It is a known result that the photo-voltage generated in photovoltaic system (such as a simple p-n junction) is directly related to the quasi Fermi levels splitting at the contact - as illustrated by Wurfel in Physics of Solar Cell (Willey, 2005, p99):

Assuming Fermi levels to be flat enough for simplicity, $$ qV = E_{F,n}-E_{F,p} $$ where $E_{F,n}$ is the Fermi level of electrons in the conduction band and $E_{F,p}$ that of the valence band.

Now, from a thermodynamic perspective, $qV$ is the work produced by the transfer of an electron from the conduction band to the valence band, and it can be related to the variation of the relevant thermodynamic potential.

In most derivation I could read, people simply write that the potential is Gibbs free energy $G=U-TS+pV$, such that $$ \Delta G = -S\Delta T + V \Delta p + \mu_n \Delta N_n - \mu_p \Delta N_p $$ and as the transformation is isothermal and isobaric, $$ \Delta G = E_{F,n} \Delta N_n - E_{F,p} \Delta N_p. $$ Considering $$ W=- qV \Delta N_n = -\Delta G, $$ we recover the result.

However, I don't see any reason why the transformation would be considered isobaric rather than isochore, and I understand Gibbs free energy as the thermodynamic potential for a closed system, whereas the system here is open. Why is Gibbs free energy the correct thermodynamic potential for such transformation ?

  • $\begingroup$ 'quasi Fermi levels splitting at the contact' - what do you mean by that? $\endgroup$ – Jon Custer Feb 27 '18 at 16:05
  • $\begingroup$ @Jon Custer: I added a figure for clarification. I mean the difference between the Fermi level for the extraction of electrons and the injection of holes. $\endgroup$ – Pen Feb 27 '18 at 16:28

First of all, I think you already know this, but quasi fermi level is a synonym for chemical potential, i.e. by definition: $$E_{f,n} \equiv -\mu_n , \qquad E_{f,p} \equiv \mu_p.$$ (hope I got the signs right)

Now, I don't see why Gibbs free energy would come up at all here. A typical context for Gibbs free energy is: you have a beaker full of chemicals in a laboratory, and the laboratory is full of air, and Gibbs free energy accounts for the fact that if the liquid in the beaker changes volume, it has to push against the air in the room, which has an ambient pressure.

In our context, the ambient pressure has no relevance. All the electrons and holes are in the crystal, and the crystal's volume is fixed. So it should be Helmholtz free energy, right? We're also not changing the temperature here, so it ends up being something like: $$\Delta H = \mu_n \Delta n + \mu_p\Delta p$$

  • $\begingroup$ I think to things remain unclear to me here. We are not changing the crystal volume, but we are not changing the pressure either, are we ? And is Helmholtz free energy still relevant as thermodynamic potential for an open system ? All together, my question probably comes down to: how do you determine the correct thermodynamic potential for such a transformation ? $\endgroup$ – Pen Mar 4 '18 at 13:07

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