Bob Smith wrote:

>> If you too can't grasp the principle that photons are converted to
>> electrons
>> proportionate to the intensity of light, then what can I say?
> 
> That's true, but as I understand it, the analog to digital conversion
> is not achieved by counting electrons or photons and that's the crux of
> our disagreement.
Well maybe I didn't express myself too well, however, you're absolutely
correct. In a previous post I thought I had made it clear that the stored
electrons were read out as voltages?

> If it we did simply count electrons or photons, then
> there'd be no need for an a/d convertor.
Correct.

>  This would simply be a purely
> digital device and we'd be capturing numbers at the sensor.   I don't
> profess to be an expert in the intricacies of sensor workings but its
> my understanding that we are measuring an analogue signal produced by
> the sensors response to light.
Yes!
 
> The more electrons produced, the
> stronger the signal, but we aren't literally counting electrons.
No we're not, well not exactly. I believe the device creates a voltage on
the basis of how many electrons it receives. More or less electrons
represent subject brightness's and hence voltages proportionately and that
is the message I have been trying to communicate!

> I'd 
> be happy to be shown otherwise.  I'm curious about how this stuff works
> and its why I participate in threads like this. I assume the strength
> of current produced by a given cell changes based on the amount of
> light hitting it.
Exactly.

> In the a/d conversion the highest number available
> it assigned to the maximum signal.  Bit depth determines how big that
> number is and thus how many potential definable steps there are between
> the lowest and highest measurable signal.
>  As someone else suggested
> its just the sampling rate.  It doesn't change the endpoints of the
> analog signal only the precision with which it's sampled.

Yes I agree, however, there are definitely more brightness levels, as
reflected from the subject, recorded more accurately in a higher, rather
than lower, bit-depth device.

Best regards,

David Kay 

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