30 March 2011

How Perfect is the Perfect Sensor?

A couple of weeks ago, the Great and Powerful Ctein published an article at The Online Photographer about how much better a sensor could be in low light using only reasonable extensions of existing technology. His conclusion was that we'd probably see a sensor capable of providing good results at ISO 64,000 in the next decade or two, if the relevant companies are sufficiently motivated to produce such a beast.

I thought it might be interesting to look at the problem from the other direction: How good could a sensor be, if limited only by fundamental physical laws?

To make the problem tractable, I'm going to assume that the limiting factor here is shot noise in the arriving photons. So, I will assume that our "perfect" sensor has a 100% quantum efficiency and is noiseless in collection, amplification, and digitization. (The sensor clearly must be made from the hide of a spherical cow.) Given that the best current low light sensor is the one found in the Nikon D3S, I'll assume that our perfect sensor will be a drop-in replacement for it (24 mm x 36 mm, 12 MP). I'll also assume that the light collection system (the lens and camera body) are lossless.

To calculate the ISO of this sensor, I'll use the "Sunny 16." This is a photographer's rule of thumb that the optimal exposure for full sunlight when using an f/16 aperture is achieved with a shutter speed equal to your ISO setting; so, if you're set to f/16 and ISO 100, you would want the shutter to be open for 1/100th of a second. 'll calculate the proper shutter speed for full exposure of our perfect sensor in sunlight at f/16, then deduce the ISO from that.

Our perfect sensor should achieve comparable dynamic range to the D3S, so we'll need 12 EV. From DxOMark's definitions, it may be deduced that the dynamic range is the ratio of the number of photons detected at 100% exposure to the number of photons detected when the signal-to-noise ratio (SNR) is reduced to 1. Given that we are assuming we are limited by shot noise, and that shot noise follows Poisson statistics, the SNR is the square root of the number of photons detected. Thus, if the SNR is 1, then only 1 photon (on average) is being detected during the exposure. So, if the lower limit of the dynamic range is 1 photon, and we want 12 EV worth of photons at full exposure. That's 2^12, or 4096 photons. The energy in each photon can be calculated by:



where h is the Planck constant, c is the speed of light, and λ is the wavelength of the photon. If we assume that the photons detected here have a mean wavelength of 555 nm (which is just about the peak sensitivity of a human's daylight-adjusted eyes), then each photon carries 3.58E-19 joules (J) of energy. 4096 photons is thus 1.47E-15 J.

The amount of light collected by a single pixel of the sensor is given by:



where P is the optical power on the pixel, A is the area of the pixel, Ω is the solid angle subtended by the lens aperture from the position of the sensor pixel, and L is the radiance of the object imaged onto the pixel. (See here for a decent discussion of this equation.) By definition, the F-number of the aperture, N, is given by



where f is the focal length of the lens and D is the diameter of the aperture. The aperture in this example is small compared to the focal length of the lens. So, in steradians, Ω is given by:



where d is the distance between the pixel and the aperture stop. We will make the approximation that f = d; this isn't strictly true, but will not affect our results here materially. Finally, the intensity of full sunlight, Isun, is about 1000 watts/m2. If we assume that this light is scattered evenly in all directions by the object being illuminated (not entirely accurate, again, but good enough for these purposes), then L = Isun/4π, and we can combine the previous equations to write:



We now have the power at the pixel in terms of known quantities. From this, we may calculate that P is 1.76E-11 W. Dividing this by the energy needed for full exposure of our perfect sensor, 1.47E-15 J, gives us a shutter speed of 12,000 s-1. From the Sunny 16 rule, this gives us an ISO of 12,000 with 12 EV dynamic range. Not bad, given that the D3S only manages about 8.5 EV at that ISO.

Suppose we would settle for 9 EV of dynamic range, comparable to the performance of the D3S at ISO 6400, which was Ctein's standard for "good results." At this dynamic range, our ideal sensor would have an ISO of 96,000. This is getting pretty close to Ctein's predicted ISO of 64,000; I hope he's right, and we see this sensor in the next decade or so, because it might be darn close to perfect!

6 comments:

  1. Hi, Nicholas,
    I believe that in your estimation you can get rid of Sunny 16 rule in favor of ISO value defition by DxOMark:
    http://www.dxomark.com/index.php/Learn-more/DxOMark-database/Measurements/ISO-sensitivity

    ReplyDelete
  2. I'm assuming I'm missing something here, but it seems like you are describing a monochrome sensor. Am I missing something or are you contemplating a foveon style sensor where the wavelengths of the photons* are determined within each receptor.

    Sorry about "the wavelengths of the photons", but I couldn't think of a more graceful way of putting it.

    ReplyDelete
  3. POD adnin, I am, indeed, neglecting the entire issue of wavelength determination here. (And your phrasing wasn't particularly nonstandard, FWIW.) I figured that, as long as I was imagining a perfect sensor, I'd imagine one that doesn't need a Bayer layer.

    Of course, since I do a fair amount of B&W photography myself, a sensor without a Bayer layer may have some appeal for me on its own...

    ReplyDelete
  4. And Sergei, I'm planning to do a follow-up post here with a bit more detail and information in the next few days. I'll include a comparison of my (admittedly hamfisted) trick for computing the ISO with the more rigorous definition to which you linked. I was looking for just such a definition when I first came up with the idea of the article, and I somehow didn't think to look for it on the DxO website. Thanks for the link!

    ReplyDelete
  5. Not to be nitpicking but wouldn't it make more sense to approximate scattering of an illuminated object into 2 pi space only? It's only illuminated by at best 2 pi incoming radiation and won't scatter from its back either.

    ReplyDelete
  6. organicdev, the issue of whether to use 2-pi or 4-pi steradians here is not always clear. Is the light that is scattered into the surface (not into the ~2-pi steradians a given section of surface presents to the outside world) lost, or does it eventually find its way back out of the front face of the object? In the limit where all of the into-surface scattering is lost, then the light should be considered to be scattered into 4-pi sr. In the limit where all of the into-surface scattering is reflected or rescattered back out, then the 2-pi sr value is more appropriate.

    ReplyDelete