I'm not sure if this is the right place to ask, but I'd like to
find out what the effective number of bits is out of the ADC's in
the AD6655 IF Diversity Receiver.
The ENOB on the AD6655 can be approximated from the following equation:
ENOB (bits) = (SINAD - 1.76)/6.02
For the AD6655, the SINAD is approximately SNRFS - 1.0dB for IFs up to 220MHz.
So, for example, on the AD6655-150 operating at 220MHz, the ENOB is given by:
ENOB = (72 - 1.76)/6.02 = 11.67 bits
I hope this helps answer your question, if not just let me know.
I'm moving this question on AD6655 to the High Speed ADC Community.
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