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Should be a simple question but I've never seen a clear answer to this.

I know that capacitors remove DC bias on a signal, like with an AC signal being output from an amplifier; just couple the signal through an HPF and you're left with the AC signal oscillating around 0 V.

Does the same principle apply to two grounds that might have some DC offset? I know that a real capacitor will have some leakage current, but if we ignore that for a moment, conceptually does the capacitor remove DC offset between two grounds?

I ask because in a power supply I know that the isolated side and input side are sometimes bridged with a cap. If we assume for the moment that the output side is floating, wouldn't connecting the two sides with a capacitor set them to the same potential? Or does this just remove any DC offset for HF noise, so now the HF noise is oscillating around 0 V with respect to some other reference (like earth???)?

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  • \$\begingroup\$ DC current cannot flow through a capacitor, so putting one between two nodes at a different DC voltage will not being them to the same potential. \$\endgroup\$
    – user1850479
    Commented Nov 23, 2021 at 3:35
  • \$\begingroup\$ The capacitor appears invisible to the DC, thus it will not make any difference. However, it will provide an impedance to AC currents. This is why you’ll commonly see cable shields tied to ground via a capacitor - high frequency AC currents get shunted to ground. If the shield was tied directly, DC currents might flow creating a ‘ground loop’ \$\endgroup\$
    – Kartman
    Commented Nov 23, 2021 at 3:36
  • \$\begingroup\$ @Kartman well then in this case, it would remove an AC offset and not a DC offset, correct? \$\endgroup\$
    – user224284
    Commented Nov 23, 2021 at 3:38
  • \$\begingroup\$ It won’t necessarily remove, but it will attenuate the AC signal. \$\endgroup\$
    – Kartman
    Commented Nov 23, 2021 at 3:40
  • \$\begingroup\$ @Kartman sure that makes sense, but going back to the DC offset for a moment, suppose there is some DC offset between two grounds... Connecting them with a cap would provide infinite impedance between these grounds, so any DC offset between them is entirely dropped across the cap. I suppose this means that the DC offset between them still exists, correct? EDIT: Meaning if we connect a voltmeter across them we would still measure a voltage... I suppose I'm answering my own question here... \$\endgroup\$
    – user224284
    Commented Nov 23, 2021 at 3:44

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No it does not remove DC offset - it allows there to be a DC offset. A capacitor blocks DC because a capacitor does not pass DC and it allows there to be a DC bias over the capacitor. It has infinite impedance at DC. And so it passes AC as it allows AC currents through and has low impedance at high frequencies.

It preventing EMI, it allows the voltage between primary and secondary to have for example 1000V of DC difference so there is isolation, but it will pass high frequency noise so it shunts the output noise of a floating DC supply to the ground-referenced mains input side.

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  • \$\begingroup\$ Thanks, I was looking at it differently but ended up realizing I had equated "remove DC offset" with "prevent DC current". The former is not the right way to describe this function of the cap \$\endgroup\$
    – user224284
    Commented Nov 23, 2021 at 6:32

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