PCB Impedance Control

Hi

If I am designing a pcb using impedance controlled layers can I treat th power planes as a reference layer as well as the gnd layers?

Cheers

Jon

Reply to
maxascent
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Hi Jon, Yes. But with a caveat. When your signals switch reference layers, make sure there is a path for the reference current.

E.g. Take a 6 layer board, layer 2 ground, layer 5 power. If the signal goes from layer 1 to 6 through a via, you should have a bypass cap bewteen power and ground near this via. Think of your signal as differential, the complementary signal being the reference.

It's for this reason that I long ago ditched power planes and use multiple ground planes instead. I route and/or copper pour powers. Then the bypass cap in the example I gave can be replaced by a ground via, because layer 5 is ground in my PCB.

HTH., Syms.

Reply to
Symon

Since without a ground one has a dipole antenna, could you qualify what you mean? The impedance of the plane is measured in units of picoHenry per square so it's not a solid ground, certainly, but without a ground we'd be in a world of hurt at these high frequencies.

Reply to
John_H

I have a number of pretty pics I made to illustrate the issue. I'll put 'em on a.b.s.e tomorrow sometime.

As an aside, at high speeds (which I define as having a track longer than 1/4 rising/falling edge, YMMVG), there is no such thing as 'ground'. The ground plane is a signal layer insulated by copper ;) [Although it helps to keep large power currents away from the high speed return paths].

Cheers

PeteS

Reply to
PeteS

I think Pete means that at high frequencies the skin effect means that the current in the plane is all on one surface of the plane. Thus the bulk of the plane might as well be an insulator. Or something like that? Cheers, Syms.

Reply to
Symon

I think he means the return current is mostly localised directly under the signal track; the copper under the "space" between tracks carries relatively little current and could *almost* be replaced by insulator.

- Brian

Reply to
Brian Drummond

Seems to me you're both right: the return current is bunched up under trace, on the side of the ground plane adjacent to the trace. This "path" is surrounded by copper. To paraphrase Doug Smith, a local EMC consultant, the other side of the plane might as well be on the moon.

Bob Perlman Cambrian Design Works

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Reply to
Bob Perlman

Thanks to everyone for devining what PeteS means. I understand skin effect. I understand the confinement of return current for high frequencies. I don't understand the statement that "there is no such thing as 'ground'."

I was hoping Pete could conjecture what Pete meant.

- John_H

Reply to
John_H

The fact that return current is confined tightly around the signal for high frequencies doesn't make the ground plane cease to be nearly equipotential. The confined return current will not raise the voltage of the ground plane significantly but instead flow the current through a low impedance path due to any induced voltage. If the voltage changes under the high frequency trace due to the return current and the finite HF plane impedance, the current will flow out around that area to the lower potential - a voltage difference on a ground plane doesn't last long; this is why the return current path is several widths of the distance to the conductor above. If the ground reference were not a ground but a strip that was the same width we identified for the return current, the characteristics would probably be a bit different since the voltage difference from under the signal to the edge of this region isn't tied down to the same potential as the rest of a plane, but isolated. Current and voltage are two parts of the power flow. Impedance (resistance, capacitance, inductance) affects the overall picture. The confinement of the return path the way we've defined it is because of the planes - because of the impedances involved - in addition to the other high frequency issues such as skin depth.

I'd love to see the difference between a plane reference and a strip reference that *should* contain the entire return current. It may be the two solutions are the same. But it may present very different results. I haven't done studies on the return current issue, but my gut says that if we ditch the ground plane, the assumptions we make at high frequency also go out the window.

- John_H

Reply to
John_H

Wow. Didn't know I would set off such a discussion :)

At high speeds, return currents flow virtually exclusively on the reference layer at not much more than the width of the signal track; it also flows only in the skin area, the depth of which depends on a number of things. It won't flow elsewhere [subject to normal current distribution laws] (unless it has to because there is no return path, which was alluded to in the first response) because of the high impedance involved. (At ~4nH / inch there's a high energy barrier, although that value is highly topology dependent).

As such, if you have a signal track on a single layer point to point, then the return current will flow immediately adjacent on the nearest reference layer (so make sure it really *is* a reference layer), and for a 6 thou track, the return current will have ~95% in 6 thou on the reference layer.

As I said, at high frequencies, the notion of 'ground' as used in power, does not really exist; there are only signals and reference, whatever it happens to be.

Cheers

PeteS

Reply to
PeteS

Absolutely. The other side of the plane doesn't really exist to the return currents. There is some speculation that return currents take the lowest energy state return path, which seems intuitively true, but I'm not sure we have proper evidence for.

There is a great deal of misunderstanding in the area of reference layers, planes and high speed signalling in general. When we take a signal through the board, we have to make sure we also take the return path through the board. I spent a long time doing calculations on diff pairs of varying topologies to figure out the optimal via sizes (and spacing, annular ring size etc) to take signals from one layer to another with minimal loss (and therefore minimal radiation). I got the loss down to < 0.3dB / via on a rather dense high speed board.

