I've been researching into this and can't quite figure out where that lost voltage is going. When silicon is excited by a photon within its absorption spectrum, it will always have an internal potential of 1.1V as per the band gap. Why is the p n junction only able to extract roughly half of this?
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$\begingroup$ The open circuit voltage is not simply dependant on the band gap. It's normally determined by the recombination rate. A quick Google found this article that seems to provide a good summary. $\endgroup$– John RennieCommented Nov 20, 2014 at 8:33
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$\begingroup$ Recombination only effects the overall current and not the voltage/electromotive force created by the electrons that are able to be extracted. $\endgroup$– ScottCommented Nov 20, 2014 at 9:02
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$\begingroup$ We have to keep in mind that if we're speaking about PN junction, then we deal with transitions between defect levels. So, I'm not sure which doping atoms are used in this very type of solar cells, but it seems that that difference in energy levels of thos defects causes solar panel to produce 0.6v $\endgroup$– ChaositCommented Nov 20, 2014 at 9:57
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$\begingroup$ Silicon diodes' threshold voltage is also $0.6V$. $\endgroup$– RuslanCommented Nov 20, 2014 at 10:14
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$\begingroup$ And, perhaps not surprisingly, the diode threshold voltage and the cell open-circuit voltage are one and the same through the saturation current. Sze (as usual) has lots about both. $\endgroup$– Jon CusterCommented Nov 20, 2014 at 14:30
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1 Answer
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I'll talk about why silicon diodes usually have threshold voltage of about $0.6V$. For the potential generated by silicon solar cells the argument is much the same.
Built-in potential is what determines threshold voltage. This potential can be calculated using the formula:
$$q\varphi_{in}=kT\ln\frac{n_{n0}p_{p0}}{n_i^2},$$
where $k$ is Boltzmann constant, $n_i$ is intrinsic concentration of charge carriers, and $n_{n0}$ and $p_{p0}$ are concentrations of dominant charge carriers (i.e. holes in p-Si and electrons in n-Si).
If we take room temperature, then $kT=0.025\text{ eV}$, $n_i=10^{10}\,\text{cm}^{-3}$, and taking doping concentrations of $10^{15}\text{ cm}^{-3}$, and taking into account that at room temperature almost all the doping atoms are thermally ionized, we have $n_{n0}=p_{p0}=10^{15}\,\text{cm}^{-3}$, so we calculate the built-in potential to be
$$q\varphi_{in}=0.57\text{ eV}.$$
Taking higher doping concentration, e.g. $10^{16}\text{ cm}^{-3}$, we'll get $q\varphi_{in}=0.69\text{ eV}$.
See this wiki page for some more about doping and charge concentration.
Have a look at the band diagram of pn junction with doping of $10^{15}\text{ cm}^{-3}$ (from that same wiki page):
We can see that on bias of $0.59\text{ eV}$ the bands are almost flattened, so increasing the bias will just tilt the whole bands, which corresponds to ohmic conductance region of diode voltage response.
