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I've just tried making my own J pole antenna. While matching for lowest SWR on the SWR meter, I noticed that my FTM400 didn't show 5 W forward power.

I hardly saw the reflected power needle move so I increased the power to 20 W just to have a quick peek. There it showed a forward power of about 12 W.

On my normal antenna I get 5 W for 5 W input and 20 W for 20 W input, but this has me stumped.

The homebrew is a J pole for 146 MHz and SWR should be in the ballpark. At 20 W I get a SWR of about 1.3.

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  • $\begingroup$ Perhaps your SWR meter is inaccurate, and your FTM400 is reducing power to avoid damaging the final amplifier transistor? Or maybe the J-pole is more reactive than the transmitter likes. If you are only using an SWR meter, then you might think about borrowing an antenna analyzer that shows the complete impedance of the antenna, including both the resistance and the reactance. $\endgroup$ – rclocher3 Jun 27 '16 at 16:06
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    $\begingroup$ You could reasonably attribute much of the discrepancy to measurement error. Neither your radio or SWR meter is likely to be calibrated to a high standard nor designed to be accurate under all conditions. $\endgroup$ – Phil Frost - W8II Jun 30 '16 at 17:30
  • $\begingroup$ Antenna has many parameters, SWR is the simplest to measure, but not give total reality . Additionally J Pole (popular sigle band) can very easy meet low SWR, but work out of normal condition (has very effective autotransformer). Antenna impedance is a complex number, SWR meter show only real part (modulus) $\endgroup$ – Jacek Cz Jul 28 '16 at 13:29
  • $\begingroup$ 2. most popular power meters (almost all in HAM radio) measure voltage, and only assume power by 50 Ohm. If impedance is different, power measure if incorrect. $\endgroup$ – Jacek Cz Jul 28 '16 at 13:36
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Let's consider this system

Transmitter --> SWR meter --> Antenna

A (V)SWR of 1.3 means that the magnitude of your reflection coefficient $\Gamma$ between your transmitter and your antenna is

$$|\Gamma| = \frac{SWR-1}{SWR+1} = \frac{0.3}{2.3}\approx 0.13$$

which is pretty good, actually, meaning that but 13% of the field strength your transmitter inserts into the antenna is reflected by the antenna, or that 87% actually reach the antenna!

You can convert that into a relation of power inserted and reflected, simply by squaring it – then things look like this: The reflection dampening factor (i.e. how much of your energy is reflected is $$a=\frac{1}{|\Gamma|^2}\approx59\approx 17dB$$.

Now, for a power amplifier to reach maximum power output, it must be connected to a device that matches its source impedance $Z_s$. In that case, 0% of power is reflected. In all other situations (e.g. $Z_s=50\,\Omega$, $Z_a\ne50\, \Omega$, output power is below the maximum output. I don't actually know how your device displays forward power – that's something non-trivial to measure – but the fact that we don't have a unity SWR indicates your PA can't sink it's full output.

It's known that

$$\begin{align} \Gamma &= \frac{Z_s-Z_a}{Z_s+Z_a}\\ \implies\\ 0.13 &= \frac{|Z_s-Z_a|}{Z_s+Z_a}\\ 0.13(Z_s+Z_a) &= |Z_s-Z_a|\\ 0.13Z_s+0.13Z_a &= |Z_s-Z_a|\\ 0.13Z_s+0.13Z_a &= Z_s - Z_a &\vee&& 0.13Z_s+0.13Z_a &= Z_a - Z_s\\ -0.87Z_s&= - 1.13 Z_a &\vee&& 1.13 Z_s &= 0.87 Z_a \\ Z_a&\approx 0.77 Z_s &\vee&& Z_a &\approx 1.30 Z_s\text{ .} \end{align}$$ I'm going to assum $Z_s = 50\,\Omega$, so your antenna impedance is either $38.5$ or $65.0\,\Omega$.

That doesn't really explain the power loss your device is displaying, but it's common for HF measurement devices to not be on identical terms.

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  • $\begingroup$ I have a forward/reflected swr meter, that's where I noticed it. Thanks for an indepth swr Z explanation, but as you mention it doesn't explain why my meter only says I'm putting 12w into the antenna when my radio is set for 20w. This really bugs me, lol. $\endgroup$ – Hans Neve Jun 26 '16 at 18:53
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    $\begingroup$ well, the emitted power might be limited by your transmitter to make sure the reflected power doesn't damage it. Best explanation I have in mind so far. $\endgroup$ – Marcus Müller Jun 26 '16 at 18:54
  • $\begingroup$ Hmm, I can put 50w out of the radio into my other antenna (same swr meter) with about 1.4 swr where meter showing 50w forward power $\endgroup$ – Hans Neve Jun 26 '16 at 18:57
  • $\begingroup$ maybe it's me who is struggling; Forward power is the power coming from the transmitter, going into the SWR meter, not the power going out of the SWR meter into the antenna, right? $\endgroup$ – Marcus Müller Jun 26 '16 at 19:02
  • $\begingroup$ I'm just buggered by what my swr meter is telling me, so we're both buggered. $\endgroup$ – Hans Neve Jun 26 '16 at 19:04
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There are various considerations which could influence your measurements.

  • Antenna height and location: compared to your (unspecified) "normal antenna" did you test the J-pole at exactly the same height and location ?
  • Feed line: did you use the same, or same lengt/type of, feed line ?
  • Weather condition: did you test in the same weather conditions: temperature, humidity, rain, wind, snow, hale can all influence measurements

In my experiments with J-poles I have a good range of failures and some successes. This is (IMHO) to do with the fact that a J-pole is an:

  • Unbalanced half-wave radiator
  • with a balanced matching section
  • fed with an unbalanced coaxial cable

This will cause all kinds of wonderful effects, most common: common-mode. Which may or may not be influencing the measurements you are taking.

The best results I have established over the years is to use a 4:1 balun, which then needs to be fed "higher" on the matching section to get a 200 Ohm match, combined with a good number of beeds on the coax.

You can easily create 4:1 match using something like this:

4:1

This will at least transform your unbalanced 50 Ohm feed into 200 Ohm balanced, for the balanced matching section of the J-pole. It still is a unbalanced half-wave radiator, nothing you really can do about that.

Personally I have given up on J-poles, while with the above method I had some success, I really think that a standard quarter-wave GP, or a vertical half-wave dipole is much easier and I had better results.

YMMV

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  • $\begingroup$ "wind, snow, hail"? $\endgroup$ – user2943160 Aug 27 '16 at 23:45
  • $\begingroup$ @user2943160, wow, thanks. $\endgroup$ – Edwin van Mierlo Aug 29 '16 at 7:11

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