What compact digicam has the biggest CCD pixels?

Discussion in 'Digital Photography' started by Paul Rubin, Apr 22, 2005.

  1. It must be we are looking a different problems. I work from a
    photographers point of view. example: I have a camera
    lens, say 500 mm f/4 for photographing wildlife. I must have
    a 1/2000 second exposure (or faster) for certain action stopping
    photos. A real life example:

    At a given ISO: X photons are delivered per square micron to
    the focal plane. If my pixels are 8 microns square I get
    64X photons in a 1/2000 second exposure.

    Put the same lens on another camera and in the
    same light conditions and shoot at an exposure time of
    1/2000 second, and if the pixels are 4 microns square, I
    get 16X photons per pixel.

    Simple physics. Simple math. Pixel area is proportional
    to photons collected in each pixel per unit time.
    Signal-to-noise drops as pixel size decreases because
    you collect less photons per pixel per unit time.

    How do photons per pixel work in your world?

    Roger N. Clark (change username to rnclark), Apr 30, 2005
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  2. Aha, Perhaps I see were we differ in understanding what each
    is saying regarding this issue. Correct me if this is not
    your position.

    When you say "theoretical maximum" you say
    theoretical maximum is perfect everything: transmission of all
    optics is 100% and quantum efficiency is 100% at all

    I say on my web page for the Canon 1D Mark II camera:
    "Figure 1. The signal-to-noise of the Canon 1D Mark II camera
    at each of its ISO settings is compared to film and the
    theoretical maximum possible." Here I mean the theoretical
    maximum possible for this device with it's quantum efficiency.
    That means that noise in the entire electronics chain is
    negligible and Poisson Statistics due to photon (electron)
    counting dominates. Given real world electronics, this is
    the ideal one can hope for. That does not mean future
    sensors could not be better, but it does mean that for
    this class of devices, it will not get better. Note:
    this signal to noise is independent of any lens or blur
    or Bayer filter over the sensor.

    Analogy of why both our theoretical maxima arguments are valid:
    resolving two close stars in a telescope. Say I have an
    8-inch aperture telescope (I do). I can validly say on a given
    evening that I am resolving the stars at the theoretical maximum.
    You counter and say that is incorrect. The theoretical maximum
    is infinite resolution (just get an infinitely large telescope).

    Does this describe our positions?
    Roger N. Clark (change username to rnclark), Apr 30, 2005
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  3. [A complimentary Cc of this posting was sent to
    Roger N. Clark (change username to rnclark)
    Me too. All I say is:

    The smaller sensor is of correspondingly smaller ISO sensitivity (if
    one measures sensitivity via exposure needed to achieve given S/N ratio).
    Yes, it is obvious to anybody with minimal photography education: if
    you have smaller sensitivity sensor, you should use longer exposure;
    if longer exposure is not possible, your image is not going to be good.
    Easy: you get the same amount of photons per pixel with smaller pixels
    and larger exposure. What is hard to understand here? It is as if
    you compare 800ISO film with some "ancient" 100ISO film with the same
    granularity. 800ISO wins hands down; but it still not fair to say
    that it has "smaller" granularity.

    With digital, due to a possibility to trade sensitivity for noise,
    both view are possible:

    in some situations one can use 8x smaller sensor as one with 8x
    lower sensitivity;

    in some other situations one can use 8x smaller sensor as one with
    2.8x lower S/N.

    Hope this helps,
    Ilya Zakharevich, May 1, 2005
  4. [A complimentary Cc of this posting was sent to
    Roger N. Clark (change username to rnclark)
    Close, but not exactly:

    a) Since I consider the "sensor assembly" only, the optic is out of
    the equation (only IR filter, blur filter, and Bayer filter remain).

    b) The Bayer filter (by definition) should filter out some light. I
    assume that it filters as little as possible while keeping the
    demosaicing noise the same as with current sensors.

    For the remaining parts of the equation: IR filter, blur filter, and
    the photoelectric part of the sensor, the "theoretical maximum"
    assumes transparency=1 and QE=1 throughout the visible spectrum.
    (Although note that some variations from 1 are possible, since they
    may be considered as "a part of Bayer filter".)

