Is there a reason the offset of this opamp is done like this?

Question

I recently found this schematic from a project at university (drawn by our lecturer). It's a low pass filter that also adjusts the gain, and adds an offset of 1.5 V (the input signal can be positive or negative and this filter feeds an ADC).
However, I just realised that the 1.5 V offset at the positive pin of the opamp doesn't include the capacitor. Is there a reason the 1.5 V is not connected to both components?

Thanks for your help!

Andy aka · Answer

I just realised that the 1.5 V offset at the positive pin of the opamp
doesn't include the capacitor. Is there a reason the 1.5 V is not
connected to both components?

Capacitors don't pass DC current so there's no benefit in doing this. If the 1.5 volt was actually a DC voltage with some useful AC signal superimposed (people are at liberty to call nets whatever names they choose) then C44 might need to be connected to 1.5 volts. Chances are... not.

Spehro Pefhany · Answer

The input (presumably) and the output (definitely) are referenced to GND. By low-pass filtering whatever is connected to R63 and the 1.5V reference any noise on the 1.5V reference is also filtered and the impedance to ground is minimized. Many reference sources are relatively noisy and they may have relatively high output impedance at higher frequencies, far more than a typical capacitor.
It would be a rare situation where there was a reason to do this any other way.

Is there a reason the offset of this opamp is done like this?

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