Latest Kodak sensors have superior Dynamic Range! (on the paper)

Discussion in 'Digital Photography' started by RiceHigh, Mar 2, 2006.

  1. RiceHigh

    RiceHigh Guest

    http://www.kodak.com/global/plugins/acrobat/en/digital/ccd/products/fullframe/KAF-18000LongSpec.pdf

    The one above will probably be used in the coming Pentax 645D with
    cropped sensor size less than 6 x 4.5cm. DR listed in this "layman"
    (IMO) or simply marketing datasheet is 75dB.

    http://www.kodak.com/global/plugins/acrobat/en/digital/ccd/products/fullframe/KAF-18000LongSpec.pdf

    For this one, I don't know what DSLR will use it! as it has a 1.24X
    crop factor, really strange to me afterall. Will the coming PEntax 10M
    DSLR used this? WHo knows? DR listed is 67dB.

    So, the question is how we intrepret this into the usual EV units we
    are familiar with??

    dB is a *comparative* logged unit in log-base *10* multipied by 10 for
    signal level difference and 20 if given the signal levels but we wish
    to calculate the power or energy difference.

    f-stops is a *comparative* logged unit in log-base *2*.

    I don't know if the Kodak datasheet is about voltage level or energy
    level on the dB (as it doesn't define it clearly). I bet it is probably
    about energy level difference base on voltage level difference, though,
    for a typical datasheet for this type of device. Thus, a 67dB means the
    logged-10 ratio is 3.35 (derviated by divding 20). To convert back to
    the f-stops, one need
    to multiply the log-10/log-2 ratio which is 3.32 when rounded up. Then,
    the DR should be 11.13 f-stops (3.35 x 3.32) instead! which I think is
    pretty good!

    For the 645D sensor, its theoretical DR can reach 12.45 EVs!
    Amazing!(?) :))

    Congrats Pentax! But hope she don't spoil the excellent Kodak sensor in
    the 645D!

    Finally, noted that the CCD is a *linear* device and actually each
    pixel is a light sensitive capactor. The voltage in each pixel stored
    in *proportional* to the light energy received in a *linear* way..
    Thus, I have just derived to the voltage difference in "how many
    doubles" last time about the voltage and this is essential equal to the
    light energy ratio.

    RiceHigh
    http://www.geocities.com/ricehigh
    RiceHigh, Mar 2, 2006
    #1
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  2. RiceHigh

    Bryan Olson Guest

    RiceHigh wrote:
    [...]
    > dB is a *comparative* logged unit in log-base *10* multipied by 10 for
    > signal level difference and 20 if given the signal levels but we wish
    > to calculate the power or energy difference.


    Mostly. dB is defined in terms of power. Knowing the power
    ratio will also tell us the voltage ratio, because power
    increases as the square of voltage. (In measuring sound, power
    increases as the square of sound pressure.)

    > f-stops is a *comparative* logged unit in log-base *2*.
    >
    > I don't know if the Kodak datasheet is about voltage level or energy
    > level on the dB (as it doesn't define it clearly).


    Both.

    > I bet it is probably
    > about energy level difference base on voltage level difference, though,
    > for a typical datasheet for this type of device. Thus, a 67dB means the
    > logged-10 ratio is 3.35 (derviated by divding 20).


    67dB is a ratio of 10**6.7 = 5,000,000 in power, which is
    equivalent to a ratio of 2200 in voltage.

    > To convert back to
    > the f-stops, one need
    > to multiply the log-10/log-2 ratio which is 3.32 when rounded up. Then,
    > the DR should be 11.13 f-stops (3.35 x 3.32) instead! which I think is
    > pretty good!


    The ratio is even larger, but let's not get carried away. They're
    not saying that the thing can faithfully record images with a
    67dB dynamic range (which would be 22 f-stops). The figure is
    just looking at saturation versus read-noise.


    --
    --Bryan
    Bryan Olson, Mar 2, 2006
    #2
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  3. RiceHigh

    RiceHigh Guest

    Thanks for your response, Bryan.

