I'm going to assume that are able to get the readings from each sensor and that they are operational.
You will need to find a way of controlling the light into the sensor in a controllable way. This can be either through ND filters (ND = Neutral Density) a controlled aperture or through chopping. A controlled aperture can be a lens with a controlled F/# 1, stop is typically 2x the amount of light. A chopper can be a spinning disk with 1/2 or 1/4 of the disks covered/open and it needs to spin fast enough that the sensor cannot keep up, it sees the "average"
If you reduce the ratio of light by 1/2 and even better also by 1/4 and compare the readings, it will be obvious which one has the higher gain.
From the comment thread : "some more info Device A has 439counts/uW/cm2 and device B has 3 counts/uW.cm2. 22 μW/cm2 is the irradiance for White LED source"
439/3 => 146.33:1 increaased sensitivity.
Let's review what is happening. The dynamic range is the ratio of the maximum signal to inherent noise of the sensor. When you increase the sensitivity you are actually increasing the gain of the circuit this amplifies the gain of the maximum and also the noise.
Let's have a made up scenario to illustrate:
sensor X is low sensitivity and requires 2000 w/cm^2 at maximum. If you increase the light the output saturates (also known as clipping). the noise level equivalent to 4 w/cm^2. the sensor is nicely calibrated so the the digital number is 2000 in the bright and 4 in the dark.
The DR = 2000/4 = 500:1
I increase the sensitivity by 100 X, that means that sensor will put out it's maximum of 2000 at 20 w/cm^2 and the noise level will be 4*100 = 400. The sensor with the higher sensitivity will by necessity have a worse DR in digital outputs 2000/400 = 5:1 in terms of light levels, 20/4 = 5:1.
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.
Best Answer
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.