In various places, e.g. this tutorial video by ti, a circuit like the following is being recommended to drive the analog input of a SAR A/D converter:

simulate this circuit – Schematic created using CircuitLab

The general intention is to ensure that the sampling capacitor of the ADC will be charged with sufficient accuracy by the end of the sampling period. If the source has a sufficiently low impedance, it can be connected directly to the analog input, but otherwise this circuit should be used.

I understand the use of the buffer amplifier, and that it has to be fast enough to settle to the required accuracy within the sampling period. I also understand how to choose the values for the extra RC filter, but I'm not entirely sure about its purpose. So why is it there? And as part of that, why is the recommended value for C1 10x to 20x the value of the sampling capacitor?

Someone else will most likely pop in with a better answer but as far as I know, the RC circuit has two components.

1) the SAR needs to see a stiff input when it samples the input. That could be either a capacitor or an op-amp that is stable with a heavy capacitive load.

2) most o-amps are NOT stable with heavy capacitive loads. Especially fast op-amps.

So the R isolates the op-amp from the heavy capacitive load presented by the SAR. But now the input to the SAR will sag because of the resistance.

The capacitor is added so that the input signal seen by the SAR doesn't sag very much when the signal is sampled.

In other words, even though it looks like a RC low-pass filter, that is not the intended result. Just consider it to be a freebie bonus!

• I can see the second argument, although the ADCs I've seen have sampling capacitors of just a few tens of picofarads. As you said, the R would be enough to isolate the op amp from those. However, AFAIK the "stiff input" is merely needed to charge the sampling capacitor within the sampling period, and I don't see how adding C1 helps with that. The larger you make C1, the longer the RC filter requires to settle, so in that sense the input actually becomes less stiff with higher C1. At some point C1 becomes so large that it hardly loses voltage from charging CSample, but not at C1=10x CSample. Feb 23, 2015 at 17:52
• @Medo42: The basic idea is that there are two ways to get good readings: (1) ensure that the input voltage recovers quickly after transferring charge to the storage cap, or (2) ensure that it moves so little that recovery time is irrelevant. For (1), both resistance and capacitance must be very small; for (2), capacitance must be very large relative to the sampling cap. I'll confess some puzzlement about the 330pF value, since I would think that would be near the worst case. Apr 24, 2015 at 23:51
• @supercat yes, that's what I wanted to express. I think the ideal capacitor value for your case 1) would be zero, so I wonder why the cap is there at all. My best guess so far is that it keeps HF content of the switch transients from the op amp output - maybe they would prolong the settling. And in addition, the C definitely improves filtering of HF noise in the other direction. Apr 25, 2015 at 16:26
• @Medo42: Ideal for the first case would indeed be zero if the input were driven by a pure resistance fed by a pure voltage source. It's possible that using an op amp and a resistance with a small but non-zero inductance might influence things in such a way that adding some capacitance reduces the instantaneous resistor current and any transient effects that might cause with the op amp. Apr 27, 2015 at 2:38
• The source impedance of a ceramic capacitor with ESR of 0.1 Ohms or less is a lot less than the source impedance of an opamp followed by a resistor in the hundreds of ohms range. Additionally, any ADC should have some kind of anti-aliasing filter, even if the signal you sample is not intended to have high-frequency components. There will always be some EMI/noise/spikes, and reducing their magnitude makes the conversion result better. Nov 24, 2018 at 21:23

Your resistor is too large. (probably a typo).

You have a pseudo sample and hold amplifier (no analog switch). The ADS8028 can handle 8 inputs. Internally it has a S/H amplifier.

The OPA836 is working as a voltage follower. The RC combo is working as a low pass filter to remove noise AND the capacitor holds the sample so that the ADC can multiplex between the 8 channels. Allows small capacitor to charge quicker.

In A/D conversions sometimes two S/H amplifiers are used. One with a small capacitor to track input. A second with a large capacitor to hold input constant during conversion. To get accurate results we need a small capacitor, to get accurate results we need a large capacitor. This is not the case here, we have large followed by small.

• That answer makes no sense -- whether it's multiplexed or not doesn't matter. You cannot multiplex while the SAR converter is taking a measurment, only between measurements. Nov 24, 2018 at 21:20

Most people here have the right idea when it comes to charging times of the internal ADC capacitor. However, what I find missing in these answers is when it comes to the external RC filter on the output of the buffer.

You MUST filter out your frequencies above the Nyquist rate. I.E. if you're sampling at 1000Hz you need to filter out everything above ~500Hz or else your results will include aliasing. This can be done with an active filter (filter components are in the feedback loop of the op-amp), but this is a more direct and obvious solution and since the ADC input doesn't require much current it really doesn't make much of a difference.

• In cases where there is a known relationship between the sampling times and the signal being sampled, one may most emphatically not want to filter out things over Nyquist. For example, if one wants to periodically power up a sensor, measure its output, and power it down, nearly all of the useful signal content will be at multiples of the sampling rate. Eliminate everything over Nyquist and there will be almost nothing left. Jun 29, 2017 at 23:11
• The RC combination in this case really is just about allowing the opamp to drive a capacitor large enough to hold the signal steady during sampling. The LTC2440 recommends a more complicated version, and doesn't mention the sampling rate or Nyquist limit at all. The LTC2440 needs a 1microFarad capacitor on its input, and driving that can be a bit tricky.
– JRE
Jun 30, 2017 at 5:00
• In general, yes, you must consider the Nyquist limits. In practice, many ADCs have a built in filter and use oversampling.
– JRE
Jun 30, 2017 at 5:03
• The RC setup described in my question is quite inadequate for removing all frequencies above nyquist though. The SAR ADCs I have seen often have a sampling period less than 1/10th of the total conversion time, and the time constant of the RC filter needs to be something like a tenth of the sampling period, so the corner frequency would be ~15x above the maximum samplerate (30x above nyquist). And it's a puny first order filter, so even frequencies at 60x nyquist are only attenuated by around 50% (amplitude). These days I think it is really more about killing HF noise from the R and OpAmp. Jun 30, 2017 at 16:29
• A first order filter has 6 dB/octave fall-off, so attenuation is 50% at 2x the filter corner filter frequency. (Well, you really get 9 dB at 2x, because you get 3 dB at the corner.) Nov 24, 2018 at 21:25

The output impedance of an op amp follower is inductive. Without the RC filter, the impulse from the ADC would generate a large spike at the op amp output and drive it nonlinear. Then the op amp has to settle to within the ADC accuracy, which takes time. Driving the opamp nonlinear can also cause the spike to propagate to other parts of the system.

• excuse me? the output of an OPAMP follower isn't inductive. The reason a buffer is recommended is to present a low-impedance source to the ADC. an SAR draws pulses of charge at the ADC rate. The reason for the RC is to present an anti-alias filter to the ADC and equally to mitigate the fact the OPAMP will typically struggle to keep the output stiff during the SAR switching action & charge transfer
– user16222
Nov 24, 2018 at 22:50
• You need to provide some references for your statements here, or at least clarify the meaning. Nov 25, 2018 at 2:35
• @user16222, the op-amp follower output impedance IS inductive for the very reason that it has shunt feedback at the output. That is just fundamental. The source follower output impedance is also inductive for the very same reason. Apr 26, 2022 at 7:46