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

Discussion in 'Kodak' started by RiceHigh, Mar 2, 2006.

1. ### Kennedy McEwenGuest

The charge, Q, integrates on a capacitor, the charge well, C to produce
a voltage.

V = Q / C

So the voltage is linear... until it saturates.

Kennedy McEwen, Mar 4, 2006

2. ### Bryan OlsonGuest

I'm missing something. Power increases as the square of
Voltage. If the output Voltage is proportional to the
charge, which is proportional to the light power, then
the power coming out in the Voltage and current increases
as the square of the power going in as light. You have the
charge inducing the Voltage without another power source.

Bryan Olson, Mar 4, 2006

3. ### Floyd L. DavidsonGuest

Power is power, the output of one thing is merely the input of
another.

I assure you that input power signals *do* have "a current factor".
It's a simple matter of what the input impedance of a device happens
to be...
That is confused. What "input power" are you talking about?
Photons? Electrons? A CCD is a *sensor*, it does not have an
electrical input signal. It has an electrical output signal,
which is what we are talking about.
It's all the same thing, and anyone who understands one
understands the other, and can interchange them in appropriate
places. When discussing a CCD alone, absent the rest of the
processing system, a raw ratio is perhaps more useful. But when
the rest of the system is part of the discussion, dB makes a
great deal more sense; and that is particularly true when
comparing fstops of exposure with the dynamic range of the CCD.
That is simply false, and demonstrating it is easy. If a given
camera has a CCD with an S/N in the range of 30-40 dB as you are
claiming, the range of fstops that could be recorded in the
images produced would be 5 to 7 (30-42 dB of dynamic range).
Yet not only do *all* of the major camera manufacturers list the
dynamic range of the CCD's they use as in the 60-75 dB range,
the also can be *measured* to record images that record a range
of illumination from more than 7 to more than 10 fstops.

What that says is simple: They all have CCD's with a dynamic
range that is more than you are claiming the best of them would

Clearly the correct formula is 20 log (S/N), and when it is used
*all* of the calculations come out correct.
There is only one definition, but they did give different ways
to calculate it with *different* *parameters*. With
"intensity", "voltage", "pressure" the correct multiplier is 20,
and with power it is 10. That does *not* generate a different
dB, it merely says that voltage and power are not the same
thing.

Indeed E = IR and P = IE demonstrate exactly what that
difference is.

Floyd L. Davidson, Mar 4, 2006
4. ### Scott WGuest

Bryan Olson wrote:

Photocells are call square law detectors for just this reason. The
power source is the charging voltage that recharges the capacitance of
the photo sites.

Scott

Scott W, Mar 4, 2006
5. ### bjwGuest

Floyd. I think we're talking past each other. A CCD has an
input signal from the light, and it's perfectly reasonable to
given exposure time. I assumed that this is what
people like to refer to as the dynamic range of the sensor -
the range of input values that it can sense.

Apparently the dynamic range spec is the dynamic range
calculated from the output voltage. In that case, I give up,
using 20 log (voltage max / voltage min) is correct and
the dynamic range of the output power is 75 dB.

like to use dynamic range to talk about the range of
input luminances the CCD can sense, and that range is
not 75 dB in _input_.
What you're showing here is that the manufacturer's
spec sheets are all consistent in using 20 log (S/N),
not whether that formula means input or output. I have
no disagreement with the fact that the CCD under
discussion has a dynamic range of 100,000/18 - I may
have said that in my original post.

bjw, Mar 4, 2006
6. ### Floyd L. DavidsonGuest

The dynamic range of the sensor, which is what it can sense, is
a measure of the *output* electrical signal.
Nothing else makes sense.
Absolutely true.
That is *NOT* true. The sensor *output* is the only way to
determine what input range can be sensed. The input exposure
range can of course be varied from far less than is sensed to
far more than is sensed... but the definition of "sensed" is
that a change in input results in a change in the output.
It clearly shows that the sensed range matches the range they
are calculating using 20 log (S/N).
What you have been claiming is that a dynamic range indicated by
100000/18 is 37.5 dB. That is not true. It is clearly twice
that, just as everyone says. If it were 37.5 dB the camera
would not be able to record an image with more than

37.5
---- = 6.2+ fstops
6

And in fact there is no question that typically digital cameras
are recording between 8 and 11 fstops and are being described
as having a dynamic range of between 60 and 75 dB using the
*correct* formula.

Your formula simply is not the right one.

Floyd L. Davidson, Mar 4, 2006
7. ### JPSGuest

In message <>,
Not necessarily. The highest exposure level on the sensor before
saturation is recorded at the lowest ISO, whereas the lowest usable
sensor signal (whatever your standard may be) may only be obtained with
amplification; i.e., a higher ISO.
--

JPS, Mar 4, 2006
8. ### Floyd L. DavidsonGuest

Yes, necessarily. (Anything else is *impossible*.)
It is *still* being determined from the output voltage.

I'm not sure your statement means what you think it does anyway.
Keep in mind that "obtained with amplification" amplifies
*everything* the same -- highs, lows, and noise -- it does not
change the dynamic range of the sensor. It might change the
operating point of the codec that digitizes the output though,
which still doesn't change the dynamic range of either the
sensor or the codec, but certainly can result in different
results. (Granted that if the operating point for one is moved
off the upper or lower end of the other, the dynamic range of
the combination is compressed.)

Floyd L. Davidson, Mar 4, 2006
9. ### JPSGuest

In message <>,
They are on a completely different scale, though, and may not both be
possible at one level of amplification, given the limitations of
digitization hardware..
It changes the DR of the *output*, though, in a single capture.
In other words, you had to make it look like I was wrong, even though I
wasn't. That's pretty much what I expected, when I saw your name.
--

JPS, Mar 4, 2006
10. ### Floyd L. DavidsonGuest

Your statement "not necessarily" apparently applied to something

It is irrelevant to the discussion you inserted it into.
Post sensor amplification cannot change the the dynamic range of
the sensor output.
So if you aren't able to make a technical point you'll try
gratuitous insults?

Compressing the image (which is not what was being discussed) is
a very different thing than reducing the dynamic range of the
sensor (which is what was being discussed).

Floyd L. Davidson, Mar 5, 2006
11. ### Floyd L. DavidsonGuest

More pointless detail that does *not* support the false claim

And please note that any manufacturer who wants to go to 14 or
even 16 bit digitization hardware can (and does) at any time.
It does *not* change the dynamic range of the *sensor output*,
which is what we were talking about and which you claimed was
true.
You were wrong then, you are wrong now, and I can't see any
indication that you are going to change that. The fact that
some unrelated factoid is true does not change what you said to
something that is right. Even if you did see my name, and
shivered.

Floyd L. Davidson, Mar 5, 2006
12. ### Bryan OlsonGuest

Ah, thanks.

Bryan Olson, Mar 5, 2006