Current balun ( CHOKE) for EFHW system 1:1,Model WOCB1152SO CMC 1.8–54 Mhz, 1KW



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SKU: WOCB1152SO Category:


Current balun for EFHW system 1:1 for end fed half wave antenna construction–Common Mode Choke

Model WOCB1152SO

Freq. 1-54 MHZ

Power – 1kw (1.5kw peak)

DIGI – 500w (1kw peak)

Insertion Loss Less than 0.02db

Choke more than -30db


Amidon core FT240-K (K material is used primarily in transmission line transformers from 1 MHz to 50 MHz. AL Value: 458)

Silverplated wire 1.5mm

Polycarbonate box ip65

Other stuff – copper, brass, stainless steel

Additional information

Weight1.0 kg
OCFD Design

BY UR4UNR — note refer to the diagrams in the website near the product image
Asymmetric Dipole Antenna (OCFD) – or incorrect name of Windom FD4
I was prompted to write an article about this type of antenna by the fact that there is not much good information to repeat correctly. I note – the right one!
I have been using antennas of the asymmetrical dipole type for a long time. I repeated the first antennas from the designs that I found on the Internet, but they worked, to put it mildly, not always good. With the advent of compact antenna analysers, the theory was able to connect to practice in the form of accurate graphs and measurements.
Therefore – I think that we need to collect everything into one article, and you should not be afraid to repeat it, even if your first antennas like “Windows” were unsuccessful. Although it is a mistake to call it Windom, but oh well. For information, the Windom Antenna is a single-wire antenna that works directly with the ground. Our version is precisely an asymmetric dipole.
An asymmetric dipole OCF is a non-grounded dipole antenna fed by a two-wire line, either a coaxial or a parallel conductor with an open wire. But since the most common and simple design is a trx-coaxial antenna, let’s talk about such an antenna design.
What we know.
A half-wave dipole is fed to the centre where it has a minimum impedance and maximum current. Under ideal conditions, its resistance at the Power point is ~ 72 ohms. It is very close to 75 ohm or 50-ohm feeders. That allows you to use a dipole connecting a balancing device 1: 1 and a coaxial of 75-50 ohms. Such a dipole will work well also at odd harmonics, for example, a 40m dipole will work at 15m and so on.
This is in ideal conditions; the harsh reality shows that all surrounding objects and the height above the ground affect the impedance at the Power point. For example, if the height of the suspension is less than 0.13 lambda, then the impedance will shift downward closer to 50 ohms. Etc.
It is also inevitable that shifting the place of supply to the side we get an increase in impedance. Our task is to find the area (the place of supply) where the antenna would work approximately the same on more than 2-3 bends. Looking ahead, I want to say that my antenna works successfully on 80m 40m 20m 17m 12m 10m and 6m
If we start from the theory, then a shift of about 33% from the end of the dipole gives us ~ 300 ohms at the Power point, which allows us to use an unun 1:6. But since the antenna almost always hangs under different conditions and is very far from ideal, moreover, at different heights its impedance in the place of the power supply is simply unpredictable.
Running ahead I want to say that with a suspension of about ~ 8 m in the place of the balun and descent of the shoulders to ~ 2-3 m, the ideal distance of the feeding place was selected, and had an impedance of about 230-160 ohms, which required a 3:1 or 4:1 transformation.
So back to the theory, remember that the impedance of the dipole Power point will depend on its height above the ground and the dielectric constant of the earth under the antenna. Therefore, a 6: 1 balun is likely to convert impedances below 300 Ohms …, which makes the impedance transformation coefficient unpredictable. In this case, the current balun will serve only as a radio frequency choke … which is better than nothing, but still not what is needed. Therefore, you cannot do without measurements with the device and the selection of the correct transformation.
The principle of operation of an asymmetric dipole (OCFD)
Knowing that for a dipole, in this example, an 80m range, at a point shifted by a certain amount of % from the centre, there is a current, we need to figure out what will happen on the remaining ranges.
To do this, you have to consider the graphics.
Consider, for example, an 80M half-wave dipole. Based on a shift of 33%, se diagram
In general, in all ranges, such an antenna works like this:
As you can see, with an acceptable rf current close to maximum, we get 80, 40,20,17,12,10,6 m ranges 3.5 MHz (80M band)
Second harmonic (2xf0) = 40M band (7 MHz) see diagram
Fourth harmonic (4xf0) band 20M (14 MHz) see diagram.
Fifth harmonic (5xf0) band 17M (17.5 MHz) see diagram.
The fifth harmonic is 3.5 MHz (5xf0) 17.5 MHz, but the 17 M frequency distribution for the amateur section is slightly higher in frequency at 18.068 MHz. As you can see in the graph, the antenna * is physically * too long at a frequency of 18.068 MHz But overall we get a quite acceptable result
Sixth harmonic (6xf0) band 15M (21 MHz)
Third harmonic (3xf0) band 30M (10.5MHz)
The third harmonic from 3.5 MHz (3xf0) is 10.5 MHz. The 10.1-10.15 MHz section is slightly lower. Thus, the antenna becomes * physically * a little short to fully cover three full waves at a frequency of 10.1 MHz. About the same thing with the 6th harmonic at a frequency of 21 MHz It can be safely stated that the antenna will not work normally at these frequencies
Seventh harmonic (7xf0) band 12M (24.5 MHz)
The seventh harmonic of 3.5 MHz (7xf0) is 24.5 MHz, but our frequency distribution of 12 M for the amateur section is slightly higher in frequency and lies at 24.89 MHz. As you can see on the graph, the antenna * is physically * a little bit longer than necessary at a frequency of 24.89 MHz. But again, we get a fairly good result on the RF current, at the place of supply.
Eighth harmonic (8xf0) band 10M (28MHz)
14th (13th) harmonic (14xf0) band 6M (49MHz)
In the area of 50-52 MHz, we also get a good result and an acceptable rf current at the Power point.
Preliminary findings. Having studied all the graphs above, we can see that:
The feeding point with a bias of 33% does not allow the use of this OCFD 80M in the ranges 30M and 15M, where the RF current is close to zero, which leads to this inoperability and impedances of thousands of ohms. And it will behave quite successfully at 80m 40m 20m 17m 12m 10m 6m having an impedance at the Power point of about 100-400ohm (depending on the location and height of the antenna). What makes it possible to successfully transform further into the 50 (75th) of our asymmetric coaxial that we need?

Once again, I want to note that our antenna is not symmetrical!
UNUN! But! To many problems with common mode… Work not stable. TVI effect
That’s why I use system current choke 1:1 + voltage transformer 1:4 . This system works perfect with OCFD. Because have good choke more then -30db and good voltage transformation from 50 to 200 ohm
My results: 80 40 20 12 10 6m – swr less 1.4, 17m – swr 2.6


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