underexspose or up iso

I think JasonW has a good point on each 12 bit value only recording 12 bits to represent each pixel. Underexposing by 2 stops wastes half the information the pixel can record, so it'll be less accurate than it could be.

So, you would agree then that increasing the ISO setting is preferable to an under-exposure at a lower ISO? (which I believe is the consensus at this point)

I've been trying to follow this thread without my head exploding. I will try to summarize, to see if my understanding is accurate. As I understand it, this is how a digital image is created:

1) Light hits each pixel on the sensor, with its associated color filter, and voltage is generated from the energy.
2) The image processor will amplify this voltage (still in analog form), depending on the ISO setting. However, some of the voltage present is not actually generated by light, and this extraneous voltage is also amplified, resulting in what we refer to as digital noise.
3) This amplified voltage is then measured by the image processor and recorded as a numeric value from 0 to 4,095 in the RAW file. This numeric value is logorithmic, such that there are more distinctly different values in the brighter end of the dynamic range than the darker end.
4) RAW conversion software can increase or decrease this numeric value as the user adjusts the exposure adjustment slider. Increasing the values to simulate a brighter exposure will also make the digital noise more noticable.

So the basic question is, is it better to amplify the analog voltage signal coming from the sensor, or is it better to simulate this by multiplying the numeric value recorded in the RAW file?

I'm still not sure if I know the answer, but do I at least understand the question?

We're of the consensus that it's better to amplify the incoming signal - that is, use a higher ISO. Adding EC in a RAW editor is not preferred, and should only be done when necessary.


(Pardon the rest if you know it. I just got on a roll... For the technical, I know I'm off by one, and it's not colors it's light levels. I know. I just think it's easier to understand this way.)

However, to clarify part of your #3. It is not logarithmic, but more like what you're used to with Stops. A stop is a halving or doubling of light hitting the sensor (or film). What is important here is that a digital sensor basically counts the number of photons hitting a photocell. If twice as many photons hit then it's twice as bright. Simple right?

Now think about counting photons. (Simplified a bit.) A pixel will appear in the darkest stop (the left-most fifth of the histogram) whenever 256 photons or less hit that photocell. A pixel will appear in the next brighter stop whenever the number of photons is between 257 and 512. 512 photons hitting the photocell is twice as many as 256, thus it would be twice as bright- 1 stop. The next brightest stop - the middle fifth of the histogram - goes from 512 photons to 1024. The next stop is 1025 to 2048. A pixel is in the brightest stop whenever anywhere from 2049 photons to 4096 photons hit the photocell. If more than 4096 photons hit a photocell it's simply recorded as 4096, which is the most the photocell can count to, and is considered blown out.

Let's go back to that darkest stop for a moment, since small numbers are easier to understand (and type). Let's say that (pick a random number) 26 photons hit a particular photocell during one exposure. The photocell next to it got hit by 27 photons. So that second photocell will produce a pixel (ignoring Bayer Interpolation for this discussion) which is just a little bit brighter than the first. The 2 pixels will be *slightly* different colors. A third photocell got hit by 28 photons, again a little brighter, and again a slightly different color.

So if we think about that darkest stop again remember that it covers all photocells that got hit by 256 photons or less. We know that 26 photons produces a lightly different color than 27 photons. Suddenly we realize that the darkest stop can only record a maximum of 256 different colors. There's aren't many variances in 256 colors, so there's not much detail possible, since detail is just a difference between two pixels. That is, if all the pixels were the same color then there's no detail - it's flat, all one color. If a row of pixels is a different color than the next row then that's a line, a detail.

So the darkest stop can only hold 256 different colors, so 256 levels of detail.

Now look at the brightest stop. It ranges from whenever 2048 photons hit to 4096. That's 2048 possibilities of different colors. And 2048 levels of detail.

Every next-brighter stop can hold twice as many possible colors, thus twice as many possible levels of detail.

This is the big reason why people suggest to "expose to the right." You get more possible colors, more details.


Man, I hope this all makes sense. I'm in work and heading into a meeting. Feel free to correct me where I'm wrong (other than the obvious mistakes made for simplicity).

Thanks, Scottes. I think the fog over my brain is starting to clear.

In your summary of the basic question, I first translate the expression "amplify the analog voltage signal" to mean "use a higher ISO setting". Then, I translate the expression "multiplying the numberic value recorded in the RAW file" to "compensate for underexposure by sliding the exposure slider". With those two translations I would say that you have the issue in hand.

My simple answer to that question, based on simple tests, is that using the higher ISO is preferred for two reasons - it gives better shadow detail and it gives less noise. What more could you ask?

Here is the URL for Reichmann's Web article on this topic:
http://www.luminous-landscape.com …-series/u-raw-files.shtml

Here is a similar document by Reichmann:
http://www.adobe.com …transistion_fro​m_film.pdf

Yes, that's my current opinion, and it's what I do.