Reflow oven for prototyping

I don't think it depends on the skin effect at all. The field inside a long, circularly symmetric, axially directed current sheet is identically zero. You can do the math, but there's a simpler way to see it, as follows:

1. Elsewhere inside, the B field has to be tangential, because the current is axial.

2. By symmetry again, the magnitude of the field can't depend on the angular position.

1. There's zero current density inside, so curl H = 0 inside the conductor. (Faraday's law)

2. Since curl H is identically zero, the integral of H dot dS along any closed loop inside the cylinder is zero. (Stokes's theorem)

1. In particular, it's zero around any circle centered on the axis.

Facts 2, 3, and 6 together mean that the field inside must be identically zero.

Skin effect will change the field inside the metal, but not inside the hole.

There will be some small effect due to current crowding, and some very minor end effects, but filling the via itself doesn't change the inductance of the via itself.

Cheers

Phil Hobbs

Given some power pad area, and a ground plane below, there must be some optimum via size and pattern for minumum inductance/resistance/thermal resistance, but I don't know what it is. Solder leaching gets involved, too.

Even relatively simple thermal systems are hard to model.

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Hi Phil, I think you missed Fred's point. At low frequencies (where the skin effect can be ignored) the iductance of a solid rod will be every so slightly larger than the inductance of a tube with the same OD.

George H.

George Herold a écrit :

Yup, exactly.

I don't think that's the case, though. Even though B is nonzero inside the metal in the low frequency case, it also doesn't have the sharp peak at the outer skin of the via. And the inductive contribution drops pretty rapidly inside, because the fields from outer layers don't add to the ones from the inner layers.

Cheers

Phil Hobbs

Fred Bartoli a écrit :

Rereading your post I missed that sentence which is just plain wrong.

At DC and low frequencies (low enough so that skin effect does not exist) a *plain* conductor do have B field inside it, linearly rising from 0 at the center to its max value at its surface

There's no field inside the air space whatsoever, and by exactly the same argument, the B field of the outer layers cancels as you go inwards.

Adding more metal won't increase the inductance, I don't think, because inductance goes as the volume integral of B**2 over all space, and your little bit of B inside the metal loses you some at the outer surface, where the B field and the circumference are both larger.

I could be persuaded otherwise, but one would have to actually do the math to be sure. (It isn't that hard a problem, but I'm too lazy to set up the integrals.)

Cheers

Phil Hobbs

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Grin, This is just my level of integral. (A freshman physics problem.) But I don't think I have to do them. So first we assume the same current I and the same OD for the two conductors. Now Fred already wrote down the term for the B-field inside the condutor. It grows linearly as a function of r (Assuming uniform current distribution.) So there's some B field inside the solid conductor.

Now for the tube, as you point out there is no field in the central region... (no current enclosed.)

Now anywhere outside of the two conductors the field has to be exactly the same. Since we assumed the same current I. The integral of B-dot-dl is the current enclosed. (with whatever

So whatever soldering the vias did to fix Michael's circuit, it did not lower the inductance. Fun, thanks again Fred.

George H.

Ah, okay. Yes, of course the field outside is unchanged, Ampere and all that. As extenuation I plead a head full of lawyering. (I'm helping a small outfit try to enforce a perfectly good patent against a video game company who seems to be trying to steamroller them.)

Cheers

Phil Hobbs

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I was going to add that it all assumes an infinitly long conductor, which is a bit of a stretch for a via. So you could always argue about end effect.

(Nothing like lawyerese to scramble the brain. :^)

George H.

As an add note, I've just done the integral for fun but am too lazy to write the detail down here...

Anyway the outcome is: conductor internal inductance is uo/(8 pi) per meter, whatever its diameter.

And that's exactly 50nH/m or 50pH/mm.

It did seriously lower the thermal resistance.

George Herold a écrit :

Hey I can't be wrong, I just live a few 10s km from Ampere's house ;-)

Well, I live a few miles from Philo Farnsworth's lab.

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In that case it had to be lousy via/pth quality. The solder fixed that and the other effect were secondary.

?-)

inwards.

Unless i have lost track here, part of the issue was does filling the via/pth with solder change the inductance? And at what frequencies?

This presumes that the via/pth is of good quality.

Just two questions.

?-)

This one wasn't:

We paid a high price for them. They were thick, double sided boards to damp any vibrations. The original design had those holes filled with solder to hold some uncased disk caps to the board. The caps were used as terminals in the band switching circuits. Tho diodes were biased to add additional capacitance to give four overlapping bands. ALl of them were near the PC trace used for the inductor in the VCO, which was on the top side of the board. We had real fun when the vendor decided to add a patter around the mounting holes without permission. We had to send most of them back. Two were built & shipped becasue we couldn't wait for the next batch of 100 boards. I had to solder copper tape over each pad to cover something similar to this photo, except there were six thin traces between the circles.

I no longer have any photos, schematics or samples availible. The telemetry products that used it was one of the cleanest on the market for over 20 years. It was better than the VME based models where we bought someone else's low phase noise VCO.

Impedance, of course (especially if we're talking about small effects) can also be resistance. Plated-through holes aren't thick copper, and some will dissolve in solder. Resistivity of solder or solder/copper alloy is quite a bit higher than that of copper, so it's unclear whether resistance goes down (more metal) or up (less conductive alloyed copper in the outermost one-skin-depth cylindrical shell).

I'm hoping this will never be a problem I have to model accurately with only theoretical tools.