"The NOCTURN Color models are powered by the proprietary KAMELEON imaging sensor, a solid-state sensor that offers less than 4e- read noise, with SXGA (1280×1024) resolution at frame rates up to 100 Hz. The KAMELEON sensor provides large 9.72µm pixels with microlenses for optimum quantum efficiency in excess of 60%, providing high-resolution low light color images that extend the vision of the human eye."
Crazy performance!!!
ReplyDeleteIs this export controlled?
How many photons hit a 10x10u pixen in lets say 10ms when a 0.01 lux illuminated scene is captured? 25ke full well, lets say x25 gain, so a signal of 1000e- for s saturated pixel. How many photons cross a 10x10u area in 10ms at 0.01lux? Just a ballpark estimation ;-)
ReplyDelete0.9e
Delete40 photons
Deletei (i posted the question above) tried to retrace the '40 photons'
Delete1lux = 1 lumen over 1 m^2
1watt = 683lumen (for 555nm)
so 1lux = (1/683)W/m^2 = 1.5x10^-3 W/m^2 = 1.5x10^-15W/um^2
PhotonEnergy Ep = hc/lambda = (6.6 10^-34[Js] * 300 * 10^6 [m/s]) / 555*10^-9 (m) = 3.6 *10^-19
PhotonCount Pc = 1.5x10^-15[W/um^2] / 3.6 *10^-19 = 4100
so about 4000 photons/sec for 1 lux per um^2, right? or 4 per ms. For a 10x10u pixel 400 per ms or 4000 for 10ms. For 0.01 lux 40 photons. Assuming a high QE of lets say 80%, this results in about 30 electrons per pixel during integration time.
https://en.wikipedia.org/wiki/Luminous_flux
https://de.wikipedia.org/wiki/Lichtstrom
https://en.wikipedia.org/wiki/Illuminance
https://en.wikipedia.org/wiki/Photon_energy
You have forgotten that 0.01 is illumination on the scene. To get the value on the sensor, you have to multiply 1/4F2, roughly 1/10 for F=1.4.
DeleteYes, 40 photons. If QE=60%, 24 electrons.
ReplyDeleteOur worldwide patented Kameleon technology is indeed great! It is used in a lot of different applications around the world and is export free.
ReplyDeleteContact us for more détails!
PHOTONIS TEAM
Great! Do you have the patent no please?
DeletePixel Pitch = 9.7um
ReplyDeletePixel Area = 94.1e-12m2
W/lux for 550nm = 1/683
Lux on sensor = 0.01/10 (F1.4)
A/W at 550nm = 0.44A/W
QE = 0.6
Exposure Time = 10mS
==> 2.29 electrons
Where is photon noise at such low light level?
ReplyDeleteHey guys, camera lens have focusing function. let's say F1.4, it should be x10 photons not /10. But considering object have 1/10 reflectivity . It should be 40x10/10=40 photons.
ReplyDeleteSorry, but higher f-no means smaller aperture and less light. So 1/8. Otherwise all the camera phones would use pinhole optics with great depth of field and life would be easy.
Delete