Function of this circuit?

[WillHuang]

In the image attached, I see a LPF. At the output, there are some additional resistors and capacitors before feeding into an ADC. I was wondering what are these resistors and capcitors for?

[David Hess]

ADCs with sampling inputs can present a dynamic load which is difficult for a operational amplifier to drive.  The output RC network isolates the ADC input from the operational amplifier output.

[orolo]

What about this?

The LPF has a corner frequency of 10Hz. The 10K resistor just after the LPF acts as a load for the op-amp.

The 1.5k resitor has two roles:

1) protects the op-amp from capacitive loading.
2) If the test point TP? is used to inject a testing signal, the 10k/1.5k divider causes the ADC to see a mix of about 0.9 times the op-amp output and 0.1 times the injected signal. In that case, the 1.8n/1MEG RC could work as a LPF (~10kHz) for the injected test signal.

I think this could be an explanation, because the 1MEG resistor doesn't seem to be RC filtering the op-amp, the 10K+1.5K take care of that.
Edit: No. Of course, a signal injected from the test point will see an RC filter with 1.8n and 1.5k, so the 1MEG resistor does nothing in that respect. What is that 1MEG resistor for?

[Benta]

You have a Sallen-Key active low-pass filter at the input, which is the worst filter topology ever designed.
It has several deficits, one being tolerance sensitivity, which is the change of filter characteristics vs. component tolerances.
It also turns into a high-pass filter at higher frequencies, the passive LPF at the output is to keep this tendency under control.

I suggest you change to an MFB topology instead.

[Benta]

By the way, TI has a very good application note:

http://www.ti.com/lit/an/sloa049b/sloa049b.pdf

Look at page 13 and you'll see the problem.