# How does Stitching capacitor provide shortest return current path between plane splits, as we know that it blocks DC current?

I heard that Stitching capacitors are used to provide shortest return path whenever a signal is routed along the 2 power plane splits. As we know the return current path will be right under the trace. So how does the stitching capacitor provide shortest return path? As we already know that capacitor blocks DC current. So when I connect a stitching capacitor, how does return current flows through the capacitor?

We're only talking about high frequencies here. If there is a DC connection, then there would be no need to split a plane. If you split the plane there is a loop created that has some area. That means there is inductance in the return path and a voltage can appear if the current changes suddenly (as when a signal switches). That's undesirable- it upsets the signal and emits EMI, and the reason why (rule of thumb) you should not pass a signal trace over a split in a plane if you can avoid it.

If you cannot avoid it, then the next best thing is to make the plane appear to be one contiguous sheet of copper to high frequencies- by stitching capacitors across the trace near (say on either side) of the signal trace.

Similar issues will occur with signals piercing multiple internal planes.

Initially, I thought it was a bit like this circuit:

simulate this circuit – Schematic created using CircuitLab

But then I found information here: http://www.hottconsultants.com/techtips/pcb-stack-up-6.html. From what I gather, stitching capacitors are only to reduce EMI etc, so it doesn't come into play for DC. So as far as I can tell, it acts as a coupling capacitor. So something like this:

simulate this circuit

Perhaps someone can confirm? The source I linked to mentions inductances which my diagram doesn't touch on, but perhaps it's the same principle?

• +1 This is quite close- but think of the plane split as an inductor rather than a resistor. – Spehro Pefhany Apr 24 '15 at 12:18