Let us assume that we have an ideal photoelectrode with no optical losses. The attainable current density for a given bandgap is then obtained by integrating the spectral distribution. Starting value is the band gap energy and the integration runs over all shorter wavelengths. If we plot the value of the integral for different starting points we get the maximum current density for a given bandgap. This is the maximum current that the photoelectrode can deliver under short circuit conditions. Lower bandgaps yield higher currents because they absorb larger parts of the spectrum.
The theoretical maximum photocurrent density under light illumination, $J_\textrm{max} (A\,m^{-2})$, is calculated by integrating the photon flux, shown in equation:
\[J_\textrm{max}=e\times \int \limits_{\lambda_a}^{\lambda_b}\textrm{Flux}(\lambda)d\lambda\]where $\lambda_a$ is the shortest wavelength of the light emitted by the light source ($m$), $\lambda_b$ is the wavelength of the absorption edge of photoelectrode ($m$), $e$ is the charge of single electron ($1.602\times 10^{-19}\, C$), and $\lambda$ is the wavelength of the incident monochromatic light ($m$). Flux($\lambda$) is the photon flux of the light source ($m^{-2}\, s^{-1}\, nm^{-1}$), and it is calculated according to:
\[\textrm{Flux}(\lambda)=P(\lambda)E(\lambda)\]where $P(\lambda)$ is the power of incident photons ($W\, m^{-2}\, nm^{-1}$). $E(\lambda)$ is the photon energy ($J$) and it is calculated from
\[E(\lambda)=hc/\lambda\]where $h$ is Planck’s constant ($6.626\times 10^{-34}\, J\, s$), $c$ is the speed of light ($3\times 10^8\, m\, s^{-1}$).
Thus,
\[J_\textrm{max}=e\times \int \limits_{\lambda_a}^{\lambda_b}\frac{P(\lambda)\lambda}{hc}d\lambda\]Example
Figure 1 (left) AM1.5 spectrum. (right) Converted current per area and time and wavelength interval (black) and maximum attainable photocurrent for a given bandgap (red), assuming an AM1.5 spectrum.
Reference:
http://www.superstrate.net/pv/limit/