# Photodiodes: the “generation” mechanism behind photocurrent

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?

• I'm not sure if this answers your question, but the photodiode absorbs energy from the EM field (i.e. photons). This means it can have an internal emf of its own. – The Photon Sep 1 '19 at 18:16
• @ThePhoton: Hmm, well the photodiode absorbs energy but that energy creates a electron/hole pair, right? I assume that the energy is absorbed is already consumed by the process that separates the electron out of the material, so there is nothing left? – Junius Sep 9 '19 at 18:43
• The electron and hole will each typically have some excess energy beyond what's needed to separate them. A solar panel is just a big photodiode. – The Photon Sep 9 '19 at 19:08
• One last question: The current of the current source in the equivalent model (i.e. the photocurrent) is somehow proportional to the amount/energy of incident light. Since the equivalent model involves an ideal diode in parallel to the current source, that means that the forward voltage of the photodiode is limited to about 0.7 volts. So the maximum current of the diode (and therefore the max. amount of electrical energy delivered by the diode) is limited by the current of the photodiode's I/U curve evaluated at U = 0.7V ? – Junius Sep 9 '19 at 19:28
• The orientation of the diode in the model should be so that photocurrent goes out from the cathode terminal, so there's no such limit. See this old question for a comparison of the I-V curve with and without optical power applied. – The Photon Sep 9 '19 at 20:42