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I am using an ADC LTC2378 with a resolution of 20 bits.

Based on the datasheet this resolution corresponds to an LSB calculated by the equation (\$2 V_{\text{ref}}/2^{20}\$), i.e., 9.536 µV.

In order to achieve a better resolution of 24 bits as described in this document Silicon Labs Oversampling I have to oversample at a frequency

$$f_{\text{oversample}} = 4^{n_{\text{bits, additional}}} f_s$$

Assuming that the original sampling frequency of the signal is 62.5 Hz and we oversample to achieve 4 additional bits we will have to sample it at 16000 samples per second.

After oversampling the simple method of averaging is considered.

Assuming that we have an accumulator of at most 34 bits width (\$2^{20}\cdot 16000\$), from the generated result we are discarding the 10 bits by simply shifting on the right. My question is what it would be the newly achieved LSB after this process.

\$2 V_{\text{ref}}/2^{24}\$ or something else?

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    \$\begingroup\$ You have made small mistake. You need to take average of 256 samples so right shifting by 8 and not 10. \$\endgroup\$
    – Rokta
    Jun 12, 2023 at 9:36

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My question is what it would be the newly achieved LSB after this process.

It depends what you mean by an LSB?

  • A noise level LSB?
  • A 'no missing codes' LSB?
  • A spurious or harmonics ENOB?
  • Something else?

The basic LTC2378 guarantees no missing codes to 20 bits.

If you oversample to reduce the noise by 16 times, you will now be able to see missing codes. This may matter in some applications, and not in others.

Check the noise specification of the original ADC. This is what oversampling will reduce, by the '3dB per doubling' factor. It will not improve spurious of harmonics ENOB, it will not improve missing codes - unless you use large scale dither with the oversampling process, when it might.

If getting better ADC performance simply involved oversampling, everybody would be doing it. Well, let's just say that more people attempt it than succeed.

If the datasheet is correct, and you have done these sums correctly

Based on the datasheet this resolution corresponds to an LSB calculated by the equation (\$2 V_{\text{ref}}/2^{20}\$), i.e., 9.536 µV.

Then if you get another 4 bits by oversampling 28 times and take the 24 MBSs of the accumulated result, then your LSB is just (\$2 V_{\text{ref}}/2^{24}\$), i.e., about 0.6 µV.

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    \$\begingroup\$ Well, to be fair, everyone is doing it – when it's an easy win. (And I mean, at the end of the day, what is \$\Delta \Sigma\$ other than an oversampling accumulating ADC.) At > 20 bits, getting your ENOB high enough is hard enough that it's not a simple win. \$\endgroup\$ Jun 12, 2023 at 9:02
  • \$\begingroup\$ Don't think you'll see missing codes at the output of a truncating accumulator of the proposed size, by the way; when I accumulate 2¹⁴ independently noisy samples, I get close enough to be using the central limit theorem, right? the bits where that's not true are getting truncated \$\endgroup\$ Jun 12, 2023 at 9:04
  • \$\begingroup\$ I am not refering to the noise at all. My question is simple. How the LSB value in uV varies when you averaging and decimating. \$\endgroup\$
    – MrT
    Jun 12, 2023 at 9:17
  • \$\begingroup\$ @MrT I think you and Neil might be referring to different meanings of "significant" in LSB. \$\endgroup\$ Jun 12, 2023 at 11:07
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    \$\begingroup\$ @MarcusMüller ΣΔ is definitely cool. But it is more than oversampling. It is about oversampling combined with noise-shaping. And these 1-bit converters is the third attribute. With 1-bit, there is no danger of missing codes or non-monotonicity. As long as the rise-time and fall-time (and shape) are identical, no other non-linearity remains in the analog end of it. There is no saturation distortion. Then if we're careful that the digital end of the ΣΔ is clean, there really is no non-linearity other than idle tones. \$\endgroup\$ Jun 14, 2023 at 18:13
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My question is what it would be the newly achieved LSB after this process.

Technically it can be whatever you want since you get to pick the numerical format. If you choose floating point (and you should) it will probably be some incomprehensibly small number.

However this question doesn't make a lot of sense. Oversampling reduces quantization noise, usually resulting in some other noise source dominating SNR, so the new LSB you reconstruct doesn't really matter. You'll probably increase your effective number of bits to 20.5 or 21 or so, even though you'll have 32 or 64 bit sample precision on your PC.

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    \$\begingroup\$ With ΣΔ and oversampling, surely the bottom bits are going to be noisy. But don't throw them away! Truncating those bits only adds more quantization error. You want to use those bits as a natural dither in your signal that help lubricate the remaining signal processing even if that is merely scaling by a gain contant. \$\endgroup\$ Jun 14, 2023 at 16:27
  • \$\begingroup\$ by the way, if this is all unsigned values, then truncation will also introduce a bias (you're always rounding down). \$\endgroup\$ Jun 14, 2023 at 17:25

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