Digital BW, The Print

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. 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 can result in shot noise coming out of the 
senor in some cases but it maximizes the dynamic range of the imaging 
system. 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?

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

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
>