    I would say 67dB is still equal to 11.13 EVs if it is assumed that the
    dB is measured for power. Just that 67dB = 20 log-10 on (Voltage levels
    ratio between the brightest scene and the darketst scene which can be
    recorded *simultaneously). Then, the "Voltage Ratio" is simly 67 / 20 =
    3.35. That means the voltage level difference is 10^3.35 = 2239, which
    is again equal to 2 to the power of 11.13.

    The *Voltage Level" is proportional to the light energy/power projected
    on the CCD as CCD/CMOS imager is a *linear* device, unlike film. So,
    this 11.13 is simply the EV difference afterall.

    Also, a datasheet which lists it has a DR of 22 EVs will not be
    realistic anyway (and it doesn't *look* realistic too).

    Cheers,
    RiceHigh
    http://www.geocities.com/ricehigh
    RiceHigh, Mar 3, 2006
    #3
  4. RiceHigh

    Bryan Olson Guest

    RiceHigh wrote:
    > I would say 67dB is still equal to 11.13 EVs if it is assumed that the
    > dB is measured for power. Just that 67dB = 20 log-10 on (Voltage levels
    > ratio between the brightest scene and the darketst scene which can be
    > recorded *simultaneously).


    Voltage is a measure of potential difference, not brightness.
    Brightness is, at least roughly, light power. EV (Exposure Value)
    is brightness times time, so it is proportional to light energy.


    > Then, the "Voltage Ratio" is simly 67 / 20 =
    > 3.35. That means the voltage level difference is 10^3.35 = 2239, which
    > is again equal to 2 to the power of 11.13.
    >
    > The *Voltage Level" is proportional to the light energy/power projected
    > on the CCD as CCD/CMOS imager is a *linear* device, unlike film.


    I'm not an EE, but my sources disagree. CCD's are linear in that
    the *charge* (or discharge) is proportional to light energy.
    The Coulombs are linear, not the Volts.

    > So,
    > this 11.13 is simply the EV difference afterall.
    >
    > Also, a datasheet which lists it has a DR of 22 EVs will not be
    > realistic anyway (and it doesn't *look* realistic too).


    On that we agree. The datasheet at issue says no such thing.


    --
    --Bryan
    Bryan Olson, Mar 3, 2006
    #4
  5. RiceHigh

    RichA Guest

    Kodak is nuts. They should put that sensor in a true DSLR, scrap the
    stupid old fashioned
    SLR lens mount dimensions and sell it for $5000. Of course, you need
    lenses of quality to make
    it work so they'd probably have to outsource them, as usual.
    RichA, Mar 3, 2006
    #5
  6. "RiceHigh" <> wrote in message
    news:...
    > Thanks for your response, Bryan.
    >
    > I would say 67dB is still equal to 11.13 EVs if it is assumed
    > that the dB is measured for power.


    That's correct, it's a measurement of power:
    http://www.ccd.com/ccd111.html

    Bart
    Bart van der Wolf, Mar 3, 2006
    #6
  7. RiceHigh

    RiceHigh Guest

    > Voltage is a measure of potential difference, not brightness.

    Yes, you're correct. But I also mentioned that CCD converts light power
    into voltage :)

    > Brightness is, at least roughly, light power. EV (Exposure Value)

    is brightness times time, so it is proportional to light energy.

    Right again, the light energy is project onto the sensor thro the
    aperture of the lens. It is actually through the lens with the aperture
    as the limiter.

    >> The *Voltage Level" is proportional to the light energy/power projected
    >> on the CCD as CCD/CMOS imager is a *linear* device, unlike film.


    > I'm not an EE, but my sources disagree. CCD's are linear in that

    the *charge* (or discharge) is proportional to light energy.
    The Coulombs are linear, not the Volts.

    I am an EE :) but not an analog device or component specialist though.
    Actually, I have not touched analog device design nor any other low
    level applications for more than a decade.

    CCD is an analog device which will produce chains of waveform according
    to the charges stored in each pixel. To identifiy the "address" (i.e.
    which pixel) in a row, a clock is used to "clock out" the charges and
    convert the value into a certain voltage. The "farthest" away pixel
    value will become the waveform head and the waveform propagate along
    with the pixel values and forming the whole waveform. Thus, each line
    will have a waveform and than the same repeated with another row of
    pixels.

