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Reading about FPD-Link II in Wikipedia, I came across this statement for which I need clarification:

Because the signal is DC balanced the application can use AC coupling, which eliminates the ground current problem between data source and destination.

It goes on to say that this is critical in automotive applications due to potentially high transient currents. There is no diagram or reference figure provided to help explain what is meant by this. Perhaps what they mean is that any high magnitude/low frequency currents affecting the ground references to the xmit and receiver devices are "filtered" (subtracted) out by the coupling capacitor....Anyway, if anyone can explain this statement better than they are doing so in the Wikipedia page, I'd appreciate it ---I'll give you credit when I make suggestions to improve the explanation on Wikipedia ;-).

jrive
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It just means that since the useful signal consists only from AC signal components and low frequencies are not needed, the DC path between transmitter and receiver can be isolated with capacitors which remove DC potential. And the capacitors then allow the receiver and transmitter to be at different DC potential.

Assuming the capacitors can handle it, there can be many volts of DC over a single capacitor, and it can still pass a low voltage AC signal just fine.

This allows the transmitter and receiver to have a potential difference between them, and it also allows the potential difference to vary within reason, due to voltage drops in wiring resistance when varying loads cause DC or AC voltage differences.

Justme
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  • that's what I figured they meant but was not clear from the explanation they provided....thanks – jrive Apr 16 '22 at 23:25