Is all digital sensor noise thermal?

And what is the relationship between temperature and noise? For example, as we cooled a CMOS towards absolute zero, would we discover that high-ISO images would be less noisy? Would they converge towards a truly noiseless image at maximum ISO and absolute zero?


4 Answers 4


It depends: On the sensor design, the ISO setting selected, the length of exposure, the intensity of the light entering the camera, etc. Thermal noise can be anywhere from almost none of the noise in a digital photograph to almost all of the noise in a digital photograph.

If the photo is taken relatively quickly with moderate light intensity at high ISO settings most of the noise will be Poisson distribution noise (shot noise) that is caused by the random nature of the distribution of photons as they strike the sensor. Shot noise is not related at all to thermal considerations.

On the other hand, if the photo is taken with long exposure times and low light intensities at low ISO settings, most of the noise in the resulting image will be read noise. That is, it will be noise caused by the camera's electronics. All such noise is influenced by heat. The warmer the sensor, analog amplifiers, and digital processing units are, the more dark current they will generate that will be recorded as noise.

If a photo is taken with the lens cap on and the viewfinder covered all of the noise in the photograph will be read noise, which is affected by thermal conditions. This is just one among many reasons why testing a camera with a lens cap on is pretty much useless for predicting actual noise in real world scenarios in any meaningful way. To measure a signal-to-noise ratio, one must include signal (light) as well as noise (thermally induced dark current).

As a camera is cooled, the improvement would be more noticeable in low ISO images than in high ISO images. This is because low ISO images tend to have more dark current (read) noise while high ISO images tend to have more Poisson distribution (shot) noise. The improvement would also be more noticeable with images made using very weak light sources for longer exposure times, such as astrophotography, than with images made using very strong light sources for shorter exposure times.

Since shot noise is not affected at all by temperature, but by the very nature of light and the way photons oscillate in waves as they move, you'll never be able to have a zero noise image. In theory one could probably cool an image sensor and related electronics to the point that dark current noise would be undetectable.


The dominant non-thermal noise sources are photon-shot noise and cosmic rays induced noise, but this only affects images taken by professional astronomers, such as by the Hubble Space telescope. All other noise sources, which then also includes shot noise due to the discrete number of dark current electrons, is ultimately thermal in nature. In ordinary astrophotography, the photon shot noise is always hidden in the thermal noise. E.g. the number of photons from a tenth magnitude star entering a 50 mm aperture lens is about 2500 per second. So, even a 1 second exposure would only have a fluctuation of the photon number of the order of sqrt(2500) = 50, the relative fluctuation is therefore just 2%.

We need to consider a 15th magnitude star (ten thousand times dimmer than the weakest stars visible to the naked eye) to get to a 20% fluctuation due to noise at 1 second exposure. But, of course, in such a picture the star is nowhere to be seen, all you see is read noise and shot noise, the latter is then not induced by the photon number fluctuation, rather by thermal fluctuations in the dark current. By improving the exposure you can reduce these thermal noise effects to make the star visible, but then the photon shot noise will also be reduced.

Suppose that a camera is taking pictures where the noise is dominated by photon shot noise, how we would we know that this is indeed the case? The crucial test would be to take a few dark frame shots. These dark frame shots should then only contain hot and stuck pixels and apart from that almost no noise at all. So, subtracting two dark frame exposures should yield almost totally dark pictures.

A camera cooled to absolute zero would, however, still be affected by read noise and dark current noise besides photon shot noise. This is then caused by quantum fluctuations rather than thermal fluctuations. Any electronic circuit is subject to quantum fluctuations at absolute zero. The simplest example to demonstrate this is an LC circuit, in such a circuit the current and voltage oscillate at an angular frequency of 1/sqrt(LC). When we then describe this system according to quantum mechanical description, we then find that just like any oscillating system, this system has energy levels given by (n+1/2) hbar omega. In the ground state the system then still have an energy due to voltage and current fluctuations of 1/2 hbar omega.


There is dark current, as inkista wrote in a comment, and there's quantization noise at the digitizer, and there's gain noise in the analog amplifiers, and there's readout noise (slightly less than 100% of collected charge is extracted).

When people write "scene-limited" or "photon-limited" noise, what they mean is that the photon shot noise is much greater than the combination of all the electronics noise sources. You can never do better than being photon-limited. Luckily, if you read up a bit on energy per photon, you'll see that an SNR of sqrt(number of photons) is a very large number – unless you're imaging distant galaxies or some such :-)

  • So assuming we're not at the photon limit, then the dominant sources of noise are all electronic, and each of those noise sources is (I'm guessing) positively correlated with temperature? E.g., readout noise would presumably drop to zero once the conductors involved are cooled below their superconducting temperature. What about the other noise sources?
    – feetwet
    Jun 28, 2016 at 16:50

No. The noise is in the photon count, and it is the square root of the number of photons. So if you get 9 photons, the noise level is 3 (33%); if you get 10000 photons, the noise level is 100 (1%).

Photons are not perfectly distributed over space, but randomly, and even at perfect 0 K, different receiver buckets will catch different amounts of photons each time you try.


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.