Hi first time posting here. There always seems to be helpful answers and tips.

Okay my question is simply are there any glaring mistakes in my approach to the PCB layout (see image links below), I'm referring to grounding, noise and decoupling more than anything. Which always seem to be a tricky subject with varying opinions on the 'best' solution, I've read just about everything I can find.

The first bit that I'm a little unsure on is what happens to the return path on a differential signal? As I presume it must run back through 'ground' at some point?

How close do you really need to get all the ground points in a DC-DC Buck converter (bottom left of the board). Is it worth adding a 100nF capacitor beside the large electrolytic's on the 24v in-feed side of the LM2575, to help reduce high frequency switching noise on the 24v rail?

The analogue section has been through SIMetrix and looks to be doing what I'm expecting.

Eagle: Schematic PCB


To answer your first question, differential signals do not have a return path. There is no return through ground. That is a property of single ended signals. For this reason, differential signals don't theoretically even require a common ground between the two end points, though don't do that in practice unless you know what you're doing and using isolation of some sort (like Ethernet does). The whole point of a differential signal is it creates its own return. It returns through the negative line, and current flows from the positive line to negative line and never flows into ground.

As for your layout, it's ok, but you absolutely need ceramic decoupling capacitors. Electrolytics have a lot of ESL and are effectively resistors, not capacitors, for frequencies above 100-150kHz. So they will do nothing at all about ringing. Because of that, you are only decoupling the frequencies below that from the rest of your board. You must use ceramic bypass capacitors on the input and output if you want to attenuate the higher frequency ringing from being felt by the rest of the circuit.

  • \$\begingroup\$ Okay that's quite simple to add a small ceramic bypass to the input and outputs. Thought I might need to do so on the input at least, I thought I had the output covered with a ferrite bead and ceramic capacitor. \$\endgroup\$ – ben b Nov 18 '15 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.