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My textbook, Introductory Semiconductor Device Physics, by Parker, says the following in a section on energy band theory:

We are aware that conduction in most metals is by electrons but conduction mostly by holes does occur in some (divalent) metals such as beryllium, cadmium and zinc and this is due to their more complicated band structure.

I find this explanation confusing. I was under the impression that "holes" are just positions that lack an electron where an electron could otherwise exist. But, in the aforementioned description, the author implies that there is a distinction between the conduction of electrons and holes themselves. This is confusing because I don't see how it makes sense to discuss conduction of holes or electrons as separate elements (that is, to discuss the conduction of holes as to the exclusion of the presence of electrons, or vice-versa) that can conduct without the presence of the other, which it seems to me is what the author is implying.

I would greatly appreciate it if people could please take the time to clarify this.

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I think your main confusion can be summarized as follows:

Why is there a difference between conduction by electrons and by holes, if the holes are simply lack of electrons, and the electrons is what moves between these holes?

The main point is to look where the electrons are normally considered as charge carriers, and where the holes are. The electrons are the charge carriers in the conduction band – the allowed energy band, which is above the band gap – the forbidden band.

Have a look at the following figure. It schematically shows the part of band structure of a direct band gap p-type semiconductor near $k=0$. Notice that in the valence band electron energy $E(k)\propto \frac{k^2}{m^*}$, and effective mass $m^*<0$, while in the conduction band we have $m^*>0$.

Band structure near k=0

Now, if we apply electric field $\mathcal E$, the electrons with $k\approx0$ will obey (in a sense; this also relies on effective mass approximation) Newton's second law of motion:

$$\frac{p_e'(t)}{m^*}=x_e''(t)=\frac{-e\mathcal E}{m^*}.\tag1$$

Notice what happens: the expected value of position of electrons in the conduction band will accelerate, as for vacuum electrons, opposite the electric field. But in the valence band, the expected value of position of electrons will, since $m^*<0$, accelerate backwards, along the electric field!

Now consider the total energy of electrons in the valence band, $E_\sum$, and total wavevector $\vec K_\sum$. If you take one electron with energy $E_e$ and wavevector $\vec k_e$ from it, you'll have total energy $E_\sum-E_e$ and total wavevector $\vec K_\sum-\vec k_e$. We can view it as addition of a particle with $E_h=-E_e$ and $k_h=-k_e$ to the valence band. Given that momentum $p=\hbar k$, we'll have, in the same sense as the above equation,

$$p_e'(t)=-p_h'(t)=-e\mathcal E,\tag2$$

or

$$p_h'(t)=e\mathcal E.\tag3$$

Whether you view motion of electrons in a crowded "place" (band) as switching of "seats" ($k$ values) by one electron after another, or as a "bubble" (vacant energy state) moving inside this crowd, this doesn't change the effect. But due to $(3)$, it's natural to introduce a concept of vacant energy state being another quasiparticle which can move in a crystal, similarly to electron.


Now, can you actually measure the difference in conductance via conduction band vs valence band? Yes, one of the prominent effects is the Hall effect, whose sign depends on the charge of the charge carriers. Another effect demonstrating the backwards motion of electron when it reaches top of allowed energy band is Bloch oscillation.

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  • $\begingroup$ Thanks for the answer. I'm probably misunderstanding what you wrote, but it seems to me that you're explaining the 1-to-1 duality between electrons and holes; but it seems to me that what the author mentions is a conduction that occurs "mostly by holes", and so there are more holes involved than there are electrons. This is the part that I don't understand, since I understand that there is a 1-to-1 duality between electrons and holes. Can you please clarify? Simpler language is appreciated, since I'm a newcomer to this area of physics. My apologies if I'm misunderstanding you. $\endgroup$ Dec 18, 2019 at 0:26
  • $\begingroup$ Once we take this view of "by holes"="via valence band" and "by electrons"="via conduction band", the "mostly by holes" means "expect Hall effect as if from positive-charge carriers", i.e. conduction occurring through valence band. $\endgroup$
    – Ruslan
    Dec 18, 2019 at 5:36
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    $\begingroup$ @ThePointer well, that's playing with words. Conduction as a phenomenon is one thing, conduction as an adjective in "conduction band" is simply a label. Both valence and conduction bands support the phenomenon of conduction. $\endgroup$
    – Ruslan
    Dec 27, 2019 at 6:16
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    $\begingroup$ @ThePointer well, I guess initially it referred to the phenomenon of conduction by free electrons, just like in most metals. But then it was found that the band below the conduction band can also let charges move and result in Ohm's law, but the name "conduction band" was already used. So, to somehow unambiguously describe it, it was called valence band. See Wikipedia: "The name "valence band" was coined by analogy to chemistry, since in semiconductors (and insulators) the valence band is built out of the valence orbitals." $\endgroup$
    – Ruslan
    Dec 27, 2019 at 6:39
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    $\begingroup$ @ThePointer note that at some point the moderators might clear all the comments (I've seen this happen on quite some questions and answers). $\endgroup$
    – Ruslan
    Feb 9, 2020 at 13:33
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If a band is completely full of electrons,for all energy levels, there is no conduction. That could be the case of that metals with 2 valence electrons, because the Pauli exclusion principle allows 2 eletrons of opposite spins by energy level.

But some of them, occupying the highest levels, depending on the band configuration, can move to the next band (where the lowest level may have lower energy than the highest energy of the former one for instance).

In that case, the electrons have plenty of unoccupied states to migrate in the higher band when an electric field is applied, and conduction is possible.

The lower band has now some unoccupied states, and the electrons can also migrate. But if there is much more occupied than unoccupied states, that migration is a step by step process, more easily described as if the unoccupied states were moving (and they are!)

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  • $\begingroup$ Thanks for the answer. I think I understand the main point you're making. However, could you please clarify "That could be the case of that metals with 2 valence electrons, because the Pauli exclusion principle allows 2 eletrons of opposite spins by energy level."? $\endgroup$ Dec 14, 2019 at 1:42
  • $\begingroup$ Also, I found what you wrote here interesting: "* But some of them, occupying the highest levels, depending on the band configuration, can move to the next band (where the lowest level may have lower energy than the highest energy of the former one for instance).*" The way energy bands were portrayed in my textbook implies that the higher in the energy band, the higher the electron energy. But it seems that what you're implying here is that there can actually be some overlap between the bands, so [...] $\endgroup$ Dec 14, 2019 at 1:46
  • $\begingroup$ [...] that the energy of the electrons at the bottom of the highest levels of the energy band actually have lower energies than the electrons at the highest levels of the band below it, allowing these two groups of electrons to, effectively swap positions in the energy band? $\endgroup$ Dec 14, 2019 at 1:47
  • $\begingroup$ If the atoms are far way, an valence orbital 2s for example can have 1 or 2 electrons. When they are part of a crystal, all that possible states form a band. The band is half full in the first case and completely full in the second. $\endgroup$ Dec 14, 2019 at 14:50
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In a book about semiconductor devices, it goes a bit too far to write about the complicated Fermi surfaces of a divalent metal like beryllium. But it should not be too much of a surprise that the $2s$ band is nearly filled when some of the valence electrons are in $2p$-like states.

Hole conduction means that the Hall effect has the other sign than that which is expected for electrons.

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