# 1 electron = 1 photon?

Started Apr 12, 2013 | Discussions thread
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Photon rate vs Optical Power; Photon E vs bandgap
4

Jack Hogan wrote:

Simplifying, assuming a modern B&W silicon sensor without a CFA or other filtering and 100% fill factor, if we see one electron produced at the output of a photosite, can we assume that it was the result of one photon making it through to silicon - and that the probability of it dislodging an electron is the Charge Collection Efficiency (QE) of the semiconductor at the wavelength of the incoming photon, so that for a given Exposure

Electrons produced = photons(wavelength) hitting silicon * QE(wavelength)?

If the visible spectrum is between 380 and 760 nm, with a near-infrared photon having half the frequency/energy of a near-ultraviolet photon therefore burying deeper into silicon, does the following responsivity curve merely represent QE at various silicon depths?

To represent QE, you need to adjust this curve for the photon rate as a function of wavelength.  At the same optical power, photon rate is proportional to wavelength since photon energy is inversely proportional to wavelength.

Making this adjustment, we obtain a curve that is quite flat between 600nm and 900nm; then the loss above and below this range is due to QE reduction.

The reduction at long wavelengths above 1000nm is due to the photons having insufficient energy to overcome the Silicon band-gap.  The reduction at shorter wavelengths below 600nm is due at least in part to the shallower absorption depth; at 400nm the graph is showing about 33% reduction in QE relative to the 600nm-900nm range.

Relative Number of electrons Generated as a Function of Impinging Photon Wavelength

Or could the fact that the responsivity at 760nm is more than 3 times that at 380nm mean that, for instance, sometimes 1 photon produces two electrons?

It would be the other way around since short-wavelength photons are higher energy.  Photons near 400nm do occasionally produce more than 1 electron in silicon, but the rate is low enough that it isn't a significant factor in the operation of our camera sensors.

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