Coupling PCB traces

Question

In my pcb design I have 2 triangular voltage signals with the following specs:

• Peak voltage = 300V
• Rise and fall time= 7us
• Freq=25kHz
• (assume a constant current of 0.5A)

Because of the limited space available, I need to trace these signals on top of each other.

I am trying to get away with a 2 layer PCB board with a standard thickness of 1.6 mm.

My Question:

I am concerned about the crosstalk between the 2 traces due to capacitive coupling.

My attempt at an estimation:

• C= (eo.er.L.w)/d
• L=50 mm
• w=0.6 mm (trace thikness)
• er=4 (dielectric constant PCB)

Thus with these values I arrive at c=0.7 pF

Then with Ic= C*dv/dt= the induced current is 50 uA. This seems a very low current which would bring me to the conclusion that I can neglect this coupling.

But intuitively, I feel that I am missing something.

Or is the analysis good enough?

EDIT: (based on the answer by @Neil_uk) The two traces are going to 2 their loads which are capacitors. Hence the current is constant for linear slope. Thus I made the assumption that the inductive coupling could be neglected.

Answer 1

Assuming the geometry you've given, you've done the sums correctly to get to 0.7pF. However, that will be an underestimate for the true capacitance which will be dominated by fringeing fields, rather than parallel plate capacitance. Inflate your figure by a factor of 2 or 3, to get 2pF. That should be accurate enough for back-of-envelope work (until somebody cranks up a line solver and posts the correct figure). (edit - The correct figure is 1.6pF, using a microstrip calculator, half the result for half the substrate thickness /edit)

Given 0.7pF, I get 30uA for a dV/dt of 300v in 7uS. A triangular waveform of 25kHz doesn't have a 7uS rise/fall time, so I'm not sure what your waveshape is. 30uA is essentially the same as 50uA if we're on the back of an envelope.

Whether the resulting 100uA (assuming 2pF) between traces is too much depends on what else is going on in the circuit. It may be trivial in a power circuit, and a job-stopper in a measurement circuit, you don't say what yours is. If that 0.5A figure you give is a reference for what's relevant, then 100uA may be far enough below that to be negligible for your application.

There will also be inductive coupling between the traces, which could be a problem if one carries a rapidly varying current, and you're sensing the voltage across the other. You don't say whether your circuit has these features.