Recently you responded to some Email concerning via inductances. I have some questions from your book on this subject. On page 414 there is an equation for calculating the DC resistance of power planes based on the diameters of two contact points space at X amount of distance.

*1st 2 part question: Is the distance based on the centers, edges or somewhere in between the two contact points? I would expect that if based on centers or on the edges that when the outer edges touch this would be 0 ohms, but this does not seem to be the case. Does this equation have an accuracy limit based on how close the two contact points get?

*2nd question: What is the proper resistance calculation for power planes for multiple vias from a single contact area to multiple vias from another single contact area? For instance, a landing pad for a power bus bar a half inch square, with a 1/4" via electrically connected for mounting the bus bar surrounded by 68 vias on .050 centers of .013" outside diameter to another landing pad that also has multiple vias per contact area. Do you do the calculation from just one via from each contact point then use the parallel resistor formula based on the number of vias? Or do you sum the circumferences to come up with an equivalent diameter for the contact points? or am I just plain crazy for trying?

Thanks for your help!

Hi Rich,

Thanks for your interest in High-Speed Digital Design.

You've hit on a tough subject, one for which I think you will not find a satisfactory answer. It's true that my assumptions when deriving the formula on page 414 were:

(1) that the holes were widely separated, and

(2) that the plate was much bigger than the separation between holes.

The formula is simple to derive. First, assume you have a really big circular plate, and ground the outer edge. Now put current into the first contact location. The current spreads out pretty much circularly from the first contract. The current density falls off as 1/r in every direction going away from the center. The total voltage from the edge of the first contact, to the center point of where the center contact is going to be in a moment (it isn't there yet) is the integral of 1/r from the edge of the first contact (d/2) to x. That gives you the ln(x/(d/2)) term. When you put the other contact in place I assume you don't much disturb the voltage in that vicinity (that's the "widely separated" assumption at work). When you inject negative current into the second contact it basically doubles the voltage. That gives you the second ln() term.

In this simple approximation, the resistance goes to zero when the contacts overlap so the edge of one is touching the center of the other (clearly not a good reflection of reality).

In your case, I am going to recommend that you build a physical simulation of the situation. Use a flat tray filled with salt water to represent a (scaled version of) your circuit board. If you want a super-accurate result, block off sections of the tray with non-conducting material (try Lucite) so that you form a lake with an outline that matches the profile of your PCB.

Use short sections (1" length) of copper plumbing pipe with appropriate diameters to represent the contact points. Set the short pipe sections in the water so that, from a top view, you see a pattern of vias that looks like your PCB configuration. Wire the pipe sections together in groups as you expect the system to be configured, using copper wire and a heavy soldering gun.

Connect a power source so that it will force current into one group of pipes, through the saltwater, and into the other group. Adjust the power system to put about one amp of current through the system, and measure the voltage drop with an ohmmeter from one group of pipes to the other.

The current flow in your simple experiment mimics the DC current flow in your real system. All resistances measured in this experiment will be scaled versions of the resistance of your actual system. All we need to know to use these results is the scale factor.

Next step: calibrate the sheet resistivity of the water with a known configuration. Try inserting a copper foil strip at each end of the tank. That makes a simple rectangular arrangement that is easily calculated.

Now scale your results to account for the ratio of (a) the sheet resistivity of your actual copper ground plane, and (b) the sheet resistivity of your experimental tank.

Have fun!

Best Regards
Dr. Howard Johnson