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(New page: '''Common Mode Current''' what is it, and how can you deal with it? by Lou Rummel KE4UYP First lets talk about differential current and ...)
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Revision as of 03:40, 7 February 2008

Common Mode Current what is it, and how can you deal with it?

                                              by Lou Rummel KE4UYP  


First lets talk about differential current and hopefully I can make this a little less confusing. This type of current is ideally what you would want flowing inside your coax.

It is called differential because it is balanced, what this means is, the amplitude of the current flowing up the center conductor of the coax is identical in amplitude and 180 degrees out of phase, with the current flowing down the inside diameter of the braided shield surrounding the center conductor.

Unfortunately this rarely happens in the real world. You will almost always have an imbalance between these two currents.

This imbalance is what will cause, Common Mode Current. The obvious question is what caused this imbalance well there are actually three possibilities.

The first scenario: If the antenna is a balanced antenna and you connect coax to it directly then technically speaking you will always have to some degree an imbalance with this differential current. You may ask how much of an imbalance can I expect well, that depends on several different factors.

For example if the coax is perpendicular to a dipole as it is going back to the transmitter the imbalance could be negligible. On the other hand it could be excessive if the length of the coax is a particular length. I will talk more about this a little later.

The second scenario: When you connect coax to an unbalanced antenna, the situation only gets worse. The antenna or more technically speaking the load has more to do with creating this current imbalance than practically anything else. The more the load is unbalanced, the more current imbalance you will see on the coax.

The third scenario: There are lengths of coax that will develop the highest level of common mode current. These lengths are odd multiples of one quarter wavelength for example 1/4, 3/4 and 1 1/4 wavelengths. When you have coax of these lengths, they will develop a low impedance path to ground, this in turn will cause excessive amounts of this common mode current to develop on the outside of the braided shield of the coax.

Because common mode current is located only on the outside perimeter of the braided shield unlike the differential current that is located on the inside diameter of the braided shield. Its velocity factor is not the same as the center conductor of the coax. For example RG-8X has a velocity factor of .78 to .82%. But common mode current has a velocity factor of .95%.

So the $52 million dollar question is how do you eliminate this problem. One of the best solutions to this problem is to install either a current balun or an RF choke sometimes called a (Line Isolator), at the feed-point of the antenna.

My Second Choice: As I mentioned before there are lengths of coax that develop the highest level of common mode current. There are also optimum lengths of coax that develop the lowest level of common mode current. These lengths are odd multiples of 1/8 wavelength.

For example 1/8, 5/8, 1 1/8 and 1 5/8 wavelengths when you have coax that are these lengths, they develop an extremely high impedance path to ground which works as efficient as a high-quality RF choke or (current balun) Because we are dealing with electrical lengths that are odd multiples of 1/8 of a wavelength precise measurements do count.

A ½ wavelength dipole for example 468/ 3.5mhz=133.7ft. for 3.6mhz it would =130ft. Notice that there is almost four feet difference in their length. But at 1/8 wavelength the difference in these two frequencies is only 0.9ft. Here are the optimum coax lengths for 1/8 wavelength on the 80m and 60m band.


80m 60m 3.5 Mhz=33.4ft. 5.373Mhz=21.75ft. 3.6 Mhz=32.5ft. 3.7 Mhz=31.6ft. 3.8 Mhz=30.78ft. 3.9 Mhz=30.0ft.

Unfortunately the next coax length would be to excessive, for most people. For example 5/8 wavelength at 3.5Mhz=167ft. This is why using this technique is so limited on 80 or 60m but on VHF frequencies like 2m where one wavelength equals only 6ft. 5" it is quite manageable.