Assume I use a photodiode under zero-bias (that is in photovoltaic mode). How/why can a current flow in this bias condition?
As far as I have understood the working principle of photodiodes, when photons hit the depletion region of the P/N junction, the energy absorbed causes the creation of electron/hole pairs. Due to the electric field in the PN junction those charge carriers are separated and "flow out" at the terminals of the diode.
This somehow makes sense to me when a diode is used in photoconductive mode, where an external voltage source provides an electromotive force (and therefore an E field in the diode), but under zero-bias, there is no E field inside the diode that could "pull-apart" the created electron/hole pairs?
Best Answer
Actually, there is an electric field in the depletion region in a diode with zero bias. Depending on what problem you're trying to solve at the moment, it's either the cause of the depletion region, or a necessary effect of the cause of the depletion region. Either way, it sweeps the depletion region free of carriers.
So when a photon smacks into a diode junction and succeeds at creating a hole/electron pair, the usual consequence is that the hole is swept into the cathode, the electron is swept into the anode. Thus, light drives current.
In fact, a pretty good model for a photodiode of any kind, whether it's an itty bitty super-fast diode for receiving laser pulses, or a component of a solar cell, is a plain old diode in parallel with a current source. The actual photocurrent is fairly constant (I'm sure the actual number of electron/hole pairs generated per photon varies somewhat, but to a 1st-order approximation it ain't much). Most of the differences in photodiode behavior as a function of operating point have to do with the behavior of that virtual "dark diode" that's in parallel with the current source.
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