Found something that confused me and so I thought the crowd here can probably answer this one since its camera-related and technical at the same time.

How can dynamic-range be larger than sensor bit-depth?

Someone sent me the DXOMark results for the Pentax K-5 which shows 14.1 EV of dynamic-range at its lowest ISO. However, given that the sensor is 14-bits, this does not fit with my intuition... It seems strange that a linear device like a CMOS-sensor can capture more DR than it has bits. Would it have a sparse dynamic-range, skipping EVs in the middle?

  • \$\begingroup\$ The DxO Mark score for Dynamic Range under the print tab is an interpolated theoretical score, not an actual measurement. Please read the page on their site where the scores and how they are computed is explained. The DR under the screen tab is a more realistis number for a 14-bit sensor: 13.44 EV. \$\endgroup\$
    – Michael C
    Aug 23, 2014 at 19:47
  • \$\begingroup\$ See this answer and comments: photo.stackexchange.com/a/47512/15871 \$\endgroup\$
    – Michael C
    Aug 23, 2014 at 20:07

4 Answers 4


Cambridge in Colour has a very good article on this. If the sensor has a linear A/D converter, the bit depth would cap dynamic range at at 14 EVs as a theoretical limit. However, if it is non-linear, then the bit depth doesn't necessarily correlate. From that, I think we can determine that the sensor in the K-5 doesn't have a linear A/D converter.

I can say, from personal experience, that this sensor definitely has enormous dynamic range. I managed to recover an image that was close to 8 stops underexposed on the K-5.

  • \$\begingroup\$ Are you sure it wasn't ISO1600 not ISO16000 you were shooting at the rest of the time? This would make the image just over 4 stops overexposed, not 8, and correlate without fact you used exposure compensation of +4 in ACR. It's still impressive, I just want to make sure the numbers are correct. \$\endgroup\$
    – Matt Grum
    Nov 9, 2010 at 15:40
  • 1
    \$\begingroup\$ Yep, it was 16000, I have another image from the sequence (and subjects) with the same aperture and shutter speed for reference (I'd post it, but I'm not at home to get it). ACR only allows for a 4 stop adjustment on exposure, so I also had to push the fill light to 100 to get more detail. Hmm, maybe I should I update the article with some intermediate steps. I've seen a similar example with a deliberate 10 stops under and that's what triggered my upgrade now rather than in January. :) \$\endgroup\$
    – Joanne C
    Nov 9, 2010 at 15:59
  • \$\begingroup\$ +1, that Cambridge in Color article is excellent. It does a great job of explaining the value of larger photosites (deeper "wells") and how they affect dynamic range. It should probably also be noted that most sensors are not purely linear, most have an attenuated A/D conv. curve (S-curve) as you reach the extremes of shadow and highlight. In RAW, a digital sensor can capture a lot of data that can later be recovered, as your article demonstrated. \$\endgroup\$
    – jrista
    Nov 9, 2010 at 18:38
  • \$\begingroup\$ @jrista - Cambridge in Colour was one of the first photography sites I ever hit when I started into dSLR shooting. I keep going back to them, very well written and easy to follow stuff. \$\endgroup\$
    – Joanne C
    Nov 9, 2010 at 18:59
  • \$\begingroup\$ @John: Agreed. CinC is a great site, and very well written at a level that is useful to both beginners and experienced photographers. Thats a difficult thing to do. \$\endgroup\$
    – jrista
    Nov 9, 2010 at 19:28

How can dynamic-range be larger than sensor bit-depth?

Dynamic range is the logarithm of the ratio between the brightest and and the darkest intensities on the linear part of the sensibility curve. There may be other definitions, but in general it is derived from the ratio of two intensities, objective physical properties of the scene. It is a real number.

Bit-depth is the number of bits per channel used to quantize the continuous variable. More bit-depth gives more distinct shades of gray in between. It is purely a question of how an image is represented in computer memory.

Dynamic range reflects how much contrast the sensor can register. The bit depth reflects how many distinct colors the camera can “give names” to. Or into how many pieces the camera can divide the range. If a camera were a ruler, then the dynamic range would be the (logarithm of the) length of the ruler, and the bit depth would be the (logarithm of the) number of marks along its edge. And you can divide the length into as many pieces as you like. Similarly, the bit depth does not have to be the same as the dynamic range.

If dynamic range is S EV, and bit depth is n, then it means that the camera can register scenes with contrast at least as large as

E_max/E_min = 2^s

(Actually a little more if you use also the non-linear part of the sensor response curve). And you can theoretically distinguish

N = 2^n

shades of gray.

I own a compact camera which can write 12-bit RAW. Inspite of the high bit-depth, its dynamic range is very modest. You can imaging an opposite situation, when the sensor can register a high contrast scene, without over- and underexposure, but if the bit depth is low, that scene will be represented with few intermediate colors.

