4

Assuming that camera manufacturers can eventually make the perfect sensor that introduces no electrical noise to the signal, at what point (measured in ISO/area of a pixel) will shot noise be so prevalent that you cease to get useful images?

Improve this question
7
  • I'm having trouble figuring out what limits DxOMark uses to determine the ISO rating of a sensor. I think they use a combination of dynamic range and SNR.
    – Evan Krall
    Oct 16, 2011 at 22:34
  • Are you serious about the "no electrical noise", or are you really interested in what is physically possible with "as low an amount of electrical noise as is physically possible"? I don't think it is possible to produce a sensor that has zero noise generated by the electronics (which you seem to imply by the statement "mfg. can eventually make the perfect sensor"), and I think the distinction would affect answers. Even an extremely low amount of thermal noise will exhibit when amplified enough, thereby imposing a limit on maximum ISO.
    – jrista
    Oct 17, 2011 at 4:38
  • This page might be useful, as it has some mathematical formulas for calculating SNR (given you know enough other information): learn.hamamatsu.com/articles/ccdsnr.html. Its for CCD's, and I am not sure what the differences for thermal and read noise might be between CCD and CMOS sensors, if there are any (I would assume there are, given the differences between the two technologies).
    – jrista
    Oct 17, 2011 at 4:44
  • @jrista you could build a sensor out of photon detectors, instead of measuring charge released by incident photons, thus achieving zero electrical noise. Whether it would ever be commercially viable to produce such a sensor with sufficient resolution is another question...
    – Matt Grum
    Oct 17, 2011 at 12:36
  • @MattGrum: Even with photo detectors, you still have an electronic gate that is responsible for turning detected photons into something digital that a computer can process. I guess if we could find a way to directly process everything as an analog signal, perhaps with quantum processors, we might be able to eliminate electronic noise in the sensor. Regardless, I guess I was thinking in terms of what is economically viable from a consumer standpoint...but who knows, maybe fifty years from now we will be able to have entirely analog processing systems.
    – jrista
    Oct 17, 2011 at 20:38

2 Answers 2

Reset to default
8

Here's a good simulation of the 'perfect' sensor that you describe (one having zero electrical noise, thus recordning every incident photon perfectly) reacting to widely differing levels of light, from 0.001, 0.01 & 0.1 photons per pixel (top row), 1, 10 & 100 photons per pixel (middle row) to 1000, 10000 & 100000 photons per pixel across the bottom row.

Click for a larger version where you can make out individual pixels. Image by Mdf some rights reserved.

You can't specificy ISO sensitivity without knowing the saturation point of the sensor (without saturation there is no overexposure), so for your hypothetical 'perfect sensor' you'd have to chose an abitrary saturation point, making the ISO values computed for the images arbitrary too.

However to answer your question it appears about 1 photon per pixel is the limit of getting meaningful images (the top right image, with 1 photon every 10 pixels, looks unrecognisable to me).

Improve this answer
1
  • I would say that the one-to-one image was unusably poor (and that makes intuitive sense, due to randomness)
    – MikeB
    Jul 5, 2019 at 15:28
2

It's hard to provide a definitive answer to your question, because it depends on what you subjectively deem “useful”, as well as on many other factors like the strength and quality of the denoising algorithm, the output medium, etc... Thus, this is not really an answer, only some hints to help you find your answer.

First, about the parameters to consider. The area of the pixel is not really important. Although smaller pixels give more pixel-level noise, you can always reduce the noise by downscaling the image (a side benefit of antialiasing). The really important parameter is the total area of the sensor. Also, readout noise is usually only relevant in the darkest areas of the image, most of the time the main noise source is photon shot noise. Thus, quantum efficiency and fill factor are also important.

Next, I recommend you take a look at this page from ClarkVision.com: http://www.clarkvision.com/articles/digital.signal.to.noise/. It is a very well documented discussion about digital noise, together with models and lots of real-life data.

Improve this answer
4
  • I've no idea what I was talking about, now I've reread you answer it makes perfect sense so I've deleted the comment!
    – Matt Grum
    Oct 17, 2011 at 21:12
  • I disagree with the idea that "the area of the pixel is not really important". The area of the pixel determines how many photons hit each pixel, for a given amount of light coming in through the lens. As you say, you can reduce the noise by downscaling the image, but by downscaling the image 2x, you're effectively increasing pixel area 4x.
    – Evan Krall
    Oct 18, 2011 at 3:26
  • There's a reason film grain size increased with ISO.
    – Evan Krall
    Oct 18, 2011 at 3:34
  • @Evan Krall: My assumption is that most of the time the resolution of the camera is more than enough for the output medium. So you (or your display software, or your printer driver...) will end up downscaling anyway. After downscaling, the effective pixel area is the total area of the sensor divided by the effective resolution of the output medium. The resolution of the sensor and its own pixel area are irrelevant.
    – Edgar Bonet
    Oct 23, 2011 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.