1 electron = 1 photon?

Started Apr 12, 2013 | Discussions thread
OP Jack Hogan Veteran Member • Posts: 6,645
Re: Photon rate vs Optical Power: photometric or radiometric values

alanr0 wrote:

QE and irradiance both vary with wavelength.  To calculate the average QE, sum the product of QE and spectral irradiance over the wavelength range of interest, and divide by the sum of spectral irradiance over the same wavelength range.

In general, if the irradiance spectrum changes, then so does the average QE.  You will need to re-calculate the number of photons incident, and the number of electrons released.

How do you decide what constitutes an 'exposure related photon'.  Does a 750 nm photon count? The human eye can (just) detect such wavelengths, but is around 8000 times more sensitive to green light. The sensor's QE will vary strongly in this region too, especially if an IR blocking filter is installed.  If you specify a sharp spectral cut-off, the number of 'related photons' you calculate will depend on the cut-off wavelength selected.  For black body illumination, increasing the upper wavelength of the summation from 700 to 800 nm would increase the number of photons counted, without changing the number of electrons generated.

Yeah, good point.  I am interested in what the sensor would see and count, so I guess that a 750nm photon counts if it makes it through the CFA et al.

2) Intuitively I would use floating point math for this operation.  Is there a good reason why one should use integer math instead?

I can't think of a good reason.  The transmission and conversion probabilities are all fractional.

I agree, yet not everyone does, pointing for instance to the discontinuity implicit in the conversion from   photons to electrons and back, unity gain being one area where the discontinuity would be minimized.  Is this relevant to a photographer, or does the effect get swamped by other statistics?

For instance if I do this for a D4 and D50 or so light off of DxO SNR curves, I get an effective QE of 15.9%, which for 7.01 e-/ADU at ISO 100 mean 44.11 photons/ADU.  electron Unity Gain is around ISO 700 and photon unity gain is around ISO 4400 - whatever the value of these gains is.

I assume these values are averaged over all types of sensel (R, G, B).  Are these 'green equivalent' spectrally weighted photons, or are you simply counting photons over the full visual spectrum?

Are you looking for precise photon numbers for some specific purpose, or just interested in the relative efficiency of D4/D50 and other sensors?

The DxO curves are linked to ISO speed, so presumably use ISO 7589 D55 (daylight) illumination.  ISO speed is based on photometrically weighted illuminance (lux) values, rather than radiometric irradiance (W/m^2) values.  If you want pure unweighted photon numbers then you need to acount for photometric weighting as well as the illumination spectrum and the wavelength dependence of QE.


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Alan Robinson

I'd like to be able to calculate the number of arriving photons from the measured number of electrons to cross check the expected photon arrivals from Nakamura, and therefore make sure that I understand the process - but I am currently cheating, calculating average effective QE from the electron FWC and the supposed number of incident photons for the given light spectrum.

DxO says that they use daylight lights for their tests, I assume that means something close to 5000-5500K (what I referred to as D50 above).  From Nakamura, that would mean about 13366 photons per lux-s-micron^2.  Since the D4 has a pixel pitch of about 7.21 microns, and DxO measures its Hsat at camera ISO100 as 1.04 lux-s (Ssat of 75), that means that at saturation each photosite in this light sees

13366 photons/lx-s-um^2 * 1.04 lx-s * 7.21^2 um^2 = 722612 photons

From the electron end, we know that DxO calculates its Full SNR Curves combining the four channels into a single gray level.  I assume that it is an arithmetic mean (does anybody know if they weigh the channels?), and from there extract full well count at various camera ISOs from a portion of the curve supposedly affected by shot noise only.  At camera ISO100 I estimate FWC for the D4 to be around a normalized 114871 electrons.  So now we can calculate the effective average QE:

Effective QE = 114871 electrons / 722612 photons = 15.9%

Here are some of the numbers:

D4 Data Extracted from Dxomark.com Full SNR Curves

It would be interesting to break effective QE down further to understand the main contributors from our data's perspective.  For instance, if we assume that only about 1/3 of the available photons make it through the color filter (a reasonable assumption?), then we could say that charge collection efficiency and other factors other than the CFA would result in an effective QE of approximately 48%.  If we assume that charge collection efficiency is somewhere around 80%, then the effective QE for perfect charge collection efficiency would be around 60%.  We still have to account for a 40% loss of efficiency

Thank you very much for your input Alan, always spot on and quite welcome.


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