I am trying to select an LDO for powering the analog domain of a sensor that has a ripple requirement of < 50 mVpp @ 1.8 V. The 1.8 V LDO I intend to use has an output voltage accuracy of +/- 2% which means the voltage can vary by +/- 36 mV, which translates to 72 mVpp voltage. Should the output voltage inaccuracy be treated as a contributor to the ripple?

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    \$\begingroup\$ I recommend linking the data of the LDO that you are talking about so that people can be more specific with their answers. If I knew the ripple rejection of your regulator, I could have given you a more specific number for the output noise in my answer. \$\endgroup\$ Commented May 5, 2021 at 5:24
  • \$\begingroup\$ @PrathikPrashanth Link added. \$\endgroup\$
    – ashare
    Commented May 5, 2021 at 5:40

5 Answers 5


There are three contributors to output ripple on an LDO. Those are the line regulation, the load regulation, and regulator noise.

Assuming your load draws a reasonably constant current, the ripple due to this should be negligible. No practical LDO will have a noise approaching 50 mV. If you were worried about 50 uV then you may have to do some digging in datasheets.

What's the input ripple to the regulator? This can be reduced with larger input reservoir capacitors if required. This is now reduced by the line regulation of the LDO. For any reasonable LDO and input ripple, your output ripple will be waaay below 50 mV.

The 2% voltage accuracy is a DC specification, and does not contribute to ripple.


Should the output voltage inaccuracy be treated as a contributor to the ripple?


The output voltage accuracy is the deviation from the nominal output voltage and the output ripple is the amount of AC component that is present in the output.

An LDO should easily meet your requirement as its 'output noise', which is the term used for linear regulators, will definitely be less than 50mV. In fact, you can expect it to be in the 100uV range or lower.

  • \$\begingroup\$ The datasheet says that the output noise is 39uVrms. So yea, you can expect 100uVpp of ripple @ashare \$\endgroup\$ Commented May 5, 2021 at 6:30

The 1.8 V LDO I intend to use has an output voltage accuracy of +/- 2% which means the voltage can vary by +/- 36 mV, which translates to 72 mVpp voltage.

That's accuracy, not ripple. It includes effects like part-to-part variation and thermal drift, but unless you look at it on a very slow timescale, drift is not ripple.

If your sensor is ratiometric (like a resistor divider) and it outputs a voltage proportional to its supply voltage, then you have to worry about the accuracy and drift of your regulator. A simple solution is to use the same voltage reference for the ADC and the sensor, since the ADC measures the ratio of input voltage to reference voltage, that solves the problem.

If your sensor is not ratiometric, but instead outputs a specified voltage like "10mV per °C" then most likely its "maximum ripple" spec comes from its own power supply rejection. In that case you should ensure low enough ripple on the power supply, but the accuracy isn't really relevant as long as the sensor gets the voltage it needs. So what matters in this case is the product of the PSRR of your LDO and how noisy its own power supply is. If the sensor draws pulsed current, then also consider transient response.


For the voltage of a chip, ripple would mainly introduces jitter to the point of failing timing altogether and not working as expected. Lower/higher DC voltages don't cause much jitter but make the chip globally faster/slower.

If your only concern is jitter, then the inaccuracy of the voltage would be inconsequential.

If your chip is rated only for maximum allowed ripple, but not for maximum DC range, then I suppose it will work no problem with higher/lower DC and the ripple rating in its datasheet is just there to guarantee performance.

In general, the inaccuracy is the harder problem to get right from a PSU point of view though, because ripple can be substantially reduced with large output bulk capacitors. So even though ripple+ inaccuracy might exceed your tolerance, check for which output capacitance and current draw this ripple is given. If the capacitance/current is larger in your application, the ripple will be lower.


Try more to find the best way to get what you want. But, if you didn't get your suitable answer, I suggest this way:

  • Connect the output of your power supply to an oscilloscope (it's better to use a digital oscilloscope, because it gives you your P.S. output's ripple-frequency).

  • Put its channel on AC mode.

  • Look at your ripples and measure minimum and maximum frequency of your ripples.

  • Then make a band-reject filter on your output (you can find how to design it in internet).

    But you have to select your components carefully, because your pk-pk voltage is very low. For example if you use a capacitor like ...uf - 100v, maybe it's not working in your circuit because of your low pk-pk voltage.

  • Find the lowest possible capacitor's voltage of your circuit.

I'm not sure actually, I never deal with this issue before but your components must work in that voltage limit (pk-pk ripple voltage) and follow this rule for the other components.


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