Small Transmitting Loops (STL) usually require a capacitor to tune the loop to resonance for ease of efficient matching.
For an efficient STL that can handle moderate power, the capacitor must withstand extreme voltage, and must have extremely low equivalent series resistance (ESR).
(Straw 2007) describes the so-called ‘trombone’ capacitor which is attributed to Bill Jones, KD7S, originally in Nov 1994 QST.
Above is Jones’ trombone match.
There is a dearth of meaningful measurement data to demonstrate the performance of this solution.
Lets consider such an arrangement to tune a 1m diameter STL for 14.2MHz.
The inductive reactance of the untuned loop is around 200Ω depending on the diameter of the tube used, and so requires a tuning reactance of about 200Ω.
As far as possible, this analysis uses dimensions that were gleaned from the reference, dimensions are converted to metric.
The large diameter tube is copper, od=22.22mm, id=18.92mm. The small diameter tube is copper, od=15.88mm. A 0.127mm thick PTFE sheet is fitted between the tubes, the assumption is that it occupies 10% of the dielectric space and permittivity and loss tangent are factored from solid PTFE on that basis.
Analysis
The problem can be broken into three segments in cascade:
- a two wire transmission line from the main loop to the lower end of the thicker tube;
- an o/c coaxial stub formed from the two tubes and PTFE sheet; and
- a s/c two wire stub which is the extension of the inner tube with s/c link at the bottom.
Taking these from the bottom up…
3. s/c stub
Above is a solution for the s/c stub assuming negligible reactance and resistance in the shorting bar. (The impedance of the shorting bar is quite small, probably around 0.0015+j2.6Ω, and has been ignored… it could have been plugged in to this stage but the amount is small compared to other uncertainties.)
Zin=0.0040+j7.9852Ω.
2. o/c stub
Above is the solution of the o/c stub. There is one in each leg, so Zin to the cascade of the series o/c stub and s/c stub is Zin=2*(0.0044-j117.6744)+0.0040+j7.9852=0.013-j227Ω.
1. two wire transmission line
The impedance of the pair of o/c stubs in series with the s/c stub is transformed by the TL formed by the outside surface of the large tubes.
Above is the solution to that transformation, Zin=0.0182-j195.5057Ω. We have the desired reactance of around -200Ω, but with 0.0182Ω of resistance due to losses in that structure. Q of the equivalent capacitor can be calculated:
Q=195.5/0.0182=10,700.
This is a pretty good result, but remember that it is calculated for ideal new materials, no corrosion, no moisture ingress etc… and lots of good ideas fail in the execution due to practical issues like these.
In practice, there will be some resistance in the shorting bar, some resistance in the various soldered joints, so it might be a little worse in practice, but still pretty good when compared to other solutions. Note that dielectric loss is small because only 10% of the dielectric space is occupied by PTFE. It is significantly worse if most of the space was PTFE, or worse, a polar polymer was used.
Extrapolating results.
The scenario is quite complex, and extrapolating results for different conductors, diameters, insulators, bands etc is quite risky.
That said, this is less likely to be a good solution at lower frequencies / with smaller loops.
References / links
- Straw, Dean ed. 2007. The ARRL Antenna Book. 21st ed. Newington: ARRL. p24-10.