I'm using AD7190 to sample 4ch single-ended signal. The datasheet suggests ADR431 or ADR421 as voltage reference for AD7190. However, an article on Analog Dialogue gives a way to calculate whether an additional output buffer is needed for voltage reference.

The article gives an example: enter image description here

So, I did calculation myself for ADR435 and AD7190.

The output impedance of ADR435 is calculated in the article above.

The AD7190 is 24bit ADC and the Iref is 7uA/V, which gives its Ro_max=(5V/2^25)/(7uA*5)=0.00425 Ohm.

The Ro_max(AD7190) < Ro(ADR435).

Is that means a output buffer is needed or my calculation is wrong?


The link of the article is down below.



2 Answers 2


The ref has 3.5µVpp noise (on 2.5V that's 19.5 bits).

This noise is higher than voltage drop due to ADC current (7uA*5)*0.075R = 2.62µV.

However, note your ADC is a sigma delta, so it doesn't draw current pulses from the reference voltage, unlike the SAR ADC mentioned in your article. So the article doesn't apply for this ADC.

Also the ADC datasheet says:

Average Reference Input Current 7 μA/V typ

Average Reference Input Current Drift ±0.03 nA/V/°C typ

This implies the reference current is constant, which makes sense for a sigma delta ADC, it's just going to end up on a comparator input.

So your reference current is just going to cause a constant DC offset which will be negligible compared to the tolerance on your reference voltage.


Short story: the accuracy of the AD7190 is not 24 bits; it's less than 18 bits.

Let's put this problem into context. The AD7190 (B grade version) has these errors: -

enter image description here

  • INL of 5 ppm of full scale maximum
  • Offset error of 75 μV typically
  • Gain error of ±0.005 % (50 ppm equivalent)

The ADC has a conversion resolution of 24 bits and that is far lower than the accuracies implied from above. 24 bits is 1 in 16.77 million or 0.06 ppm of full scale.

Hence, using \$2^{24+1}\$ in your calculations is pointless because the basic errors due to INL alone are equivalent to 1 in 200,000 or a shade under than 18 bits. If you factored in the uncalibrated gain error, it's equivalent to 25 ppm at half scale and that is 1 in 40,000 or a little less than 16 bits.

So, don't bother using 24 bits in that formula because it misses the point; realize that choosing a reference with adequate output impedance has no point beyond an accuracy of 18 bits.

Bottom line: the ADR435 output impedance is not going to be a problem.


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