Myth busted: smaller pixels have more noise, less dynamic range, worse diffraction, etc.

[Daniel Browning]

You know, Daniel, I'm pretty good at math, but I don't really get a couple of passages -maybe out of inexperience in the field- could you please send me -via email if you prefer (i'll send you my address)- all the calculation with a legend of the symbols and everything written out plainly?

Certainly. I did use a lot of shorthand. You can get a very thorough explanation of all the points here:


Noise, Dynamic Range and Bit Depth in Digital SLRs


But I'll add my own brief explanation:

by which I mean: "um = micrometers" (in reality the "u" should be the greek letter mu)

Yep: µm.

and for example the explanation why S/N of the 5DII is taken to be 1:1

The noise of 23.5 electrons per pixel was taken from Roger Clark's measurements posted on his clarkvision.com web site. Signal was chosen to be 1:1 (23.5 photons : 23.5 electrons) because that is traditionally the lower bound of dynamic range. It shows the effect of read noise greatly. The calculation can be repeated with any other signal that is smaller than the full well capacity (e.g. 10,000 photons), but then only photon shot noise will affect the image and not read noise. In other words, 1:1 was chosen to show the effect of read noise.

• 6.4 µm is the size of the 5D2 pixel.
• 23.5 electrons per pixel is the read noise of the 5D2 pixel.
• 23.5 photons is the arbitrary signal chosen to demonstrate the effect of read noise.
  
• 2µm S is the Signal (in photons) of the 2-micron LX3 pixel.
• Scale factor is how signal scales with pixel size.
• 2µm N is the read noise (in electrons), taken from Emil Martinec's measurements.
• 2µm S/N is the signal-to-read-noise ratio when given the same signal as the 5D2 pixel (23.5 photons).

The demonstration is incomplete because it doesn't demonstrate the effect of photon shot noise (which is always sqrt(S)) and how that contributes to total noise:


total noise = sqrt(photon shot noise squared + read noise squared)