Gound plane resistance calculator

Just installed it. That software seems powerful but also seems to have quite a learning curve.

Try it first for a disk with a finite diameter contact in the middle, and grounded around the periphery. I just did some Real Live Calculus and I'm getting* a resistance equal to

(rho/(2*pi)) * ln(r1/r0)

where rho is the resistivity of the material, r1 is the outer radius, and r0 is the inner radius. IOW, it goes to infinity as r0 shrinks to a point.

I suspect that for _two_ points of a given radius on a big infinite sheet that the resistance vs. radius will be monotonically decreasing, starting at infinity, going down to some fairly constant number having to do with the spacing of the points, then going down rapidly as the perimeters of the contact circles approach one another.

If so, you could find a lot more wrong ways to solve your problem than by choosing that constant plateau and calling it "the resistance".

• Consider an annular ring of radius r and thickness dr, made of material whose resistivity is rho ohms/square. This ring is made up of (2*pi*r)/dr squares, and thus it's resistance is rho/(2*pi*r) dr.

Now, note how convenient it is that PURELY BY CHANCE we chose dr for the ring thickness -- because now we can integrate this quantity from r0 to r1:

R = int{rho/(2*pi*r) dr}_r0^r1.

And this, of course, evaluates to R = rho/(2*pi) * ln(r1/r0).

I actually made low value resistors on-chip (diffused) back in the early '60's that way... easiest way to get 50 Ohm resistors when all you have is 300 Ohms/sq. ...Jim Thompson

It helps if you have a problem that matches a tutorial.

Yes, it sure would but mine doesn't :-(

Now where's the stores with that Teledeltos paper ... ?

Cheers

Ah, thanks, now I remember Bob Pease's hint where to get it. But too late for this time, I'd need it in two days. I guess we'll just wing it then :-)

I don't think so. I've chased it a few times, but without success. As John says, you can get a 6-digit DVM for $50 from eBay, so using aluminum foil is probably a win. Though the contact resistance to Al foil is pretty variable--it's even worse than nickel.

Cheers

Phil Hobbs

Black anti-static foam?

A large swath of foil would work. On a hard surface so you can press a flat kind of contact onto it. Once when we did stuff like that at a large semiconductor company one of the big shots came by in his expensive suit. "Dudes, we are not on kindergarden here"

I thought about that but it's too irregular.

Just for fun I tried this out in FEMM, with a 4" square board with a 1" notch out of one corner, with terminals at opposite corners and 0.001" copper. I got different conductivities for different diameter contact points: 684 mho for 0.02" diameter. 898 mho for 0.1" diameter, and 1040 mho for 0.2" diameter.

So it varies with diameter.

Use copperclad FR4. Measure a strip to calibrate the sheet resistance. Solder brass spacers or whatever to make contacts.

John

Um, won't the resistance always be the same?

Ohms per square and all.

Doesn't matter per square what.

You could do it in Spice. Make a grid of resistors... lots of resistors. Some clever cascaded cutting and pasting might work. Then short out the resistors that represent contacts.

There used to be a thermal analysis program called SAUNA that worked that way. Structures were modeled as grids of resistors, and a circuit solver was sic'd on it.

John

That's a comparatively miniscule variation for such huge diameter changes. Or in RF-speak just a few dB :-)

Can you send a copy of that file (reply-to works)? Then I can use that as an example or starter. Well, maybe. I am not very good with software that uses pictograms. Probably I won't get to it anymore this week but it would be something worthwhile to learn.

I like John's idea of using regular old copper clad and scaling it. You can even solder to it. Put a few amps through it and the voltages become quite reasonable. If you've ever searched for a short between power planes on a multilayer board..

Best regards, Spehro Pefhany

I should refuse, on the grounds that I don't know what the heck I'm doing.

But I sent it anyway.

That's basically what finite element analysis is, but with a triangular mesh that's automagically constructed and operated on, instead.