why does the fermi level is situated in the middle of the valence band and conduction band in an intrinsic semiconductor at $T = 0$ Kelvin? what is the physics behind it?
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The Fermi level lies approximately in the middle of the bandgap for an intrinsic semiconductor, independent of temperature.
At $ T=0K $, all states in the valence band are occupied by electrons and all states in the conduction band are empty. At temperatures above $0 K$, the occupation of states with electrons is governed by thermodynamics and is described by the Fermi-Dirac-Dirac distribution function, which minimizes the free enthalpy for a given temperature.
$$f(E,T)=\frac{1}{e^{\frac{E-E_{F}}{k_{B}T}}+1}$$
It provides the probability that an energy level is occupied by an electron in a semiconductor
At $ T=0K$, the probability of an electron to have energy below $E_{F}$ is 1 and above $E_{F}$ is 0. At higher temperatures, the step function is "smeared out", as you can see in the image.
At temperatures above $0K$, some states in the valence band are occupied and not all states in the valence band are occupied anymore. Note that the semiconductor is still electrically neutral. For each negative electron in the conduction band, there must be a positively charged hole in the valence band, i.e. every electron that leaves the valence band to the conduction band leaves an uncompensated positive charge in the valence band, which means that there are equally many electrons in the conduction band as holes in the valence band.
Since the Fermi-Dirac distribution is symmetrical, the Fermi energy must be in the middle of the bandgap.
Note that this is only an approximation and only valid if the density of states n the conduction and valence band are equal.
If there is a difference in the densities of the valence and conduction band, it is given by:
$$E_{F}=E_{C}-\frac{E_{g}}{2}+\frac{k_{B}T}{2}ln(\frac{N_{V}}{N_{C}})$$