    Afterall, all the pixel values are represented as voltage level at the
    different parts of the waveform. I supposed that this should be
    *linear* to the charges stored in each pixel. Just a guess though as I
    am not a CCD specialist here.

    Another example is that charged rechargeable batteries are full of
    charges but then it will have its voltage potential according to the
    chemical used. The current is dependent on the voltage and the charge
    flow per second is the current.

    > this 11.13 is simply the EV difference afterall.


    >> Also, a datasheet which lists it has a DR of 22 EVs will not be
    >> realistic anyway (and it doesn't *look* realistic too).


    > On that we agree. The datasheet at issue says no such thing.


    I think in more detailed technical manual, it will tell. But I still
    think that it does not make any sense at all to tell the potential
    developer that the sensor has a DR of 22 EVs which is impossible for
    any photographic recording media, including B&W films!
    RiceHigh, Mar 3, 2006
    #7
  8. RiceHigh

    RiceHigh Guest

    RiceHigh, Mar 3, 2006
    #8
  9. RiceHigh

    rafe b Guest

    On 3 Mar 2006 05:45:06 -0800, "RiceHigh" <> wrote:


    >I am an EE :) but not an analog device or component specialist though.
    >Actually, I have not touched analog device design nor any other low
    >level applications for more than a decade.
    >
    >CCD is an analog device which will produce chains of waveform according
    >to the charges stored in each pixel. To identifiy the "address" (i.e.
    >which pixel) in a row, a clock is used to "clock out" the charges and
    >convert the value into a certain voltage. The "farthest" away pixel
    >value will become the waveform head and the waveform propagate along
    >with the pixel values and forming the whole waveform. Thus, each line
    >will have a waveform and than the same repeated with another row of
    >pixels.



    I'm an EE, and I work with CCDs quite a bit.
    I've looked at CCD waveforms on a scope and
    written firmware for processing and calibrating
    image data acquired by CCDs.

    The sensels acquire charge, that's correct,
    but it's a voltage waveform you're looking at
    outside the chip.

    The A/D measurement is generally done using
    "correlated double sampling." Within each
    pixel's period, there is a baseline
    measurement, which is captured by a sample/
    hold. Some time later, the final value is
    sampled in a 2nd S/H. The differential
    measurement is what's then fed to the A/D.

    Typical period for measuring 3 channels
    of one pixel is about 250 nanoseconds.

    There are lots of CCD data sheets available
    on the web, if you're interested.

    CCDs are pretty crude devices. You'd be
    surprised at how much "massaging" of the
    data is involved in a typical flatbed or
    film scanner. That's why I kind of smirk
    when folks make such a big fuss over 8-bit
    vs. 16-bit scans.

    My LS-8000 can return 14-bit scans, but
    in reality the low-order four or five
    bits are mostly noise. In a typical
    flatbed scanner it's far worse than that.



    rafe b
    www.terrapinphoto.com
    rafe b, Mar 3, 2006
    #9
  10. RiceHigh

    Guest

    In message <>,
    rafe b <rafebATspeakeasy.net> wrote:

    >My LS-8000 can return 14-bit scans, but
    >in reality the low-order four or five
    >bits are mostly noise. In a typical
    >flatbed scanner it's far worse than that.



    There is absolutely nothing wrong with sampling at a depth that results
    in mostly noise in the lowest bits. It should *always* be done, IMO, to
    limit posterization. Adding noise to break up posterization after the
    fact is just that; more noise.

