questions about antennas and sleeve baluns

[Agent Smith]

Hello, I am interested in antenna construction. I have purchased some inexpensive collinear antennas, but I have serious doubts about their validity. Over the past few months, I have tried to delve into this topic and have read hundreds of documents, but there are still some things I don’t fully understand. Can you help me?

For instance at http://www.w8ji.com/sleeve_baluns.htm i've found an explanation of sleeve baluns, but it's not clear if the sleeve connected at the center feed point and running backwards to the line is a radiating element or is radiating when its open end is toward the whip.


I appreciate any help.

[Solder_Junkie]

Rather than relying on guesswork and speculation, a common issue for those discussing common mode currents on coax cables, why not measure it?

A common mode current meter that works up to 2m and is within 20% from 1.8 to 70 MHz, is really easy to make. This site shows you details, including calibrating the meter:
http://www.ifwtech.co.uk/g3sek/clamp-on/clamp-on.htm

Ferrite sleeve chokes are effective, those in the UK can buy useful sleeves from CPC, item 559-568, cost is £1.61 + vat for a pack of 5. These slide over RG213 size cable and provide a measured 19 dB of common mode current attenuation on 2m (using 5 ferrites).

Antennas-Amplifiers use a small coil of coax on their 2m Yagi antennas, Tonna used an aluminium tube grounded to the boom at the end away from the driven element, and Innov use ferrite sleeves. On none of these antennas can I measure any common mode current. Incidentally, the sleeve on the Tonna Yagi is shorter than a 1/4 wave in free space, as it will be effected by the coax through the sleeve.

SJ

[Agent Smith]

(sorry for my language, i don't speak english very often)
First, thank you for your reply.
I love multimeters and i'd like to have another one more, but i'm sure there is a common mode current because the measure with nanovna is very unstable and looks like a fine V shape when i touch the line with my hand. I just received some toroids from ebay, anyway my answer here is just to understand, to know what to do in the future, and so.

My only operating frequency is 869.5 MHz, i have three enclosures and i should fit them with different elements combinations to have a choice among three different gains or radiating lobes. If the sleeve balun is a radiating element i won't hide it under the enclosure aluminum base, if not i can save 8 valuable centimeters. Unfortunately i have only 0,15W of power, everything should be optimized.

[Solder_Junkie]

150mW at 860 MHz is not going to show on a simple common mode current meter.

Your reference to W8JI appeared to refer to amateur radio antennas and power levels (up to 1500 Watts, depending on frequency and country/licence). Clearly this is not the case.

At 150mW, I wouldn’t bother about common mode currents.

SJ

[pienari]

Is it LORA or some other system?

[Agent Smith]

Yes, the network is meshtastic over Lora. I'd like to understand for first, it's a matter of knowledge, or curiosity. Sometimes i need to know that i can't sleep 🙂

[antenna]

I will try to answer your question, but I have to make a couple important notes about antenna and transmission line theory first to help you understand it.

1- To get useful power into a half wave radiator, the source impedance must match the feed point impedance, and the feed point impedance depends on the feed point location along that radiator.

2- If the feed point is at the center (quarter wave on both sides), the source needs to have a low impedance

3- If the feed point is at the end of the half wave radiator, the source needs to have a high impedance.

4- Quarter wavelength transmission lines will invert the impedance seen at the other end, so if one end is shorted, the other end is a high impedance.  If one end is open, the other end appears as a short.

5- If a shorted quarter wavelength transmission line is used as a series element, it acts like a parallel LC tank circuit shifting the signal by 180°, so the signal applied to one leg will be coming out the other leg 180° out of phase.

Referring to the middle antenna in figure 22.5 on w8ji's web page, the low impedance transmission line  is feeding the quarter wave radiator at the top and the quarter wave sleeve "skirt" going down.  The two form a half wave dipole.  At the bottom of that first skirt, there is a signal with a high impedance (since it is at the end of a half-wave dipole).  The inside of that skirt sleeve will act like a parallel LC tank in series with that high impedance shifting the signal by 180°.  So what comes out of the inner leg of the sleeve (the signal riding on the coax shield) is a high impedance signal with current traveling in the same direction as the first half wave element.  Looking at antenna C, that outer part of the shield acts as the tip of the next half wave element.  And since we are driving a half wave element at the end this time, that high impedance signal coming out of the sleeve is perfect.  The signal phase matches the top radiator, and the impedance is proper for the driving location.

