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I was going through this question: Why does the Fermi-level decrease with temperature increase?

As the temperature increases though the semiconductor becomes intrinsic where number of holes equals number of electrons, the number of carriers will still increase. In other words, the resistivity decreases as temperature increases as shown below.

enter image description here Source: https://eng.libretexts.org/Bookshelves/Materials_Science/Supplemental_Modules_%28Materials_Science%29/Electronic_Properties/Resistivity

The resistivity also decreases as doping level increases as shown below.

enter image description here

As the temperature increases, the fermi level moves down more below the conduction band for an n-type semiconductor as shown below.

enter image description here

I had thought that more closer the Fermi level is to the conduction band, the more conductive the semiconductor will be; in other words, less resistivity. I cannot still convince myself that why this is not true. If fermi level moves away from the conduction band, it means that 50% probability of finding an electron is quite below the conduction band and therefore it's more difficult for an electron to jump to the conduction band. Where am I having it wrong? Could you please help me?

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    $\begingroup$ An intrinsic semiconductor has the Fermi level at mid-gap - equal numbers of electrons and holes. So, you just need a thermally generated carrier density greater than the doping levels and the Fermi level will be at mid-gap. $\endgroup$
    – Jon Custer
    Commented Mar 24, 2023 at 1:58
  • $\begingroup$ Thank you! So, at higher temperatures, it's the thermally generated carriers which contribute more to conductivity compared to the contribution from doping levels. I hope I have it correct. $\endgroup$
    – PG1995
    Commented Mar 24, 2023 at 16:27

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If I understand you correctly, you are asking why conductivity increases as temperature increases. As the temperature of an extrinsic semiconductor increases, its conductivity first goes down, and only starts going up once the semiconductor becomes intrinsic.

For an extrinsic semiconductor $N_D \gg n_i$, that is $E_F$ is over at least a few $k_B T$ above the intrinsic level $E_i$. In this regime, $n\approx N_D$ is fixed, as required by charge neutrality. As you increase temperature, electrons can more easily hop into the conductance band, so in order to retain the same electron density, the Fermi level must come down. Conductivity is proportional to the electron density and mobility. The former stays constant while the latter decreases with increasing temperature due to increased scattering with the vibrational modes of the lattice, so conductivity decreases.

For an intrinsic semiconductor, $n_i \gg N_D$, which roughly means $E_F$ is within about $k_B T$ of $E_i$. In this regime, $n\approx n_i$, which is a strong function of temperature. As you increase temperature, $E_F$ doesn't change much because it's already very close to $E_i$. With $E_F$ roughly constant, electrons can more easily hop into the conductance band, so the density of conduction electrons increases. While mobility still decreases, this effect is outweighed by the increase in electron density, so conductivity increases.

I had thought that more closer the Fermi level is to the conduction band, the more conductive the semiconductor will be; in other words, less resistivity.

This is true if you are keeping the temperature fixed, controlling the Fermi level by adjusting the doping density. But you need to remember that the probability of a carrier accessing the higher energy conduction band states increases at higher temperatures, which manifests itself as an increase in the intrinsic carrier concentration.

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  • $\begingroup$ Thank you for the help. Relater to your last para. So, if doping level is kept constant and temperature is increased, the Fermi level moves away from conduction band for an n-type semiconductor. At one point the semiconductor will become intrinsic and Fermi level will lie between conduction and valence bands. So, is it fair to say that at higher temperatures once the semiconductor becomes intrinsic, it's the thermally generated carriers which contribute more to conductivity compared to the contribution from doping levels? $\endgroup$
    – PG1995
    Commented Mar 25, 2023 at 3:36
  • $\begingroup$ @PG1995 That's correct, an intrinsic semiconductor has $n \approx n_i \gg N_D$, meaning the contribution of donors to electron concentration (and hence conductivity) is negligible. $\endgroup$
    – Puk
    Commented Mar 25, 2023 at 3:39

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