    The deeper you digitize, the more random the lowest bits get, but you
    get slightly more subject detail by going fairly deep into the noise. I
    have exposed high-contrast text subjects with my 20D 10 stops under at
    ISO 1600, and could read the text after removing some coarse banding.
    The biggest enemy to seeing the text was not random elements of sensor
    noise, but more banding, left over from the influence of abnormal pixels
    used in determining banding strength. If I had a slider for each
    horizontal and vertical line, to control the blackpoints independently,
    (and a lot of time), I could probably go a couple stops more in
    under-exposure, and read the text.
    --

    <>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
    John P Sheehy <>
    ><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
    , Mar 3, 2006
    #10
  11. Re: Latest Kodak sensors have superior Dynamic Range! (on thepaper)

    "Bart van der Wolf" <> wrote:
    >"RiceHigh" <> wrote in message
    >news:...
    >> Thanks for your response, Bryan.
    >>
    >> I would say 67dB is still equal to 11.13 EVs if it is assumed
    >> that the dB is measured for power.

    >
    >That's correct, it's a measurement of power:
    >http://www.ccd.com/ccd111.html


    The calculations were also correct, though that was the hard way
    to do it! Since fstops or EV (same basic difference) are
    logarithmic, and so is the dynamic range (which is the signal to
    noise ratio)... the ratio between the two is

    20 log 2

    for each time one is doubled (1 fstop or EV). That works out
    to this,

    SNR_in_dB / 6.020599

    Hence a dynamic range (SNR) of 67 dB is a range of 11.128459
    fstops.

    Obviously dividing by 6.02 is very accurate (11.13) and
    just dividing by 6 is easily close enough (11.17).

    --
    Floyd L. Davidson <http://www.apaflo.com/floyd_davidson>
    Ukpeagvik (Barrow, Alaska)
    Floyd L. Davidson, Mar 3, 2006
    #11
  12. Re: Latest Kodak sensors have superior Dynamic Range! (on thepaper)

    Bryan Olson <> wrote:
    >RiceHigh wrote:
    >> I would say 67dB is still equal to 11.13 EVs if it is assumed that the
    >> dB is measured for power. Just that 67dB = 20 log-10 on (Voltage levels
    >> ratio between the brightest scene and the darketst scene which can be
    >> recorded *simultaneously).

    >
    >Voltage is a measure of potential difference, not brightness.
    >Brightness is, at least roughly, light power. EV (Exposure Value)
    >is brightness times time, so it is proportional to light energy.


    The voltage from the sensor is an indication of brightness
    detected by the sensor. Regardless, 11.13 is the correct value
    within about 2/1000th or so. Seems close enough...

    > > Then, the "Voltage Ratio" is simly 67 / 20 =
    >> 3.35. That means the voltage level difference is 10^3.35 = 2239, which
    >> is again equal to 2 to the power of 11.13.
    >> The *Voltage Level" is proportional to the light energy/power
    >> projected
    >> on the CCD as CCD/CMOS imager is a *linear* device, unlike film.

    >
    >I'm not an EE, but my sources disagree.


    Disagree with what?

    >CCD's are linear in that
    >the *charge* (or discharge) is proportional to light energy.
    >The Coulombs are linear, not the Volts.


    The voltage will be linear if the charge is linear, as voltage
    is directly proportional to charge. Voltage is 1 Joule per Coulomb.

    > > So,
    >> this 11.13 is simply the EV difference afterall.
    >> Also, a datasheet which lists it has a DR of 22 EVs will not be
    >> realistic anyway (and it doesn't *look* realistic too).

    >
    >On that we agree. The datasheet at issue says no such thing.


    --
    Floyd L. Davidson <http://www.apaflo.com/floyd_davidson>
    Ukpeagvik (Barrow, Alaska)
    Floyd L. Davidson, Mar 3, 2006
    #12
  13. RiceHigh

    Alfred Molon Guest

    In article <>,
    says...
    > Thanks for your response, Bryan.
    >
    > I would say 67dB is still equal to 11.13 EVs if it is assumed that the
    > dB is measured for power. Just that 67dB = 20 log-10 on (Voltage levels
    > ratio between the brightest scene and the darketst scene which can be
    > recorded *simultaneously). Then, the "Voltage Ratio" is simly 67 / 20 =
    > 3.35. That means the voltage level difference is 10^3.35 = 2239, which
    > is again equal to 2 to the power of 11.13.


