• Removed user
    April 17, 2023, 4:37 p.m.

    This is very interesting. Check my understanding of this. Setting aside ISO for the moment, in order for the exposure for your "A" example to be equal to "B", the f-stop would need to increase or open up (proportionally?) because A's shutter speed increased. So it would be something like this...

    A. f/5.6, 1/400s
    B. f/8, 1/200s

    Have I got that right?

  • April 17, 2023, 4:59 p.m.

    That is my understanding; but what do I know? :) :) :)

    David

  • Removed user
    April 17, 2023, 5:03 p.m.

    I'm not sure what I know either. LOL

  • April 17, 2023, 5:03 p.m.

    And here we don't have 150 posts limit either :)

  • April 17, 2023, 5:08 p.m.

    Actually, I havent used flash since getting a Canon R6. I dont do any "studio" photography (portraits, etc), and apart from the fuschias on my windowsill and shots of equipment that I build, my photos use available light deliberately. I guess I am a "purist"! :)

    David

  • Members 976 posts
    April 17, 2023, 5:10 p.m.

    ISO can be set aside indefinitely for this, but the light can't ;)
    For the same light, ...

  • April 17, 2023, 5:12 p.m.

    The problem that analogies can be misleading. I suspect that David isn't aware of photon shot noise and its effect on imaging.

  • April 17, 2023, 5:14 p.m.

    No -- I am not a Physicist, just a humble engineer.

    Sorry! You have lost me completely here...

    David

  • Members 1737 posts
    April 17, 2023, 5:19 p.m.

    Certainly, in most photographs the main source of noise is photon noise.

    But I was talking about read noise. What I said about read noise is true in the absence of photons, and therefore the absence of photon noise, although the traditional way to make PTCs requires photons.

  • Members 1737 posts
    April 17, 2023, 5:21 p.m.

    Does this help?

    Screenshot 2023-04-13 100652.png

    Screenshot 2023-04-13 100652.png

    PNG, 86.9 KB, uploaded by JimKasson on April 17, 2023.

  • April 17, 2023, 5:23 p.m.

    Bob,

    The problem is that the people who understand all this explain it in terms that others like me cannot follow. Your last sentence is true, because photon shot noise is unfortunately far too high above my head! As I said, I am not a Physicist and have not studied theories of light and its propagation beyond A-level (lenses and prisms, etc!) :) (I have a BSc in Electrical and Electronic Eng and do have some idea of how lasers work, but that is a different subject.)

    Am I unique among the members of the forum?

    Best,

    David

  • Members 457 posts
    April 17, 2023, 5:41 p.m.

    Not long ago, my knowledge of this topic was probably less than yours at the moment. With the help of people who were on DPR and (mostly) moved to DPReviewed, I learned a lot.
    Here is a good (and correct) introduction to noise in digital photography:
    www.photonstophotos.net/Emil%20Martinec/noise.html

  • April 17, 2023, 5:42 p.m.

    Hi David, if you have a BSc in Electrical and Electronic engineering I'm really surprised that shot noise wasn't covered somewhere in the curriculum. Still, that appears to be the case. I think I for one, have tried to explain photonshot noise in layman's terms many times. As have many others. The problem in this type of discussion is knowing the level to pitch the discussion. Trying to explain these things to beginners I would always give a brief explanation of what shot noise is - simply the noise caused by the randomness of arrival of quantum particles - in the case of photon shot noise, photons. In electronics it tends to be electron shot noise that is taught, but it's the same thing apart from the particle responsible. However, when one is engaged with someone who is not a beginner, it seems to be disrespectful to assume a level of ignorance of the concepts, and very much gets treated like that by some people. I blame the people who spread the myth of 'ISO noise' when they didn't know anything (it seems ) about the sources of noise in imaging.
    Anyhow, for what it's worth, the dominant source of noise in imaging is photon shot noise, which is the noise caused by the randomness of the photons that build up the image. The SNR of shot noise is the square root of the number of photons counted in a nominal area of the image - for instance, if you're measuring noise per square millimetre of the image you count the photons over a square millimetre. If you want to get your head around how it works, look out the window at the ground when it's raining softly. At the beginning, when there are few raindrops, you see a random, spotty pattern (noisy). As more drops get collected that pattern becomes smoother (less noisy). When the ground is saturated, you just get a smooth layer of water, and you can no longer differentiate differences between the amount of raindrops in different patches of the ground.

  • Members 976 posts
    April 17, 2023, 5:46 p.m.

    Walter Schottky?

  • April 17, 2023, 5:47 p.m.

    Yes, I suspect that David was assuming that read noise is the dominant source of noise, and therefore assumed that the additional noise that appears to come with high ISO settings must be generated in amplifiers, because he couldn't see anywhere else it would come from. For an electronic engineer, which he is, he should be able to understand the effect of variable gain as a case of stage matching, for the same reason that RF systems generally have an AGC.

  • Members 273 posts
    April 17, 2023, 5:47 p.m.

    Bachelors and Masters in EE here. Shot noise was not covered in my curriculum. In fact, the majority of what I learned about, well, everything, I learned outside of any school curriculum.

  • April 17, 2023, 5:50 p.m.

    I'd have thought it was rather important in a whole range of EE topics. I'm amazed.

  • Members 976 posts
    April 17, 2023, 5:51 p.m.

    That's amazing.