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Even when there are holes, the holes move only because electrons are jumping from one hole to the next. So why doesn't it make sense to say that electrons are the majority carriers in p-type semiconductors? Also, since only the electrons move, shouldn't there be a buildup of electrons on one side of a p-type semiconductor due to Hall effect? But instead there is a buildup of holes instead. Why does this happen?

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    $\begingroup$ To understand semiconductors you will need to embrace the concept of holes being on the same footing as electrons. Positive charge carriers are not unknown even in metals, but in a semiconductor the detailed balance between holes and electrons is absolutely fundamental to understanding device operation. $\endgroup$
    – Jon Custer
    Sep 15, 2021 at 14:22
  • $\begingroup$ @JonCuster Yes I understand that holes are a fundamental concept, and that if I just think of the Hall effect case with only holes moving in the direction of the current, I would get an explanation of why there is a hole buildup. However, that doesn't really answer my question as to why they are called the majority carrier, and why there is no electron buildup. Because the electrons are still moving in the opposite direction, unless I am understanding this wrong $\endgroup$ Sep 15, 2021 at 14:28
  • $\begingroup$ They are the majority carrier if $p > n$. If there are very few free electrons, there aren't any (many) moving. $\endgroup$
    – Jon Custer
    Sep 15, 2021 at 14:45
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    $\begingroup$ As JonCuster said, you need to embrace the concept of holes. If one hole moves, one hole moves. Ignore the electrons that moved to make room for it in its new state. They aren't helpful to understanding the transport behavior in the semiconductor. $\endgroup$
    – The Photon
    Sep 15, 2021 at 15:06
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    $\begingroup$ The thing you need to understand is that they don't affect the transport behavior and you need to focus on the holes, not the valence band electrons that "surround" them. $\endgroup$
    – The Photon
    Sep 15, 2021 at 15:54

3 Answers 3

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I think to understand why holes are the preferred "thinking tool" for p-type semi-conductors one must understand the following concepts:

  1. In a crystalline materials there are so-called energy bands which define the possible relationships between an electron's energy and its (crystal) momentum "k".

That is, there is a function $E(k)$ which defines the set of possible energy states which may be present in a crystal.

  1. In a semiconducting material there is a gap in the band structure such that there are energy levels which electrons may not possess (they are forbidden in a sense).

  2. The energy-momentum relationship can be approximated as parabolas just above and below the energy gap (i.e. the band gap).

  3. The parabola approximating the energy band just above the gap is pointing up / concave up. The parabola just below the gap is pointing down / concave down.

  4. The average velocity of an electron can be expressed as:

$$v(k) = \frac{1}{\hbar}\frac{\partial E(k)}{\partial k}$$

  1. The current density associated with a single band is then calculated by adding up the contributions from all electrons.

$$ j = (-e)\int_{occupied}\frac{dk}{4\pi^3}v(k)$$

Here the flag occupied means to integrate over only those states which electrons occupy.

  1. In general, the relationship $E(k)$ is a complicated function and thus not very straightforward to manipulate mathematically. However, it is possible to show that the current density integral is mathematically equivalent to the following integral:

$$ j = (+e)\int_{unoccupied}\frac{dk}{4\pi^3}v(k)$$

The reason this simplification is important is that in the valence band (i.e. the band below the band gap) electrons fill almost all the states except those near the top of the band (i.e. the unoccupied states). However, as previously mentioned the function $E(k)$ can be represented by a downward facing parabola at the top of this band. Thus a previously intractable integral becomes "trivial" as one only needs to integrate a quadratic equation. This is a very key point: it is mathematically much simpler to consider that the current density is due to fictitious positively charged particles which 'occupy' the top of the valence band instead of considering the mathematical equivalent, but non-fictitious, fact that the electrons present in valance band occupy almost all the valence band states except those near very near to the top of the band.

Finally, to answer the posed question: why are the unoccupied states (i.e. holes) treated as the predominant charge carriers.

In a p-doped material, the doping atoms such as boron "accept" electrons present in the crystal. These electrons, generally speaking, are taken from the valence band thus leading to a state where more holes are present in the valance band than in the pure semiconducting material.

Also in this conception of semiconductors, it is incorrect to say that only electrons move. When an electron moves, if it moves to a previously unoccupied state (i.e. a hole), the hole must have moved as well. Actually, as @Matt indicated, the motion of electrons and holes under an electric field follow identical trajectories. This occurs because electrons in the upper part of the valence band move as if they had negative mass and thus in the opposite direction of an applied force.

