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My professor just uploaded the following two foils:

figure 1

and

figure 2

Now I understand that if temperature increases we have some electrons in the conduction band and some holes in valence band, which would (I think at least) lead to a rise of chemical potential, which is what you see in figure two.

Now what I don't understand is how these two foils do not contradict each other, as in one $E_f$ is rising and in the other $E_f$ is getting smaller. Can somebody explain maybe, what actually happens to the chemical potential when we raise the temperature in a undoped semiconductor? Also I do not understand why, in a doped semiconductor $E_f$ will tend exactly to the middle of the gap again when temperature is raised, that is to say behave like an undoped semiconductor.

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  • $\begingroup$ I agree the plots are confusing as to just what was meant. For your last question, when the number of (thermally generated) intrinsic carriers is greater than the number of dopants, the material looks intrinsic and the Fermi energy will be mid-gap. $\endgroup$
    – Jon Custer
    Commented Feb 1 at 16:31

1 Answer 1

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The first diagram seems to correspond to an n-doped semiconductor, because the chemical potential (Fermi level) is close to the conduction band. The second diagram seems to correspond to a p-doped semiconductor because the chemical potential is close to the valence band. In both cases, with rising temperature, the chemical potential (Fermi level) will move towards the intrinsic level (near the middle of the band gap) because the electron or hole concentrations due to doping become comparable and in the end negligible compared to the intrinsic electron and hole concentrations at high temperatures.

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