 # Shot Noise - A question for the wise ...

Started Jul 21, 2012 | Discussions thread
Photoelectron Shot Noise and QE

Hello Detail Man,

QE has something to do with it. In order to understand this, suppose you illuminate a monochrome sensor with uniform, essentially monochromatic light (monochromatic enough so that it's QE is constant), such that you send in an average of n photons/pixel in one exposure time. By QE here, I'm using the usual definition of external quantum efficiency (i.e. = average electrons out/ average photons in). Let's call this average electron out = m, so that m = QE*n.

Photoelectron shot noise variance, V, as measured at the output, is given by:
V = QE^2 * n + QE * (1 - QE) * n

It is useful to look at these 2 parts in the sum, before adding them, in order to see what's going on. The first term is due to the fluctuations in the input light field itself, for normal light. This normal light is Poissonan, such that the variance in this photon count = the mean, i.e. = n, on input to the pixels , but when we measure this on output, it gets multiplied by QE^2. The second term is due to the imperfect conversion to electrons, i.e having a QE less than 1. This can be seen by regarding the incoming photon as having a probability = p = QE of generating an electron, as well as having a probability = (1-p) = (1-QE) of not generating an electron. This existence of the 2 probabilities is described as the so-called Bernoulli choice, and when you have n photons with this choice, this generates a Binomial distribution, whose variance is given by p(1-p)n (see for example Wikipedia on the Binomial distribution and its variance).

When the QE is low, say .1 or so, the first term, the "in the light" noise is much lower than the second term. As we go higher in QE, the first term becomes more important, such that as QE nears 1, then there is almost no uncertainty in generating the electron, and the first term dominates.

The first term can be reduced if we could generate this sub-Poissonian light, and this is an active area of research in the field of Quantum Optics.

After discussing all this, we can go ahead and do the sum:
V = QE^2*n + QE*n - QE^2 * n = QE*n = m

which is our old frield that says that our output noise variance, V, is equal to the output average electron count, m.

If you want to see some references, I list some here: