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[filmscanners] RE: Density vs Dynamic range



Peter,

> > CCDs, in and of them selves, don't have anything to do with number of
> > bits,
> > as they are analog devices, but their dynamic range does...  If the CCD
> > has
> > a dynamic range of 5000:1, it will require a 13 bit A/D to be able to
> > extract the full dynamic range of the CCD.
> >
> > Regards,
> >
> > Austin
>
> Austin, I have kept out of this up to now, but I have step in to disagree
> with this as it's just plain wrong.

Well, I know it's just plain right.  And, funny enough YOU agree with what I
said in your supposed disagreement!  I don't mind a misunderstanding, or
questioning what I wrote, but if you're going to blurt out that something is
"just plain wrong" you ought to read carefully, and understand, what was
said before saying that.

I know exactly how these things work.  I design with them for a living. Even
most of the new CCD specs give a number of bits!  Go read the Kodak spec for
the Kli14403 sensor.  It says "Dynamic range - 12 bits" and later goes on to
say "Dual mode operation is provided to increase dynamic range to 14 bits".
They base this on the dynamic range which is based on noise...exactly what I
said above.

> The range of the CCD can be represented by any number of bits.

Yes it can (as I've said many times, if YOU read my posts), but that has
nothing to do with what I said.  READ IT AGAIN.  "To extract the FULL
dynamic range"...  NOTE THE WORD FULL.  There IS FOR A FACT an optimal
number of bits, as I said, as Kodak also says.  If you use less, you will
NOT get the FULL dynamic range, and if you use more, you will not get any
better data.

> What will
> vary is the degree of resolution of different values within the range.

Yes, but it gets to a point where more bits does you NO good because you are
simply reading noise.  That is what limits dynamic range.

> The dynamic range of the scanner is determined by how much noise is
> generated in the CCD.

Correct, and if you understood what I wrote, I said the EXACT same thing.  I
fail to see where I was "just plain wrong" as you say the EXACT same thing I
said.

Plain and simple, do you agree that a dynamic range of 5000:1 REQUIRES 13
bits to represent every integer value between 1 and 5000?  If so, then
where's the problem?  If not, then plain and simple, why not?

Austin

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