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ADS1232: Calculating PSRR regarding PGA gain.

Daisuke Higa
Genius 3665 points
Part Number: ADS1232

Hi support team.

I have been asked by my customer  that how to calculate and measure PSRR.

Could you provide the information that how to calculate or measure PSRR.

My customer is thinking about how to calculate below, is it correct?

PSRR = 20log(Vin_niise_rms / Vout_noise_rms) 

Vout_noise_rms is PGA output noise.

Could you please tell me regarding above question?

Best regards,

Higa

  • 0
    TI__Guru** 113765 points

    Hi Higa,

    The customer may find the following TI Precision Labs helpful:

    https://training.ti.com/ti-precision-labs-adcs-ac-dc-specifications-offset-error-gain-error-cmrr-psrr-snr-and-thd?context=1139747-1140267-1128375-1139102-1128657

    For the ADS1232 there are two potential PSRR errors.  One relates specifically to the DC shift of the power supply source (often as a form of supply drift) and the second is an external AC noise component (EMI, RFI or power line-cycle as examples).  The PSRR measurement needs to take into account the complete ADC path to include the PGA and modulator/filter stages.

    For the DC PSRR you would need to take into account the delta or change of voltage.  With the DC measurement, you need to have a very stable reference and the DC source needs to be a precision and low noise source.  So instead of 'noise' voltage you would look at the change or delta of the input source relative to the change in the ADC output code response (which you would translate the codes back to volts).  Note here that you would need to provide an input voltage for determining the delta between readings and any potential noise source (which includes the noise of the reference and noise of the input source) will affect the results.

    The AC measurement may be a bit easier, at least at frequencies within the passband.  The reason being that you can potentially view the delta as a waveform to compute the change in voltage.  At higher frequencies the shift may appear less like an AC waveform and more like a DC shift.  Again, any noise in the input source and reference will appear as noise in the output which may make it difficult to ascertain the delta in the measurement.

    Best regards,

    Bob B

  • 0 in reply to Bob Benjamin
    Genius 3665 points

    Hi Bob-san

    Thank you for your prompt response.

    I have some questions.

    -First question is about ⊿V in datasheet of ADS1232.

     Is it ⊿V = ⊿Vps (in training material below)? 

    -Second question is about relationship between PSRR and Ripple frequency on power souce.

    I could not find any graf or spec regarding it in datasheet.

    -Second question is about relationship between PSRR and PGA gain.

    Is PSRR affected by PGA gain?

    If no, can I consider that PSRR is 100dB(min) /120dB(typ) which written in datasheet of ADS1232?

    If yes, please tell me how much is PSRR affected by PGA gain?

    -Third question is about how to calculate PSRR_error.

    Is the calculation below correct?

    ⊿Vps_error  = ⊿V * 10(-PSRR[dB]/20)

    For example;

    ⊿Vps_error = 1*10(-100/20)

                        = 10uVp-p 

    Best regards,

    Higa

  • 0 in reply to Daisuke Higa
    TI__Guru** 113765 points

    Hi Higa-san,

    See my responses below.

    Best regards,

    Bob B

    Daisuke Higa said:

    Hi Bob-san

    Thank you for your prompt response.

    I have some questions.

    -First question is about ⊿V in datasheet of ADS1232.

     Is it ⊿V = ⊿Vps (in training material below)? [Bob] Yes, this would be equivalent.

    -Second question is about relationship between PSRR and Ripple frequency on power souce.

    I could not find any graf or spec regarding it in datasheet. [Bob] Unfortunately we do not have PSRR data for the ADS1232 showing the AC (or ripple frequency) change in supply. We only have information with respect to DC changes.

    -Second question is about relationship between PSRR and PGA gain.

    Is PSRR affected by PGA gain? [Bob] The current datasheet is not very clear on which supplies (AVDD or DVDD) are being given in the table.  It should be showing as AVDD nominal at 5V.  There was some additional tests done a couple of years ago showing that the best PSRR was at the highest PGA gains.

    If no, can I consider that PSRR is 100dB(min) /120dB(typ) which written in datasheet of ADS1232?  [Bob] At 5V AVDD and at gains of 64 and 128 you should expect to see the 100dB min and 120dB typical PSRR performance.

    If yes, please tell me how much is PSRR affected by PGA gain? [Bob] The analog input is bypassing the PGA at gains of 1 and 2.  When using the PGA, the performance is better than by bypassing the PGA.  If the measurement uses gain, such as would be the case in a load cell or pressure sensor application, you should see the performance I just noted.  At a gain of 1 or 2, it becomes more difficult to capture the same results as shown in the datasheet and the limit could lower in measurement results.  This is dependent on measurement capability and operating temperature.  Also remember that a 1V shift in AVDD is considerable and the device is also affected by the operating voltage and the reference.

    -Third question is about how to calculate PSRR_error.

    Is the calculation below correct? [Bob] This is the inverse calculation to determine the level of change as voltage error at the ADC output.  To actually calculate the error in dB, you would use the 20 * log (⊿Vps_error/⊿V).  ⊿V is known by the test, but ⊿Vps_error must be measured.

    ⊿Vps_error  = ⊿V * 10(-PSRR[dB]/20)

    For example;

    ⊿Vps_error = 1*10(-100/20)

                        = 10uVp-p [Bob] Yes, that would be the correct result for a 1V shift in AVDD supply at a gain of 1.  This would be highly unusual for an AC ripple.  It might be more like 10mV.  If you were to adjust the AVDD by 10mV you would see a 100nV shift and if you follow Table 1 in the ADS1232 for a gain of 1 this change would be in the level of noise of the ADC at a gain of 1.  That is why it is difficult to get meaningful data at a gain of 1.

    Best regards,

    Higa

  • 0 in reply to Bob Benjamin
    Genius 3665 points

    Hi Bob-san

    Thank you for your kind support.

    I understand.

    Best regards,

    Higa