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I'm building a trap dipole for the high HF bands, 10 / 12 / 15 / 17.
So far what I've got in the air is:

  • The 10 meter section (4.92 meters, 16.14 feet, resonant at 29Mhz)
  • The 10 meter traps (8pF cap in parallel with a 4µH inductor, resonant at ≈28Mhz)
  • The 12 meter section (80 cm, 31.5 inch)
                8pF/4µH                             8pF/4µH
             ----{10M}------------{Balun}------------{10M}----
            0.4M        2.46M        ║       2.46M          0.4M
                                     ║
                                     ║ 30.4M coax

Now take a look at the output from my antenna analyzer: What my SARK110 sees

Whhaaaat... Well, I'm happy with the 10 meter resonant point (given by M1) but what the heck happened next? I read somewhere that traps would electrically shorten all subsequent dipole sections, but i'm not sure what to make of this. Why would the 12 meter section be so far out of tune. Any ideas?

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    $\begingroup$ Not completely sure but don't think the TL parameter should be set for this measurement. Doesn't that add a non existing piece of transmission line for simulation? $\endgroup$
    – s3c
    Feb 11, 2015 at 10:10
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    $\begingroup$ Is that diagram really accurate? Just looking at the 10m section, you have 9.84m of wire, which is about a full wavelength (not a half wavelength, as I'd expect...) $\endgroup$
    – Phil Frost - W8II
    Feb 11, 2015 at 12:27
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    $\begingroup$ The answer, briefly, is that at frequencies lower than the trap's resonant frequency, the trap looks like an inductor, and functions like a loading coil. It's effect is somewhat diminished by being near the end instead of at the base, but I'm a bit unclear on the exact math. $\endgroup$
    – Phil Frost - W8II
    Feb 11, 2015 at 16:20
  • $\begingroup$ s3c: your completely correct, I will correct the TL parameter. $\endgroup$
    – Turingrad
    Feb 12, 2015 at 1:07
  • $\begingroup$ Phil Frost, The diagram was not accurate. The 4.92 meters was the length of the entire 10 meter section. I have corrected it in the diagram. So I just need to shorten the 12 meter section until it's resonant at the desired frequency. I'll get back to you. You should put that comment down as an answer for the metrics. $\endgroup$
    – Turingrad
    Feb 12, 2015 at 1:16

1 Answer 1

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Traps require that the sections that follow them to be shortened because they look like loading coils. We can demonstrate this by calculating the impedance of your trap at 28MHz and 24.9MHz.

The impedance of a parallel LC circuit is given by:

$$ Z(\omega) = -j \left( 1 \over C \right) \left( \omega \over \omega^2 - \omega_0^2 \right )$$

Where

  • $\omega = 2 \pi f$
  • $\omega_0 = 1 / \sqrt{LC}$

So for your trap:

$$ \omega_0 = {1 \over \sqrt{4\:\mathrm{\mu H} \cdot 8\:\mathrm{pF}}} = 176776695 $$

This is the resonant frequency in radians per second; if we divide by 2π we see that it's 28.13MHz as expected. We can then calculate the impedance at the relevant frequencies:

$$ Z(28\:\mathrm{MHz}\cdot 2 \pi) = j73.6\:\mathrm k\Omega \\ Z(24.9\:\mathrm{MHz}\cdot 2 \pi) = j2.89\:\mathrm k\Omega $$

In the case of 28MHz, the trap is a high impedance, and this serves to reduce the current on the section past the trap to some negligible level.

At 24.9MHz the impedance is much lower, but it's not nothing. Additionally, it is no longer a quarter-wave down the antenna, but something less. Because this impedance is purely imaginary and positive, it's an inductive impedance. We can divide the reactance (the imaginary part of impedance) by ω to get an effective inductance:

$$ {2.89\:\mathrm k\Omega \over 2 \pi \cdot 24.9\:\mathrm{MHz}} = 18.5\:\mathrm{\mu H} $$

Thus, at 24.9MHz (and just that one frequency), the trap looks just like a 18.5μH inductor. Just like a loading coil.

Because the coil is not at the base of the antenna but closer to the tip, its effect on the feedpoint impedance somewhat diminished. I don't know the exact math, although you can find loading coil calculators around the net. Normally we use a loading coil when we don't have room for a full-sized antenna, but here we are reasoning in the opposite order: because you have a loading coil, your antenna can't be full-size.

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  • $\begingroup$ Further pruning suggests that I will need a negative length of wire to me resonant at 24.9 as desired. Does anyone have some anti-copper I could borrow? $\endgroup$
    – Turingrad
    Feb 13, 2015 at 6:34
  • $\begingroup$ @Turingrad Thats....interesting. Have you verified the resonant frequency of your traps? If its off a bit, then on 10m there can be significant currents on the 12m sections, and depending on which way the trap is off, that can make the 10m resonant point go up or down. Is the 10m section, with no traps, resonant at approximately the same point as it is with traps? (the hypothesis here being the trap is off, and you've made the 10m sections too long to compensate, and now they are too long for 12m also) $\endgroup$
    – Phil Frost - W8II
    Feb 13, 2015 at 12:43
  • $\begingroup$ "the hypothesis here being the trap is off, and you've made the 10m sections too long to compensate, and now they are too long for 12m also" This rings true and it's just the kind of mistake I would make. I'll let you know as soon as I do. $\endgroup$
    – Turingrad
    Feb 13, 2015 at 16:51

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