I try to estimate the open-circuit voltage of an ideal solar cell (no non-radiative loss) with perfect absorption, illuminated by a black body at temperature $T_{\odot}$ and at full concentration. I found myself facing incoherent results.
I first tried to use the standard Boltzman approximation to estimate the absorbed and radiated currents
$$J_{abs} =\frac{\pi}{4\pi\hbar^{3}c^{2}}\stackrel{\infty}{\underset{E_{g}}{\int}}dE\,E^{2}\exp\left(\frac{E}{k_{B}T_{\odot}}\right)$$
$$J_{em} =\frac{\pi}{4\pi\hbar^{3}c^{2}}\left(\stackrel{\infty}{\underset{E_{g}}{\int}}dE\,E^{2}\exp\left(\frac{E}{k_{B}T}\right)\right)\exp\left(\frac{qV}{k_{B}T}\right)$$
and the open circuit voltage is given by
$$\exp\left(\frac{qV_{OC}}{k_{B}T}\right) =\frac{\stackrel{\infty}{\underset{E_{g}}{\int}}dE\,E^{2}\exp\left(-\frac{E}{k_{B}T_{\odot}}\right)}{\stackrel{\infty}{\underset{E_{g}}{\int}}dE\,E^{2}\exp\left(-\frac{E}{k_{B}T}\right)}$$
For $E_{g}=1$ eV, the open volage is larger than the gap $V_{OC}=1.05\,E_{gap}$, which makes no sense. This probably shows that Botlzman approximation is not valid in this case. So I tried to carry out the calculation with the generic expression
$$J_{abs} =\frac{\Omega_{\odot}}{4\pi\hbar^{3}c^{2}}\stackrel{\infty}{\underset{E_{g}}{\int}}dE\,\frac{E^{2}}{\exp\left(\frac{E}{k_{B}T_{\odot}}\right)-1}$$ $$J_{em} =\frac{\Omega_{em}}{4\pi\hbar^{3}c^{2}}\stackrel{\infty}{\underset{E_{g}}{\int}}dE\,\frac{E^{2}}{\exp\left(\frac{E-\Delta\mu}{k_{B}T}\right)-1}$$
Here, $J_{em}$ presents an upper bound $\frac{\pi}{4\pi\hbar^{3}c^{2}}\stackrel{\infty}{\underset{E_{g}}{\int}}dE\,\frac{E^{2}}{\exp\left(\frac{E-E_{g}}{k_{B}T}\right)-1}$ reached for $\Delta\mu=E_{g}$. If the source is hot enough (for instance $T_{\odot}=24\,000$ K), this upper limit is lower than the absorbed current $J_{abs}$, which means that the cell can not be at equilibrium.
I am doing a stupid mistake somewhere ? I tend to blame the perfect absorption, which is certainly disturbed at some point by the large population changes...