    The datasheet says saturation signal of 100000 electrons, read noise of
    18 electrons. 100000/18 = 5555 which is 2^12.4, which corresponds to
    75dB.
    --

    Alfred Molon
    ------------------------------
    Olympus 50X0, 7070, 8080, E300, E330 and E500 forum at
    http://groups.yahoo.com/group/MyOlympus/
    Olympus E500 resource - http://myolympus.org/E500/
    Alfred Molon, Mar 3, 2006
    #13
  14. RiceHigh

    Guest

    Alfred Molon wrote:

    > The datasheet says saturation signal of 100000 electrons, read noise of
    > 18 electrons. 100000/18 = 5555 which is 2^12.4, which corresponds to
    > 75dB.


    This is correct when decibels are defined equal to 20 log (voltage).
    The voltage is produced at the readout amplifier and is proportional
    to the charge in the CCD pixel (electrons). But the number of
    electrons detected is equal to the number of photons detected
    and thus proportional to the light energy falling on the pixel
    in the exposure time (the power).

    The use of dB = 20 log (voltage), I think, comes from EE
    measurements of electrical signals, where because
    power = voltage*current, power is proportional to voltage squared
    (for Ohmic devices anyway, like wires and resistors).
    For a CCD, power is proportional to readout voltage, and
    properly one should use 10 log (voltage).

    IOW, when CCD manufacturers use dB = 20 log (voltage) for
    dynamic range, they are telling a little white lie. The real
    DR of that chip is 37.5 dB.
    , Mar 3, 2006
    #14
  15. Re: Latest Kodak sensors have superior Dynamic Range! (on thepaper)

    wrote:
    >Alfred Molon wrote:
    >
    >> The datasheet says saturation signal of 100000 electrons, read noise of
    >> 18 electrons. 100000/18 = 5555 which is 2^12.4, which corresponds to
    >> 75dB.

    >
    >This is correct when decibels are defined equal to 20 log (voltage).
    >The voltage is produced at the readout amplifier and is proportional
    >to the charge in the CCD pixel (electrons). But the number of
    >electrons detected is equal to the number of photons detected
    >and thus proportional to the light energy falling on the pixel
    >in the exposure time (the power).


    dB is a ratio of *power*. That is true even if it is calculated
    with voltages, currents, or something else.

    >The use of dB = 20 log (voltage), I think, comes from EE
    >measurements of electrical signals, where because
    >power = voltage*current, power is proportional to voltage squared
    >(for Ohmic devices anyway, like wires and resistors).
    >For a CCD, power is proportional to readout voltage, and
    >properly one should use 10 log (voltage).


    If power is proportional to the readout voltage, then the
    same formula applies, as above: SNR = 20 log (S/N).

    >IOW, when CCD manufacturers use dB = 20 log (voltage) for
    >dynamic range, they are telling a little white lie. The real
    >DR of that chip is 37.5 dB.


    More like 74.89 dB.

    Regardless of what the circuit impedance is, the voltage *is*
    going to cause current to flow, and therefore the correct
    formula multiplies times 20, not 10.


    "Dynamic Range = 20 x Log(Nsat/Nnoise)"
    http://micro.magnet.fsu.edu/primer/digitalimaging/concepts/dynamicrange.html

    "Dynamic Range = D(dB) = 20√ólog10(full well capacity/read noise)"
    http://www.du.edu/~rmatson1/AstrCh3outline.htm

    "20 x log (full well capacity / readout noise) [dB]"
    http://www.du.edu/~rmatson1/AstrCh3outline.htm

    A quick search with google turns a many many such references.


    --
    Floyd L. Davidson <http://www.apaflo.com/floyd_davidson>
    Ukpeagvik (Barrow, Alaska)
    Floyd L. Davidson, Mar 3, 2006
    #15
  16. RiceHigh

    Guest

    Floyd L. Davidson wrote:

    > If power is proportional to the readout voltage, then the
    > same formula applies, as above: SNR = 20 log (S/N).
    >
    > >IOW, when CCD manufacturers use dB = 20 log (voltage) for
    > >dynamic range, they are telling a little white lie. The real
    > >DR of that chip is 37.5 dB.

