Are smaller sensors more efficient...

Started Nov 2, 2013 | Discussions thread
Jack Hogan Veteran Member • Posts: 6,982
SNR: 3dB or 6 dB per octave/stop?

Steen Bay wrote:

Jack Hogan wrote:

Steen Bay wrote:

Jack Hogan wrote:

Anders W wrote: You and Steen are talking past each other Jack. Steen already knows what you are telling him above. But what he is trying to say is that you need to increase exposure by 2 stops in order to double the SNR, i.e., make it increase by 6 dB. The relationship between exposure and SNR is nonlinear (quadratic) whereas that beween exposure and DR is linear.

I know that. But since the curves above are displayed according to 'Measured ISO', and Measured ISO is actually an absolute exposure (Hsat=78/Ssat in lx-s), if for a given , unchanged Measured ISO (Exposure), say 400, one curve is higher than the other by 2 dB that's a difference of 1/3 of stop in SNR, not 2/3 of a stop, correct?

Don't ISO 100 have 1 stop better SNR than ISO 200, even though the difference only is 3 dB?

Yes because that's a doubling in exposure. In the graph above however we are comparing SNR at the same exposure, that's the whole point of Ssat.

OK, what is the difference in SNR then between mFT and FF compared at the same exposure (assuming same QE)? Wouldn't you say that the difference is two stops, even though there only is a 6 dB difference?

You know, I gotta remember that we are not trying to answer the straightforward question

'How many dBs is a doubling of SNR?'

for which the earlier wikipedia article and this page on noise at DxO are appropriate as they are - and will result in the correct conclusion that a doubling of SNR corresponds to 6 dB = 20log(2)

SNR = mean/stdev

As photographers we are implicitly trying to answer a different question altogether:

'How many stops would I need to change Exposure (ss/aperture) to obtain this new vs old SNR ratio?'

This question is harder to answer because while the mean signal is linear with Exposure (i.e. double one the other doubles), standard deviation (noise) is non-linear with Exposure and so it depends. For the purposes of the SNR18% data, we can make the simplifying assumption that noise around 18% of Full Scale should be mostly shot noise which varies with the square root of the signal - and so will the SNR. Therefore if the Exposure (ss/aperture) doubles, the relative SNR will decrease by 3dB = 20log(1/sqrt(2)).

So we are both right: a doubling of SNR corresponds to 6 dB; and when one changes ss/aperture by one stop shot noise SNR should change by 3dB.

But for a photographer the latter is more relevant, so you are righter: in order to match the SNR18% performance of a future FF sensor full of BSI photosites, the owner of a D610 would have to increase shutter speed or aperture by 2/3 of a stop. As a landscaper I am starting to take notice

Jack

PS Incidentally, about Effective QE, assuming two SNR18% shot noise dominated curves with the caveats above, when the Exposure is the same (as indicated by reading the graph at the same Ssat/'Measured ISO') the ratio of the two SNRs is simply the ratio of the two standard deviations.  Squaring it will give us the ratio of the number of electrons generated by the two sensors. If the Exposure (Ssat) is the same, we know that the number of impinging photons per equal sized photosite should be the same, so the difference in e- generated must be due to a difference in Effective QE, which therefore doubles/halves for a 3dB change in SNR18% aotbe.

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