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Say there's a camera sensor, in a given time, a pixel sensor (abstract a group of subpixels from Bayer filter into one indivisible pixel) can only be activated when more than 10 photons hit it, after that, the pixel sensor is able to record photons by a scale of 1, like 11 photons, 12 photons, etc. And at the maximum, it can record 16,000 photons.

What, then, is the minimum signal level? Would it be 10 or 1?

Or would it be impossible for the sensor to have smaller steps (1, in this example) than the activation requirement (which is 10)?

Another example (which might not fit in Photography) is a thermometer, if a thermometer can record from -30 degrees celsius to 70 degrees celsius, what would be the "dynamic range" of this thermometer?

Amarth Gûl
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  • Does this answer your question? What is the lowest level of luminous flux a camera can detect? – Saaru Lindestøkke May 09 '22 at 21:31
  • @SaaruLindestøkke I don't really think so. My question is less about detecting light, but more about the definition of minimum signal level for calculating dynamic range, and the situations when the scale of measurements varies. – Amarth Gûl May 09 '22 at 21:45
  • What photographic problem are you attempting to solve? That is, what photo do you want to take that answering this question will allow you to take? This question is more for Electrical Engineering SE or Physics SE (particularly since you also bring temperature measurements into it), not Photography as defined in this community's description. – Michael C May 10 '22 at 05:30
  • Ironically, "digital" sensors aren't really digital. They're devices that accumulate analog electrical charges and output analog voltages when read out. The information from a "digital" sensor is analog until those voltages are converted from analog signal to digital information by an analog-to-digital convertor. In contrast (see what I did there), "analog" film does have a 'cutoff' number of photons that must strike a film grain within a specific time interval before the molecular structure of the film grain will be altered and thus "has been exposed". Pretty cool huh? – Michael C May 10 '22 at 05:45
  • @MichaelC I am not asking the question to take a specific photo, simply trying to get a clearer view of dynamic range. In point of fact, I have considered asking this question in Physics or ECE section but figured that they probably would say the same (that I should go to Photography), plus, they would likely answer in decibels rather than stops. Anyways, thanks for the additional info – Amarth Gûl May 10 '22 at 15:50

2 Answers2

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I think the notion of a sensel being "activated" may not be the right way to think about it. Rather, it's a matter of reading the level out of each sensel, in which case the minimum signal read out would be zero, plus the possibility of various types of noise.

Whether you get a different digitized level for 10, 11, 12 photons, vs for 10, 20, 30 photons and so on, is more a matter of resolution or precision than minimum/maximum signal levels. And this is going to depend on sensor technology (and more specifically, the post-readout analog to digital conversion) - older sensor subsystems had e.g. 10- (1024 discrete values) and 12-bit (4096 values) precisions while a lot of more modern ones today have 14-bit precision (16384 values).

In your thermometer example, the dynamic range would be 100 degrees (in base-10, anyway - there are other units used depending on the context). But if you can only measure e.g. -30, -20, -10, 0, etc., you have less precision than one that can measure a single degree difference (or even smaller) over the same range.

twalberg
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  • Sensors don't have 12-bit or 14-bit values. Analog-to-digital convertors that convert the analog voltage signals from the sensor have 12-bit or 14-bit resolution. – Michael C May 10 '22 at 05:34
  • @MichaelC Valid point. Reworded a bit to clarify that... – twalberg May 10 '22 at 12:01
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There is no minimum... every photon of visible light releases an electron (e-).

But there is a minimum usable, which is electrons above the noise floor of read noise (RN)... this read noise is just the random packet error of photosite readout, and a typical number would be in the range of 5e- RN. So the sixth electron would be readable above the noise floor (measured as accumulated voltage).

And for a sensor with photosites with a max capacity (FWC) of 20,000e-, and a noise floor of 5e-, the dynamic range would be 20,000/5; 4000 discrete values or ~ 12 stops (log2(FWC/RN)).

Steven Kersting
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  • Just to reaffirm: by this calculation, if the RN goes to 10e- then the dynamic range would be dropped to 11 stops, and if the RN slips to 2e- then the dynamic range would be raised to around 13 stops? – Amarth Gûl May 09 '22 at 21:03
  • @AmarthGûl, yes (10.97 and 12.7), but that also assumes a 14bit ADC for linear raw files to go above 12 stops. – Steven Kersting May 10 '22 at 01:08
  • "... every photon of visible light releases an electron (e-)" Not exactly. Photons vibrating at shorter wavelengths of light release slightly more energy and thus add a slightly higher analog charge to the photosite than photons vibrating at longer wavelengths do. When the photosite is read out, the total cumulative analog electrical charge is read out, not the total number of electrons captured. EEs use the term e-, but it's really only shorthand for an approximation of the energy released by a single photon from within the middle of the range of the visible spectrum. – Michael C May 10 '22 at 05:40
  • Si photodiodes have a very even sensitivity/response uniformity (less than 2% across the spectrum). I am aware that it is the cumulative charge that is measured/converted; but I guess my reply didn't make that clear. – Steven Kersting May 10 '22 at 13:07
  • @StevenKersting The variable I'm addressing above isn't about the difference in sensitivity from one sensel to the next, it's the difference in energy yield between a photon oscillating on the UV end of the visible spectrum vs. the IR end of the visible spectrum. Red photons of light carry about 1.8 electron volts (eV) of energy, while each blue photon transmits about 3.1 eV. – Michael C May 11 '22 at 10:34
  • @MichaelC, I think you are convoluting the energy of the wavelength of light itself and the reactivity of an Si photodiode (which depends on the filtering/doping/Si thickness/wavelength). https://www.hamamatsu.com/content/dam/hamamatsu-photonics/sites/documents/99_SALES_LIBRARY/ssd/si_pd_kspd9001e.pdf – Steven Kersting May 11 '22 at 12:13
  • I think you mean you think I'm conflating the two, but I'm neither conflating nor convoluting ( to coil up; form into a twisted shape. ) them. Photons oscillating at "red" wavelengths/frequencies yield less energy when they strike the same sensel than photons oscillating at "blue" wavelengths do. Filtering only determines what percentage of each make it through the filter. But those "red" photons that make it through all have the energy of a "red" photon. Those "blue" photons that make it through all have the same energy as a "blue" photon, – Michael C May 11 '22 at 12:46
  • Filters do not shift the oscillation of photons. They may absorb or reflect their energy, but they do not shift the frequency at which they are oscillating. The only way to shift the wavelength/frequency of photons is to apply relativistic speed differences between the frame of reference of the source emitting the photons and the observing position's frame of reference. – Michael C May 11 '22 at 12:51