Cheers

PeteS

Reply to
PeteS

Hi John

Let's first define what 'ground' means. For power, it's the 'zero potential' path for return currents. For RF radiation, it's literally the ground (where we get the name) which has a nominal '0' potential.

For high speed signalling, there isn't really a thing that has zero potential across it's length, be it signal or reference. The return currents in a nominal ground plane for high speed returns will induce significant potential gradients across the plane. As a 'ground plane' is usually used as a description of an area of equipotential, the term ceases to have meaning when it is no longer equipotential - as is the case when it is a high speed signal reference/return path.

That's really what I mean by the comment that at high speeds, ground doesn't really exist.

Cheers

PeteS

Reply to
PeteS

From my experience, a strip reference is the same as a plane. There are a number of issues here, not least that we are returning low speed high currents on the same plane that we are returning high speed currents.

For low speed (DC or close to it) currents, the entire plane is the return path. As the speeds increase, the return path becomes more and more constrained, primarily due to inductance of the return path. The return path would be identical [in width] to the signal path at infinite frequencies, for the above postulate. For reasonably high speed signals (at 1.25GHz fundamental, for instance) 90% or so is constrained within those bounds.

When looking at a plane, one must ask what the reference problem is; for a 1/4 wave antenna, the entire zone is required to make the reflected antenna operate properly (because it is reflected - it's not really the antenna); for a signal return, a plane is not necessary. This is borne out (when one thinks about it) by the principle of coplanar waveguide.

So - is a plane necessary? Usually, yes; and it's usually a very good idea, but at high speed it's not a 'plane' in the ordinary sense.

Cheers

PeteS

Reply to
PeteS

That's not necessary. There's already so much plane-plane capacitance that the planes are already equipotential as far as the tiny charge injected by the signal trace can affect them.

Really, on a board with, say, 3000 vias, are you going to put a bypass via near every signal via? I've heard of people asking for two!

John

Reply to
John Larkin

Hi John, I wish that was always the case! There are problems relying solely on the interplane capacitance. There will be an impedance discontinuity at the via no matter what, but using solely the interplane capacitance increases this discontinuity. Clearly the inductance of the signal in this case is higher. Also, if a bus does the transistion, all the even mode effects add up. Finally, the high Q of the planes means you are vulnerable to plane resonances.

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Clearly putting a bypass cap at every via is impractical, and if you re-read my post, that is not what I suggested. However, I stand by my recommendation that in places where many nets and/or critial nets change reference planes, a bypass capacitor nearby is important. Clearly, a single capacitor, perhaps connected with multiple vias to the planes, can be shared by many signal vias, as it will have a lot more capacitance than, for example, the planes can provide.

HTH., Syms.

Reply to
Symon

If that was true, would we have a problem crossing plane cuts?

My handwave says the plane cut would be C/4 relative to the via case. /2 because the caps for the 2 planes are in series, and another /2 because each cap is only half the size.

This sounds like something that should be in (one of) Lee Ritchey's books. Anybody got the page number handy?

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These are my opinions, not necessarily my employer's.  I hate spam.
Reply to
Hal Murray

Then I should be able to compute/estimate the value of "nearby" and the size of the cap needed.

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These are my opinions, not necessarily my employer's.  I hate spam.
Reply to
Hal Murray

I've experimentally TDR tested plane cuts, and they are generally invisible as long as they are thin as compared to board thickness, or if the ground plane slit is shorted out by an adjacent power plane.

All this popular "return current" theorizing is madness. There's so much dielectric shorting a small slit that a signal trace cruises over it and never sees it.

Spend s few minutes with a hunk of copperclad, some edge-launch SMAs, an xacto knife, and a 20 GHz TDR scope. It's enlightening. The other fun thing to do is to add some "classicly bad" trace/slit/via structures into spare bits of production boards and TDR them.

John

Reply to
John Larkin

Wow, that's about as incoherent as papers get!

Incredible, dangerous nonsense. Absolutely moronic. Quote:

"Referencing the Top and Bottom of the Same Plane. Whenever a signal switches layers and references first the top and then the bottom of the same plane we must still ask the question, how does the return current get from the top to the bottom of the plane. Do to the "skin effect" the current cannot flow through the plane, it can only flow on the surface of the plane.

This is some sort of joke, right?

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That last one is academic and inconclusive, and the board he uses is not at all realistic.

Except in extreme situations, it's safe to assume that parallel planes in a multilayer pcb, bypassed with scattered caps, is an equipotential structure. It's that simple.

John

Reply to
John Larkin

The way I've heard it done is that you can you a power plane as a reference if it is the actual power used for the driver of your impedance controlled net.

Reply to
kayrock66

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