If I hold a semiconductor sample at a certain temperature $T$, its lattice temperature $T_l$ will equalize: $T_l=T$. But how is this lattice temperature related to the carrier (electron or hole) temperature $T_{e/h}$ if we assume a Fermi-Dirac distribution of carriers in thermal equilibrium $f^{e/h}_k=\frac{1}{\exp( [E^{e/h}_k-\mu_{e/h}]/[k_BT_{e/h}]) + 1}$? The carrier temperatures are not the same as the lattice temperature, are they? Is there a relation between $T_l$ and $T_e$ or $T_h$? Or does this depend on further knowledge of the system?


1 Answer 1


Most of the time (e.g. at room temperature), the temperature of the lattice $T_\ell$ is equal to the electronic temperature $T_e$. The simpliest argument to give in favor of this is that electrons are strongly coupled to phonons via the so-called electron/phonon coupling.

In that way, the thermal equilibrium of the lattice is directly linked the thermal equilibrium of the electrons, so that $T_\ell=T_e$.

But electron/phonon coupling cease to be efficient for temperatures under approximatively $10\,\mathrm{K}$ (depending on your sample). So if you are able to cool your semi-conductor (or metal) under this temperature, you will actually only be able to cool the lattice but not the electrons : $T_\ell<T_e$.

That's why people who are working in low temperature physics and cryogenics are having such an hard time to actually cool a sample below a few $\mathrm{K}$.

  • $\begingroup$ Is there a good estimate how the hole temperature differs from the electron temperature? $\endgroup$
    – DaP
    Nov 14, 2014 at 15:46
  • $\begingroup$ I actually don't know any case where the hole temperature is different from electron temperature. I would say they are always equal since holes are just mathematicals tricks to describe electrons. $\endgroup$
    – dolun
    Nov 15, 2014 at 11:11
  • $\begingroup$ But, consider the Fermi distribution I described in my question. If you do not lose carriers, than the number of electrons and holes must be the same. You get the number of carriers by integrating the distribution and then you obviously need a different temperature for electrons and holes since they have a different mass and chemical potential. $\endgroup$
    – DaP
    Nov 15, 2014 at 11:40
  • $\begingroup$ I do agree that there exists some cases where effective mass of electrons and holes is different, especially when you do the separation between light and heavy holes in the valence band. But most of the time, $m_h=-m_e$ still valid. And clearly, there is no such a thing as different chemical potential for electrons and holes. And it's very doubtful that one can define temperature for holes as a real thermodynimical measurable quantity since electrons and holes in semi-conductor can not be separated in two well independant systems. $\endgroup$
    – dolun
    Nov 15, 2014 at 12:36

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