Why do silicon solar cells only produce ~0.6V when the band gap of silicon is ~1.1V? 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? 
 A: 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.
