# A quick question about thermal voltage

Joined Aug 30, 2018
24
I was skimming around G. D. Mahan's Condensed Matter in a Nutshell and came across a pretty interesting nugget: $\varepsilon (k) = \frac{\hbar^2 k^2}{2 m^*} \sim k_B T \,.$ (It's important to draw a distinction between $$k$$, which is the lattice's wavenumber, and $$k_B$$, which is the Boltzmann constant.) Basically, the total energy of a semiconductor lattice is related to its wavenumber—and it approximately scales up to the thermal properties of the system.

So, provided that we know this, does that mean that the thermal voltage of a semiconductor would be more accurately described as
$V_T = \frac{\varepsilon (k)}{q_0} = \frac{\hbar^2 k^2}{2 m^* q_0} \,?$ The answer probably seems trivial to people that have spent more time reading about semiconductor theory than I have.

#### ZCochran98

Joined Jul 24, 2018
179
The term $E\equiv\epsilon(k)=\frac{\hbar^2k^2}{2m^*}$ is exactly equivalent to kinetic energy of a particle and, in quantum mechanics (considering you're reading a condensed matter book you likely know or are at least familiar with this), is frequently used when writing momentum in terms of energy or vice-versa. Now, the kinetic energy of an electron in a material is due to a large number of things (take a look at the Hamiltonian for the Standard Model!) However, thermal energy (for things sufficiently above absolute zero) is one of the biggest contributors to kinetic energy. So, with that in mind, $E_{therm}=\frac{\hbar^2k^2}{2m^*} \approx k_BT$ is a good enough approximation that either statement is technically correct, though the Boltzmann's constant form is a lot easier to use (you don't need the lattice wavenumber or the reduced mass of the electrons).

To properly answer your question: for most situations, neither formulation is technically "more accurate" (one formulation is derived from quantum mechanics/solid-state physics and the other from statistical mechanics), though the form with Boltzmann's constant is more "instructive" as to the origin of the thermal voltage (and the originator of its name as a thermal voltage). At really low temperatures the thermal voltage will also be very small, but you may have other electric potentials to deal with due to other sources of kinetic energy in the electrons, so the formulation using lattice wavenumber will not be technically an appropriate description of the thermal voltage. The total potential? Quite possibly - just not for the thermal voltage.

• The term $E\equiv\epsilon(k)=\frac{\hbar^2k^2}{2m^*}$ is exactly equivalent to kinetic energy of a particle and, in quantum mechanics (considering you're reading a condensed matter book you likely know or are at least familiar with this), is frequently used when writing momentum in terms of energy or vice-versa. Now, the kinetic energy of an electron in a material is due to a large number of things (take a look at the Hamiltonian for the Standard Model!) However, thermal energy (for things sufficiently above absolute zero) is one of the biggest contributors to kinetic energy. So, with that in mind, $E_{therm}=\frac{\hbar^2k^2}{2m^*} \approx k_BT$ is a good enough approximation that either statement is technically correct, though the Boltzmann's constant form is a lot easier to use (you don't need the lattice wavenumber or the reduced mass of the electrons).