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ADS1291 ENOB and Noise-Free-Bits Calculation

Aaron Hill43
Prodigy 50 points
Other Parts Discussed in Thread: ADS1291, ADS1298

There seems to be an error in the data sheet for the ADS1291. Starting on page 7 there are 3 equations and tables that provide Vrms and Vpp values and then calculated ENOB and Noise Free Bits (NFB). However, when using the data in the table (Vrms and Vpp values) and the provided equations, I do not get values remotely close to values given in the table for ENOB and NFB.

Could you enlighten me on the error in the data sheet or the possible misunderstanding I have with how the table data was generated?

I have reviewed other TI ADC data sheets that provide ENOB/NFB equations and they are not the same as the ones given in the ADS1291. Also in the ADS1298, only the Vrms and Vpp values are given and no equations to calculate ENOB or NFB from the data. I am doing an ADC comparison study; your timely response is greatly appreciated.

Thank you,

Aaron Hill

  • 0
    TI__Guru**** 173726 points
    Hi Aaron,

    I see what you mean regarding the tables in the ADS1291 datasheet. I'll look into that and get back to you as soon as possible.
  • 0
    TI__Guru**** 173726 points

    Hi Aaron,

    Please accept our apologies for the confusion here. For the effective number of bits, please use this equation: ENOB = log(VREF / (sqrt(2) * Gain * VNrms)) / log(2). For 'Noise Free Bits' please use NFB = log((2 · VREF/ (Gain * VNpp) / log(2). I'll see what we can do to get the equations in the datasheet updated in the next revision of the document.

  • 0 in reply to Tom Hendrick
    Prodigy 50 points

    Tom,

    No worries. Thank you for the prompt response. Amending the equations as you suggest does remedy the inconsistency with the tabular data. So thank you for that. I do have a follow on question that is directly related to this forum thread.

    Equation 1 status that SNR = ENOB x 6.02. However, in my readings I have uncovered a different formula. Namely, SNR = 6.02(N) + 1.76. Or more commonly stated N = (SNR - 1.76)/6.02, and it is common for sources to say that N is ENOB. Further some sources will replace SNR with SINAD. I understand the confusion around SNR vs SINAD and I understand the difference. But for equation 1 to state that SNR=ENOBx6.02 seems wrong as the 1.76 term is left out which has an entangled mathematical relationship in the derivation of the equation. Could you maybe clarify if eq 1 is wrong as well?


    As a follow on to that question, the datasheet has THD (total harmonic distortion) measurements in it. Are the noise values provided in the table taking into account the THD?

    Thank you,

    Aaron Hill

  • 0 in reply to Aaron Hill43
    TI__Guru**** 173726 points
    Hi Aaron,

    Yes, the 1.76 factor is normally in the equation, so the SNR numbers in the table should be higher than what is stated there, we'll fix all of that with the next datasheet revision. The THD numbers in the Electrical Characteristics table are taken with a 100Hz sine wave input with PGA gain of 1. The noise numbers are taken from measurements with the inputs shorted over a 10 second period.
  • 0 in reply to Tom Hendrick
    Prodigy 50 points
    Thanks again Tom. I would like to point out that if Eq 1 includes the 1.76 factor then Eq 1 and Eq 3 will no longer be consistent( based on the amended formula you gave). To resolve this, Eq. 3 should not define ENOB, but SNR instead. Then Eq. 1 with the 1.76 factor will define ENOB (or some form of it since THD is not accounted for in this calculation).

    Thanks again for your help in looking into this matter, it has resolved a lot of confusion I have been having with the mathematics behind these formulations.

    Thanks!

    Aaron Hill