Open circuit photodiode energy

Question

Where does a photo-diode use the energy it collects from light and other EM waves when not loaded? Does it dissipate it as heat, reflect it or something else?

Answer 1

Where does a photo-diode use the energy it collects from light and other EM waves when not loaded?

Short answer: it doesn't.

Voltage is potential energy - energy that is there, ready to do work. It only does work if a load is attached. If no load is attached, then no work is done.

Does it dissipate it as heat, reflect it or something else?

Certain wavelengths of photons striking the PD junction cause electrons/holes to migrate and polarize, which we perceive as electrical voltage potential.

There will be some small amount of heating from the incident photons and conversion. Those photons are absorbed. Some input photons may incite localized recombinations, resulting in a very small amount of fluorescence. This intensity is much less than the irradiating intensity so would be challenging to measure. More input photons = more heat.

There will be no heating from current flow since there is no load.

When the photon energy source is removed, the migrated electrons/holes will recombine via the leakage resistance of the junction. This will create a miniscule amount of heating.

As far as I know, there has been no mention of "increased reflectivity" or other physical changes between loaded and unloaded junctions. Perhaps there could be some slight change in self-emission, but this is speculation.

Also, a PD will emit light in the wavelength of it's bandgap voltage (self-fluoresce) if externally forward-biased like a typical LED; although this is a very inefficient process because they were not designed for emission.

Answer 2

It's simple: given no external circuit to flow through, the electron-hole pairs created by the photoelectric effect recombine, giving up their energy to lattice vibrations (heat). Thus, an unloaded solar panel runs hotter than a loaded one.

Answer 3

First of all, not all wavelengths can generate a current flow. Depending on the physical properties, PDs can be "transparent" to some wavelengths.

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The answer to your main question is simply where the Open Circuit Voltage, V_OC, of solar cells (or PDs in general) comes from.

V_OC is the maximum voltage that a PD can generate when illuminated. It's related to temperature, dark current, illumination, and some other factors. The equation is the same with Shockley's diode equation:

$$ V_{OC}=K\cdot \ln\Big(1+\frac{I_L}{I_s}\Big) $$

\$K\$: Includes temperature and some other physical parameters

\$I_s\$: Dark current or reverse saturation current

\$I_L\$: The light-generated current

As the light-generated current increases the voltage developed across the PD increases.