So to conclude:

  1. It is more mathematically convenient when calculating the current density to treat the valence band as being filled only with positively charged particles called "holes" which have energies near the top of the band. Though, in reality, the valence band is almost completely filled with electrons except near the top.

  2. For p-doped semiconductors, atoms which accept electrons are added. The electrons picked up by these atoms are typically taken from the valance band and thus more "holes" are present than would be otherwise. Thus there are more holes in the valence band than electrons in the conduction band. Therefore upon the application of an electric field, the holes will be the dominant charge carriers.

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  • $\begingroup$ This is nearly correct but gets an important point wrong. Under an electric field a given state will move in the same direction whether or not it is filled. When a hole moves in the valence band you dont have an electron in the valence band moving in the opposite direction. Also, you hint at it but dont come out and say it directly: states at the top of the valence band have negative effective mass. Without this you wouldnt have holes acting as positive carriers for things like the Hall effect. $\endgroup$
    – Matt
    Sep 15, 2021 at 17:20
  • $\begingroup$ I see, I forgot about this. I updated my answer to include a hopefully more correct description. $\endgroup$
    – gigo318
    Sep 15, 2021 at 18:10
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why doesn't it make sense to say that electrons are the majority carriers in p-type semiconductors?

When we talk about "electrons" being charge carriers in semiconductors, we're referring to conduction band electrons only.

If the number of conduction band electrons is less than the number of valence band holes, we say that holes are the majority carrier because there are more of them than there are of the type of electron that can transport charge.

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  • $\begingroup$ I think I get it. So the limit on the conduction is placed by the number of holes, not the number of conduction band electrons, right? When $p>n$, that is. But what about in the case of Hall effect? For a particular setup, suppose there is a buildup of electrons in the positive z direction in an n type. For the same setup, there will be a buildup of holes in the positive z direction for a p type. But if the conduction electrons feel a Lorentz force in the positive z direction, then there should be a deficit of holes there instead. $\endgroup$ Sep 15, 2021 at 15:37
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    $\begingroup$ Because of ...quantum mechanics and stuff... the holes really behave like a positively charged carrier, and not just a "bubble" in a sea of negatively charged carriers. And they produce a Hall effect as positive carriers. Again, you can ignore the valence band electrons that are "moving out of the way" to allow the holes to move; they do not help you understand transport in semiconductors. $\endgroup$
    – The Photon
    Sep 15, 2021 at 15:53
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You noticed something that a lot of people seem to overlook with their simplified mental model of holes. If holes were "just" missing electrons then a Hall measurement would tell you the charge carriers were negatively charged, but in semiconductors we can see both negative and positive Hall coefficients. So what is going on? Well, holes are more than just missing electrons, and dont settle for answers from people who insist they are. (There are many such people. In my experience, holes are widely misunderstood. I blame professors who tell their students that holes are like a bubble in water. Its a horrible analogy.)

There are a few counterintuative effects in play. Semiconductors have two "bands" of interest where there exist allowed energy states with a band gap in between which has no allowed states. What makes semiconductors interesting is that both of these bands can have a non-negligible contribution to current.

The energy-momentum (E-k) relationship for each band defines the allowable states, and actually defines the bands themselves. At the bottom of the conduction band things behave in a reasonable intuative way, but at the top of a band, things get weird.

First, lets talk about filled bands. A filled band contributes exactly zero current. This is the first counterintuative thing we have. This is because of symmetry within the bands. For every state moving in one direction, you have another state moving in the opposite direction, and these exactly cancel out. This means when we have a nearly filled band, such as the valence band, we find it easier to talk about the missing states, and compute total current as the current from a filled band (which is 0) minus whatever those empty states would have contributed had they been filled.

Okay, but we still have electrons, right? Yes and no. The E-k relationship at the top of a band is approximately an upside down parabola. This means those states have a negative effective mass. This makes them move the "wrong way" under an electric field. This is another one of those counterintuative things about semiconductors. Add up these effects and you have particles that act like they have a positive charge.

So while you may decide its easier to think about electrons with negative effective mass and counting up all the contributions from a mostly filled band, everyone else likes to think about "holes". They act exactly like positively charged particles and it will serve you well to just accept them as particles that are equally as real as electrons.

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