If you have a microstrip trace above a ground plane, and put a voltage step in at one end, there will be a traveling wave front that moves down the trace. In the moving "risetime" region, the capacitance of the trace must be charged, so there is a current into whatever constitutes that capacitance. In microstrip, some current will be into air and some into the FR4, ideally into a ground plane, layer 2.
What is the return path for the current into the air? How about the current that winds up in the ground plane? What happens when the signal jumps through a via to layer 4 and suddenly gets "referenced" to a layer 3 power pour?
The short answer is: don't worry much about it. Do NOT pay HoJo $600 to attend one of his silly seminars.
I wonder what HoJo would say about the return path of a Goubau line.
--
John Larkin Highland Technology, Inc
lunatic fringe electronics
I found the thread--it was from December 2005, and titled "Interesting experiment". The pages on the Wayback Machine don't have the video link that I can see. Really really funny video, though, if you know anything about corona.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
"There ya go" (in the voice of Ron Reagan) "thinking in the time domain again." :^)
Well not current. I'm thinking of the E-field, but at high enough frequency you have to think of the entire E&M radiation. I guess I see the field lines in the air terminating back on a ground plane somewhere. But I'll freely admit my picture is fuzzy, and could be worng.
Doesn't matter, power and ground are the same except at some low frequency. (You know that, of course)
Never planned to. The ground current concept helps me a lot in laying out 2-layer boards.
Never heard of it....
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(Is that balun right in the launcher circuit? Seems like it needs another wire? to the launcher maybe This is a bit better,
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It's kinda like waveguide, but the fields are on the outside. What carries the current in a waveguide?
Sure. For one thing, the higher the frequency, the higher the inductive reactance felt by currents that go too far from the trace. Johnson isn't altogether wrong about the phenomenon, as you say, but his explanations are ridiculous. His making a career out of being vaguely right (in spots) for the wrong reasons is sort of amusing.
I mean, the whole thing is an undergraduate fields problem, at least for straight conductors or simple curves. Even the mathematically challenged could fire up one of several free method-of-moments simulators and see the problem very clearly. I expect that FastHenry would do a pretty good job of showing what slots in the ground plane do, at least in the low-frequency limit (where radiation from the slotline antenna isn't important).
Just visualizing the fields (pretty easy to do with the method of images) will give an intuitive feel for the problem.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
OK, you're using the same language as JL, and I'm not understanding. What current is leaving the trace? Polarization current?
Thanks.. I fixed at least one layout (not by me) with a piece of copper tape soldered across ground planes under the signal trace. (It's hard to argue when it works.)
Yeah... Well I'll just think more then.. there's this energy idea floating around...
Sorry, I was unclear. I'm talking about ordinary current in the ground plan e. It spreads out more at low frequency. HoJo ignores the EM mode issues es sentially completely, at least in the book and in what I've read of his oth er stuff (what you can get for less than $10k).
And that now-vanished video from 2005 showed that he had no idea whatsoever of the orders of magnitude involved.
Some of it continues into space as radiation (swapping periodically between E and M fields). This is encouraged if the trace on the PCB is folded back on itself in a suggestive manner, so that constructive interference leads to lobes of strong antenna gain.
(Obviously, patch antennas are optimized for such purposes. The substrate presents effective loading capacitance that doesn't radiate, so these can never be very wide band devices. But that's plenty for wifi and stuff!)
Most of the D-field lines arc through space, back to the dielectric surface (which is ~ground), eventually returning to ground proper.
A via is a discontinuity, where current flows over one plane, then stops; while on another layer, it begins, and continues.
The current flows need to be balanced between the planes. So on the plane-to-plane surfaces, there will be a current that diffuses out from around the via. The wavefront from this hole expands radially, so that inductance goes as ~log(time).
For most average via shapes and sizes, the plane inductance looks ~constant (in the same way a diode drop is ~0.6V), and the via inductance (that is, the part due to the barrel size and length, and clearance to planes) will be dominant.
At high frequencies (~ps steps, say), you'll see peaks due to the planes being relatively tight around the via (capacitive), where the via penetrates them; and dips due to the via and plane inductances (where the wave crosses the gap). These can be balanced up to a modest frequency, but not beyond, because you're making a CLC lowpass kind of structure. The LC cutoff frequency will be on the order of 1/4 wave = via height.
