Why the Fermi level determines the conductivity of the material?

Considering electronic band structure in solid state Physics, the Fermi level is defined as the chemical potential appearing at the Fermi-Dirac distribution

$$f(\epsilon)=\dfrac{1}{\exp[(\epsilon-\mu)/k_BT]+1}.$$

In that sense, it is just one specific value of energy that we chose to give a name to it.

Now, the Fermi level is quite important because of the following points, according to Wikipedia's article:

1. In an insulator, $\mu$ lies within a large band gap, far away from any states that are able to carry current.

2. In a metal, semimetal or degenerate semiconductor, $\mu$ lies within a delocalized band. A large number of states nearby $\mu$ are thermally active and readily carry current.

3. In an intrinsic or lightly doped semiconductor, $\mu$ is close enough to a band edge that there are a dilute number of thermally excited carriers residing near that band edge.

In other words, it seems that the conductivity properties of the material are determined by $\mu$.

But why is that? Why $\mu$ has all these properties?

How can we actually find out these properties of the Fermi level? How can we find out that the Fermi level determines the conductivity according to these points?

• As to point 3, the position of μ in doped semiconductors depends on temperature. Only at low temperature it is close to the edge of a band. But at high temperature, when the material behaves like an intrinsic semiconductor with equal concentrations of electrons and holes, it will be near the middle of the band gap. – Pieter Nov 28 '16 at 6:59
• @Pieter - In a doped semiconductor (e.g., n-type Si, $N_D$ larger than $10^{16}cm^{-3}$) there is a wide temperature range (approximately from $T=150K$ to $450K$) called extrinsic region where the electron density is practically constant $n≈N_D$ due to the full ionization of the dopant (ionization energy typically 0.05eV) and the Fermi level is below the donor level and closer to the conduction band than to midgap. At (very) low temperatures, the donors "freeze out" and the Fermi level rises above the donor level but $n$ drops sharply because the FD-distribution approaches a step function. – freecharly Nov 28 '16 at 19:16
• @freecharly: yes, carrier concentration constant with rising temperature but the Fermi level moves towards the center of the band gap. The Boltzmann factor increases. Here is a figure for silicon: ioffe.ru/SVA/NSM/Semicond/Si/Figs/123.gif – Pieter Nov 28 '16 at 19:56
• @Pieter -Thanks, I know this temperature behavior, you find it in many textbooks, like, e.g. the one by Sze. My last comment had this in mind. The relation between increasing electron concentration and decreasing distance of the Fermi level from the conduction band holds, of course, at a given temperature when you change the (quasi-) Fermi level by doping or by charge injection and don't change the shape of the FD-distribution. – freecharly Nov 28 '16 at 21:02

As you have already indicated, the Fermi level $\mu$ determines the Fermi-Dirac energy distribution of electrons (and holes) in the solid. Thus in a semiconductor at a given temperature, if the Fermi level is close to the conduction band (brought about by doping with donor atoms), a large electron concentration appears in the conduction band together with a low hole concentration in the valence band. Conversely, if the Fermi level is close to the valence band (by acceptor doping), you get a large hole concentration there and a low electron concentration in the conduction band. In an intrinsic semiconductor (no doping) the fermi level is close to the midgap energy and you have equal electron and hole concentrations in the conduction and valence band, respectively. Similarly, in an insulator with a large bandgap, the Fermi level is in the middle of the band gap and you have very low electron and hole concentrations. In metals the Fermi level lies in a partly filled band and it can be shown that conduction is due to the electrons close in energy to the Fermi level. Thus the conduction properties are determined by the energetic position of the Fermi level with respect to the conduction and valence bands or by the position in an energetically allowed band of a metal or degenerate semiconductor.
• @Pieter - In metals only electrons near the Fermi surface in k-space are contributing to electrical current. Electrons in fully occupied k-space with energies below the Fermi level do not contribute to conduction because the current contributions of all individual electron states cancel each other. Solid state quantum theory results in a conductivity formally proportional to $n$ only for a parabolic spherically symmetric $E(k)$. In general, the electron effective mass and scattering probabilities entering the conductance formula are determined by the electrons at the Fermi surface. – freecharly Nov 30 '16 at 0:19