    The "practical" estimate assumes transparency*QE = 0.8.
    As I said in a private email, from my point of view, this statement is
    a logical fallacy. Please correct me if my understanding is wrong:

    You deduce the "theoretical maximum" from measured data (by fitting
    the theoretic curve against the measured data). Now you look at the
    fit, and see a good one. From this you assume something about
    having a "theoretical maximum".

    All *I* can deduce from the fit is that the model of noise

    readout noise + Poisson noise + dark current noise

    is a good fit for the noise of the sensor. (Actually, you did not
    write what the "theoretic curves" are for. If they are for Poisson
    noise only, then the additional conclusion is that readout noise and
    dark current noise are low. But "we" know it already from your tables
    of readout noise and dark current; we do not need all these curves...)
    Well, I doubt that you will find more than a handful of readers of
    your site who got this impression from reading your statement. And
    how can you explain your own ardor in attacking my calculation of the
    photon count if you insist on this interpretation?
    Sorry, I have no idea what do you mean by "this class of devices"...
    By what your wrote above I'm forced to think that you mean "devices
    using exactly the same sensor assembly, but different ADC and/or
    firmware", but this can't be so...

    Ilya Zakharevich, May 1, 2005
  5. That isn't how ISO sensitivity of a sensor is determined. Please pick
    another term for what you're trying to say (e.g. "constant S/N
    sensitivity"), and make it clear that it is not equivalent to what most
    photographers think of as ISO-rated sensitivity.

    Now, when you talk about a digital camera as a system, it can vary the
    ISO by changing amplifier gain (or the equivalent on the digital side).
    In that context, it may make sense to talk about varying ISO, as long as
    everyone realizes that the sensor isn't changing at all.
    Here, you're talking about a 2.8X sensor size ratio, which gives a
    factor of 8 difference in pixel area. An "8X" smaller sensor would have
    1/64 the area per pixel. In photography, "X" usually refers to linear
    magnification, not area, unless you make it clear that you're comparing
    area. Having cleared that up:
    What sensitivity measure are you talking about? Normal ISO sensitivity
    is based on reaching some fraction of full well capacity, and since this
    capacity scales with pixel size, available camera sensors have similar
    ISO sensitivity whether the pixels are large or small. So you're
    comparing against a mythical lower-sensitivity sensor that isn't in use.
    Or if you're talking about "constant S/N sensitivity", the larger
    sensors have much *higher* sensitivity. Either way, this doesn't make
    sense when comparing existing sensors.

    Maybe you meant to say that in some situations you can use the 2.8X
    smaller sensor instead of the one with 8X higher sensitivity being
    operated at 8X its natural ISO because you had to stop down much more to
    get the same depth of field.
    Again, the larger sensor will have the *higher* S/N operated at its
    natural ISO, or the same S/N operated at it's "constant S/N
    sensitivity". It won't ever be used with a lower S/N. Do you really
    mean "higher S/N", or "lower noise" instead? Then your statement makes

    In that case, you're right about this tradeoff. But you *also* have to
    look at the useful range over which the tradeoff can operate.

    With the large (~8 um) pixel pitch of DSLR sensors, the useful tradeoff
    range is perhaps ISO 100-1600, limited on the low-ISO end by full well
    capacity and on the high-ISO end by increasing noise. This range of
    choice is good.

    When you reduce the pixel area by 8X, you get a sensor where the lower
    ISO limit is *still* 100 due to full well capacity issues, but where the
    upper ISO limit is 200 because it gives you the same noise as 1600 in
    the large-pitch sensor. The small physical size of sensor has its
    advantages, but you now have a very narrow range of useful ISO, and you
    basically can't use the camera in dim light or at high shutter speeds.

    If you were to reduce pixel area by another factor of 2, you'd end up
    with a sensor that has only one usable ISO: 100. And with even smaller
    pixels, the two limits actually cross, leaving no usable ISO at all.
    The sensor can't be used below ISO 100 without overflowing the small
    pixel wells, and even at ISO 100 the noise performance is worse than the
    DSLR sensor at 1600.

    You may say that more noise is acceptable. But no matter how much you
    reduce your S/N standards, there comes a point when shrinking CCDs that
    you cannot achieve an acceptable S/N.

    In other words, you can trade small sensor size against effective ISO
    while keeping S/N constant *only over a limited range*, and you have to
    pay attention to how the limits move as the pixel size changes.

