ea82c3de   
   From: noreply@invalid.invalid   
      
   "eryer"  wrote in message   
   news:4357c36f-498f-4349-8172-76522242dcaa@f6g2000yqa.googlegroups.com...   
   Thanks for your answer   
      
   On 29 Set, 07:22, "Bret Cannon"  wrote:   
   > If you know the quantum efficiency or   
   > responsitivity at a particular wavelength then the conversion gain can be   
   > used to calculate the number of photons or joules of light that were   
   > incident on a pixel to give the observed digital output.   
      
   So i should remove dark current to obtain the real number of photons   
   incident on a pixel...if i have this camera chip   
      
   Conversion Gain 0.072DN/e   
   Fill Factor 80%   
   Responsivity 23.82 DN/nJ/cm^2 at 550nm   
   QExFF 52% at 550nm   
      
   How can i do it?   
   Conversion Gain=1/0.072=13.88   
   QE=52%/80%=0.65=65%   
   So, on 13,88 electrons, only 13,88*0,65 are generated from external   
   light?   
   Thanks   
      
   Your approach is not correct.  Dark current is not related to quantum   
   efficiency and to a first approximation it is not related to the light   
   exposure.  Only very expensive cameras that control the temperature of their   
   array detectors specify their dark current levels, since it is usually not   
   an effective marketing tactic.  For an accurate measurement of the light   
   level, the dark current needs to be measured under similar conditions to the   
   measurement of the "bright" scene.  To do this record a series of exposures   
   with different exposure times, with the lens cap on the camera and other   
   precautions to make certain that no light enters the camera.  Perform a   
   linear least squares fit for each pixel versus exposure time to obtain an   
   intercept, B (as a digital number) and a rate, M (as a digital number per   
   second) for each pixel in the camera.  This is most easily done in an array   
   based software tool like MatLab or SciLab.  Typically, there is an offset   
   built into the electronics, which may be subtracted off within the camera   
   but this is not universal.   
      
   Now for each pixel in a image recorded for an exposure time t_exposure,   
   calculate the dark current for this exposure as B + (M x t_exposure), and   
   subtract this dark current contribution from the pixel value for this image   
   to get a Net_DN for this pixel for this image.   
      
   The corresponding number of photons incident on the pixel to produce this   
   Net_DN is (Net_DN x 13.88 e/DN)/52%.  Of this number of photons, only 80%   
   will actually reach a sensitive part of the pixel with the other 20% being   
   lost to metallic conductors or areas on the edges of the pixel where the   
   electrons are not collected.  Of the 80% of incident photons, only 65% will   
   produce electron-hole pairs that are collected in the capacitor of the pixel   
   with the rest being lost to reflection or direct conversion to heat.   
      
   Bret Cannon   
      
   --- SoupGate-Win32 v1.05   
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