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It is well known that the voltage $V$ across, say, a p-n junction can be related to the electro-chemical potentials (usually and improperly called Fermi levels) at the borders of the junction $$ qV = \mu_{n}-\mu_{p}$$

I'm trying to find a way to understand this in thermodynamic terms. This formula means that the work provided by one electron going from one end of the junction to the other is the difference of electro-chemical potentials. But I'm struggling to recover that from a pure thermodynamical perspective.

When removing dN carriers from the n side of the junction, the energy of the system changes by $$dU_n = -p_n dV_n +T_n dS_n -\mu_n dN $$ when adding dN carriers to the p side of the junction, the energy of the system changes by $$dU_p = -p_p dV_p +T_p dS_p +\mu_p dN $$

The total energy change of the system equals the work and heat provided by the system, ie $$ dU_n + dU_p = \delta W + \delta Q = qV dN + \delta Q $$

Now, how do I get from there to $ qV = \mu_{n}-\mu_{p}$ ?

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    $\begingroup$ Considering that the variation of free energy $F$ gives the maximal available work, we consider two processes on the carrier population Removing dN carriers, resulting in $dF_n = -p_n dV_n -S_n dT_n -\mu_n dN - q \phi_n dN$, where $\phi_n$ is the electrostatic potential. Considering that this process does not change the temperature or pressure in the system, $dF_n = - (\mu_n + q\phi_n)dN$. In the same way, when adding $dN$ particles on the p side, $dF_p = (\mu_p + q\phi_p)dN$. The work provided by the system is $$\delta W = qVdN = -dF_n - dF_p =((\mu_n - \mu_p) + q (\phi_n - \phi_p))dN$$ $\endgroup$
    – Pen
    Commented Oct 5, 2017 at 6:12
  • $\begingroup$ Is this derivation correct ? $\endgroup$
    – Pen
    Commented Oct 5, 2017 at 9:26
  • $\begingroup$ Take the case when no heat is provided to the system, and there's no volume change (i.e., all the work actually is due to the potential.) $\endgroup$
    – Roger V.
    Commented Feb 25, 2023 at 8:48

2 Answers 2

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There is a distinction between the built-in voltage and the externally applied voltage in a p-n junction. In thermodynamic equilibrium, the p-n junction maintains a single chemical potential across its entire structure. When you apply the filed the expression $eV=\mu_p - \mu_e$ holds, where V is applied voltage.

The ionization potentials for the left and right ends remain unaffected by the contact. However, the ionization through p-n junction will be different thanks to the vacuum level bending.

Edit: The fact that you have positional-dependent chemical potential indicates that you are out of equilibrium.

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I would suggest going through Howard Reiss' Chemical Effects Due to the Ionization of Impurities in Semiconductors paper in J. Chemical Physics 21(7) 1209-1217 (1953). In particular, look through section V which is titled 'The Relation of the Fermi Level to the Total Free Energy of an Electron Assembly'. There, Reiss makes the point that 'although the Fermi level always represents the chemical potential of a weakly coupled electron assembly, it is hardly ever the Gibbs free energy per electron' (in italics in the original).

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  • $\begingroup$ I'm quite puzzled by the argument... Sure, $\mu\neq F/N$ in the general case, but why would you need such a relation ? As long as you consider small variations, $\mu= \partial F/ \partial N$ should be enough ? $\endgroup$
    – Pen
    Commented Oct 5, 2017 at 6:04
  • $\begingroup$ @Penangol - the paper goes in to great detail, which is way too long to detail. Good reading. $\endgroup$
    – Jon Custer
    Commented Oct 6, 2017 at 2:36

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