    Dave Martindale, May 1, 2005
  6. Ah, here is a major difference. The digital camera manufacturers do not
    define ISO this way. I have not seen an actual specification,
    (anyone out there still reading know about this definition?),
    but it is clear that it is some percentage of electrons in the full
    well for a given exposure time with a given light intensity.
    A percentage of full well is the closest analogy for getting to a
    given density on film.

    In my opinion, we should use the same method of defining ISO as
    the manufacturers are using.

    Side note: the astronomical CCD suppliers seem to be using a definition of
    CCD ISO that is based on signal-to-noise compared to film. I have
    seen numbers of ISO 32000 and higher. This is not what the
    digital camera manufacturers are using.
    Nothing, except you changed the definition of the experiment, and in
    general are changing the definition of ISO with pixel size.
    I do not understand how this is relevant.
    It helps to see that different definitions are being used for ISO.

    The sensitivity of CMOS and CCD detectors do not change with pixel
    size. The quantum efficiencies of the devices is constant as a
    function of pixel size. The active area could change, but
    making small pixel less area (edges and in cmos, the on pixel
    transistors), but manufacturers have gotten beyond that issue by
    using micro lenses to focus light onto the active areas, giving
    essentially 100% active area.

    The original example I set up was keep the lens constant and just put
    in detector (CCDs or CMOS) with different pixel sizes. This is no
    different than putting a 50mm f/1.4 lens on a Canon 1D mark II,
    A Canon 20D (6 micron pixels), a Nikon D70, or some point and shoot
    camera. The lens delivers X photons per square micron per second.
    Smaller pixel cover less area so collect fewer photons. This is
    basic fundamental physics that will not change.

    Your response, correct me if I'm wrong, is to change the exposure
    time on the small sensor to collect more photons.
    You change the definition of ISO for the small sensor
    to give it an advantage. That is incorrect the way I see it,
    and was not the premise of the original problem, and is not
    how ISO is defined.

    Let's try and agree on one thing: Do you agree that given
    X photons per square micron per second incident on a surface,
    and given two "buckets" to collect photons, one Y microns
    on a side and the other 2Y microns, the 2Y bucket collects
    4 times as many photons per second as the Y bucket?

    Roger N. Clark (change username to rnclark), May 1, 2005
  7. Gee, Figure 2 is labeled "Canon 1D Mark II Signal to Noise
    compared to theoretical photon statistics."

    If they are for Poisson
    Yes, but many people do not read tables. A graph showing how the
    data fits is much more instructive.

    Digital cameras of a few years ago had much more noise. The reason is
    not a change in the QE of the basic detectors. It is due to improvements
    in the electronics processing the data that the CCD/CMOS chip collects.
    The QE of CCDs has not changed significantly in over 20 years. Yet digital
    cameras have improved steadily in the last few years. Why? Improvements in
    electronics. So one can analyze the noise in the electronics
    and the sensor. The best one can possibly do is get to Poisson
    statistics. Once Poisson statistics dominate the noise, you can no
    longer improve the results from that device by making further
    electronics improvements. You are at the theoretical maximum possible
    for that device. from the Canon 12D Mark II pro DSLR, to the canon
    S60 point and shoot show that they are at the maximum possible for
    the device, and by numbers others are deriving for other cameras,
    including CCD and CMOS cameras, they have all reached this limit.
    Electronics engineers could not improve on the noise in the camera
    by making further improvements to the electronics. That is when
    it is said the system is performing at its theoretical maximum
    possible. That does not mean by changing the system, like
    changing the QE of the detector, one couldn't do better. But given
    the technology in use today, CCDs or CMOS sensors, we have
    reached a fundamental limit.

    I say on the web page:
    "The only way to improve on the signal-to-noise is to acquire
    more photons." I can see you have a different idea, and I
    have already rewritten the page to include the different
    ways to collect more photons. I also changed to working on
    theoretical maximum to qualify it, limiting the result
    to the device being tested.
    The computation of system quantum efficiency if quite difficult,
    and you used many many approximations.
    I responded quite aggressively because of your attacks,
    such as "However, my back-of-envelop calculations suggest
    that your conclusion of "working at theoretical maximum" is more than
    an order of magnitude off mark."
    Then after some discussions about approximations:
    Roger said:
    You've been all over the place with your numbers. But I thought
    you were trying to derive quantum efficiency of the sensor.
    I got confused with you using QE for the whole system
    including optics transmission (this is not standard nomenclature).
    This would be like giving the elevation of a mountain from the
    center of the earth, not from sea level. Valid, but because
    not the normal use of elevation, so people just get confused.

