I would like to split a coaxial feed line 3 ways, maintaining 50 Ω impedance.

I know how to do this 2 ways; you use 1/4 wavelength 75 Ω coax as matching stubs.
I know how to do this 4 ways; you do 2 way twice.

2 way Matching

But how do I do this 3 ways, ideally using 50 Ω and 75 Ω cables (as I have them in stock)?

I want to use the 3-way split to feed 3 dipoles in an end-fire-array configuration, on the 70cm band, with a maximum power of 50 W.

The reason for 3 is that there is no room for 4 (4 would be handy), and I would like more gain then 2. Furthermore, three is easily switchable in regards to direction: the middle dipole feed will stay static, while the outer dipoles reverse their phase offset.

  • $\begingroup$ How much power are you talking about? Sometimes it's just simpler to use a 4-port splitter and terminate the unused port with a resistor. $\endgroup$ – Dave Tweed N3AOA May 17 '16 at 13:31
  • $\begingroup$ Good question, 70cm band, 50 watts $\endgroup$ – Edwin van Mierlo May 17 '16 at 13:34
  • $\begingroup$ Do you need isolation between the output ports? $\endgroup$ – Phil Frost - W8II May 17 '16 at 13:37
  • $\begingroup$ Thanks for the answers, comments and references. I am going to build this array, for me to experiment with... and because it is fun to do so. :-) $\endgroup$ – Edwin van Mierlo May 18 '16 at 18:43

The picture in your question is almost a Wilkinson power splitter. Missing is a 100 ohm resistor between the outputs:


simulate this circuit – Schematic created using CircuitLab

Without that resistor there's no isolation between the output ports. You say you are using this to feed an antenna array. The consequence of not having isolation between the ports is that if any of the antenna elements is not a perfect match, that reflected power will show up on the other antenna elements, messing up your phasing.

With the resistor, if there's a perfect match everywhere, each output is in phase and so there's no voltage across the resistor.

When there's some reflected power from one of the antennas, half the power goes back through the input, and the other half is dissipated in the resistor. Ideally, none goes out the other output. In practice you can expect 20dB of isolation.

But you have three antennas. Fortunately, Wilkinson power splitters can be made with any number of ports. This article by Microwaves101.com illustrates the concept well:

enter image description here

This requires a different impedance 1/4 wave section. Z0 is probably 50 ohms, so:

$$ \sqrt{3} \cdot 50\:\Omega = 86.6\:\Omega $$

If you happen to have some 87 ohm coax sitting around, you are good to go.

If not, an equivalent circuit can also be realized with lumped elements. See:

Keep in mind that with an end-fire array there is significant coupling between each element of the array which can make things messy in practice. See the answer by tomnexus.

  • $\begingroup$ now that was the answer I was hoping for! Thanks! $\endgroup$ – Edwin van Mierlo May 18 '16 at 14:51
  • $\begingroup$ @EdwinvanMierlo Just keep in mind what tomnexus said, also. I think he raises some good points. $\endgroup$ – Phil Frost - W8II May 18 '16 at 14:57
  • $\begingroup$ absolutely, good points made in both answers. $\endgroup$ – Edwin van Mierlo May 18 '16 at 15:24

Go carefully when feeding an end fire array. Unlike a vertical collinear array, the coupling between ports is substantial. Even with a perfect well isolated splitter, you won't get the current distribution and hence patterns that you expect.

What you need to do is simulate the array in NEC or something, and experiment with the voltage magnitudes and phases on the elements, until you have the current distribution or radiation pattern that you want. Then only can you start to design the feed network - three different transformers and delay lines to achieve the feed voltages you want.

Some end fire arrays are designed for feeding simplicity. The LPDA has a simple transmission line running between elements, they each take some current and the rest flows on down the line. The yagi generally has a zero feed point voltage on all but one of the elements, which is of course easy to feed. There are others, like a yagi with two or three driven elements, the W8JK and the HB9CV with two, but the common theme is that the feedline design is an integral part of the antenna design.

I tried a quick simulation of four cases:

  1. Simple $\lambda/2$ dipole.
    • Result: Well matched at 445 MHz. Gain=2.1 dBi.
  2. Three dipoles $\lambda/4$ apart. Fed with $-90^\circ, 0^\circ, +90^\circ$ current sources.
    • Gain = not calculated for current sources, sorry. F/B = 7 dB.
    • Currents on the three are equal, and phases are as requested.
  3. Three dipoles $\lambda/4$ apart. Fed with $-90^\circ, 0^\circ, +90^\circ$ voltage sources.
    • Gain = 8 dBi, F/B = 6 dB.
    • Impedance at ports: Insane. One port 1.5:1, one 3:1, last port has power flowing back from it, SWR is negative, hard to describe.
    • Currents on the back two are equal, almost no current on the front dipole.
  4. Three dipoles $\lambda/4$ apart. Fed with transmission lines: first each dipole has a $\lambda/4$, 86.6$\Omega$ line, which would transform 50$\Omega$ into 150$\Omega$. Then each has a 150$\Omega$ line back to a common source, one zero length, one $\lambda/4$, one $\lambda/2$. This would be the naive phasing to get the end fire effect.
    • Gain = 5.6 dBi, F/B = 5 dB.
    • Impedance at port: SWR of 1.5:1.
    • Currents quite mismatched, middle dipole has about 60% of the current of the side two. Phases are about right for end-fire.

The next step, which I can't easily automate, would be to play with the transmission line lengths and impedances to achieve equal current, correct phasing. As a guess, it might pay to split half the power to the middle dipole, and 1/4 to each side one, keeping the -90, 0, 90 phase relationship.

In summary, N-port equal power equal phase splitters are for collinear arrays and perhaps broadside arrays, but for end-fire things get a lot more complicated.

  • $\begingroup$ This is good information, but completely tangential to the main question on this page. You should consider creating a separate question and answering it with this information, like Phil Frost did with his information about lumped simulations of transmission line sections. $\endgroup$ – Dave Tweed N3AOA May 17 '16 at 21:53

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