What does it mean that the Fermi level for some semiconductors lie in the band gap?

Is Fermi level definition different from what is know as usual?

We define the Fermi level as the highest level of energy which is occupied by electrons.

Then what is missing here?


5 Answers 5


No the definition of the "Fermi level" doesn't change, it's only its relative position in the band structure of a material changes.

Definition: The Fermi energy or Fermi level $E_F$ is the chemical potential of electrons at $T=0.$ The states that are filled at $T=0$ are called the Fermi sea.

In a solid the energy eigenvalues are filled in accordance with the Fermi-Dirac distribution, which says that the probability of a state of energy $E$ being occupied by an electron is given by: $$f(E)=\frac{1}{e^{(E-E_F)/k_BT}+1}.$$

Now what does it mean when the Fermi level lies in the band gap or instead in one of the bands? Here's an overview using a nice diagram from wikipedia:

enter image description here

The y-axis is the energy and x-axis the density of states(of electrons) available for different band structures of different materials.

  • Metals: are very good conductors because the Fermi level lies inside one of the bands, meaning there's no energy gap to overcome for an electron in the valence band to pass on to the conduction band. Which explains why metals are populated by quasi-free electrons in the conduction band.

  • Semi-metals: Very similarly to metals, $E_F$ lies in one of the bands, but unlike metals, here the overlap between valence and conduction bands are very small. But still no gap!

  • Semiconductors: For semiconductors, there's a small energy gap between the two valence and conduction bands, the states in the gap are not available to the electrons. So in order to have moving charge carriers, electrons and holes, the system first has to be heated up (excited, also possible with an applied E-field) in order to overcome the gap $E_g$ and be able to occupy the available states in the conduction band. Now because for semiconductors the gap is usually very narrow, about $\leq 4 eV$, meaning it requires a specific minimum amount of energy for the transition. Because this gap is so small, the properties of semiconductors lies between those of conductors and insulators, but unlike metals, semiconductors increase their conductivity with temperature, because the overlap between the two bands becomes greater with increasing T. Finally as you see in the above diagram, $E_F$ lies in the gap for semiconductors, which means occupied and unoccupied states in valence and conduction band respectively are energetically separated. The neat thing about semiconductors is that they can be n and p doped, which respectively causes the fermi level to shift towards the conduction band and the valence band.

  • Finally for insulators, the energy gap between the two bands in so large, that it becomes very difficult to excite electrons towards the conduction band, hence their poor conductivity.

A final point: one may ask where do these gaps come from anyway? For solids, when you take the nearly free electron model and add a periodic potential $V(\mathbf{r+R})=V(\mathbf{r})$ to the picture ($\mathbf{R}$ the lattice vector), and solve the Schroedinger equation using degenerate perturbation theory, we see that a gap opens up between the energy eigenvalues at the zone boundary in momentum space $\mathbf{k}$, i.e. $$E_{\pm}=\epsilon_0(\mathbf{k})\pm |V_G|$$

$V_G$ being the value of added periodic potential at zone boundary. In conclusion when electrons are subject to a periodic potential, gaps arise in their dispersion relation. Thus the electron spectrum breaks into bands, with a forbidden energy gap between the two bands.

Clarifications added in regard to the question asked by Calmarius in comments:

Question: From this answer it's still not clear to me why would the fermi level lie in a forbidden gap. Based on what I have found on the internet they say fermi level is the maximum energy level occupied by electrons on 0K "the top of the Fermi sea". For semiconductors this mean all electrons are in the valence band. So why don't the fermi level lies right at the top of the valence band instead? It's not clear to me.

