I am not sure what really the term "sensitivity" means when characterizing image sensors, so I will describe what I am looking for in the sensor at the following conditions:

  • the exposure time is very short
  • the intensity if light is very low

Therefore, I need a sensor that will allow me to "see" very dim light at short exposure times. To compare the sensors (assuming equal spectral responses, exposure times), I decided that the most sensitive one will be the one that will "see" the dimmest light.


One problem is that different manufacturers provide different information/specs for their product, so it is difficult to compare them and choose the best one for application. I took one example and tried to calculate it's minimum amount that must reach the pixel before it can be detected. Please, correct me if I am wrong:

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  • I use "Output due to dark current" and "Conversion Gain" values to determine how many electrons converted from light are needed to reach the threshold above which they will become "detected": 6mV/3.4uV/e = 1765 electrons.
  • I find conversion efficiency at a desired wavelength (let it be 700nm): Given that the QE(675nm) is 60%, I find that at the 700nm QE=60% * 0.87 = 52.2%
  • Thus, the minimum amount of photons that must strike the pixel before they become detected is 1765e/.522e/ph= 3381 photons

Now, I wish to compare this sensor to S11639, however, there is not information to calculate the number of photons as in the first example: - The amount of electrons needed is: 0.4mV/25uV/e= 16 electrons! That is 110x less than for the first sensor. Now, there is no QE plot for this sensor. The only thing I can do is assume that at 700nm it has a low QE value, say 20%. But even with this bad QE value the number of photons needed to reach detectable range is 16e/.2e/ph= 80 photons - muuuch less than in the first case. Thus is it right to assume that the second sensor is much more suitable for my application with very low light intensities and exposure times because it needs less photons to reach the detection threshold? Are my calculations correct? Anything that I am missing?

  • \$\begingroup\$ Note that in the Hamamatsu sensor, the dark current specification ("Dark output voltage") has a footnote that says that the integration time is 10 ms. This means that to compare it to the first sensor, you need to multiply the number of electrons by 100, because the first sensor is giving you a 1-second reading. So the two sensors are actually rather comparable, 1765 e-/second vs. 1600 e-/second. \$\endgroup\$
    – Dave Tweed
    Commented Sep 26, 2014 at 23:01
  • \$\begingroup\$ @Naz Your calculations are incorrect, but I can't answer now. If you don't see an answer in the next 24 hours please remind me. \$\endgroup\$ Commented Sep 27, 2014 at 4:55

2 Answers 2


For your conditions what is the most important factors will be NEE (Noise Equivalent Exposure, QE = Quantum efficiency).

The ratio \$\dfrac{SEE}{NEE}\$ is your Dynamic Range (DR) where SEE = saturation Equivalent Exposure.

You need to understand what is meant by exposure, this is the integral of the photon flux over time. In other words the # of photons collected in a time period.

Unfortunately, most sensor manufacturers Quote NEE and SEE in electrons (after the conversion of Photons to carriers - here electrons) rather than actual photons so you will need to bring in the QE to calculate the actual light levels. These numbers are often implied with a saturation level being quoted, in that case the QE is implied.

In your low light high speed application you need a sensor with as small a NEE as possible, and you will need to see in the data-sheet some mention of CDS (Correlated Double Sampling) or kTC noise removal.

After update with datasheet: *****

Using nominal Vsat with conversion gain:

\$FW = \dfrac{V_{sat}}{G_{conversion}}=\dfrac{2.7}{3.4*10^{-6}} [\dfrac{V}{\dfrac{V}{e^{-}}}] = 274,118 [e^{-}]\$ FW= Full Well

This is close enough to the 800 \$ke^{-}\$ in the data-sheet. SO the SEE = 800 \$ke^{-}\$.

Dynamic range is 71 dB which is 3548:1.

\$ NEE = \dfrac{800,000}{3548}=225 [e^-]\$

Using your dark current calculation of 1765 electrons being generated in 1 second, the noise associated with that is:

\$ \sqrt{1765} = 42 [e^-] \$

Ideally the dark current contributes a variable baseline with temperature and the noise associated with that baseline shift is the shot noise of the leakage current.

