For one camera, the black level of the sensor will make the non-zero output when it is covered. For example the output of the sensor is 10-bit and black level is 63, if the output is reduced by black level, the range of the output will be 0~960. So the dynamic range is reduced from 0~1023 to 0~960. Am I right?

From one document, there is one figure:

Analog offset is added to pixel voltage

I am not an electronic engineer, so I don't understand why the analog offset is added to pixel voltage. The output of ADC is reduced by digital offset, do this will reduce dynamic range greatly?


1 Answer 1


The thing is that you want on one hand to make the best calibration you can, but on the other hand you don't want to insert nonlinearity or othe artifacts. So you have some analog compensation, in order not to amplify anything that is not actual signal. But still after amplification some offset may remain, so you have to calibrate digitqlly too.

And you are right, the dynamic range does not span over the full ADC scale. Still, if you have like 90% of your scale it's fine. Hope linear calibration will be enough. I once had a sensor that required polynomial calibration, and even with that the performance was not perfect.

  • \$\begingroup\$ Thanks for your timely response. Why is black level calibration added to pixel voltage? In my view, the value should be reduced from pixel voltage. I don't understand how nonlinearity or othe artifacts is brought to the output? \$\endgroup\$ Commented Jun 13, 2015 at 17:55
  • \$\begingroup\$ Add and subtract is the same. Artifacts, no matter how small, are amplified by the gain component. \$\endgroup\$
    – user76844
    Commented Jun 13, 2015 at 17:57

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