous positive charge causing a massive violation of the charge neutrality of the semiconductor.

Added answers to you specific questions:

Q: "For example, for the n-type semiconductor, there is a slight probability of finding electrons in the conduction band. But you will still find more electrons in valence band compared to the conduction band? "

A: Yes, in the n-type semiconductor there is sizable number of electrons in the CB. But there is a much higher number of electrons in the VB.

Q:"Again, for the p-type, the conduction band is filled with holes, but there is still a slight probability of finding holes in the valence band? "

A: For the p-type semiconductor, there are very few electrons in the CB. There is a large number of holes in the VB, but still much less than the number of electrons. In this picture, there are no holes in the CB.

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  • $\begingroup$ Can you please write that again. I'm not sure what you mean. $\endgroup$ – user367640 Apr 15 '18 at 14:22
  • $\begingroup$ @user367640 - I have added an explanatory paragraph to my answer. $\endgroup$ – freecharly Apr 15 '18 at 18:24

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