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More exactly 0.65 - 2.65 for the present circuit. Because the input is biased to 1.65 V with an ADC input range of +/- 1 V. According to the AD9248, single ended operation with 0 to 2 V input range can be achieved by biasing VINn to 1 V. The voltage divider has to modified respectively.
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Thanks for the explanation. The span is clear now, but where does the 1.65V come from? Which voltage divider do you mean? Unfortunately I also don't understand from the datasheet what is the native input range of the ADC. With 2V span I would admit that it's -1 to +1V biased to 0V. But why is an input biased at 1.65V required? Sorry for the probably dump question.
Edit: Does it have something to do with VCCAD33 (which is probably 3.3V) which is divided by the voltage divider (on the right of the schematics) by a factor of 2? Does this mean that I have to connect VINn to (0V) ground to get an input range of -1 to 1V?
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