  • \$\begingroup\$ +1, great answer. One tip: I believe the word you need in place of "discretize" is "quantize": Quantize -verb: Math,Physics . to restrict (a variable quantity) to discrete values rather than a continuous set of values. \$\endgroup\$
    – jrista
    Nov 9, 2010 at 18:35
  • \$\begingroup\$ Thank you. My English is far from perfect, but it seems in the world of computing and mathematics discretize is more appropriate when the continuous space is replaced with an equivalent discrete space for the purpose of calculation en.wiktionary.org/wiki/discretize (e.g. a real number with an IEEE floating point value or an integer). Discretization is a software engineering decision. Quantized variable for me, instead, is a variable for which all values are forbidden except some. So “quantized” sounds like a physical restriction for me. But may be you are right. \$\endgroup\$
    – sastanin
    Nov 9, 2010 at 19:13
  • \$\begingroup\$ Technically speaking, a sensor does "quantize" light into specific, physically limited "buckets". If we assume a 12-bit RAW image, there are 4096 discrete "quanta" that you can 'discretize' into. Where as discretize would mean that you could concert a real space into a variable number of discreet spaces, with a sensor, the discrete space is fixed, and there are only 4096 specific discrete values you could convert the analog space into. It may be a moot point, but I do think quantize is more applicable here. ;) \$\endgroup\$
    – jrista
    Nov 9, 2010 at 23:01
  • \$\begingroup\$ OK. I'm convinced. \$\endgroup\$
    – sastanin
    Nov 10, 2010 at 11:29
  • \$\begingroup\$ @jrista While we're discussing English, the word you want is "discrete", not "discreet". \$\endgroup\$
    – coneslayer
    Nov 11, 2010 at 2:56

Firstly to be clear, dynamic range has an inverse relationship to noise - low noise (all else equal) leads to a greater dynamic range. Noise comes primarily from the sensor electronics (read noise, dark current noise), from the discrete nature of light (photon/shot noise) and from conversion from analogue to digital (quantisation noise).

DXO mark dynamic range scores are based on the difference between the light intensity required to saturate the sensor and the light intensity at which the SNR hits 1:1 (i.e. the point at which the signal equals the noise)

You would expect that in the absence of shot noise and readout noise that the DR of a sensor with a linear response would equal the bit depth. Given the K-5's score in the presence of these sources of noise indicates to me that the image pipeline has a moderate degree of nonlinearity (all sensors have some inherent nonlinearity), probably engineered that way to increase the dynamic range.

Nonlinearity helps escape the bit depth limit, bit what you gain in gradations in the shadows you lose elsewhere in the tonecurve (albeit probably somewhere less important). There is no such thing as a free lunch!

With regard to the K-5, it's class leading at low ISO sensitivity, which is determined mainly by read noise. It's really great to see the manufacturers turning their attention to this area and it thoroughly deserves the attention, however dynamic range at higher ISO sensitivities is dominated by photon noise which is only countered by capturing more light, so large sensors will always have an advantage here. As some people predominantly shoot ISO400 and above so it's worth bearing this in mind!

  • \$\begingroup\$ I agree, at ISO80, the K-5 is stunning and stacks up well with some medium format and full frame for the lower ISO ranges. As the ISO starts to jump, it starts to lose the lead. However, it still manages to stay quite close, so that's quite the achievement for Sony (who makes the sensor) and Pentax (who implemented it). The D7000 has very similar characteristics given that it's a variant on the same sensor and Nikon did a very good job on their implementation. \$\endgroup\$
    – Joanne C
    Nov 9, 2010 at 18:57

The "dynamic range" (DR) is not an absolute characteristic.

The most coarse definition of DR is "ratio between the brightest and darkest gray intensities which sensor may record just fine".

The DR of a digital sensor is derived from two measurements:

  1. clipping intensity [for the most sensitive channel] at a given colour temperature (DxO is most probably using D65);
  2. intensity which produces boundary amount of noise (i.e. if it is any darker the noise is unacceptable).

Then, you have two ways of computing DR of digital image.

  • Dumb way is using pixel data to compute the noise ("screen" measurements on DxO site). If you compute the DR of the linear sensor with X bit ADC this way it can in no way be bigger than X EV.
  • Smart way (which is the only possible way of comparing photos from cameras with different resolution) is taking the resolution into account when computing the noise ("print" measurements on DxO). The DR is not limited by ADC this way, one can potentially make a camera with a bigger sensor and the same ADC and it will have bigger percepted dynamic range.

So, you won't find any camera the "screen" DR of which expressed in EVs exceeds the ADC resolution expressed in bits.

Comments on other answers:

From that, I think we can determine that the sensor in the K-5 doesn't have a linear A/D converter.

There is not a single digital sensor with nonlinear A/D conversion developed. Every tonal conversion which camera does (including the special output modes of cinema cameras and Sony A7 series in particular) is done using the discrete data.

The Kodak DCS Pro 14n has dual slope ADC operation mode in which the output is piecewise linear.

Given the K-5's score in the presence of these sources of noise indicates to me that the image pipeline has a moderate degree of nonlinearity

K-5 has perfectly flat response (as any other camera with probably the only exclusion being Kodak DCS Pro). I have measured it myself.

Note: DxO Labs are not resizing or printing anything for "print" measurements, they rather use resollution coefficient in the formulae. Sidenote: in this post "linear" is not "logarythmic".


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