7
\$\begingroup\$

If a via gets plugged with solder, does this destroy its ability to pass high frequency (RF noise) currents? Since there is no longer a surface along which the high frequency currents can travel due to the skin effect?

Somewhat related, with regards to via-in-pad technology, I read that conductive materials are usually undesirable because thermal mismatching between the fill and surrounding PCB causes thermal fatigue. But I've not found any mention of high frequency effects of conductive fills. The only other effects ever mentioned are higher thermal and electrical conductivity (presumably for low frequency currents only).

EDIT: I've been trying to make drawings that show current flow on both sides of every PCB trace, layer, and via in a full closed loop but have not been able to do it in a clear, coherent way so for now I've simplified it down to the drawings below

According to the discussion the RF currents don't actually flow on the inside surface of the via anyways since since it is acting as a waveguide (I don't know anything about waveguides so I'll just take your word on that). This somewhat changes my question since now the question has turned into how high frequency via currents flow in general, unfilled or filled. Accordingly, these are two of the scenarios I had in mind.

This image below is the typical trace to via to trace. So according to the discussion, the current travels on the outside of the via. Since current enters the trace from the top (by means of a component lead). Then is the drawing below accurate in that the current must wrap around the edges of the trace to get underneath it in order to reach the outside surface of the via? Trace-via-Trace

Assuming the above is correct...traces tend to be thin so current does not have to take a large detour to get to the edges of the trace in order to wrap around to the underside. But then consider the scenario below where current is dropped into the middle of a large plane via a wire or component lead. Current dropped into the middle of a plane

I usually run into this in power circuits when I have large current carrying traces that are essentially planes. The planes often connect to planes in other layers to carry the current in parallel. Large through-hole snubbing components end up dropping down into the middle of the plane due to the physical size of the component. If current cannot travel down the inside of the vias, whether conductively filled or unfilled, doesn't that mean they must take the long trip to the edge of the plane in order to get underneath it so that they can access the outside of the vias in order to continue traveling?

Should I be adding little "cutout holes" in the plane so that the high frequency currents don't have to travel to the edge of the plane in order to get from top to bottom?

I think you also run into this if you have two ground planes (power planes, or any plane) and stitch them together with vias. The via drops current directly into the middle of the ground plane. The current entering the plane travels on the outside of the via and so ends initially ends up on the inner surface of the plane. Then it must find it's way to some edge along the plane so it can wrap around to the outer surface. Presumably, in this case the "cutouts" in theplane I was referring to in the paragraph would accomplished by anti-pad of vias from the signal layers passing through the plane? That would mean that via stitches for planes should be done close to signal vias?

Like so: enter image description here

\$\endgroup\$
14
  • \$\begingroup\$ There is still a surface, on the outer diameter of the via. Is there an effect because there's no surface on the inner diameter? Very interesting question. \$\endgroup\$
    – The Photon
    Apr 11, 2019 at 22:07
  • \$\begingroup\$ There is, but since we mount our components on the outside surface of the layer, for the currents to get to the inside of the layer they have to travel all the way to the edge of the trace (maybe not so bad) or the edge of the plane (really bad) to get to the outer surface of the via. Then they have to do the same thing again to get back to the outer surface again at the destination. \$\endgroup\$
    – DKNguyen
    Apr 11, 2019 at 22:10
  • \$\begingroup\$ If you're talking about top layer microstrip geometries, at high frequencies the current is concentrated on the bottom side of the trace, close to the return current path on the next plane layer below. \$\endgroup\$
    – The Photon
    Apr 11, 2019 at 22:14
  • 1
    \$\begingroup\$ Actually I think that's the answer. Just like the inner part of a power transmission line doesn't matter in the skin-effect dominated frequency regime, and we can replace it with steel (or air) without affecting the effective resistance, same thing with the via. It doesn't matter if the center is air or solder or swiss cheese, current will flow mainly on the outer surface of the via because the return currents are flowing outside the via, not inside it somewhere (like in coax). \$\endgroup\$
    – The Photon
    Apr 11, 2019 at 22:21
  • 2
    \$\begingroup\$ @Toor The gap in your reasoning is that a via has an anti pad in whichever layer it penetrates, and current can switch from top to bottom of a plane using that hole. The place this sometimes screws you if you are not paying attention is in situations where you have to change reference layers (not just reference planes) as stitching the layers near the signal via becomes critical. \$\endgroup\$
    – Dan Mills
    Apr 11, 2019 at 22:41

1 Answer 1

5
\$\begingroup\$

If a via gets plugged with solder, does this destroy its ability to pass high frequency (RF noise) currents?

