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One can create a very cheap and easy to build capacitive galvanic isolator for low freq (audio)? 

In+ --||-- Out+

In- --||-- Out-

Adding resistors does not help much (in fact it increases the noise).

Obviously the two sides are galvanically isolated!!!!

There is no DC path between the in and out. Yet why doesn't this work or is used?

(I thought adding a load resistor on both sides would help with DC discharge if necessary)

Basic analysis says this should work! (cap open for DC == block signal DC and ground DC, cap closed allows signals only)

My guess is that capacitive isolation works but since it it still forms an electrical loop rather than an electrical break that one gets with opto's and magnetics, that ground loops still exist. Although it isn't quite as satisfying since there is still galavanic isolation and that, technically, should break any ground loops.

Ricardo
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AbstractDissonance
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  • What exactly is the question? – horta Oct 05 '15 at 02:21
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    Or perhaps a better question: why would you want to galvanically isolate audio? Are you running an amp from a high voltage power supply? – horta Oct 05 '15 at 02:33
  • It is used. But you need to account for the HPF created. – Ignacio Vazquez-Abrams Oct 05 '15 at 02:39
  • @IgnacioVazquez-Abrams. im not worried about the response. Ive never seen it used before. Any references to your claim? – AbstractDissonance Oct 05 '15 at 04:49
  • We use this in the back-channel of our UART (device is powered by a galvanic isolated flyback, but has to send data back). It's cheap, low power but has the drawback that no static level can be transferred, so additional stuff is needed to get the UART working correctly and you have to worry about the frequency of the signals. – Arsenal Oct 05 '15 at 06:51
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    Capacitors do not provide galvanic isolation. Therefore, given that the premise of your question is incorrect and there are no apparent "other reasons" mentioned in your question I'm voting to close. – Andy aka Oct 05 '15 at 08:29

2 Answers2

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There is no DC path between the in and out. Yet why doesn't this work or is used?

With galvanic isolation you usually want protection from AC mains and not just from DC. But AC voltage passes right through your capacitors - the voltage would still be lethal!

You can see these caps e.g. in audio amplifier outputs. You only want the AC component on your speaker or headphone coils.

Turbo J
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I answered a somewhat similar question here.

Common mode noise is a problem with this approach.

If two circuits are perfectly galvanically isolated, they can be referenced to different voltages, and those voltage references could change relative to each other without affecting the signal being transferred. That's not the case with this approach. If your input reference vs. your output reference was, for example, a 1 kHz square wave, that noise is going to get coupled into your signal.

A perfect magnetic isolator doesn't have that problem. Of course, any real magnetic isolator also has a capacitance associated with it, but it's usually quantified somewhere on a datasheet.

Your circuit also may fail an AC hi-pot test, depending on the caps involved.

Community
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Stephen Collings
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  • Capacitors can work as a quasi-isolated coupling if the signal to be conveyed by the capacitors is of a much higher frequency than the signals that need to be blocked. While transformers or optos are often better, capacitors can be smaller and cheaper. – supercat Jan 10 '17 at 23:00