# Why does the Fermi-level decrease with temperature increase?

On temperature increase there should be an increased probability of finding an electron in the conduction band, this corresponds to the Fermi level rising up towards the conduction band so that there is a greater possibility of finding an electron in the conduction band.

But the plot shows the opposite. Why is this so?

The picture seems to be describing the behavior of an n-doped material as a function of temperature.

1. For an intrinsic semiconductor, one where $$n=p$$, the Fermi-level is approximately at the middle of the gap (if the effective masses in the conduction and the valence band are very similar) almost independently of the temperature.

2. In an extrinsic material, one that has $$n\gg p$$ or $$p\gg n$$, the Fermi-level is closer to one of the bands; closer to the conduction band (CB) if $$n\gg p$$ and to the valence band (VB) if $$p\gg n$$.

I think the confusion here comes from thinking that a doped material (with donor impurities in OP's case) behaves extrinsically no matter the temperature. This is not true. As we increase the temperature we expect the material to start behaving intrinsically at some point (vid. point 2 in my Physics.SE answer here). As this happens, the Fermi-level will approach asymptotically, for very high temperatures, the intrinsic Fermi-level at $$\approx E_g/2$$.

For an n type semiconductor, Fermi level is given by $$E_f=E_c-k_BTln(N_c/n)$$ where symbols have usual meanings. From this you can see why.

• Thank you, I wanted some conceptual understanding as to why it happens Commented Dec 27, 2020 at 11:22

The decrease in Fermi level with temperature is a consequence of charge conservation. The net charge per unit volume in the material is $$n_d$$, the concentration of donor electrons. At $$T = 0$$, the Fermi-Dirac distribution is a step function and all states below $$E_f$$ are occupied (specifically, the donor states and valence band states). When $$T$$ is large, the Fermi-Dirac distribution varies over a huge energy range extending beyond the band edges. If the Fermi level were still close to the conduction band edge, the total electron density would be much higher than $$n_d$$, which violates charge conservation. Therefore, the Fermi level has to lie deep in the band gap.