# Electronic – Photon count with ADC, am I doing it right

I have this LED and I want to characterize this photodiode.

The LED should produce light of 370 nm, but I measured it and it is around 358 nm, and at that wavelength the quantum efficiency ($$\\eta\$$) of the photodiode is 0.5141.

Now, I feed the photodiode signal to this ADC, I got data and I turned it into a histogram.

If you are familiar with photon statistics then you know that for poissonian light the mean of those histograms should be equal to their sigma squared, for example in one case I have:

$$\\mu\$$ = 19,462.2

$$\\sigma\$$ = 135.785

$$\\sigma^2\$$ = 18,437.56

close enough I guess.

If I understood correctly this book(eq. 5.56 and 5.57) then the number of photons should be given by this equation:

$$\n_{photons}\$$ = $$\\frac{\mu}{\eta}\$$

which in this case would give me some 37,856.83 photons, which seems like a very low number…

Then I started decreasing the voltage and getting things like this:

$$\\mu\$$ = 7123.51

$$\\sigma\$$ = 135.7

$$\\sigma^2\$$ = 18961.29

This is called super-poissonian light. Here to get the photon count I need the general formula (eq. 5.57):

$$\\sigma_{counts}^2\$$ = $$\\eta^2 (\sigma_{photons})^2 + \eta(1+\eta)\mu_{photons}\$$

in this case I know 2 of the values but I also have 2 unknowns and as far as I can see no other equation or formula that I could use.

So basically I'm wondering:

1.- If I'm using the formula for the amount of photons correctly

2.- What should I do when the light becomes super-poisonian?