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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???)?

sparaps
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  • 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. – user1850479 Nov 23 '21 at 03:35
  • 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’ – Kartman Nov 23 '21 at 03:36
  • @Kartman well then in this case, it would remove an AC offset and not a DC offset, correct? – sparaps Nov 23 '21 at 03:38
  • It won’t necessarily remove, but it will attenuate the AC signal. – Kartman Nov 23 '21 at 03:40
  • @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... – sparaps Nov 23 '21 at 03:44
  • Capacitors do not remove DC bias on a signal. An inductor might because it will short out the DC. – Andy aka Nov 23 '21 at 03:44
  • @Andyaka Then why do we call series capacitors DC blocking caps if they do not block DC? In fact they would block DC because they will drop all DC voltage across them when placed in series with a load, so of course they are DC removing caps when measured from the perspective a load placed in series with the cap. – sparaps Nov 23 '21 at 03:47
  • They do block DC but they don't remove DC from a signal. – Andy aka Nov 23 '21 at 03:52
  • DC blocking capacitors *do* block DC--they sustain a DC voltage across themselves, preventing any DC coupling across them. – Hearth Nov 23 '21 at 03:56
  • @Kartman another question that comes up, is if the secondary side in an isolated power supply is floating, doesn't the use of a capacitor across grounds eliminate the radiated emissions from the floating conductor by sourcing or sinking some AC current? This would be equivalent to grounding out a ground plane to its enclosure to sink RFI currents. – sparaps Nov 23 '21 at 04:00
  • @sparaps You mean like this? https://electronics.stackexchange.com/questions/592908/what-is-the-purpose-function-of-these-bypass-capacitors-in-this-isolated-suppl – DKNguyen Nov 23 '21 at 04:07
  • Actually yes that would make sense. I'm also wondering does it remove the DC offset. My thought now is that the answer is "no" because the two sides are not being shorted at DC, they are only (nearly) shorted at high frequency AC. – sparaps Nov 23 '21 at 04:11
  • @DKNguyen Also from here: https://electronics.stackexchange.com/questions/368986/understanding-topology-for-emi-supression-in-a-power-supply "...a Y-capacitor is added between ground on the primary side and secondary side. This is one typical method for reducing the common mode noise caused in the secondary side by primary side switching noise via the capacitance across the windings of an isolation transformer" In this way, isn't the "ground offset" problem on the secondary side I'm referring to just a common-mode noise problem as seen from the primary side? – sparaps Nov 23 '21 at 04:37
  • If by "removing DC offset" you mean "changes the DC offset to zero" then no, it does not remove DC offset. The DC offset that was there when the two nodes are isolated will still be the same when you connect the capacitor between those two nodes. – DKNguyen Nov 23 '21 at 04:48
  • @DKNguyen Yes I see that, if you measured across the grounds with the cap in place, then you would just be measuring the voltage across the cap, which is equal to the DC offset. I suppose what I mean is "block any current flow from pri to sec due to DC offset so that the DC offset won't affect anything on the secondary side" but that seems trivial and no cap would be needed if that were the only reason. – sparaps Nov 23 '21 at 05:10

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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.

Justme
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  • 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 – sparaps Nov 23 '21 at 06:32