How is exposure mapped to 12-bits?

[Russ Lasson]

This article that I found referenced in a thread here on RED User says that brighter exposure levels have more numerical values in a 12-bit linear RAW file.

http://www.luminous-landscape.com/tutorials/expose-right.shtml

If the first f-stop has 2048 values, but the fifth f-stop only has 128, that sounds more exponential or logarithmic than linear.

If it was linear, wouldn't each f-stop have the same number of values to capture the data? Isn't the noise generated from lower light levels hitting the sensor thus creating a lower signal to noise ratio?

I don't really want to start an advanced math thread here. So my real question is, with the RED One, does it really have more values for highlights than shadows like this article suggests with RAW cameras?

Thanks,

-Russ

[jbeale]

It is the f-stop scale that is logarithmic. Going up or down one f-stop doubles or halves the amount of light going through the lens. The camera converts whatever light it sees to digital counts in a linear way, which is why half the A/D range is within the brightest full f-stop, because that is fully half the light there is.

Log units like f-stops and EV (exposure value) are often used in photography because your eye responds to light in a logarithmic way.