When we have the desired number of elements, how do we prevent the high-Z signal from the inside of the last sleeve from going down the shield as common-mode current?  Again, it comes down to feed point impedance looking into the section you don't want power on.  if the combination of coax and radio enclosure is a half wavelength, or multiple, it will gladly accept power from the sleeve above and become part of the antenna.  You will have common-mode currents. 

Back to your question of which way the sleeve faces.  Lets say you put a quarter wavelength sleeve on the shield facing the other way.  Any common-mode signal looking into the open end will see a high impedance.  Take a look at the next illustration below, section titled "decoupling impedance".  In this case, a sleeve with the open end facing the other way, towards the load, prevents common-mode currents.  This is because there is a low impedance feed point driving a low impedance load, and since there is a low impedance load, the low impedance signal goes into the load only and the high impedance formed by the sleeve (which is in parallel with the load) has little effect on anything, just like a 10kΩ resistor across an 8Ω speaker terminal.  All of the current goes into the matched load

This is often called a bazooka balun.  Although mathematically identical to the other sleeves facing down the coax, we are taking advantage of the high impedance to tell those low impedance signals they can't come this way. This only works if there is a mismatch, so the open end of the bazooka needs to be at a low impedance point or it will simply do what the other sleeves did and pass the signal down the coax shield.  In figure 22-5 (c), the low impedance point is at the bottom of the antenna where the high impedance coming from the inner part of the sleeve is transformed to a low impedance by a quarter wavelength of shield to where the drawing ends.  That would be the ideal location for the open end of the bazooka sleeve.

The bazooka balun can technically be inside the enclosure, but I fear you have missed one of the most important parts of that paper, the choke impedance.  The sleeve and the coax shield forms a transmission line and the choking impedance is a direct function of that transmission line impedance, so if the sleeve diameter is too small, your choking impedance will also be small.  The ratio of sleeve diameter to shield diameter must be large for any of this to work properly, and a large sleeve will most likely not fit inside your enclosure.  If you are going to try this with a sleeve that is tight against the coax jacket, you may as well just abandon the whole idea.

Edit:  and don't forget to mind the velocity factor!  A tight sleeve with mostly PVC or whatever the jacket is made of filling the gap will have a very different VF from a sleeve full of air (or the outer surface of the sleeve doing the radiating).  So remember, if you want this to work, you need a sleeve much larger than the coax diameter, and the space between the sleeve and shield must be mostly air, otherwise the outside of the sleeve (radiator) and the inside of the sleeve (phasing element) will be different lengths electrically (due to the dielectric in the sleeve) and everything goes to shit.  Something else worth considering is placing quarter wavelength radials at that low impedance point right above the mouth of the bazooka sleeve so the current at that low impedance point has a place to go instead of seeing just the sleeve and reflecting back up the chain.

[Agent Smith]

I will try to answer your question

Thank you very much for your helpful, well detailed answer, i just saved it in my notebook, translated in my language.

Now i want to read it several times while drawing some schematics to be sure i'll understand it all. When ready i'll try to practice.

[Geoff-AU]

I have purchased some inexpensive collinear antennas, but I have serious doubts about their validity.

That is wise.  Collinear antennas are a little difficult to do well.

For instance at http://www.w8ji.com/sleeve_baluns.htm i've found an explanation of sleeve baluns, but it's not clear if the sleeve connected at the center feed point and running backwards to the line is a radiating element or is radiating when its open end is toward the whip.

One very, very useful mindset is to always consider coaxial cable to have 3 available "conductors" for RF to use:
1) The outside skin of the inner conductor (core)
2) The inside skin of the outer conductor (shield)
3) The outside skin of the outer conductor (shield)

Skin effect splits the coax shield into an inner conductor and an outer conductor.

Differential current will only flow on conductors 1 and 2.  Common mode current will flow on conductor 3.

So for the figure 22-6 of the collinear, the differential current is easy.  The current in conductor 1 goes right to the top of the exposed element.  The current in conductor 2 turns around and flows along the sleeve, thus forming the second half of the dipole.

