# What is the relationship between dark current and the contribution of dark noise in a photodiode?

For example, if I have a datasheet for a photodiode that lists a dark current of 1000nA. I have a transimpedance amplifier with 1kohm transimpedance and 1 mV voltage output voltage noise.

Presumably the specified dark current is the current mean value. Clearly the dark current will result in a mean output of 1000ohms*1000nA = 1mV. Dark current is also Poisson distributed, but does that mean that the noise variance also 1 mV such that the voltage noise from the amp and photodiode are equal? Or is the relationship more complicated?

• Poisson processes, with large enough numbers of them, integrate into Gaussian distributions. Saying it is Poisson-distributed should allow you to infer that the statistics of large numbers of random variations behave in ways predictable using the standard deviation and mean mathematics. Dark current is Poisson because there are no long-range forces felt by charges across a PN junction, so the random quantum events will be Poisson in nature. This is called shot noise for photodiodes (and any PN junction with a current.) It's not complicated. – jonk Apr 28 at 20:45
• You can reduce the dark current by zeroing out the voltage across the photodiode (use a virtual ground and perhaps add a DAC and some circuitry to null it more precisely) and by cooling the junction. You can reduce the shot noise by cooling it and by limiting the bandwidth. Photodiodes used at low enough photon fluxes will experience a new source of noise from "photon flocking" (a tendency for bosons to clump up into the same exact quantum state.) But you have to get down below femptoamps of generated current for it to become a new, significant issue. – jonk Apr 28 at 20:49

$$darknoise = \sqrt{2*q*Id*B}$$