Wire over ground transmission line

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New for March 2021.  This page will provide some background on our wire-over-ground calculator which calculates characteristic impedance, located here. Where did the impedance formula come from?

You can find that equation in the ITT "Reference Data for Radio Engineers", but it has no attribution. Let's venture a guess as to why: it is solvable with a closed-form equation, just like coax. You don't see any attribution for the coax formula, do you? It turns out that the most-often used equation is an approximation of a the fully-solved equation, which we will explain here By the way, the IEEE Reference Data for Radio Engineers has more than a dozen impedance formulas for different wire and ground geometries. A full reference is provided at the bottom of this page.

There are various web sites that show different geoemetries for the calculation.  Don't believe everything you see, the true geometry is that the "height" is measured to the center of the wire (think about how you would use radial coordinates to solve this...) The relative dielectric is assumed everywhere, not just under the wire. In this case, the transmission is purely TEM.

Wire-over-ground geometry and coordinate system

This Japanese web site provides a comprehensive, closed-form solution to the problem, and shows that the formula we use is an approximation.  But the approximation is so small in most cases that the full sulution seems almost esoteric.

Perhaps the worst-case situation you might encounter would be a 1 mil wire, 20 mils above a ground, representing a long wirebond. That inpedance works out to 263 ohms, shich is pretty close to a pure inductance. That is why wirebonds are usually modeled with an inductor, or a series R-L if you want to capture the resistive loss.

Before we get started there are at least four different solutions that are correct.  First, you can use the diameter or the radius of the wire. We chose the diameter. And second, you can use natural or base 10 logarithms.  We kind of like base 10, but if you want to use "ln" all you need to do is change the "138" in the final equation to "60". Becaase this formula starts out as a calculus problem, the derivation should at least start with the natural logarithm. And we will use the diameter of the wire, since that is what we used in our wire-over-ground calculator, and that is how ITT published it.

There are two intermediate steps to creating the impedance formula.  First, you have to find the capcitance per unit length (C') and ithe induatcne per unit length (L'). Both of these are useful, we will try to add them to the calculator soon.

The capacitance calculation comes from Gauss's Law.  We will skip the integration, but if you want to see it, we refer you to the aformentioned Japanese site. Here is the final C' result.

The inductance calculation comes from Ampere's Law. Here is the final L' result.

Now, every good boy and girl knows that characteristic impedance is calulated from the square-root of L'/C'. After a little simplifcation, we end up with:

It is easy to see that out that the stuff inside the SQRT function can be reduced to (4h/d)^2, when h/d is large. Even at h/d=2, the error is less than 2%. So here is the simplified, assproximate formula.

Now, let's gather together all the constants and simplify it further. 

And last, for those that prefer log in base 10, here is the formula that appears on our wire-over-ground calculator.

We wish to point out that if you enter a crazy value for h/d, you will get a crazy high impedance, hundreds or even thousands of ohms. Whether that is useful in real life is another matter.

Q.E.D.* and get back to work.

* quit eating Doritos, they are bad for you.


Howard W. Sands & Co., Inc., A Subsidiary of International Telephone and Telegraph Corporation, "Reference Data for Radio Engineers", Sixth Edition, 1983, Section 24-22.





Author : Unknown Editor