Now what happens with conductor 3?  Think about the distance any RF has to travel to make it from the open end of the sleeve to the same point on the outside of the shield.  The sleeve is 1/4 wavelength long so the round trip is 1/2 wavelength.  Therefore, any common mode current coming up the coaxial cable will flow onto the sleeve (it has nowhere else to go) and when it reaches the end of the sleeve it is exactly out of phase - this will act to cancel out the common mode current!

The next figure with the balanced load can be thought of in the same way.  The only difference is that the coax shield continues on both sides of the sleeve connection, so the differential current wants to stay on the inside of the coax shield so it can be closer to the opposing current on the inner conductor.  Therefore the sleeve does not radiate differential current, but it suppresses common mode just the same.

[Agent Smith]

One very, very useful mindset is to always consider coaxial cable to have 3 available "conductors" for RF to use:

Thank you for your answer, i stored it too in my notebook, It's another piece of knowledge to keep.
Now i understand the difference between forward and reverse sleeves, i already tried to build my first 4 x λ/4 collinear, after placing 3 ferrite cores at the bottom it resonates seamlessly. Not tested on the field yet: i don't know what kind of lobe it will radiate.
Now i'm only trying to make the ferrite part more lighter with some specific toroids.
To be continued...

[Agent Smith]

Hello again
Currently the step is to understand why the collinear just made shows two vertexes at the bottom of the swr chart. There is also a double change in the phase chart never seen before in my other simple dipole. It looks like one vertex is from some elements like the sleeves and the other spike comes from the whip, one of them not correctly sized or something like that. I think i have to move the vertexes into one, mainly to get a narrower band: i have only one frequency, no channels, everything outside is noise.
As i can't measure the radiating lobes i wonder if a wrong tuning like now may affect the performances.
Do you have any suggestions?
Thanks

[Agent Smith]

Collinear antennas are a little difficult to do well.

That antenna is operating three meters over the roof, i'm still doing some tests. After many attemps i finally was done.
Now i'm managing to make some other antennas, i purchased some ferrite cores, please one question about that: the inner diameter of the cores is a little larger than the outer diameter of the element i have to mount it on, there is a little free play. Is it a problem or the ferrite should only be around the brass pipe, like the way power supplies have them around the cables?
Thank you for your time.

[Geoff-AU]

it'll be fine.  ideally you don't have a large gap, it does reduce the performance but not by much.

[Agent Smith]

Great thanks

[radiolistener]

just buy NanoVNA and you can measure your antenna and check if it works good or bad. It also allows to see how other things around antenna affects its performance in realtime...

[Agent Smith]

just buy NanoVNA and you can measure ...

Right, i could remove the working cores and put on these ones just to check the difference. As soon as i lower the mast.
Thanks

[Agent Smith]

I have one more question just out of curiosity: now the collinear antenna project is completed, working on my roof, i took a note of the measures with nanovna, it looks fine tuned:
Swr 1.016:1
R jX 49,96  0,78Ω
Phase alost vertical line within 5 MHz
The only thing i can't understand is why i should have made the whip 94 mm long, while the working frequency is 869.5 MHz, so the wavelength is 345mm, quarter wave 86,25mm, multiplied by the velocity factor, let's say 0.84, should be at least shorter than 86,25mm. Maybe there is another combination between elements length?
Should i guess an element electrical length different to a quarter wave means lower efficency?

[Geoff-AU]

Element length different to 1/4 wave will change the radiation pattern.

Velocity factor applies to the surrounding media, so a coax cable has a velocity factor of 0.66-0.85 on the inside (ie centre conducter, and the inner surface of the shield) and your velocity factor is 1 otherwise (whip, and the outer surface of the shield).  VF of air is close enough to a vacuum.

An element longer than resonance is inductive, it may be compensating for extra capacitance in another part of your antenna.  Or, if you are measuring at the end of a feedline, then the situation is different depending on the length of the feedline.  R+jX impedances change continuously with feedline length, performing a complete inversion of the feedline's characteristic impedance every 1/4 wavelength.

Measure with an exact 1/2 wavelength (or multiple) of feedline, or do a port extension adjustment on your VNA so you're seeing the correct result and not the effects of the feedline.