Don't worry, your confusion is well placed and your question is rather common, note that some of the things I describe here may or may not be already clear to you, but I have to make sure they're mentioned.. First things first, the statement that "fermi level is the maximum energy level occupied by electrons at 0K" is completely wrong (this would be actually a more valid definition for the valence band energy), and for some reason seems to be a common misunderstanding, as you have realised yourself by looking through the web. You could add a correction to that definition, by saying that "the maximum energy level occupied by electrons in the valence band always lies below the fermi level", and this would be sort of say something more accurate about the fermi level. The most general definition though, which is always correct, is: "the chemical potential at 0 Kelvin". Frequently though the Fermi level is wrongly defined as the energy of the most energetic occupied electron state in a system, this definition leads to errors as soon as you have discrete energy eigenstates, in other words when there's a gap between the most energetic occupied state and the least energetic unoccupied state in the system. Whereas the correct definition is the chemical potential at $T=0 K$, and will be halfway between the minimum of the conduction band and the maximum of the valence band. It is exactly in the middle, i.e. $E_F = 1/2(E_c-E_v)$ when you have e.g. an intrinsic semiconductor, meaning equal density of conduction band electrons $n_0$ and free valence band holes $p_0$, or usually written as $n_0 = p_0 = n_i,$ (i for intrinsic) this equation should already resolve some of your confusion about what defines $E_F.$ As you dope the semiconductor, the latter turns into an inequality, e.g. $n_0>n_i>p_0$ for n-doped.

To simplify the picture, imagine a metallic hulk which is made of a periodic array of atomic cores, without its electron cloud, necessary for electrostatic neutrality. Suppose now electrons are added to this array of cores until neutrality is reached. The first electrons added go into the lowest possible kinetic energy states, or as seen in k-space, states of smallest wave-vectors. The next electrons, due to the Pauli exclusion principle, must build up occupancy of successive shells in k-space, of ever higher energy (the arrangement of this levels depends on both the dimensionality of the system and boundary conditions). As soon as enough electrons are supplied to reach neutrality, a chemical potential is established (assuming this neutrality was reached for 0 K), which corresponds to the Fermi level, note again this chemical potential will not be equal to the highest occupied k-state of electrons. Now an important remark: if you kept trying to continuously add electrons to the system, the available momentum states allowed in the system do not form a continuum, (because of the periodic arrangement of the cores, in simple terms) and when you solve the Schrodinger equation for the simplest such model, you immediately come across an energetic separation, the bandgap, between the valence (where the electrons are locked in fixed k-states) and conduction band, which should be overcome for any further addition of electrons to the system. In a nutshell, in any system the density of free holes in the valence band and electron is conduction band defines the fermi level. The fermi level for semiconductors is always between the two bands, its exact position determined by doping concentration.

Recommended references:

The Oxford Solid State Basics, by Steven H. Simon.

Fundamentals of Semiconductors, 4th edition, by Peter Y. Yu and Manuel Cardona.

  • 3
    $\begingroup$ Semi-metal - you learn something new everyday !! +1 $\endgroup$
    – Gowtham
    Commented Dec 29, 2014 at 17:19
  • $\begingroup$ Thanks. At T=0 is the identity which i did't pay attention to, I think. $\endgroup$
    – P.A.M
    Commented Dec 29, 2014 at 18:08
  • $\begingroup$ From this answer it's still not clear to me why would the fermi level lie in a forbidden gap. Based on what I have found on the internet they say fermi level is the maximum energy level occupied by electrons on 0K "the top of the Fermi sea". For semiconductors this mean all electrons are in the valence band. So why don't the fermi level lies right at the top of the valence band instead? It's not clear to me. $\endgroup$
    – Calmarius
    Commented Jul 18, 2015 at 18:28
  • $\begingroup$ @Calmarius Thanks for your question, I've added some notes for you in the answer, hope you find this helpful. $\endgroup$
    – Ellie
    Commented Jul 18, 2015 at 20:15
  • $\begingroup$ @Phonon I wanted to clarify your last example, the metallic hulk. If I start without any electrons and start filling up the valence electrons, do I reach electrostatic neutrality when I add the last electron that fills the highest valence band, despite being the last band being below the Fermi level? It seems to me that the Fermi Level dictates how I can keep adding electrons, but when there's a band gap (insulator) those extra states that I would fill up if considering a metal are not filled. Which leads to the question - have I reach neutrality or is there a mismatch due to the bandgap? $\endgroup$
    – Mr G
    Commented Sep 18, 2019 at 12:14

The root of the confusion seems to be that the forbidden gap should have no electrons. However, Fermi level which by definition is the maximum energy level that may be occupied by electrons at 0K lies in this gap.

