I've been having a discussion with a colleague about ENOB (effective number of bits) calculations for DACs and ADCs. We both come across it from different directions (he being more analog, and me being more digital).
My understanding of ENOB is that it is an indicator of the bit depth you can reliably detect (for an ADC) given the noise of the system. So with a given noise floor, you can't use the bottom 100 ADC codes for example so you your ENOB is number of bits required to generate the remaining number of codes. The limiting factor on the ENOB value is always going to be actual bit depth of the ADC itself. This is a very digital perspective on things.
His understanding is that ENOB calculations are based on analog measurements, and there is no limit to the maximum ENOB - it is entirely dependent on the noise characteristics of the ADC. A rather analog view on things.
I agree in that the ENOB calculations we do in the office are completely derived from analog measurements with no prior knowledge of the ADC's bit depth. However, I can't understand how an 8 bit ADC could have an ENOB greater than 8.
Would the quantisation noise of the ADC be the limiting factor on the ENOB? If measuring the ENOB of a perfect sine wave with only 8bit quantisation noise present, would the ENOB be a perfect 8 or would it be higher?