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Inspired by this question's accepted answer, I've been studying how PCB trace impedance can be affected by the trace's surroundings. However, something in the accepted answer is confusing me:

Having copper pour all over the place close to my signals!!! That would cause impedance discontinuities everywhere.

Why does this cause impedance discontinuities?

As I understand it, copper pour near a trace on the same layer causes an impedance discontinuity because it offers an capacitively coupled path as it runs adjacent to a signal trace. The coupling changes the capacitance for adjacent regions of the signal trace but not the remaining portion. This coupling occurs even if the copper pour is electrically isolated (for reasons I don't understand, but that is a separate question).

Is the above paragraph accurate? If so, are there other reasons for the impedance discontinuity?

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A track surrounded by a copper pour makes a Coplanar Waveguide:

https://en.wikipedia.org/wiki/Coplanar_waveguide

This is a form of transmission line, and you can make it with a controlled constant impedance provided there is enough copper on both sides. As with any transmission line, current traveling in the center conductor is mirrored by its return current which travels in the two copper "sides" sitting next to it.

Any interruption in the copper "sides" breaks the transmission line and creates an impedance discontinuity.

Component-side copper pours tend to be discontinuous and interrupted in many places by traces. They quickly turn into a mess of bits of copper strewn all over the place. In this case, your traces are made up of bits of coplanar waveguides with the side grounds haphazardly connected to a ground node somewhere, sometimes by very long bits of copper.

The top copper pour will also allow signals from one trace to couple into nearby traces.

A ground plane placed below is a much better option if you can have it.

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  • \$\begingroup\$ Thank you for the insightful answer. For clarification: Does a copper pour adjacent to a trace on one side of a trace but not the other also form a coplanar waveguide? \$\endgroup\$ – cr1901 Apr 2 '17 at 21:44
  • \$\begingroup\$ Kind of, yes it will create a transmission line. Return current will want to flow in the closest copper to the signal line, because this is the lowest inductance path. \$\endgroup\$ – peufeu Apr 2 '17 at 21:58
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You must consider the fact that all PCB material has a very small DC leakage current. And as such is a weak conductor.

When you place copper on that PCB. The distance that two traces are apart from each other becomes a capacitor. Now you have two modes of current conduction. each of which can be made to be beneficial or if ignored very problematic for the designer. In the case of most RF frequency circuits a 50 Ohm trace is required. So the width of the trace is the series Inductance matched to the shunt capacitance of the distance to ground tracks. Included in the equation is the caricaturist impedance of the PCB material itself.

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