I believe that there is an analog portion to the circuitry before it
gets to the A/D converter.  And this analog portion includes a
programmable amplifier.  So the A/D is always giving you all the bits,
its just that at higher amplifications for higher ISO, the noise
factor is much higher.


On Wed, May 18, 2016 at 2:03 PM, John <sesso...@earthlink.net> wrote:
> I'm pretty sure you're over-thinking this.
>
>
> On 5/18/2016 2:37 PM, Larry Colen wrote:
>>
>> I expect that the answer to this question is "Yes, but it makes no
>> practical difference".
>>
>> Let's say a camera has a 12 bit D/A and at base ISO of 100 has 10 stops
>> of dynamic range.  In theory the chart would look something like this:
>>
>> DR ISO
>> __ _________
>> 10       100
>> 09       200
>> 08       400
>> 07       800
>> 06      1600
>> 05      3200
>> 04      6400
>> 03     12800
>> 02     25600
>> 01     51200
>>
>> Lets say that you are doing a macro photo of a grey card. It's lit from
>> the side, to show some texture, so you end up with 3 stops of DR coming
>> off the card. Lets's also assume that you expose to the right, so those
>> three stops are represented by the three MSBs.
>>
>> At ISO 100, your data will vary in range from  0x200 to 0xFFF.  However,
>> at ISO 12800 your data will vary from 0x001 to 0xFFF, so you will have
>> all 12 bits of data to express those three stops of light.
>>
>> I might be making some critical mistakes involving noise floor, or just
>> from not looking at the difference in the numbers of photons represented
>> by a least significant bit. However, there is a good chance there there
>> is someone on this list who is nerdy enough to have already worked out
>> the math.
>>
>>
>
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