1 electron = 1 photon?

Started Apr 12, 2013 | Discussions thread
Senior MemberPosts: 1,260
Average QE without integer arithmetic
In reply to Jack Hogan, Apr 15, 2013

Jack Hogan wrote:

alanr0 wrote:

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?

Well, an individual photon either gets converted or it doesn't, but what happens to a single photon is not a very precise predictor of what happens to 10,000 photons.  You can run QED calculations for each photon, effectively rolling the dice 10,000 times (probably needing floating point calculations along the way) for an uncertainty of a few percent in the total number of electrons.  I may be missing something, but I don't see any advantage in such Monte-Carlo calculations over a straightforward floating point calculation of average conversion efficiency.

So I don't see anything special happening at unity gain.  With a few dozen electrons or more, electron statistics (shot noise) will dominate.  At lower signal levels, the 'electron counting' argument is only valid if ADC levels align precisely with the sensor black point for every pixel.  I agree that higher ADC gains will deliver diminishing returns from this point, but there is nothing magic about unit gain ISO.

Without digging deeper, I am happy to follow John Sheehy, who seems to prefer prefer measurement and sound analysis to vague arm-waving arguments about the way the world ought to work.

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).

Oops, so this rather than Nikon's D50.

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

I have not been through your calculations in detail, but does this help with the missing 40%?

For 5500K black body radiation, energy is distributed more or less equally into bands 400-500, 500-600, 600-700 nm (give or take 10% or so, see table in Kaye & Laby, or image here).

Assume 48% QE for 555 nm light incident on each green sensor, in line with your analysis, and only a little better than the KAF 8300 data sheet.  The green filter width at 50% of maximum transmission is approximately 100 nm, so roughly 1/3 of the available light is captured (compared with monochromatic green light at 555 nm).

Hence effective QE for green sensels exposed to 'white' light is 48%/3 = 16%.

Blue QE will be less (fewer photons per Joule), Red QE could be similar or higher (more photons per Joule + additional 700-750 nm energy captured).  I have also ignored 'out-of-band' sensitivity (spectral crosstalk) which will increase the total number of photons collected for all channels.

Keeping it simple, QE for each colour is around 16%, so average QE for sensor is also 16%.


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Alan Robinson

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