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From: Manfred Mornhinweg <>
Date: Fri, 06 Dec 2013 17:23:54 +0000
List-post: <">>
Hi all,

If one is to believe Philip H. Smith in 'Electronic Applications
of the Smith Chart', McGraw-Hill 1969, page 6, Fig 1.3, the maximum voltage
appearing on a lossless transmission line with an SWR of infinity is twice
the voltage when matched.

This is only a half truth. It's valid when the line is fed from a signal source having the same impedance as the line's.

For example: Imagine a source that provides a stiff 100 volts of RF, then has a series resistor of 50 ohm. That signal is fed into a piece of 50 ohm coax cable, which is terminated in a 50 ohm dummy load (SWR 1:1).

In this case there will be 50 volts of RF, anywhere along the cable, regardless of its length. The RF current will be 1 ampere, everyhere.

If you now remove the dummy load, you will get the situation described by Mr. Smith: The voltage along the coax cable will cyclically vary along its length, between zero and 100 volts.

To understand this, imagine two extreme cases: In the first case, the cable is an exact half wave long. So the signal source is connected at a high impedance point, where the cable shows essentially infinite impedance. No current will flow into the cable, so there is no voltage drop on the 50 ohm resistor in the signal source. So we get 100V at the cable's input, and the same 100V at its other end. At the mid point of the cable there is no voltage, but 2 amperes of RF current.

And the other extreme case: The cable is one quarter wave long. Without the dummy load, the cable's input will be at essentially zero impedance - a short circuit. So the 100V source with the 50 ohm series resistor will put 2 amperes of RF current (and zero voltage) into the cable. At the open end, we get 100V and no current.

In intermediate cases, with odd cable lengths, there will be some voltage and some current at the input of the cable, with 90 degree phase relationships, but always there will be 100V and no current at its output.

Now let's see where Smith's truth ends: Consider another case, in which the signal source is just 50V, but has zero internal resistance. If we connect this to the same cable, with a dummy load at the end, the conditions are just like our first case: 50 volts (and 1 ampere) everywhere along the cable. But if we now remove the dummy load, things get interesting: The voltage at the output depends absolutely on the length of the cable!

When the cable is a half wave long, presenting an infinite impedance to the signal source, we get 50V at the input, the same 50V at the output, and 1A of RF current at the midpoint. But if the cable is 1/4 wave long, presenting a zero impedance to the source, our zero resistance signal source would inject an infinite current into the cable, resulting in an infinite voltage at its other end! So here we have a case where the voltage doesn't double at infinite SWR, but instead rises to infinity.

Of course signal sources with zero internal impedance don't exist in the real world. Lossless coax cables don't exist either. But a signal source with an internal impedance much lower than 50 ohm is totally plausible - many RF amplifier stages behave like that, even if their nominal load impedance is 50 ohm. And even with a practical, lossy coax cable, such a setup can result in far more than twice the voltage, when the SWR is very high. And it can also result in excessive current flowing in the amplifier's components, damaging them. That's why so much equipment has protection circuits against high SWR built in.

Mr. Smith and his chart are entirely correct, but only for the specific situations to which a certain graph applies!


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