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I'm trying to understand why the ground plane in PCB is preferred to be as wide as possible. Mathematically the self-inductance in a free space conductor has a relationship as shown in the uploaded picture:

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But I want to understand the physical analysis of how is the increase of width will reduce the self-inductance?

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    \$\begingroup\$ Think of the increase in width as putting two in parallel. Their mutual inductance means the drop isn't twofold, but there is a drop. \$\endgroup\$ – Neil_UK Dec 20 '16 at 15:06
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When two small wires are bifilar wound around a magnetic core, they exhibit the same inductance as one single wire. The emphasis in the first sentance is "small" meaning that any flux produced by one of the small wires is perfectly coupled to the 2nd wire.

In effect, each wire (in the pair) acquires an inductance that is twice the inductance of a single wire and, when there are two inductors are in parallel, the net inductance IS the same as if a single wire were used.

The magnetic core helps achieve flux coupling between those two wires in this little thought experiment.

Now, remove the core and coupling is not the same; current flowing in one wire (despite being closely wound with the other) does not 100% couple to that other wire and, as you spread the gap between those two wires, coupling gets even smaller and ultimately, the inductance of the pair starts to look like L/2.

It's the same effect with a wide copper conductor - the current is largely spread evenly across the width but flux generated by a part of the current along one edge does not 100% couple with the flux generated by a similar part of current flowing along the other edge.

So progressively coupling starts to reduce as the copper conductor gets wider and you start to move away from L to L/2.

The limit of L/2 does not take into account frequencies where the applied signal wavelength starts to be significant in terms of track thickness of course. That creates a whole bunch of other things to consider!

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