    I think I just made a breakthrough in trying to understand
    your emails.

    The SIGNAL-TO-NOISE achievable with the detector the
    way it is derived on my web site is independent of
    how many photons are incident on the focal plane: the signal-to-noise
    is controlled by the full well capacity. The full well
    capacity determines the number of photons that can be counted
    and thus the noise due to Poisson statistics. In the case
    of the 1D mark II, that is 52,000 electrons/photons. If you
    magically increased transmission of all optics to 100%
    and quantum efficiency of the sensor to 100%, the full well
    capacity is still 52,000 electrons and the maximum signal-to-noise
    achievable is still square root 52,000 = 228. That is the
    theoretical maximum under any scenario. To increase the
    signal-to-noise, you mist find a way to count more than
    52,000 photons.

    To repeat, the theoretical maximum signal-to-noise curves
    derived on my web site at:
    are independent of optics transmission, quantum efficiency,
    blur filters, pixel colored (GRGB) filters. The maximum
    signal-to-noise is purely a function of how many photons you
    can count in a Poisson statistics limited system.
    No, CCDs and CMOS detectors that are front side illuminated.
    No improvements in ADC, firmware, analog gain, etc will
    improve the signal-to-noise.

    If more photons could be delivered to the sensor, e.g. by improving
    optics transmission or improving quantum efficiency, the camera
    would achieve its maximum signal-to-noise at a higher ISO, but
    that signal-to-noise would still be the top limit set by the
    full well capacity.

    Roger N. Clark (change username to rnclark), May 1, 2005
  8. [A complimentary Cc of this posting was sent to
    Dave Martindale
    The ISO definition of "digital sensitivity" is not freely available.
    And, actually, did you see it? I expect it is as irrelevant to
    practical photography as one for sensitivity of film...
    What *most* photographers think of what is sensitivity of digital
    sensor is, obviously, pure junk. You probably mean "most
    photographers who have a clue"; anyway, I bet you can hardly find two
    who think the same. ;-)
    One which determines exposure. So read it:

    in some situations one can use 8x smaller-in-area sensor as one which
    requires 8x larger exposure.
    The notion of "signal" as an absolute number in digital world makes
    little sense (unless you measure in electrons, which is far from "art
    considerations"). S/N is what makes better sense (at least as far as
    you do not mix different pixels, and do not change gamma; sigh...).

    And yes, I meant exactly what I wrote.
    Sure. This is part of the "equations" I outlined above.
    50 on contemporary cameras.
    Again: this is a very big oversimplification of the situation. Small
    sensors come with different depth of field (for the same f-stop); they
    also come with *much better* lenses (if you measure two lenses per
    MTF-of-this-lens vs MTF-of-diffraction-bound-lens-at-samef-stop).

    Additionally, smaller sensors are much easier to image-stabilize.
    These 3 factors create a lot of situation when the difference in
    sensitivity is fully compensated. In some (rare?) situations it may
    be even ever-compensated: consider situation when you must shoot
    hand-hold, AND need a large depth of field. In this case smaller sensor
    will give better results (unless you use image-stabilization with
    larger sensor too; this is yet more rarely applicable).
    Please show which S/N of 18% gray you consider acceptable, and which
    is the limit for full well per micron^2 you use. And another thing
    you forgot: QE is also a part of the equation; and it may yet be
    improved about 5x.
    Both statements are wrong. Just look how cameras of '04 perform in
    these regards. Roger says that QE of small sensors is on par with QE
    of larger sensors; this is not what I see, but still, at ISO100 you
    get much better noise than ISO1600 of cells with 8x the area.

    Hope this helps,
    Ilya Zakharevich, May 2, 2005
  9. [A complimentary Cc of this posting was sent to
    Roger N. Clark (change username to rnclark)
    Roger, what I was talking about was art of photography, not what
    manufacturers or ISO committees say in their booklets. You choose the
    exposition to achieve some result; what I was talking about is the
    relation of exposition to the result you get.

