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For RF pcb design, it's common knowledge there shouldn't be any right-angle traces, because capacitance increases in the corner, changing the impedance and causing signal reflections. Less clear (to me at least) is what happens when a right-angle is formed by a passive component, like so:

PCB trace

Pad 10 is the output of a voltage-variable attenuator--the signal is anywhere 50MHz-1GHz and can be up to 20dBm here, and it is traveling from right to left, then up through C8 on to U2. Notice that capacitor C8 is oriented at 90 degrees with respect to the trace.

Is this even a problem? Should I move that cap up and bend the trace to it? Or am I being overly-fastidious and this doesn't matter?

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    \$\begingroup\$ Making the transition from PCB trace to component terminal is already a (relatively) huge lump of capacitance. The direction that the trace leaves the pad has no significance whatsoever. \$\endgroup\$ – Dave Tweed Jan 3 '15 at 1:22
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    \$\begingroup\$ @DaveTweed that sounds like most of an answer... \$\endgroup\$ – Phil Frost Jan 3 '15 at 3:26
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Strictly speaking, the AC signal current through the capacitor is exactly the same current as if the capacitor was bridged - this means it acts just like you'd expect a right angle pcb trace to act but, don't beat your self up on this - at 1GHz maximum it's not a major problem. The problem occurs when you have the track folding back on itself and forming a capacitor and an inductor - this forms a parallel tuned circuit in series with the signal and can lead to anomalies but, at 1GHz this isn't going to be significant.

Try calculating the effective inductance of the trace round the corner and the capacitance formed - this will be well above 1GHz.

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