2
\$\begingroup\$

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?

\$\endgroup\$

2 Answers 2

2
\$\begingroup\$

I think this depends on where you draw the imaginary box around the 'ADC'. If the ADC is 8 bits and you see only 8 bits at the interface, then ENOB can't be better than 8 bits.

If you draw the box around an 8 bit or 1 bit or 16 bit converter than has internal oversampling and digital filtering and you're presented with 24 bits at the interface to that box, then ENOB certainly can be more than whatever is used internally. A Delta-Sigma converter in its simplest form is a 1-bit converter, yet you can get 19 or 20 bits ENOB (out of 24 bits presented).

ENOB can be calculated from SINAD (Signal to noise and distortion ratio), which includes quantization noise, distortion, reference noise and thermal noise. All the noise sources add, at best they add in quadrature, so you can never have lower noise than the quantization noise, but there are many other possible sources of noise and distortion (especially in high resolution converters).

\$\endgroup\$
2
\$\begingroup\$

It would be 8 bits, period.

ENOB is mainly defined by the nonlinearities in the converter — the fact that the "steps" in its transfer function are not all the same height. This creates distortion products in the output of the converter that limit its effective noise floor and resolution, even if all other sources of error are eliminated. If the converter is perfectly linear, then you get the full 8 bits.

I'm not sure what your colleague means by "analog measurements", since the output of the ADC is purely digital. The digital output includes all sources of noise, including the quantization noise of the converter itself, which becomes the dominant source if all other sources are reduced to zero.

That said, it is possible to increase the effective resolution of a converter below the noise floor by adding a dithering signal to the input and doing some judicious digital filtering at the output. But this isn't what the ENOB specification is about.

\$\endgroup\$
1
  • \$\begingroup\$ Sorry I'm mixing up ADCs and DACs because we test both. The analog measurements my colleague makes are from the output of the DAC. What I should have done though is dig out the ENOB and quantisation noise formulas. Max. SNR (and therefore SINAD) of an 8bit converter is 49.92dB. ENOB = (SINAD - 1.76)/6.02 => ENOB = 8 As SINAD drops (because of noise), then ENOB drops. Simple. \$\endgroup\$
    – Oliver
    Aug 28, 2014 at 7:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.