Video Digitization Answer

  1. Full scale is 62500 electrons, so even if there were no noise, the biggest useful number would be just short of 65536 = 216. The question is: what is the minimum useful number of electrons to count? Dark current and read noise come to 45 e-, the square root (shot noise limit) of which is ~ 7 electrons. Thus, each digitization has an uncertainty of 6-7 electrons. If we want to ensure that we can see the noise, choose 1 bit amplitude = 6 electrons, which means the highest count we can use is 62500/6 = 10411 counts. That is between 213 = 8192 and 214 = 16384. If we chose 1 bit amplitude = 7 electrons, then the highest count would be 62500/7 = 8930, still greater than 8192, so we might as well say: 14 bits.
  2. This is tricky only because it's so straight forward. Since the optimum clocking speed of the array is 40 MHz, each channel must digitize at that speed! Each channel looks at 80 × 1024 pixels = 81920 pixels. That means we can digitize the output of the array 4 × 107 / 81920 = 488 Hz.
  3. This is harder than it looks. We have to find EITHER a 14 bit converter that can operate at 40 MHz OR a different resolution converter that is this fast OR we have to slow down the ouput. Amazingly, there are several such ADCs, for example the AD10465 from Analog Devices (why not check Texas Instruments and other vendors for competitive products?).
  4. If the signal is 0, S/N = 0. If the signal is full scale, then we're looking at 62500 electrons, and the shot noise is S1/2 = 625001/2 = 250. Note that if we were only interested in high intensities, 1 part in 250 would mean using a flash 8 bit ADC! If we were really rigorous, we'd include the influence of read noise, but since 45 electrons is <1/250 of full scale, that noise is irrelevant to high intensity measurements. If we did want the overall S/N, since the read noise and dark current are additive offsets, σtotal = (σfull scale2 + σdark/read2)1/2.
  5. Uncertainty for a single, full scale read is 250 electrons. We get 488 reads per second. 4881/2 = 22, and t for 95% confidence for large numbers of reads (>30) is about 2. Thus S/N is approximately 250 * 2/22 = 23 electrons, or an improvement in S/N vs. a single read of about an order of magnitude.