The highest ISO camera that I'm currently aware of is the Canon ME20F-SH, which goes up 4.56 million ISO (correct me if I'm wrong). In camera, it measures light sensitivity in dB gain instead ISO. Why is it measured in this unit instead of traditional ISO? Ive never heard of using dB gain to measure light sensitivity before this, so what exactly does a dB measure in photography?


2 Answers 2


The sensitivity measured in “ISO” is realized in digital cameras by an amplifier with a variable gain. The amplifier gain can be measured in dB as well, which may be more convenient for some applications. The “ISO” is basically a legacy measure mapping the experience from film photography where each film had a given, non-changeable, sensitivity which was standardized and given the well-known numbers (100, 200, 400, etc.). While the ISO keeps the relation to the light value by utilizing this standard, the amplification factor is based on the native sensitivity of the sensor and therefore makes a comparison between different cameras less convenient. If you know the “base ISO” of a camera (mapping to the sensitivity at amplification factor 1, or 0 dB, resp.), you can compute the dB value from the ISO and vice versa.

Edit: I just checked the data given at the Canon website and it seems that the native ISO of this camera is around 800. A rough calculation: They give a maximum amplification of 75 dB which translates to a voltage amplification factor of approx. 5600. This means a difference of approx. 12 ISO steps. Starting from ISO 800, this would bring you in the range of 4000000 ISO (3280000 exactly. 4560000 is one third of a “stop” more, meaning the base ISO would be 1000, but on the canon website they just claim “above 4000000”, no exact number).

  • \$\begingroup\$ So does one stop of ISO equal 1 dB, 2dB, 3dB, 4dB, etc.? \$\endgroup\$
    – Michael C
    Apr 7, 2017 at 21:33
  • 3
    \$\begingroup\$ 1 “stop” in voltage amplification means doubling the voltage value or 6 dB. 1 stop more light means doubling the power or 3 dB. We are calculating voltage amplification since the light power translates to a voltage level at the sensor, which will then be amplified and digitized. Probably I should not call the first one a “stop”, but it makes sense in this example. \$\endgroup\$
    – Chris
    Apr 7, 2017 at 21:43
  • 2
    \$\begingroup\$ To make it more clear, we have to assume here that the nonlinear sensor will cause an output voltage proportional to the incoming light power, which is then amplified and linearly digitized. The actual implementation may differ slightly. \$\endgroup\$
    – Chris
    Apr 7, 2017 at 21:59

In camera, it measures light sensitivity in dB gain instead ISO. Why is it measured in this unit instead of traditional ISO?

Unlike film, digital cameras use electrical circuits to capture images. Pretty much everything in that world is expressed in logarithmic terms to mirror the logarithmic behavior of the real world.

Ive never heard of using dB gain to measure light sensitivity before this, so what exactly does a dB measure in photography?

Decibels themselves don't actually measure anything absolutely; they have to be relative to something. For example, in wireless circles, radiated signal strength is often expressed in dBm, or decibels relative to a milliwatt. Sound pressure levels are expressed in dBA, or decibels relative to the threshold of human hearing on the "A-weighted" scale. None of these are recognized by any standards body.

Photography has no such convention, but it does have a less-used logarithmic standard enshrined in ISO 2240 based on a German standard called the DIN number. That and the linear ASA standard were incorporated into ISO 2240. The latter is what most people call the ISO Number. Pedants will point out that the ISO number technically includes both figures. I'll use ASA and DIN to avoid confusion.

DIN uses degrees (18°, 21°, 24°, etc.) to express its differences. Being logarithmic, every change of 3° represents an approximate doubling or halving of sensitivity depending on the direction. DIN 0° is equivalent to ASA 0.8 and DIN 3° is equivalent to ASA 1.6. Continue adding threes and you get into more familiar terrirory, with DIN 21° being equivalent to ASA 100 and DIN 24°being equivalent to ISO 200. This is exactly the same behavior as decibels, so it wouldn't be out of line to establish a convention called dBS (decibels relative to whatever sensitivity is defined as ASA 0.8*) and say that the sensor in a camera has a base sensitivity of 21 dBS.

This hasn't happened because photographers aren't clamoring to express these things in decibels. ASA numbers are preferred because there weren't many of them for a very long time and, being linear, they're easy for most people to think about.

As you noted, sensors are improving and the numbers are only going to get larger. I don't figure there will be a wholesale switch to DIN or decibels because, again, most people don't think well in logarithmic terms. If I had to guess, at some point somebody's going to say that ASA 128,000,000 compared to ASA 4,000,000 is a little much and we'll start using the SI suffixes and we'll talk about ASA 128M vs. ASA 4M.

*Interestingly, I had a difficult time finding a description of what exactly the ASA numbers represent in terms of amount of light reaching film. I've read other ISO standards, and while I'm sure they describe it in gory detail, I'm unwilling to part with US$60 just to find out.

  • \$\begingroup\$ But is using a linear unit like ASA really easier to think about in the context in which it's actually used? Or is it that the manufacturers like publishing exponentially larger numbers with each new model? ;) \$\endgroup\$ Apr 8, 2017 at 16:05
  • \$\begingroup\$ @junkyardsparkle The irony in that first question is that logarithms exist so we can think about things that behave exponentially in linear terms. :-) \$\endgroup\$
    – Blrfl
    Apr 8, 2017 at 17:58
  • \$\begingroup\$ I would suggest to invent a new unit that bases on a quadratic/square root scale. I think photographers might love it ;-) \$\endgroup\$
    – Chris
    Apr 9, 2017 at 20:59

Your Answer

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

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