# How does this antenna match work?

I had a matching technique described to me by a professional broadcast engineer. Here's the problem: you have a vertical antenna with a low resonant impedance1, let's say $38\Omega$, but you need a better match to a $50\Omega$ feedline.

Solution: put a coil between the base of the aerial and ground. Then, you feed off a tap in the coil:

How could this work? I can understand how adding a shunt inductor could improve the match if the antenna presents a capacitive load. But, why feed off a tap in the middle of the coil? If it's working as an autotransformer, wouldn't it be making the antenna impedance appear to be less to the feedline (the opposite of what we are trying to accomplish)? Also, if the antenna is already resonant, wouldn't the addition of shunt inductance make the match worse? Or is this why the feed is off a tap? Please help me understand how this works.

1: in my particular case, the impedance is even lower (I measure $14\Omega$) due to shortening of the antenna.

• autotransformer?
– user16324
Commented Jun 3, 2014 at 17:14
• @BrianDrummond that's my guess, but I'm not sure. Commented Jun 3, 2014 at 17:17
• @Andyaka No. There is a loading coil in that picture (in the middle), but that's an irrelevant detail. I'm asking about the coil at the bottom. Also note important difference: the feedpoint is a tap on the coil. Also, a loading coil is used to make an antenna resonant at a frequency lower than its physical length, that is, to electrically lengthen the antenna. The matching coil here is used to match an already resonant antenna to a feedline of a different impedance. Commented Jun 3, 2014 at 20:21
• @PhilFrost It's probably me being thick but i don't understand the question! Commented Jun 3, 2014 at 20:30

The only way I can see it working is if the lower impedance antenna (38 ohm) were connected to the "tap" of the auto transformer. A 38 ohm tap and a 50 ohm feedline means the tap is at 87% of the way up i.e. $(0.87)^2 \times 50\Omega =$ 37.8 ohms.

An 18 ohm tap is at 60% i.e. $(0.60)^2 \times 50\Omega =$ 18 ohms.

I can't see any other way round this other than the broadcast engineer got in a muddle or your ears need washing out LOL.

NB These calculations assume the auto transformer windings are 100% coupled to each other.

• Interestingly, what you describe is in another image adjacent to the one from the question. Though, the schematic shows a core and suggests a higher inductance, where maybe with the "autotransformer" solution there isn't enough inductance for it to be anything like an ideal transformer, or maybe parasitics are relevant. I think the text mentions an inductance on the order of $1\mu H$. Still, possible it's just plain wrong, and good to have a more experienced review. Commented Jun 5, 2014 at 20:20
• @PhilFrost - it's still an auto transformer that new image is - think about the righthand coil just being placed below the 12.5 ohm tap and it should make sense dude. That 12.5 ohm tap is exactly at 50% hence it gives 0.25 x 50 ohms. Commented Jun 5, 2014 at 20:45

Autotransformers can be used for both step-up and step-down impedance matching:

[Source]

• "Here's the problem: you have a vertical antenna with a low resonant impedance". I'm not missing phase: it's specified as 0°. Also, don't you think the mutual inductance of the coils is somehow relevant? Commented Aug 23, 2014 at 11:38
• @PhilFrost Yes, I do. See my edit. Commented Aug 23, 2014 at 18:27