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?

  • \$\begingroup\$ 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. \$\endgroup\$ – The Photon Sep 1 '19 at 18:16
  • \$\begingroup\$ @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? \$\endgroup\$ – Junius Sep 9 '19 at 18:43
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – The Photon Sep 9 '19 at 19:08
  • \$\begingroup\$ 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 ? \$\endgroup\$ – Junius Sep 9 '19 at 19:28
  • \$\begingroup\$ 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. \$\endgroup\$ – The Photon Sep 9 '19 at 20:42

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.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ To make them more efficient, they are often (usually?) PIN diodes with a much wider depletion region. \$\endgroup\$ – Peter Smith Sep 1 '19 at 16:02
  • \$\begingroup\$ In which case it would act like a current source in parallel with that diode kept in a dark room. \$\endgroup\$ – TimWescott Sep 1 '19 at 19:35
  • \$\begingroup\$ I think that answers my question although the explanations in the wikipedia section you referenced in your answer is a bit too complicated for me to fully understand. The cause for the depletion region and the E-Field caused by the depletion region (at any PN junction) is the "tunneling/diffusing" of electrons into the p doped region and holes diffusing into the N region, right? \$\endgroup\$ – Junius Sep 9 '19 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.