Dynamic Range -- what it is, what it's good for, and how much you 'need'

Started Oct 17, 2011 | Discussions thread
FrankyM
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Re: Wrapping it up.
In reply to Great Bustard, Oct 22, 2011

Thanks Joe, now I understand better what you mean. Really it boils down to how read noise scales with pixel size (since shot noise is invariable for a given area). Does it scale linearly with pixel area as in your example?

Great Bustard wrote:

FrankyM wrote:

What I'm interested in is mathematical proof of your statement above (academic interest only).

".. I can tell you, for a fact , that more pixels for a given sensor size and efficiency results in more IQ all the way around."

Sorry, didn't notice your post until just now. OK, let's consider two sensors of the sames size that are equally efficient (same QE, same read noise / area, same saturation / area), where Sensor A has four times the number of pixels as Sensor B.

Without loss of generality, assume QE = 1 and the read noise for a pixel of Sensor A is 4 electrons. Let's also assume that 9 electrons fall on each pixel of Sensor A, and thus 9 x 4 = 36 electrons fall on each pixel of Sensor B (just to make the computations more simple).

The photon noise for one pixel on Sensor A is sqrt 9 = 3 electrons. This makes the total noise sqrt (3² + 4²) = 5 electrons / pixel. The total noise for four pixels on Sensor A is sqrt (5² + 5² + 5² + 5²) = 10 electrons, giving us an NSR of 10 / 36 = 28%.

Now let's consider one pixel of Sensor B. The photon noise will be sqrt 36 = 6 electrons, and the read noise is 4 x 4 = 16 electrons. This gives us a total noise of sqrt (6² + 16²) = sqrt 292 = 17 electrons, resulting in an NSR of 17 / 36 = 47%.

So, more pixels on an equally efficient sensor result in less noise / area. It is interesting to note that if the read noise were to scale with the linear dimension of the pixel, rather than the area, then the noise would be the same.

I guess we've hashed this to death.

Pretty much!

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