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I noticed that image sensor manufacturers provide some optical characteristics in different formats. I wish to know how to interpret that information.

  • Sensitivity is characterized in \$\dfrac{V}{lx \cdot s}\$ for S11639 and TCD1201, which is \$\dfrac{V}{J \cdot s/m^2}\$. At the same time, RL1024P and S10077 report sensitivity in \$\dfrac{V}{J/m^2}\$. Now time 's' is gone from the characteristic. How do I interpret this characteristic in terms of light intensity and exposure time?

  • Conversion efficiency seems to be obvious - voltage increase per every additional electron. Is this related to sensitivity in any way, or it just gives the resolution and range of the output voltage?

  • Dynamic range - is it also the 'output' characteristic and has nothing to do with optical characteristics of the device?

It would be nice to have an extensive explanation for each term and how it is related to each other. What set of characteristics is preferable, what are the compromises. Any additional parameters that are worth mentioning?

EDIT:

Also, how do I convert Aviiva M2CL sensitivity that is reported in digital format to analog format, form \$\dfrac{LSB}{nJ/cm^2}\$ to \$\dfrac{V}{nJ/cm^2}\$

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  • \$\begingroup\$ I can see the value of an explanation of terminology, and how attributes relate to each other. However, I don't understand what you mean by "What set of characteristics is preferable". I would expect the set of characteristics to depend on the application. For example the sensitivity half angle, wavelength, etc. is chosen to best match the project. Finally, and this may be no more than a use of English issue, but I don't necessarily think of the practical or available sets of properties of a component as "compromises" as much as trade-offs. However, you might have something else in mind. \$\endgroup\$
    – gbulmer
    Aug 7, 2014 at 15:27
  • \$\begingroup\$ @gbulmer You are right about choosing devices for particular application. However, I am not asking which of those imagers will will work best for my project. Rather, which of the mentioned parameters should be prioritized over the others and in which cases. Disregard the wavelength. If you could, please explain about sensitivity half angle. \$\endgroup\$
    – Nazar
    Aug 7, 2014 at 15:42
  • \$\begingroup\$ I apologise. Ignore me. I was thinking of single photosensors, and had assumed those image sensors had optics, but AFAICT they don't. \$\endgroup\$
    – gbulmer
    Aug 7, 2014 at 16:43

2 Answers 2

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Lux is a photometric unit, energy units are radiometric units and the two are related through the response of the human eye. The Lux will nominally have the same units as the radiometric units but it will be weighted by the "typical" eye response. To make things easier to understand/compare when ever you see Lux think the radiometric terms in equivalence.

Sensitivity in \$\dfrac{V}{Lx \cdot S}\$: \$Lx\$ is a per unit area measure so the \$m^2\$ is implied. Your TCD1201 must be wrong, the units should be \$\dfrac{V}{W \cdot s/m^2}\$ to be equivalent.

That now explains your confusion in that \$W \cdot s\$ is Joules. So a reception of a given amount pf photon energy integrated over time will be a measure of Energy which corresponds to a Voltage signal.

Conversion efficiency: Is the conversion of photons to electrons. From energy of light and wavelength you can derive the photon flux and energy per photon. With Quantum efficiency and the transfer efficiency (moving the charge to the output node) these two terms combined give your your conversion efficiency, you get the number of charge carriers generated per photon.

Dynamic range: Has everything to do with the optical properties of the sensor. Dynamic range ideally is limited by the shot noise of the sensor which tracks as the \${N}^\frac{1}{2}\$ with N being the number of photons.

SNR is also important as is it the ratio of SEE to NEE - Saturation equivalent exposure to noise equivalent exposure.

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  • \$\begingroup\$ So, assuming the parameters are given for the same wavelength, it seems obvious to choose one that has higher V/(lx*s) number, highest conversion efficiency, dynamic range and SNR. But which priority order/ trade-offs should I follow? \$\endgroup\$
    – Nazar
    Aug 7, 2014 at 16:54
  • \$\begingroup\$ Unfortunately, Stack exchange is not a good place to write a book in. :) a lot will depend upon what you lighting conditions are, are they controllable (can you add light) do you need to freeze the action (i.e. is is a fast moving object)? Do you need to be in the shot noise limited regime? Etc. etc. I wouldn't look or care too much about sensitivity as that is easily changed by an amplifier, the limits are imposed by SEE and NEE. the sensitivity is only useful for matching the sensor output to an ADC. \$\endgroup\$
    – placeholder
    Aug 7, 2014 at 21:17
  • \$\begingroup\$ Oh, I see... I have a very dim light and short exposure time, so I thought that higher sensitivity/responsivity will give me better results, meaning that small light intensity will spread better over the dynamic range of the imager. Then, to get the best performance I would choose the imager with the smallest saturation exposure. In this case less light is needed to reach saturation output. Right? \$\endgroup\$
    – Nazar
    Aug 8, 2014 at 16:32
  • \$\begingroup\$ If you had to choose from S10453 and S11639 assuming the same spectral response, light source, and integration time, which one would have better optical characteristics and how much would the performance be affected when interchanging one with the other. I wish to use S10453 because of its size and readout rate, but I am afraid that it might be not as good as the S11639. \$\endgroup\$
    – Nazar
    Sep 25, 2014 at 17:03
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Here are some materials I found after internet search:

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