How to MATCH a 50 Ohm antenna/coax to 75 ohm Coax/hardline

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**K1JJ**:

With the popularity of surplus 75 ohm aluminum hardline, knowing how to match it to a standard 50 ohm antenna or feedline is important. There's no need to convert the antenna over to 75 ohms. The simple technique below also works for straight 75 to 50 ohm coax transformations, etc, or vice versa.

Simply make up a little known, "1/12 wavelength" coaxial transformer. It's a 1/12 wavelength piece of 50 ohm coax and 1/12 wave of 75 ohm coax in series connected between the antenna and 75 ohm feeder. Also find below a technique to match TWO 50 ohm antennas to a single 50 ohm line using ALL 50 ohm coax... no need for a 75 ohm coax matching section.

Formula: 1/12 wave coax section length = 984/FreqMhz * 0.0815 * Velocity Factor

Example on 6M for RG-213 coax: 984 / 50.125 * 0.0815 * 0 .66 = 12.67"

[Same for the 75 ohm RG-11U coax]

All these coaxial techniques are good for reasonably wide bandwidth, but for only ONE BAND at a time. Not for multiband/broadband use. If multi-band is desired, then use a broadband toroidal unun or balun.

HERE'S AN ARTICLE THAT DESCRIBES THE MATCHING SECTION IN DETAIL:

http://ourworld.compuserve.com/homepages/demerson/twelfth.htm

And, in case that page link gets changed, here's the meat of the technique here in print:

MATCHING 50 ohms to a single 75 ohm feedline:

(1) A twelfth-wave matching transformer for 50 MHz

A transformer for 50 MHz, matching 75 to 50 ohms: From the figure, or the equation, the required length of matching section is 0.0815 wavelengths. At 50 MHz (=6-meter wavelength) this becomes 0.489 meters. Allowing for a velocity factor of 0.66, the physical length bcomes 0.323 meters, or 12.7 inches. The figure below illustrates the complete transformer. [The picture shows the two pieces in series and connected to the 50 ohm and 75 ohm lines. SEQUENCE: 50 ohm antenna > 1/12 wave 75 ohm coax > 1/12 wave 50 ohm coax > 75 ohm feedline]

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Matching TWO 50 ohm antennas to a single 50 ohm line using ALL 50 ohm coax:

Matching several equal impedances Z0, in parallel, to Z0 impedance cable

Suppose we wish to match two elements of a phased array, individually fed by a matched 50-ohm cable. Putting the two feeders to the individual elements in parallel gives a combined impedance of 25 ohms. To match this 25-ohm impedance to conventional 50-ohm cable, we need a length of 50-ohm, and a length of 25-ohm cable. The 25-ohm length can be constructed by putting two 50-ohm lengths in parallel. The figure below shows the general arrangement.

In the above figure, since we are matching an impedance ratio of 2:1, each length L is (from the above equation or the figure given the required length of matching section for a given transformation ratio) 0.0781 wavelengths. If this were to be used at 28 MHz (10.7 meters) the length becomes 0.836 meters. Allowing for a velocity factor of 0.66, this becomes a physical length of 0.552 meters or 21.7 inches. All lengths L would be 21.7 inches, and all cables are of the same (e.g. 50 ohm) characteristic impedance..

Note that this is a perfectly general way of solving the matching problem of putting N feeders, each of impedance Z0, in parallel. The required matching section impedances are Z0, and Z0/N. It is always possible to make a feeder of characteristic impedance Z0/N simply by putting N sections of impedance Z0 in parallel.

73, Tom, K1JJ

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