I would like to route my USB 2.0 (12MHz) signal on the bottom layer of a 4-layer PCB. However, the edge-coupled microstrips would then be referenced to the power plane, not the ground plane. The signals are going to the USB pins of an STM32F105 microcontroller.

The signal path should be 90-Ohm differential, but I don't know how important the characteristic impedance is since I won't be using the High-speed (480MHz) rate.

My board's stackup is:

  • Top (signal, components)
  • Ground (unbroken)
  • Power (3.3V, unbroken)
  • Bottom (signal)

I have radio transceivers on the top layer, using differential, edge-coupled microstrips for the rf signalling. All the IC's are decoupled fairly well.

I see a few options:

  1. Route the traces relative to power plane.

    I would prefer this. Will it work? Does it couple noise onto the rest of the PCB's power plane?

  2. Remove a section of power plane, exposing the ground plane underneath the microstrips.

    The ground plane is now so far away that my traces would be quite wide...

  3. Isolate an area of copper in the power plane under the microstrips, and stitch it to the ground plane. Kind of a "ground mezzanine" :)

    I think this would work well, but the top layer is too thick with traces for me to do a good stitching job. I don't want to use blind or buried vias. Would this solution require good stitching?

  4. Don't concern myself with the characteristic impedance. The traces will be about 600 mil long, and I'm only signalling at 12MHz...

Here's the section in question. I intend to run the traces to the far left, underneath the microcontroller, then bring them up to the top layer. For scale, the small components are 0402 (1005 metric).


(those traces aren't actually crossing on the same layer; it just looks like it because I highlighted them in red)

Are any of these good options? Is there anything I'm missing?



3 Answers 3


The short answer is you're going to be fine, route it above VCC and make sure you have some VCC to GND decoupling caps near your chip. Plus your route is pretty short at 600mil, I've seen some people do terrible things to USB routes that still end up working :)

I think the best way to understand this is to consider where your return current will flow. Current is going to follow the path of least impedance. In a microstrip's case, when you consider the high frequency current of your USB edges this current will flow back in the reference plane directly below the trace. It doesn't matter if that trace is GND or VCC, it's the path of least impedance so that is where current must flow.

Now some interesting things, current always flows in a loop. When that return current, happily flowing along on your VCC plane gets back to your chip it must find a way to GND to complete the circuit. It's going to do that by the path of least impedance again which hopefully in your case will be the VCC decoupling capacitors you placed nearby.

That's for the AC portion once your signal has moved to the DC portion your current loop will go back to following the path of least resistance.

Also a lot of people will refer to USB as differential and then cite the fact that most of the current will return in the pair itself. But in USB 1.0 it's usually just two single ended drivers referencing GND so the current should travel the way I described above.

Finally even when you have a differential pair there will almost certainly still be common mode return current unless you can somehow guarantee that from the source to the destination, through connectors, and routing, that the D+ and D- lines are the same length and never hit any single ended discontinuities etc.

Hope that helps. There are some good books out there on signal integrity if you want to learn more about this kind of stuff such as Johnson's handbook of black magic, and Eric Bogatin's books.

  • \$\begingroup\$ "in USB 1.0 it's usually just two single ended drivers referencing GND", what, and they don't reference Vcc? Whatever driver you use, it has an equally low impedance connection to Vcc and GND, and it can source or sink current from either equally. It's also not true that the return current has to make it back to ground to complete the circuit. It has to complete the circuit, yes. But if the current came from ground in the first place, then it has to find its way back to Vcc to complete the circuit. \$\endgroup\$
    – Phil Frost
    Dec 3, 2014 at 13:26
  • \$\begingroup\$ Fair enough I was just trying to point out that they don't reference each other without being confusing but you are correct. \$\endgroup\$ Dec 3, 2014 at 14:18
  • \$\begingroup\$ You should tie the power plane you are referencing against to ground with a capacitor at both ends (chip and connector) \$\endgroup\$ Apr 7, 2016 at 11:41

I would like to route my USB 2.0 (12MHz) signal on the bottom layer of a 4-layer PCB. However, the edge-coupled microstrips would then be referenced to the power plane, not the ground plane.

Simple solution: redraw your schematic.


simulate this circuit – Schematic created using CircuitLab

Now you can feel better about making a microstrip relative to that other layer, which is now the ground plane.

Of course this makes absolutely zero difference to the operation of your circuit, and your microstrip doesn't care either. Ground is a power plane. It's just the one with a funny symbol attached to it in the schematic.


Just consider the power plane 'ground' when you route it. The DC voltage of the reference plane for a controlled impedance doesn't really matter, so long as it is a continuous plane. Since it is a differential trace, you shouldn't have any major problems with crosstalk as differential traces are designed to cancel out both received as well as emitted crosstalk (the average potential on the pair should be approximately constant and the average current should be approximately zero).


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