    So when I say ISO sensitivity I mean approximately this: the number
    you enter into your lightmeter to get the exposition for the scene
    (assuming, e.g., Adams system). So *this* notion of sensitivity is
    affected, e.g., by exposure compensation; a 100ISO film used with
    exposure compensation +1 "becomes" a 50ISO film in this sense.

    If you know a better term for this, please advise. I did not see a
    good one...

    [Of course, what the ISO sensitivity dial is doing on digital cameras
    is not much different from exposure compensation. Of course, a
    particular firmware can implement these two dials in slightly
    different ways.]
    One can discuss the work of digital cameras on different levels;
    physical, electronical, mathematical; however, the final judgement is
    the impression made by the resulting image. It was this level which I
    used in this (sub)thread.

    Hope this helps,
    Ilya Zakharevich, May 2, 2005
  10. Paul Rubin

    Paul Rubin Guest

    That's EI, the exposure index. It has not much to do with the actual
    sensitivity of the material. You can shoot Tri-X at EI 6400 but that
    doesn't make it an ISO 6400 film. EI is just whatever speed you shoot
    the film at. ISO actually means something.
    Paul Rubin, May 2, 2005
  11. [A complimentary Cc of this posting was sent to
    Roger N. Clark (change username to rnclark)
    The calculation of the photon flow is extremely elementary. If a
    freshman in my class would not be able to do this, their grade would
    suffer (unless the errors are in arithmetic ;-). [The Bayer matrix
    stuff is another question.]
    Yes, I did not intend it to be more than 10% precise. All the
    approximations are clearly documented.
    This is not what you can find on Usenet.
    Absolutely not. *You* were all over the place with your refutations;
    the calculation done on Usenet uses absolutely different methodology,
    and gets a very similar result. This result is confirmed by the
    calibrated spectral curves for the sensors you provided later. While
    if *any* of your refutations had any relation to reality ("a factor
    tof 2 here, a factor of 2 there, another factor of 2 in that place"),
    the answer would be very different.
    To the contrary; full well of about 8K electrons is quite enough for
    photograhy purposes. So increasing QE 5 times, and decreasing full
    well leads to much smaller sensors with similar performance.
    Again, you are mixing counting electrons with counting photons here.
    5x times improvement is possible with technology of today (QE=0.8
    CMOSes are available). Back-side illuminated sensors are available
    You assume same size of sensels.

    Hope this helps,
    Ilya Zakharevich, May 2, 2005
  12. No. The question I asked has nothing to do with photography.
    It is a simple physics question. Please try and answer the
    physics question without any preconceived notions about
    photography. Ignore quantum mechanical tunneling. Change photons
    to rain drops, or whatever it takes to get any photographic
    ideas out of the question. And no photons jump out of the
    bucket once inside.

    The question is:
    Do you agree that given X photons per
    square micron per second incident on a surface,
    and given two "buckets" to collect photons, one Y microns
    on a side and the other 2Y microns, the 2Y bucket collects
    4 times as many photons per second as the Y bucket?

    Possible answers: A) Yes, or B) No.

    Roger N. Clark (change username to rnclark), May 2, 2005
  13. Dave is correct. try search the ISO standards organization
    site. i did but came up with too many things to wade through.
    Perhaps someone else will do better.
    Dave and I think the same on this issue.
    I have used several dozen digital cameras
    of all sizes from pro DSLR to small P&S. All operate ISO the
    same way: a percentage of the maximum signal the camera can
    record, i.e. the full well. It is clear the definition is X amount
    of light in Y seconds gives Z percentage of full well. This
    is exactly analogous to film recording a certain density.
    We just do not know what X, Y and Z are in the formal definition,
    but it will be the same for all cameras, regardless of sensor
    size, just like it is for film and different film sizes.

    Roger N. Clark (change username to rnclark), May 2, 2005
  14. This is pointless. I other threads I tried to get
    this discussion on a non-personal track. I even admitted
    that perhaps some of my writings were not perfectly clear.
    Perhaps you should look at your own writings and see how confusing
    they are to others, especially when you use a term that has
    industry standard meaning (like QE, ISO) but you use it in a
    different context.
    There have been a similar number of emails you sent to me personally
    as are seen in this newsgroup. I put all my messages in one my
    system so I can find things (including usenet). As the discussions
    have been blurred between private emails and usenet perhaps
    I have gotten confused on which came from where. Sorry.

