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

http://www.kodak.com/global/plugins/...00LongSpec.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/...00LongSpec.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 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

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

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.

"RiceHigh" wrote in message
oups.com...
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

Kodak did this with Olympus as their new 4-3 system, namely for the
E-1, E-300 and E-550 (but not the E-330).

RiceHigh
http://www.geocities.com/ricehigh

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

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

"Bart van der Wolf" wrote:
"RiceHigh" wrote in message
roups.com...
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)

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)