5
\$\begingroup\$

In microscopy we often can sacrifice spatial resolution to collect more photons per pixel. Using sCMOS sensor and binning by factor of 2, we can get bump (up to 4x) in our signal-to-noise ratio (SNR). There is a bit of difference between CCD and CMOS here, but since all novel DSLRs I know use CMOS, it doesn't matter.

Now question. Does taking underexposed picture (say, 1 stop: 1/60s instead of recommended 1/30s) with modern DSLR and binning (downsampling) image afterwards will give same result in terms of image quality as at recommended exposure?

What about shooting at ISO 1600 and downsampling, how close result will come to image taken with ISO 800 and 2x longer exposure?

Is it realized in this way or similar in any practical cameras as a function?

PS: 1) I will experiment myself as soon as get hands on my fancy new dslr 2) googling referred me only to some shady forums without much useful discussions 3) I know that if you underexpose, you sacrifice dynamic range, essentially putting all those intensities into lower part of histogram, which effectively results in 6-8 bit images instead of 12 or 14.

\$\endgroup\$
5
  • \$\begingroup\$ What is limiting your shutter speed, why go 1 stop faster than the optimal? \$\endgroup\$
    – Iliah Borg
    May 2, 2015 at 13:15
  • \$\begingroup\$ no, i can shoot even at 1/8000. Problem is that exposure like that in dim light will fill only fraction of censor's dynamic range. For, say, portret you want your model's face to fill almost all dynamic range. \$\endgroup\$ May 2, 2015 at 23:11
  • \$\begingroup\$ I'm trying to ask what is the limiting factor to go for a slower shutter speed. \$\endgroup\$
    – Iliah Borg
    May 2, 2015 at 23:36
  • \$\begingroup\$ @IliahBorg motion blur, for example, more pronounced at slower shutter speeds. \$\endgroup\$ Aug 12, 2015 at 11:00
  • \$\begingroup\$ This interesting question has two mutually exclusive answers, both with an equal score... who's right? \$\endgroup\$
    – pipe
    Mar 2, 2020 at 0:12

2 Answers 2

2
\$\begingroup\$

No, it won't. As modern DSLRs doesn't actually do analog binning, reducing resolution won't make up for the signal-to-noise ratio loss from underexposing.

Underexposing

When you underexpose one stop, the signal-to-noise ratio roughly halves. When the singal halves, the signal induced noise is reduced with the square root of that (ref), but the readout noise is unaffected.

Binning

Modern DSLRs doesn't do analog binning in the chip (ref). The resolution reduction is done after reading the data from the pixels, which means that the binning doesn't increase the signal-to-noise ratio.

High ISO

Using the ISO setting instead of underexposing will actually give a slightly better result than underexposing. The reason is that the analog signal is amplified before readout, so this means that the signal induced noise is also amplified, but the readout noise stays the same.

\$\endgroup\$
2
  • \$\begingroup\$ Also, using high ISO and downsampling allows you to do more noise-reduction in software increasing the final image quality. \$\endgroup\$
    – user22723
    May 3, 2015 at 13:23
  • \$\begingroup\$ Why the downvote? If you don't explain what it is that you think is wrong, it can't improve the answer. \$\endgroup\$
    – Guffa
    Aug 5, 2015 at 14:01
4
\$\begingroup\$

Yes, binning will always increase dynamic range. There's no need to underexpose. Binning (i.e. adding) 4 (presumably 2x2) input pixels to create 1 output pixel gives you 4x signal but only 2x the shot noise. Shot/Poisson noise is proportional to the square root of signal level, so this noise source is only halved not quartered. Other noise sources may also exist, but for a well exposed pixel the shot noise is predominant. So your final signal-to-noise ratio will be doubled after a 4 pixel binning.

\$\endgroup\$
1
  • \$\begingroup\$ The purpose of underexposing and then restoring exposure later is to gain the additional dynamic range at the bright end of the histogram, if that's where you want it. \$\endgroup\$ Jan 28 at 21:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.