# How does a 40/15m dipole with capacitance hats work?

I followed an article Antenna Here is a Dipole [PDF] by James Hearly, NJ2L to build a dual band 40/15m dipole:

The antenna works more or less as expected. My 40m dipole was resonant at 7.030 MHz and 21.600 MHz. Adding capacitance hats according to the scheme almost doesn't affect 40m and shifts the second resonance to 21.300 MHz.

Although I don't understand how it works. Why capacitance hats make the dipole electically longer on 15m, but don't affect 40m? Also why the third harmonic resonance turns out to be ~2.5% higher than 7.030 x 3?

I realize this might be a complicated question. In this case maybe there is a book on this subject you could recommend? "The ARRL Antenna Book" doesn't give such details. Should I try "Antenna Physics: An Introduction" also by ARRL?

• Congratulations! I can't find any other measured results on this antenna. Please describe the antenna construction, including the feedline and, if possible, include feedpoint impedance measurements. Commented Oct 2, 2020 at 9:02
• @BrianK1LI The balun is 7T RG58 on FT140-43. It has > 1000 Ohm active impedance on both 40m and 15m bands. The feed line is 10m of RG58. The antenna was measured with FAA-450 (EU1KY) antenna analyzer with OSL-calibration for the feed line. The impedance @ 7.030 is ~34+0j Ω, ~63+0j Ω @ 21.330. Usually I get closer to 50 Ohm, but I've been using quite a thick insulated wire this time. Capacitance hats were made and placed according to the article, I've been using thick enameled copper wire, 1.5mm in diameter. The mast is only 7m. I had a few DX contacts (9000+ km) on 15m in FT8 nevertheless. Commented Oct 2, 2020 at 9:13

As you know, the 40m half-wave is three half-waves on 15m. The two end half-wave sections are 180$$^o$$ out of phase with the center section:

Overlaying the current distribution after adding "capacitive hats" spaced $$\lambda$$/2 on 15m:

The size and position of the "hats" establishes the resonance point of the center half-wave section on 15m. If one extrapolated to zero the current distributions at the ends of the center half-wave section, the result would be consistent with a physically longer element.

As noted by others, the resonant feedpoint impedance of this antenna tends toward 100-$$\Omega$$ on 15m:

An electrical $$\lambda$$/4 section of 75-$$\Omega$$ coax is often used to match a 50-$$\Omega$$ feedline.

Increasing the size of the "hats" tends to raise the resistive part of the feedpoint impedance; moving the "hats" toward the feedpoint lowers it. Judicious position and sizing of the "hats" delivers a better match to 50-$$\Omega$$ without the need for a transformer:

The "hats" have little to no effect on 40m because their small capacitance is a large reactance that adds little coupling between the small potential differences on either side of each "hat."

• Would it be accurate to say the hats are more relevant on 20m than 40m since they lie near a high-voltage node in the former, but not in the latter? Commented Oct 2, 2020 at 12:46
• Many thanks for the answer and charts! But I'm afraid I'm still having difficulties to understand the main mistery - why the hats affect 15m band and don't affect 40m band. Your explanation is a set of words to me that don't have any sence when put together. I'm afraid I'm not such an expert in antenna theory. Could you please explain the last sentence as if I'm 10 years old? Commented Oct 2, 2020 at 13:11
• @PhilFrost-W8II I believe so, Phil. Since Q=CV, there can be more stored charge - hence, greater effect - when the "capacitance" is placed where the potential difference between the two sides is large. The voltage node is such a point because of the phase reversal there. Commented Oct 2, 2020 at 20:01
• @BrianK1LI I would like to clarify if I understood all correctly. Can I consider "hats" as terminals of a single large capacitor? And since they are placed in voltage nodes on 15m larger voltage means larger charge and hence larger effect? Commented Oct 2, 2020 at 21:46
• @AleksanderAlekseev-R2AUK When I considered your question, Aleks, I asked myself the same question. I modeled lumped capacitors in place of the "hats," but the results are completely different. It's not too surprising: the "hats" are distributed elements that couple energy between lengths of the antenna on either side. Commented Oct 2, 2020 at 21:57

Also why the third harmonic resonance turns out to be ~2.5% higher than 7.030 x 3?

This is because of "end effect", caused by the discontinuity where the wire stops at the end of the dipole. I'm fuzzy on the theory, but the ends act like capacitors, which adds a little bit of loading, and makes the wire's electrical length about 0.02 WL longer than it would be otherwise.

Assuming 1.0 velocity factor (to avoid distraction), that means that if we want a dipole to resonate, instead of cutting it to be 0.5 WL long, we cut it to be 0.48 WL physically so it acts like 0.5 WL electrically.

But if we operate this 0.48 WL dipole on the third third harmonic, it will be 0.48*3 = 1.44 WL long physically, and end effect still adds 0.02 WL at the new frequency, not 0.06, so the antenna will be 1.46 WL long electrically — about about 2.5% short of resonant.