    Gee, as least I continued to work with you trying to give you
    more accurate information. But then you use that to twist
    it into more personal attacks.

    8,000 full well electrons gives a signal to noise maximum
    of 89, and an 18% gray card of 38. This is OK, but not great
    and not comparable to any DSLR, and not even as good as most
    point and shoot digital cameras. It would be very restrictive in producing
    quality images in various environments. With read noise of
    about 15 electrons, the maximum dynamic range is 9 stops,
    compared to DSLRs around 11 stops.
    Yes. That is the way it works. one electron = one photon
    in CCDs and CMOS sensors. And that is perfect, because it means
    you can get to Poission statistics being the limiting factor
    in noise.
    But if you magically increased the QE to 1.0, the maximum signal to
    noise is still fixed by the full well depth. You can't improve signal
    to noise beyond this, period, for the given well depth.
    It matters not what the sensel size is. What matters is the
    full well depth.

    Look at full well depth like an integer number. Lets say
    your counter is 16 bits. 2^16-1 = 65535. No matter how much you
    count, you can never get beyond 65535. If your noise is limited to
    the square root of what you can count, then your signal-to-noise
    ratio can never be greater than 65535/sqrt(65535) = sqrt(65535) = 256.

    Now lets say your counting balls. Someone is throwing balls at you
    and you are throwing them in a bucket to be counted by your 16-bit counter.
    You miss some of the balls. You drop some of the balls.
    Some of the balls you throw miss the bucket.
    Those don't matter. Only the ones in the bucket get
    counted and the noise is the square root of the ones that make
    it into the bucket. The maximum signal-to-noise ratio you can
    ever achieve is 256 because you can only count a maximum of 65535
    balls. You can improve your catching ability so more balls make
    it into the bucket, but all that does is you fill the bucket
    faster. You can still never get a better signal-to-noise ratio
    than 256. The only way to improve the nose is get a bigger bucket so
    you can count more balls.

    Can everyone out there still reading understand this?

    How to improve getting more balls into the bucket is a DIFFERENT and
    UNRELATED problem to the maximum signal-to-noise ratio that can be achieved
    by counting.

    The maximum signal-to-noise ratio for different cameras on my web pages,
    are correct and independent of quantum efficiency, transmission of
    lenses, filters, the time of day and the stock market, as well as
    all the things Ilya is talking about.

    Now Ilya is talking about some interesting stuff and he does have interesting
    ideas about improving throughput. But that just fills the electron wells

    Roger N. Clark (change username to rnclark), May 2, 2005
  15. You make perfect sense and your notes are, as always, worth reading. (Even
    though I still think you (slightly) overestimate the noise in film<g>.)

    But Illya's been in my killfile for quite a while now. It improved the S/N
    in this newsgroup significantly.

    David J. Littleboy
    Tokyo, Japan
    David J. Littleboy, May 2, 2005
  16. [A complimentary Cc of this posting was sent to
    Roger N. Clark (change username to rnclark)
    Hmm? It is trivial to find. And it is $69 for an electronic copy,
    IIRC. Too much for the values of X, Y, Z.
    Pardon my misunderstanding... Full well is a property of a sensor,
    right? How much charge it can keep before I-do-not-know-what
    (leakage?) occurs. To change it, you need to disassemble the camera
    and physically switch the sensor ;-).

    Or do you mean the charge which saturates the ADC? These two deserve
    different names; it looks like people use "full well" in both

    Anyway, all this started with the following remark (by Dave):
    I have no idea what he meant by this remark ("in that context"). Is
    he thinking what "in normal context" ISO number is not changing with
    the change of gain?

    In any case, my point of view is very simple: if you have two cameras,
    both this ISO dial going from 100 to 1600, how to determine which one
    is more sensitive to light? (This is very similar to "pushing the
    film".) Obviously, you need some way to compare ISO100 on one camera
    with ISO100 on another one. The only remaining variable is the noise
    level. So it makes sense to compare two cameras *in the settings
    which provide identical images*.

