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)
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).
The LSB {not to be confused with LSB for byte, lsb for bit} is always the average analog level between all consecutive binary encoded voltages. Thus full analog scale is used, not from midscale, nor ENOB.
\$\text{LSB=Vref}/2^N\$ for N bit ADC and Vref is the max. analog unipolar value.
Otherwise for bipolar voltage input , one uses the full-scale range(FSR) ADC's \$\text{LSB=FSR}/2^N\$
other info
Alternatively, SNR = 6.02N + 1.76dB so for N=8 bits, ideal SNR~50dB. The ENOB indicates the binary number of bits in resolution, limited by noise.
The ENOB dynamic range over asynchronous noise, distortion and ADC error sources measured in binary exponent bits and is best-case near full-scale. This value must be de-rated by the analog (full/actual ratio minus 1) for much smaller signals and is reduced due to ADC errors including non-monotonic, gain & offset errors and noise errors.
For Analog data ENOB is defined as \$\text{ENOB}=\dfrac{\text{SINAD-log}(1.5)}{\text{log}(2)}\$ where the 1.5 is the ideal ADC quantization error and \$\text{SINAD}=10\log \left(\dfrac{\text{(signal+noise+distortion) power}}{\text{(noise+dist.) power}}\right)\$
ENOB helps to define the logarithmic dynamic range.
When using a serial data link with synchronous noise from overshoot or frequency dependent of data pattern dependent precompensation equalization, the Link term LENOB is used which uses Vpp levels instead.
thus \$\text{LENOB=log}_2\text{(SNR[Vpp])} \$
When calculating 1LSB, take 1LSB=Vref/(2^N).
When considering spurious and other unwanted signals, take their voltage to be estimated by Vref/(2^ENOB).
Note that when you've calculated 1 LSB, you might hope it means that when you make that increment to the voltage into the ADC, the output number increases by 1. It doesn't. Read the data sheet carefully for what is guarranteed. A 'no missing codes' spec will guarrantee you see all the possible output numbers at some time or another. A monotonic spec will guarantee that as the input voltage goes up, the output code will never go down.
