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1.  A pixel at the centre of the imaging area of a CCD contains 1000 e-. The CCD has 2048 x 2048 pixels and a CTE of 99.999%. How many electrons will the pixel contain at the output of the CCD?

Assuming that the CCD is a three-phase device, moving the charge by one pixel will require 3 transfers.

To transfer the charge down the parallel register will therefore require 3 x 1024 = 3072 transfers.

To transfer the charge along the serial register will require another 3 x 1024 = 3072 transfers.

Assuming the CTE is the same in both the parallel and serial registers, the number of electrons in the pixel at the output is given by 1000 x (0.99999)(3072 x 2) = 940 e-.

This is only 94% of the original charge. To increase this to the more acceptable value of 99% would require a CTE of 99.9999%.

2.  A CCD pixel contains 80,000 e-. If the full-well depth of the pixel is 100,000 e-, the gain is 1.2 e-/ADU and a 16-bit ADC is used, will the pixel be saturated?

The pixel contains only 80,000 e-, which will therefore not saturate the 100,000 e- full-well depth of the pixel.

However, on readout, the 80,000 e- will be converted to 80,000/1.2 = 66,667 counts, which is above the 216-1 = 65535 saturation limit of the ADC. Hence the pixel will be saturated.

©Vik Dhillon, 12th December 2010