    >
    > More like 74.89 dB.
    >
    > Regardless of what the circuit impedance is, the voltage *is*
    > going to cause current to flow, and therefore the correct
    > formula multiplies times 20, not 10.


    If you want to talk about the _output_ power, then yes current
    flows. But for the input power, the signal you get doesn't have a
    current factor in it.

    My point is that the CCD has a dynamic range of 3.75 decades
    (factors of 10) in input power. If one redefines decibels so that it
    takes a step of 20 to make a decade, 75 dB is correct. I just think
    that's confusing.

    > "Dynamic Range = 20 x Log(Nsat/Nnoise)"
    > http://micro.magnet.fsu.edu/primer/digitalimaging/concepts/dynamicrange.html
    >
    > "Dynamic Range = D(dB) = 20√ólog10(full well capacity/read noise)"
    > http://www.du.edu/~rmatson1/AstrCh3outline.htm
    >
    > "20 x log (full well capacity / readout noise) [dB]"
    > http://www.du.edu/~rmatson1/AstrCh3outline.htm
    >
    > A quick search with google turns a many many such references.


    I accept that this formula is commonly repeated, although nobody
    I know who talks about CCD signal to noise actually uses decibels -
    they just talk about S/N ratio. (That includes users of Howell's
    book.)
    However, I think the formula is misleading, because normally decibels
    mean a change by 10 for every decade in power, while this formula
    turns it into a factor of 20. (Since the signal in CCDs is linearly
    proportional to the input power.)

    The Wikipedia article on decibels,
    http://en.wikipedia.org/wiki/Decibels

    suggests that these definitions co-exist - the sentence beginning
    "Note that in physics, these [20 log V/Vo] equations are considered
    to give power ..."

    Great, a unit with two definitions that differ by a factor of 2!
    , Mar 3, 2006
    #16
  17. RiceHigh

    Bryan Olson Guest

    Floyd L. Davidson wrote:
    > Bryan Olson wrote:

    [...]
    >>CCD's are linear in that
    >>the *charge* (or discharge) is proportional to light energy.
    >>The Coulombs are linear, not the Volts.

    >
    >
    > The voltage will be linear if the charge is linear, as voltage
    > is directly proportional to charge. Voltage is 1 Joule per Coulomb.


    But that formula has Voltage inversely proportional to charge
    (for constant energy).

    Is there a good source showing that the conversion from charge to
    voltage in CCD readout is linear?


    --
    --Bryan
    Bryan Olson, Mar 3, 2006
    #17
  18. RiceHigh

    RichA Guest

    But that is Kodak working as a parts source. They should re-introduce
    their own
    DSLR.
    RichA, Mar 3, 2006
    #18
  19. RiceHigh

    rafe b Guest

    "Bryan Olson" <> wrote in message
    news:lQ3Of.18564$...

    > But that formula has Voltage inversely proportional to charge
    > (for constant energy).


    Then the formula is wrong. The correct formula is: Q = CV.
    Q is charge, C is capacitance, V is voltage.


    > Is there a good source showing that the conversion from charge to
    > voltage in CCD readout is linear?



    As the sensels are dumped, each in turn, the charge
    from each creates a current waveform. The current
    waveform is "converted" to a voltage waveform
    by dumping it to ground through a simple resistor.

    E=IR, so the current-to-voltage conversion is
    completely linear.


    rafe b
    www.terrapinphoto.com
    rafe b, Mar 3, 2006
    #19
  20. RiceHigh

    Guest

    RichA wrote:
    > Kodak is nuts. They should put that sensor in a true DSLR, scrap the
    > stupid old fashioned
    > SLR lens mount dimensions and sell it for $5000. Of course, you need
    > lenses of quality to make
    > it work so they'd probably have to outsource them, as usual.\


    WTF. It's a 35x46mm sensor. It isn't aimed at the 35mm lens mounts.
    They don't need to try to convert photographers to some new standard
    when Pentax and Mamiya already make good 645 lenses that will
    cover it. But my guess is that the camera built around this sensor
    will cost more than $5000.
    , Mar 3, 2006
    #20
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