    In ideal conditions, when you can focus well, and stabilize the camera
    (and the subject ;-) enough, the only parameter to compare is the
    noise. So it makes sense to say

    "Camera A is 8x more sensitive than Camera B".

    This means that to produce identical noise, you need to turn the ISO
    dials differently (and, correspondingly, use different exposure).
    And, as I demonstrated by many examples, this is applicable not only
    in "ideal conditions", but in some "real life situations" too (due to
    better lenses, larger depth of field, and better stabilization of
    smaller sensors).

    Hope this helps,
    Ilya Zakharevich, May 2, 2005
  17. [A complimentary Cc of this posting was sent to
    Roger N. Clark (change username to rnclark)
    Right. All it has to do is politeness. The only purpose of such
    questions is to insult. Do you agree that sqrt(5) is larger than 2? ;-)

    Hope this helps,
    Ilya Zakharevich, May 2, 2005
  18. [A complimentary Cc of this posting was sent to
    Roger N. Clark (change username to rnclark)
    Roger, just look on how the discussion goes: when you say that I was
    "all over the place", it is "a non-personal track", right? When I say
    anything, it becomes personal, right?
    Look at your graphs. 8K full well *is* what you get from current dSLR
    on certain ISO settings. Actually, it may be that these ISO settings
    are what people use most. Using smaller ISO numbers allows much more
    postprocessing done. However, if you use smaller sensels, the need to
    postprocess somewhat decreases, since you got much more information
    without postprocessing.
    I still have no idea what you mean here. What I see is that most
    photons equal 0 electrons, only very few of photons equal 1 electron...
    I do not argue with this. If you *need* S/N ratio of 250, you need a
    lot of electrons, so a lot of photons. So: do you have "actual
    photography" examples where S/N ratio above 100 is beneficial?
    AFAIU, maximum achievable S/N number is irrelevant to photography;
    what is relevant is whether noise decreases the artistic value of the
    final image, or not. And the noise one cannot see should not, IMO,
    decrease the value. So if you know many situations where "high" S/N
    matter, please let us know.

    Ilya Zakharevich, May 2, 2005
  19. Paul Rubin

    Chris Brown Guest

    AFAICS, Ilya is bordering on being a full-blown Usenet kook. He's obsessive
    about one topic, a topic on which he seems to feel he has some special
    insight. Dresses his arguments up in tehcnobabble, a lot of which makes very
    little, or no sense at all, and is extremely reluctant to address questions
    that use real hard information and science when he finds them inconvinient,
    apparently prefering to resort to attacking the poster's intelligence and/or

    Roger, you've been more than patient and reasonable, but I really think
    you're wasting your time engaging with him - it seems that his point about
    pixel size and noise, if anyone can work out exactly what it is, is almost
    an article of faith to him, and nothing anyone says, however well backed up,
    is goping to change it.
    Chris Brown, May 2, 2005
  20. This is incorrect. What happens when you boost ISO does not
    change the full well capacity. Let's consider the Canon
    1D Mark II. The chip's full well capacity is ~52000.
    At iso 100 you can collect electrons up to the full well.
    At iso 400, the signal is multiplied by 4 times in the
    analog conversion process, so 1/4 of the full capacity saturates
    the 12-bit A-to-D converter. At ISO 400 the full well
    capacity is still 52000 electrons, but you can only count
    52000/4 electrons, and your noise is sqrt(52000/4).
    Right. But it doesn't matter how many you do not count.
    It only matters what you ACTUALLY count.
    Yes. Every day. I find noise in my 1D Mark II images I do
    not like. Try boosting the signal in the shadows to produce
    a clean image. You couldn't do it with film. You can do a
    lot with DSLR images, but often the shadows still have low
    S/N. But your question relates to the high signal. No,
    the S/N of > 100 is fine. But interesting images have more
    than just bright stuff. You need high signal to noise
    in the darker parts of the image. To get that you must collect
    a lot of photons, which means you collect even more photons for
    the bright areas.
    I disagree. But that is a technical issue.
    Yes, I agree.
    Yes. It is important whenever you are using the image data to
    restore an image. Example:
    You trade S/N for spatial resolution in image restoration
    algorithms. Even simple unsharp mask shows this trade.

    Roger N. Clark (change username to rnclark), May 2, 2005
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