No. In the high frequency regime, the current tends to flow on paths that minimize the inductance of the loop formed by the signal current and the return current. This means that current through the via will be flowing on the outer surface of the via, not the inner (hollow) surface.

As eye candy, I offer this HFSS simulation of a via structure I did a while back, with surface current density plotted (f = 5 GHz):

![enter image description here

I was going to use this simulation to make my point, but realized I have modeled the vias as being solid metal anyway. This is common practice in signal integrity EM simulation as far as I know.

Here's a view of the ground currents in the same simulation as above: enter image description here Again the currents are concentrated on the outer surface of the via, near the signal currents that they return.

And here's a view that shows the underside of the ground planes: enter image description here This shows the return currents doing just what you imagined in your drawing: flowing around the edge of the antipad for the signal via and then to the nearest ground via.

One thing that is not quite right in your drawing is you seem to show the current flowing on the top side of the signal trace, where in a high frequency situation it will mainly flow on the bottom side of the trace.

To respond to the specific questions you added,

This image below is the typical trace to via to trace. So according to the discussion, the current travels on the outside of the via. Since current enters the trace from the top (by means of a component lead). Then is the drawing below accurate in that the current must wrap around the edges of the trace to get underneath it in order to reach the outside surface of the via?

Yes, but it will do so near the component lead, and flow mainly along the bottom side of the trace, so as to minimize the inductance of the trace-return loop. It won't wait until it gets to the via to move around to the bottom of the trace.

But then consider the scenario below where current is dropped into the middle of a large plane via a wire or component lead. ... I usually run into this in power circuits when I have large current carrying traces that are essentially planes. The planes often connect to planes in other layers to carry the current in parallel.

In a power circuit, you hopefully don't have much RF current running around to begin with. At 3.5 MHz, a typical 1 oz copper layer is only one skin depth thick, so substantial RF currents will be able to penetrate the plane. At lower frequencies, skin depth won't play much of a role. In any case the inductance of the wire will prevent it being very useful for supplying high frequency currents. You'll want to bypass your higher-frequency currents near where they're generated rather than try to supply them from a wire back to your power source.

Should I be adding little "cutout holes" in the plane so that the high frequency currents don't have to travel to the edge of the plane in order to get from top to bottom?

I'd use adequate bypass capacitors rather than rely on a wire from far away to deliver high frequency currents to a power net.

That would mean that via stitches for planes should be done close to signal vias?

Yes, if you are running a high frequency signal through a via so that its return current must flow from one plane to another, you will want to place ground vias nearby to give the return current a path to do that.

\$\endgroup\$
11
  • 1
    \$\begingroup\$ How does this work if the via is connected to a plane when it is not so easy to get from one surface of the trace to the other? Because the surface that a component is mounted to is the same surface that would be on the inside of the via. Does this mean that the high frequencies would need to travel all the way to the edges of the plane to get to the outside surface of the via? Because that seems like it would be really bad for emissions. \$\endgroup\$
    – DKNguyen
    Apr 11, 2019 at 22:32
  • \$\begingroup\$ @Toor, usually the plane isn't perfectly unbroken from where some component connects to it all the way to the edge. For example, there has to be a break in the plane to not short to the other pad of your component. Maybe it would help if you share some example layout, and we can examine it in detail. \$\endgroup\$
    – The Photon
    Apr 11, 2019 at 22:37
  • \$\begingroup\$ I guess that's true. Give me time to make a drawing that has all the scenarios in my head. \$\endgroup\$
    – DKNguyen
    Apr 11, 2019 at 22:49
  • \$\begingroup\$ I added drawings to my original post. \$\endgroup\$
    – DKNguyen
    Apr 12, 2019 at 13:38
  • \$\begingroup\$ With regards to the scenario where you answered "adequate bypass capacitors", the wire dropping into the plane is not a power delivery or lead. Those enter from the edge of the plane. The leads I am talking about are are not power delivery leads. They are snubber leads so I don't think your answer is applicable for that scenario. \$\endgroup\$
    – DKNguyen
    Apr 12, 2019 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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