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I'm confused about why the Fermi level is located inside the band gap in semiconductors because it's defined to be the energy level where an electron has a 50% chance to be, but there can't be electrons in the band gap. I now think they mean it's the energy where the Fermi-Dirac distribution reaches 50% but now I'm wondering what this distribution even means when there is a band gap. It's a distribution that gives the occupation probability for a certain energy (given a certain temperature). But why isn't the occupation probability zero in the band gap? Or does it give the occupation in case there was no band gap?

In any case, I was wondering how to interpret the picture below. On the left, the valence band is completely occupied. In the middle, there are a lot of states with a lower occupation probability in the valence band compared to the left, but very few in the conduction band with a higher (but still very small) occupation probability. But it can't be the case that there are suddenly less states being occupied because there are still as many electrons as before. So what am I getting wrong? And how can I interpret the Fermi-Dirac distribution better?

Thanks in advance.

enter image description here

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    $\begingroup$ There are many similar questions. Perhaps start with physics.stackexchange.com/q/351374 or the other questions under the 'Related' tab to the right. Essentially, the Fermi-Dirac distribution has to be combined with the actual band structure to determine occupancy. $\endgroup$
    – Jon Custer
    Commented May 5, 2020 at 20:35
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    $\begingroup$ Does this answer your question? The concept of Fermi level $\endgroup$
    – Jon Custer
    Commented May 5, 2020 at 20:36
  • $\begingroup$ Yes the diagrams look a bit sloppy to me. $\endgroup$
    – Andrew Steane
    Commented May 5, 2020 at 20:37
  • $\begingroup$ @JonCuster thank you, the thread helped me a little. I don't have a good background in modern physics but I'll read up on the origin and meaning of the Fermi Dirac distribution. It's just strange that the shape of the distribution doesn't change when some states are not allowed: when you know certain states aren't allowed, doesn't that tell you other states must have a higher occupancy probability? $\endgroup$
    – Sudera
    Commented May 6, 2020 at 10:12
  • $\begingroup$ The distribution and the available states are two very different things. Actual occupancy of some state or another is a convolution of them. $\endgroup$
    – Jon Custer
    Commented May 6, 2020 at 13:09

1 Answer 1

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Fermi energy vs. Fermi level One has to distinguish between the Fermi energy and the Fermi level (=chemical potential). The former is the energy of the topmost energy state, whereas the latter is the energy needed to add/remove an electron (and also the parameter of Fermi-Dirac distribution). In a continuous spectrum at zero temperature the two are the same, which is why the two terms are often used interchangeably. In semiconductors they are not identical... and usually both mean chemical potential, since speaking of a Fermi energy in a semiconductors doesn't make much sense (if we define it as the difference between the highest and the lowest occupied energy states, it is just equal to the width of the valence band).

Fermi function and density-of-states So here we are dealing with the chemical potential: adding the last electron to the valence band costs zero energy, whereas adding the first electron to the conduction band costs the gap energy $E_g$, which is why the Fermi level is chosen to be in the middle of the gap. More mathematically strict: the total concentration of electrons is given by $$n =\int dE \rho(E)f_\mu(E)=\int dE \frac{\rho(E)}{e^{\frac{E-\mu}{k_BT}}+1},$$ where $\rho(E)$ us the density-of-states. This concentration should not change with temperature, i.e. all the electrons removed from the valence band should be found in the conduction band. Note, that the density-of-states is zero in the gap region: the fact that the Fermi function is not zero there does matter, since there are no states that electrons could occupy.

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    $\begingroup$ Thank you. I now understand that you have to multiply the probability of occupancy of the state at a given energy level with the density of states available at that energy level. I think I just don't really understand what the Fermi Dirac distribution is or where it comes from, and what it means in the case of semiconductors. $\endgroup$
    – Sudera
    Commented May 6, 2020 at 10:01
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    $\begingroup$ @Sudera The figure that you posted is not quite clear: the electrons removed from the valence band are at the bottom of the conduction band, but in the figure the Fermi function is almost zero in the conduction band. $\endgroup$
    – Roger V.
    Commented May 6, 2020 at 11:39
  • $\begingroup$ @Vadim Can you clarify the definition of Fermi level you provided? "to add/remove an electron" Where are we adding to or removing the electron from? Are we adding the electron to the lowest unoccupied state and removing electron from the highest occupied state? $\endgroup$
    – physu
    Commented May 13, 2020 at 11:30
  • $\begingroup$ @Sudera Fermi level is the same as chemical potential in statistical physics. One adds/removes electron to/from the lowest unoccupied/occupied level. Fermi energy, on the other hand is just the energy corresponding to the Fermi wave number, e.g. $\epsilon_F = \hbar^2k_F^2/{2m}$ in an isotropic free electron gas. $\endgroup$
    – Roger V.
    Commented May 13, 2020 at 11:36
  • $\begingroup$ Talking about adding/removing and electron in semiconductor just doesn't make sense, since the energy is too different. The correct meaning of the Fermi level/chemical potential is that it is a parameter in the distribution function. $\endgroup$
    – Roger V.
    Commented May 13, 2020 at 11:39

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