The dark shot noise and the amplifier noise being independent of each other add in quadrature:

\$ Noise_{total} = \sqrt{225^2 + 42^2} = 229[e^-]\$

Using your QE calculation from above, the NEE is \$ \dfrac{229}{0.522} = 439 \gamma\$

You can do the same with the Hamamatsu S11639.

However, you still cannot directly compare the two because you have omitted an very important datapoint. What is the area of a pixel?

What is important is that to compare these two sensors in the same conditions. You need to understand the irradiance required to meet NEE, which has units of \$\dfrac[W][m^2]\$ but \$\dfrac{\gamma}{m^2}\$ is comparable if you are using a single wavelength. Here \$\gamma\$ means photons.

Your next step in the comparison is look at the optical setup, f/#, resolution etc.

  • \$\begingroup\$ Could you, please, check if my calculations are correct? \$\endgroup\$
    – Nazar
    Commented Sep 26, 2014 at 22:06
  • \$\begingroup\$ Yep, that makes sense. However, I thought that all the info in the datasheet is given for a single/average pixel of a given sensor. Thus, if I wanted to keep the resolution constant (assuming they both have same # of pixels), I would only zoom in/out the image to cover the entire sensor. This would not change the amount of photons that strike each pixel, regardless of the pixel size. So, in this case, there is no need to compare pixel sizes, is that right? \$\endgroup\$
    – Nazar
    Commented Sep 29, 2014 at 14:00
  • \$\begingroup\$ That is very astute! Most people say that the bigger the pixel the lower the noise because of increased photons ... etc. etc. But if you change the focal length of the lens to match the field to view to the image plane then the photon irradiance should stay the same ... AS LONG AS you keep the same f/# and you don't run into NA issue in the pixels. But to a first approximation that is right. \$\endgroup\$ Commented Sep 29, 2014 at 14:50
  • \$\begingroup\$ Nice. +1 more word to my vocabulary. Speaking of the Hamamatsu sensor... I calculate FW=2V/25uV/e=80ke. I also get number of dark electrons generated as 0.4mV/10ms * 100 = 40mV/s / 25uV/e = 1600 e/s; whereas the noise is 40 electrons. But how do I calculate the amplifier noise? The specksheet gives me two dynamic ranges, which one do I use? \$\endgroup\$
    – Nazar
    Commented Sep 29, 2014 at 15:34
  • \$\begingroup\$ readout noise is 1.0 mV/ 0.025 = 40 electrons. 80Ke/40 = 2000:1 the Vd based measure is meaningless, no point in putting that in the datasheet. Withe your dark current calculation (which is correct) that is sure some leaky PD. \$\endgroup\$ Commented Sep 30, 2014 at 5:01

Sensitivity in image sensors can also be expressed in terms of "photon efficiency", which at a fundamental level, is a measure of how many of the photons striking the area of a pixel actually get converted into detectable electron-hole pairs in the photodiode junction. These accumulate over the duration of the exposure to become the total charge for that pixel.

The other factor you need to consider is the "dark current", which essentially are electron-hole pairs created by things other than the light, such as the temperature of the sensor, cosmic rays (and other ionizing radiation) and leakage. This sets a lower limit on the level of light you'll be able to detect.

Datasheets for sensors will specify these parameters, which will then allow you to determine whether they're suitable for your particular application.

  • \$\begingroup\$ The sensitivity is the conversion constant from electron to signal level or ADU, like volts/electron or ADU/electron. The QE (Your "photon efficiency") comes before the signal chain - In my books the QE should be wrapped up in the whole thing, but most manufacturers separate them out. The conversion constant (i.e. sensitivity) is dominated by the sense node capacitance. \$\endgroup\$ Commented Sep 26, 2014 at 16:57
  • \$\begingroup\$ Dave, Please, check out my edit. I tried to calculate the minimum amount of light, but I am not sure if I did it correctly. \$\endgroup\$
    – Nazar
    Commented Sep 26, 2014 at 21:29

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