You'll be lucky to find a 16 bit ADC where you can have a lower Vref (input range) and get better performance compared to sticking with a higher input range and choosing a better input amplifier.

For instance, the AD7687 has a specified signal-to-noise ratio of typically 95.5dB with a reference voltage of 5V - if the 5V reference is lowered to 2.5V, the typical SNR drops to 92.5dB i.e. it gets 3dB worse.

The AD7685 has a similar story, so does the AD7988-5 et cetera....

My advice is find the best ADC you can afford and operate it with Vref set to the best possible value to maximize performance (usually the highest value permissible) - then design your front-end amplifier to deliver the best performance (usually by trading-off current consumption to reduce noise.

Sampling rate - this affects perceived quantization noise, for instance, if you sample at say 30 kHz, all the relevant q-noise will be contained in that bandwidth and so if you sample at 100 kHz, the noise is spread over a wider bandwidth meaning you can use process gain to reduce the noise in your digitized 10 kHz signal - average several samples and decimate in software. Process gain is the same if you converted to analogue and used filters after the DAC - the faster the sampling rate the more noise you can remove by filtering.

The statement:

- The noise density created by ADC quantization will be \${{2.11\times 10^{-4}}V\over \sqrt{10,000Hz}} = {{2.11\times 10^{-6}}V} \$

is incorrect. The analog bandwidth is going to be no more than half the sampling rate. This calculation is not necessary anyway, since you already have the RMS value for this noise.

What you need to do is compute the corresponding RMS value for the analog noise at the ADC input, which is \$5\times10^{-4}\frac{V}{\sqrt{Hz}}\times\sqrt{5000 Hz} = 3.5\times10^{-2}V\$. It will be less if you can band-limit the input signal to something less than the Nyquist bandwidth.

But this gives you a worst-case scenario. It basically says that you have roughly a 100:1 (40 dB) SNR (relative to a full-scale signal) at the ADC input, which would suggest that anything over about 7 bits will be enough.

To address the broader issues you raise: The real question is what is the probability distribution that each source of noise introduces into the stream of samples. The quantizaiton noise is uniformly distributed, and has a peak-to-peak amplitude that's exactly equal to the step size of the ADC: 3V/4096 = 0.732 mV.

In comparison, the AWGN over a 5000 Hz bandwidth has an RMS value of 35 mV, which means that the peak-to-peak value is going to be less than 140 mV 95% of the time and less than about 210 mV 99.7% of the time. In other words, your digital sample words will have a distribution of ±70 mV/0.732 mV = ±95 counts around the correct value, 95% of the time.

**EDIT:**

- The measurement precision will corresponds to \$ 3V/0.05mV = 2^{16} \$, which has 16 bit resolution.

Be careful — you're comparing a peak-to-peak signal value to an RMS noise value. Your actual peak-to-peak noise value is going to be about 4× the RMS value (95% of the time), so you're really getting about 14 bits of SNR.

- Now let us come back to the real case. When a 12-bit ADC is to be employed, could the 12-bit resolution be simply treated as quantization noise? If this is the case, 12 bit ADC can also lead to a 16 bit resolution result.

The 12-bit resolution *is* quantization noise. And yes, its effects are reduced by subsequent narrow-bandwidth filtering.

- What bothers me is "Can I really get a more precised result than ADC resolution WITHOUT oversampling?"

Yes. Narrow-bandwidth filtering is a kind of long-term averaging. And the wide-bandwidth sampling *is* oversampled with respect to the filter output. Since the signal contains a signficant amount of noise prior to quantization, this noise serves to "dither" (randomize) the signal, which, when combined with narrowband filtering in the digital domain, effectively "hides" the effects of quantization.

It might be a little more obvious if you think about it in terms of a DC signal and a 0.01-Hz lowpass (averaging) filter in the digital domain. The mean output of the filter will be the signal value plus the mean value of the noise. Since the latter is zero, the result will be the signal value. The quantization noise is "swamped out" by the analog noise. In the general case, this applies to *any* narrowband filter, not just a low-pass filter.

## Best Answer

A general rule of thumb is that is you want something to not contribute to your noise budget, that it must be at least a factor of 10 higher SNR than the dominant noise source in your signal chain. As an example, if you have a signal source that is at 300 :1 SNR, run your ADC at 3000:1 and for all intents and purposes you can ignore the ADC.

The only way to do this properly is to do a noise analysis.

Post processing (via in DSP for example) has the potential to extract out salient features from above the noise but you have to be careful. You have to have sufficient bit depth so you don't introduce rounding/truncation errors. You have to ensure that you are conserving the nature of the noise (gaussian/poisson pdf) or else the noise floor may rise in an unpredictable way and may not be amenable to DSP techniques. These sorts of steps (matched filters etc.) typically at best can improve the SNR by factors of \$ \sqrt{N} \$ and often the processing cost (# of operations) often follows \$ N^2 \$ so these sorts of steps often become rapidly very expensive. But agains a proper analysis will show this.

I would caution you against

assumingthat a DSP technique will automatically reduce your noise. It is very important that you lot at your noise sources via histogram analysis to ensure that the PDF (Probability Density Function) is amenable to processing. I.e. it appears well behaved, Gaussian or Poisson, is not multivariate and is stationary