Digital BW, The Print

Truman,

> Actually the sensor see's photons and electrons or a potnetial
> difference (voltage) or a current flow (depending on the particulars of
> the sensor and the A/D ) results which is what is recorded.

I don't quite understand what you are saying here...  Of course the sensor
"sees" photons, as in light, it does not "see" electrons, but during the
integration period, an image is obtained by gathering electrons generated by
photons incident upon the photodiodes.  The output is a voltage based on the
number of electrons generated.

> In most
> imaging systems I have worked on, there is a circuit that adjust the
> output the voltage of the image sensor to maximize the dynamic range of
> the sensor into the A/D.

That's exactly what I just described in my other post.  It is typically not
an adjustable circuit, it's simply a voltage gain (or attenuation) depending
on the output voltage of the CCD and the input voltage requirements of the
A/D that simply matches them to, as you say, maximize the "range" (I'm not
quite clear that it actually maximizes the "dynamic range", I'd have to
think about that, but it certainly matches the two ranges) of the sensor
into the range of the A/D.

> So my question is do the scanners on the market have such a
> circuit or is the photon range coming through the negative mapped
> linearly (or almost linearly) onto input range of the A/D?

The ALL have to have a circuit to do what I described, unless the voltage
from the CCD is the exact as the input requirement of the A/D.  That means
that "the photon range coming through the negative" is mapped linearly (yes,
almost linearly ;-) into the input range of the A/D.  I believe that is what
I said in my last post, and provided an example.

There are some scanners that do not do it that way, as I believe the
Leafscan actually adds more gain in the higher density regions, though I'm
not convinced that does much in reality, as you are still limited by noise.

> That is pretty critical to the question if a 16 bit A/D really a
> 16 bit A/D.

No, there are no scanners of the typical type we use that use actual 16 bit
A/Ds, and if they do, they are reading noise in the low bits if they do use
a 16 bit A/D.  Typically, scanners today are 12 or 14 bits at best, as
that's all they need at this point in time with current CCDs.

I think the issue you may not be getting here, is the minimum increment of
discernability is noise in the CCD, and that's what dictates what happens
downstream  If you know the noise level of the sensor, say, it's got a noise
floor of 35 electrons, and it has a saturation of 230,000 electrons.  That
basically gives you  230,000/35 or 6571 discernable steps.  You only need a
13 bit A/D to be able to represent those 6571 different values, plus one bit
for good measure ;-) (and they don't make 13 bit A/Ds that I've seen), or a
14 bit A/D.  That will give you a TRUE density range of 3.8 (not one based
on the number of bits in the A/D), which is actually pretty good.

Regards,

Austin

> Truman
>
> Austin Franklin wrote:
>
> > Hi Truman,
> >
> > >
> > No, not at all.  How did you arrive at that conclusion?
> >
> > The CCD will output a voltage, and that voltage will pretty linearly
> > correspond to the number of electrons the CCD sensor "sees".  The
> > lower the
> > number of electrons it sees, the lower the output voltage, and
> the higher
> > the density that it saw...  The converse is true as well, the higher the
> > number of electrons it sees, the higher the output voltage, and
> the lower
> > the density that it saw.
> >
> > Austin
> >