If so then what one needs to understand is that the fermi distribution function $$ f(E)=\frac{1}{1+\exp[(E-E_f)/kT]} $$ is not the same as the actual population of electrons in a certain energy. This population is a product of the distribution function and the density of states. Since there are no states to occupy in the forbidden band gap, there is no electron population there.

However, at 0K the highest energy that may be occupied is still the Fermi energy level. If it lies in the forbidden energy gap, there is nothing at this energy, all electrons are below it.

  • $\begingroup$ $g(E) = 0$ for $E_v < E < E_c$. So $$n = \int_{E_v}^{E_c} f(E) g(E) dE = 0$$ $\endgroup$
    – Shub
    Commented Sep 2, 2023 at 5:14

The Fermi energy does not define a sharp cut off between unoccupied and occupied states. There is a continuous distribution in the electron density with energy. At non-zero temperature there are always some electrons above the Fermi energy, and holes beneath it.

Rather, the Fermi energy defines a level relative to which electron density is specified. For a simple band gap, if the Fermi energy is exactly in the middle of the gap, the electron density at the bottom of the conduction band is the same as the hole density at the top of the valence band. If the Fermi energy is biased toward the conduction band the electron density at the bottom of the conduction band is higher than the hole density at the top of the valence band.

Often the band gap is large compared to the range over which the Fermi energy varies, so the electron density at the bottom of the conduction band and the hole density at the top of the valence band are small and can sometimes be approximated as zero.

  • $\begingroup$ I couldn't get the visualization of the 2nd and 3rd paragraphs on the basis of the energy bands structure. I Would appreciate if you make it more clear, please. $\endgroup$
    – P.A.M
    Commented Dec 29, 2014 at 18:11
  • $\begingroup$ Your first statement begs the requirement of non-zero temperature BEFORE making it. As far as I understand, metals do not have a tangible metric to quantify holes. Holes in metals are dissipated almost immediately by delocalization throughout valence bands. Finally, as I understand, due to differences in chemical potentials (or effective masses) of electrons in valence and conduction bands, the position of the Fermi energy can depend on temperature; it is only exactly in the middle of the gap at $T = 0$ K. $\endgroup$ Commented Feb 17, 2021 at 2:38
  • $\begingroup$ @JeffreyJWeimer All true, thank you. The OP was about semiconductors, so I didn't think to consider metals. $\endgroup$
    – garyp
    Commented Feb 17, 2021 at 12:53

Fermi level lies in middle of forbidden gap .energy required to be at fermi level can called fermi level energy (say Ef). so if an electron (excited) acquired some energy which airs a tendency in it to jump from valence band to conduction band (say energy acquired is E ).

Now if this energy is E > Ef & electron reaches fermi level or lands somewhere between fermilevel & Conduction Band , the probability that it would cross fermilevel and land in Conduction band is more. on the other hand if E < Ef & electron doesnt reach or cross fermi level the probability of it dropping back into Valence band is more.

The key is that it should atleast attain fermilevel energy such that the probability of more electrons reaching the conduction band is high .So to be on the safer side we give more energy than required such that greater fraction of total electrons emitted lands in conduction band despite deflection & rebounding.

The landing of electrons though may be influenced by several other factors like

  • orientation of energy source
  • presence of foreign matter etc.

$E_F$ is the highest occupied energy level, however in the context of semi-conductors $E_F$ should be taken as the chemical potential of the semiconductor. The chemical potential is in the band gap of a semiconductor in most cases and thus it is incorrect to say that there are occupied energy levels at the Fermi level.

I understand this goes against the highest rated answer, but according to Solid State Physics by Ashcroft and Mermim (page 573), the above is correct (though please bare in mind this is quite an old and despite being a classic, this information may now be outdated, it was out in 1976).

  • $\begingroup$ Ashcroft and Mermin is still a standard textbook, it is NOT outdated! $\endgroup$
    – Sha
    Commented Sep 24, 2022 at 3:50

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