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I have researched and looked at several papers describing the effect of temperature on the $I_{sc}$ and $V_{OC}$, and seen many numerical explanations as to why the voltage drops, with reference to I-V curve and other equations, but none actually described why this physically happens. What causes voltage to drop at increased temperatures. Is it the decreased bandgap? It would be appreciable if you could link any related papers that explain this phenomenon physically.

Thanks

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    $\begingroup$ The band gap of silicon certainly decreases with temperature. I’ll leave exact numbers to a Google search. $\endgroup$
    – Jon Custer
    May 19, 2021 at 20:41
  • $\begingroup$ @JonCuster yes, but in what way does that contribute to a decreased voltage? $\endgroup$
    – Saumya Shah
    May 20, 2021 at 9:17

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We can begin with the equation for solar cell current

$I=I_0(e^{qV/kT}-1)-I_L$

The open circuit voltage is the voltage when current ($I$) is 0. Solving the above equation gives us

$V_{oc}=\frac{kT}{q}\ln\left(\frac{I_L}{I_0}+1\right)$

A temperature increase causes an increase in the intrinsic carrier concentration which in turn increases the diode saturation current $I_0$. Since $I_L\gg I_0$ the $+1$ can be ignored and this equation can be rearranged as

$V_{oc}=\frac{kT}{q}(\ln I_L-\ln I_0))$

The change in $V_{oc}$ can clearly be seen here as a function of temperature, both in the $\frac{kT}{q}$ term and in the $I_0$ term. These have opposing effects, and it turns out even inside $\ln$ the change in saturation current ends up dominating the temperature effect.

Part of the change in intrinsic carrier concentration comes from the reduction in bandgap, and part of it comes from the smearing of the Fermi distribution.

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    $\begingroup$ I understood the mathematical basis, but I am not yet clear on the physical basis for this phenomenon. Why does an increase in the diode saturation current result in a decrease in VOC? $\endgroup$
    – Saumya Shah
    May 21, 2021 at 11:07
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    $\begingroup$ Let me have a go at putting words to the math, which is the correct answer here. The cause of the saturation current is recombination, due to holes and free electrons meeting each other. Higher T causes faster diffusion which results in more collisions. The effect is a recombination current. In a simplified view, this can be understood as a shunt, which lowers the voltage. $\endgroup$
    – W_vH
    Jan 28 at 23:43
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The saturation current is in a direction opposite to the illumination current and therefore an increase in temperature increases the saturation current and reduce voc.

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The Fermi-Dirac statistical distribution expresses how carriers populate energy levels $$\hat n(E) = \frac{1}{e^{{\epsilon-\mu}/{k_BT}}+1}$$ And this is how it looks like at different temperatures (taken from Wikipedia): https://en.wikipedia.org/wiki/Fermi%E2%80%93Dirac_statistics And this behavior represents the fact that electrons can be excited spontaneously by thermal energy (e.g., received from phonons), therefore they have higher chance of populating higher energy levels. This smears the distribution and weakens the interatonic bonds (based on valence electrons), which effectively decreases the bandgap. Since the open-circuit voltage is directly related to the bandgap (as the potential drop across a p-n junction is related to the band alignment), this also decreases the open-circuit voltage.

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  • $\begingroup$ But there are no states in the gap to ‘smear’. $\endgroup$
    – Jon Custer
    Aug 27, 2022 at 23:09
  • $\begingroup$ It does not matter, the Fermi-Dirac distribution is still smeared, even if in certain energy bands the density of states is zero. The overall probability per energy level is their product. $\endgroup$
    – Omer Luria
    Aug 29, 2022 at 7:36
  • $\begingroup$ The band gap is defined by where the top of the valence band and the bottom of the conduction band are. Fermi-Dirac says something about the occupancy of those state in thermal equilibrium. It says nothing about where those states actually lie. $\endgroup$
    – Jon Custer
    Aug 29, 2022 at 12:40
  • $\begingroup$ You are right, the location of the bands is more related to weakening of the valence bonds (which decreases $E_g$, and but that also decreases $V_{oc}$). The smear of the Fermi-Dirac distribution increases the saturation current through unwanted minority carrier diffusion, which decreases $V_{oc}$. This is because more carriers can be spontaneously excited to higher energy levels. $\endgroup$
    – Omer Luria
    Aug 31, 2022 at 19:42

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