The plane diffusion inductance can of course be short-circuited, by placing ground-stitching vias around signal vias. In a 4-layer board, this requires a bypass capacitor as well, which incurs more inductance and may not be worthwhile.
ane. It spreads out more at low frequency. HoJo ignores the EM mode issues essentially completely, at least in the book and in what I've read of his o ther stuff (what you can get for less than $10k).
er of the orders of magnitude involved.
So let me try this as a hand-wavy understanding of why the return current concentrates under the trace at high frequencies. This could be totally "off the wall". In which case I hope you'll tell me a s much.
I wanted to look at this from an energy point of view... What current confi guration has the lowest energy.
Now at low frequency there will be a tendency for the charges to spread out as much as possible, due to the coulomb repulsion. I'm then going to propose that at higher frequencies it's the energy in the electric field that is going to work against the coulomb repulsion. So th e energy density in the E-field goes as E^2 times the frequency. And at high er frequency there will be lower energy with a more concentrated field, but ov er a much smaller volume under the trace. (Geesh, that last statement could b e a total crock. There's that E^2 term! What I really need is put in some numbers... not sure how.)
OK I have no idea if that is in anyway feasible. If you have any better ideas, please share.
The actual distribution is biased towards the edges of the trace (indeed, skin effect works in foils as well, where it becomes edge effect; the characteristic length depends on R/sq instead of rho, so depends on foil thickness), and therefore the image current in the ground plane tends to appear there as well.
If the substrate height is greater than the trace width, then the image currents from the edges of the trace are blurred out by distance, and the current density vs. offset from centerline looks roughly Gaussian. (Fields don't 'want' to be sharp -- a consequence of E&M being a Lagrangian system or something like that*.)
If the trace is close, but the frequency is low, then the center of the trace will carry largely the same current density as the edges, and again the image-current distribution is fairly smooth (not having two local maxima, anyway).
If the trace is close, and the frequency very high, there will be two local maxima, under the edges; but the current in the center won't go to zero, it will remain nominal.
A consequence of "foil edge effect" is the sometimes erroneous construction of foil windings in transformers and inductors. In transformers, if the image current is over the breadth of the foil winding, it looks just like traces-over-ground-plane, and works nicely. If foil overlaps foil, with no opposing winding inbetween, then very little current can flow on the facing sides of the foils, and their edges burn up. Some companies claim to make something that's like flat Litz wire; West Coast Magnetics comes to mind.
(*With other consequences like size vs. performance tradeoffs in antennas, difficulties of making very intense fields (see Halbach array?), or very sharp fields (you can't do the physics textbook problem of a "particle entering a constant B field region"). Also, electron optics, diffraction limits in conventional optics, uncertainty principle (spacial resolution versus aperture), etc. etc.)
HoJo is crazy about return currents and "reference planes." In real life, on a decent multilayer board, power planes and ground planes are AC equipotential; the whole PCB is one big brick of ground. The signal in a trace doesn't remember what "reference plane" it was born on, and doesn't care where the rising-edge capacitive current is going in its wake; it's headed off at most of the speed of light.
Vias are invisible at 1 ns, and barely visible to 100 ps edges.
A chip that drives, say, a long 50 ohm microstrip can't tell if its load is a transmission line or a 50 ohm resistor. After it drives the rising edge high, it sees a DC load. The "return current" is just DC coming in on the Vcc pin and exiting the signal pin.
--
John Larkin Highland Technology, Inc
picosecond timing precision measurement
jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com
Well some picture in you brain.. maybe it's got springs and balls. You then see if the model come's close to fitting reality. If not, find a new model.
Well it doesn't at DC where the current spreads over most of the ground plane, I guess to keep the resistance the lowest.
Huh, Thanks I hadn't thought of that... I was mostly thinking of skin depth in terms of the trace thickness and not the width. But let's forget all about the width and thickness... assume a very thin wire above a ground plane.
OK That's all very nice. But there is some effect that doesn't really depend on the trace width. That's OK don't worry. I'll let it stew for a while.
If that means you're considering the B^2 energy storage, I think you're on the right trail. Consider that the part of the ground plane nearest the RF hot line is transformer-coupled to the nearby RF currrent... and apply Lenz's law.
"F HoJo!!!" (sorry for shouting) I regret having mentioned him.
I'd just like a handle on the mechanism for ground current distribution. (leaving off the complication of power planes.) It's probably really simple and stupid and I'm just not seeing it.
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