Power Dividers (revisited)



Power Dividers seem to be in demand lately and OAL has had its share of orders for them. Not a complaint, glad to see the work orders come in. Through the years we have found that the more of a product we produce the better we get at it and learn new methods (less labor intensive) and economical ways of producing things.

As a review of Power Dividers, the configurations vary depending on available material and costs of new material. Believe me folks, material costs are not getting any cheaper the inverse if anything, but those of us that like to build will stand the costs, to a point.

The configurations: Square tube with round center conductor, Round tube with round center conductor, Square tube with a flat plate center conductor and Round tube with a flat plate center conductor. The last two mentioned you will probably never see, have to or care to build. OAL prefers the first configuration; the square tube w/round center conductor because of available materials, minimal cost and ease of construction, providing one has the proper tools.

Constructing a 2 X power divider, the materials required are; a length of 1” square aluminum tube, a length of 0.5” diameter (O.D.) copper pipe, 3 – 4 hole flanged bulkhead mount coaxial connectors, either “N” female or UHF (SO-239), and a pair of square plastic end caps.

The two meter power divider may be of the most interest to hams wishing to stack a pair of like yagis (beams). So, let’s take dimensions into consideration considering the middle of the 2 M band (146 MHz). A wavelength at 146 MHz in “free space” is 2.0548 meters or 205.48 centimeters or ~81 inches. A power divider is a quarter wavelength section so the center conductor or 1/2” copper pipe will need to be 81 ÷ 4 = 20.25” – Vp. Vp is the velocity of propagation. RF energy travels about 4% slower in a material (copper) than in air. Therefore, the center conductor will need to be cut to ~19.5”. The square tube will be cut two inches longer than the center conductor for the end caps to fit properly. Make the square tube 21.5” long. Come in an inch from each end of the square tube and mark/dimple, in the center with a center punch. Do this on one side of an end and twice on the adjacent sides, of the opposite end. Drill three ⅝” (0.625) holes, where marked or dimpled. Center a connector over the holes and mark with pencil or scribe the four holes of the flange on the square tube. Do this for the three holes. Drill these holes for #4-40 screws, #43 drill bit (0.089” dia.) Tap 12 holes with a 4-40 tapered tap. The end with the two opposing holes, mount two connectors with the 4-40 screws (8 places).

Prepare the copper tube by running a ⅛” diameter rat tail file across one end, making a slight half-moon, one on each side of the diameter. The opposite end of the copper tube will require one half-moon but 90° around from the two at the other end. Place the copper tube with the two half-moon ends inside the square tube. Fit the half-moons over the center conductors of the coaxial connectors; do not solder the tube in place to the connectors at this time. Confirm that the half-moon at the other end can be seen through the hole at the other end of the square tube. If not rotate the copper tube 180° and refit. Fit a single connector into the square tube so that the connector center conductor rests evenly on the half-moon of the copper tube. If the fit is not true, use the rat tail file to make it so. Once the fit is true, mount the connector in place using the 4-40 screws. Solder the copper center conductor in place (three places) and fit the plastic end caps into the square tube ends. This completes the constructions of the 2 X, 2 meter power splitter.


Testing can be completed in several ways; assuming one has a VSWR bridge. Place appropriate 50 ohm dummy loads on the two opposing connectors. Place the VSWR bridge between the single port of the splitter and transmitter/transceiver. Check for reflected power at both ends of the band and in the center. Splitters are broad banded devices and if construction was proper the VSWR should not exceed 1.2:1 if even that much.


The theory if you’re interested: The dimension of ½” copper tube comes from the fact that in matching a 50 ohm transmission line to two antennas of 50 ohm terminal impedance each require a transmission line of 35.35 ohms; two 50 ohm antennas paralleled = 25 ohms. The other end of the quarter wavelength section is 50 ohms to match the transmission line; therefore, the geometric mean is the square root of 25 times 50 to obtain 35.35 ohms. Now the reason 0.5” diameter copper tube is used is found by:


35.35 = 138 Log (1.08) D/d

1.8038 = (1.08) D/d

1.67= D/d

d = D/1.67

The inside side of a 1” square aluminum tube is 0.875,

0.875/1.67 = 0.5239, ~1/2”


73, Dave – W6OAL