Detail Man
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(I think) that range of maximum SNR will only increase with higher full-well-capacties and/or higher Quantum Efficiencies. bobn2 is the "guru" on that - perhaps he will see this post and comment.Thanks for the explanation. I knew most of these things, but when I asked for why raw levels of high ISO (higher amplification, shorter exposure time) images can be higher the missing link was that shot noise indeed does have a temporal component, which in turn is amplified/multiplied by the following analog gain stage. It's the fact that within half the time you can get more than half the electrons that wasn't clear to me when I asked.
How static has that limit been for the last couple of years and is there a realistic chance to increase this range anytime soon?As you can see from DxOMark's data, the SNR of photo-cells is on the order of only around 40 dB.
Not really sure quite what you are asking. The SNR of the MOSFET amplifier technology (not likely to change a great deal) is around 80 db - which is vastly higher (the linear square ) of the SNR of the photo-cells at there very best (with full illumination due to the highest exposure, lowest F-Number and lowest Shutter Speed).Interesting. Within what border does amplification of that 40 dB window happen within the 80 dB range, does it happen close to the clipping point (away from the noise) or rather somewhat below?The MOSFET amplifiers that interface to the photo-cells with each MOS image-sensor photo-site have their own noise (which is on the order of around -80 dB below the voltage signal level that they can roughly linearly process). They also add more (summed) capacitance to the photo-cells that they interface to. bobn2 tells us that it is that (input) capacitance of the MOSFET amplifiers which typically "dominates" over the capactiance of the photo-cells themselves.
As ISO Gain increases, the Exposure has to decrease (in order to retain the same "image-brightness"). As a result the SNR of the photo-cells decreasing in those cases (of higher ISO Gains), the 80 dB SNR of the MOSFET amplifiers is always vastly higher than that of the optical photo-cell transducers themselves.
The degradation of the output SNR of the system is (essentially, only) a direct result of the decreased SNR of the photo-cells in lower light levels (and the lower exposure that results from lower light-levels). The only way to improve upon that is by having lenses with lower T-numbers, or the ability to use longer Shutter Times (necessitating tripod-stabilization and low levels of image-sensor dark-noise and thermal-noise sources). With less photons for the image-sensor, those are the only options left to increase Exposure (which does not in itself include the ISO Gain at all).
It's not unlike when trying to use a high-quality microphone and pre-amplifier to record very faint low-level sounds. The SNR of the recording (even with the best equipment) will suffer more and more the lower the sound-pressure-level (SPL) to be recorded.
Thanks for pointing that out again. As you might have guessed I don't care too much for the math, because I usually never really need it. But of course this makes me less of a profound part of an discussion based on these math and the accompanying understanding. There's just not enough room in my head for everything (including remembering things I already learned), so I take the practical route and might have to ask from time to time. Also helps to keep the mind open, but I recognize the additional effort for people I ask these things about. Sorry for that.It's just the square-root function. The Photon Shot Noise varies by that relationship as the amount of light (or the intensity of light per unit area, or the total amount of light transduced by an image-sensor), varies. The square-root of 2 is 1.414, square-root of 4 is 2, square-root of 16 is 4, etc. ...![]()