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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?

enter image description here

Thanks for your help!

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  • \$\begingroup\$ What are R63 and R60 connected to ? C44 will charge to 1.5V (disregarding the effect of the input supplied at R63). \$\endgroup\$ – AJN Aug 1 at 14:59
  • \$\begingroup\$ R63 and R60 are connected to the differential output of an isolated amplifier (ACPL-790B). The offset of 1.5 V is constant. \$\endgroup\$ – Marcos Aug 1 at 15:03
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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.

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  • \$\begingroup\$ I added the schematic showing the isolated amplifier that feeds the opamp with the offset. \$\endgroup\$ – Marcos Aug 1 at 15:16
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    \$\begingroup\$ There would be no advantage I can think of in connecting the cap to the reference voltage. It's a bit less important in this case because the ACPL-790B has an internal electrostatic shield connected to GND2 (not shown in your schematic symbol) which conducts common mode input noise directly to the ground. \$\endgroup\$ – Spehro Pefhany Aug 1 at 15:44
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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.

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