What is the effect of incident light on the junction capacitance of photodiodes?

Question

My background understanding: Light incident on the depletion region of a photodiode excites electrons creating electron-hole pairs which are then swept to the edges of the depletion region by the electric field of the depletion region.

Question: Do the new charges at the edge of the depletion region make the depletion region more narrow therefore increasing the junction capacitance? Would this effect be noticeable in something like an avalanche photodiode where there are a lot more free charges?

I am just curious about photodiodes. Especially their effect on the frequency response of amplifiers.

Thanks.

Answer 1

The answer depends on the current/voltage ($I/V$) conditions you apply to the terminals of the photodiode during illumination:

(1) In an open circuit diode ($I=0$), a forward (photo-) voltage is created by the generated electrons and holes flowing to the n- and p-type side of the junction. Thus the depletion region will become narrower and the capacitance increases.

(2) In a short circuit photodiode ($V=0$), an outside photocurrent is flowing from the p- to the n-terminal while the built-in potential drop across the depletion zone is unchanged. Therefore the thickness of the depletion region and capacitance of the diode remain constant.

(3) An avalanche photodiode is operated under reverse bias in the bias region near the breakdown voltage where a high electric field in the depletion region causes charge carrier multiplication and thus a strong photocurrent gain. If a resistive load is used, a small decrease of the reverse bias voltage and depletion region thickness should occur leading to a capacitance increase.

Note: Any effects of diffusion capacitances are neglected in this answer.