We know that ADC 1LSB=Vref/(2^N) where N is the resolution of the ADC.
We know that ADC Effective number of bits (ENOB) is always less than the resolution "N".
When calculating 1LSB do I need to take 1LSB = Vref/(2^ENOB) or I need to take 1LSB=Vref/(2^N)
Best Answer
Have you read the Wikipedia page about ENOB?
The designed value of an LSB is \$1 LSB = \frac{V_{ref}} {2^n}\$
Due to non-idealities ENOB is less than n
That still means that you need to treat the ADC as if it was ideal and as designed.
An example: \$V_{ref} = 2.56 V\$ and \$n = 8\$ then ideally \$1 LSB = \frac{2.56V}{8^2}= 10 mV \$
Now assume that the \$V_{ref}\$ I'm using is very noisy and this reduces ENOB from 8 to 7 bits. Then the LSB bit is still there but it will be so noisy (random) that I cannot use it. A 10 mV change (1 LSB as designed) at the input of the ADC will not be detected in the output as it will disappear in the noise.
A (2x 1 LSB as designed) 20 mV change however should be detected, the LSB +1 bit will change. This means that 1 LSB is still 10 mV, you just cannot use it (unless you average a lot of samples so the noise averages out but that's a different scenario).