How does this antenna balun work?

[Guineafowl]

TL;DR Summary

How does a wiggly PCB track convert balanced -> unbalanced, and match impedance?

It’s this type, commonly fitted to UHF Yagis for TV signals:


Online searches, including a reverse image one of the above, haven’t returned any accounts of how it’s designed.

A folded dipole in a Yagi setup apparently has an impedance of 10-40 ohm. The signal must pass to 75 ohm coax. So there must be a step-up impedance match, and a bal-un transformation.

Someone knowledgable on another forum suggested the wiggly track is a delay line, to bring the two ends of the balanced signal in phase to pass down the coax centre conductor.

 

[tech99]

It is a delay line as you suggest, and doubles the voltage applied to the antenna terminals. In this way it provides an impedance step up of 4, so a 75 Ohm cable is match to a 300 Ohm folded dipole. The two antenna terminals are fed with equal and opposite voltages, so achieving the action of a voltage balun. There is a more commonly seen version of this balun using a piece of coaxial cable half a wavelength long.

 

[Guineafowl]

It is a delay line s you suggest, and doubles the voltage applied to the antenna terminals. In this way it provides an impedance step up of 4, so a 75 Ohm cable is match to a 300 Ohm folded dipole. The two antenna terminals are fed with equal and opposite voltages, so achieving the action of a voltage balun. There is a more commonly seen version of this balun using a piece of coaxial cable half a wavelength long.

Thanks.

Does a folded dipole not have a reduced impedance when installed in a Yagi? Here’s what sparked the question:

The cheap aerial (made in 1984) installed at my house, causing surprisingly minor glitching on one mux only. As you see, no balun. Is it really the case, for the sake of a cheaply-printed PCB, that a 300 ohm dipole is connected straight to 75 ohm coax?

What would an equivalent circuit of the PCB balun look like? I’m not quite getting how the copper track relates to more familiar diagrams of voltage and current baluns.

 

[Baluncore]

Does a folded dipole not have a reduced impedance when installed in a Yagi?

A normal dipole is about 70 Ω, the folded dipole will be close to four times that, 280 Ω. The presence of the boom and elements will not change the impedance much.

The cheap aerial (made in 1984) installed at my house, causing surprisingly minor glitching on one mux only. As you see, no balun. Is it really the case, for the sake of a cheaply-printed PCB, that a 300 ohm dipole is connected straight to 75 ohm coax?

There should have been a 300 Ω to 75 Ω balun there, but they could not be bothered. Perfection is not required with strong signals.

 

[tech99]

This is a picture I found of the same thing using half a wavelength of coaxial cable. If the parasites are closely spaced, then the feedpoint resistance will drop, but for wide band antennas, such as for television, wide spacings are used. The antenna manufacturers are not too concerned with impedance mismatch as it does not reduce gain very much. If no balun is used, the pattern can be distorted a little. But if the cable proceeds from the dipole at right angles the effects are small, especially as the cable is long.

 

[Guineafowl]

This is a picture I found of the same thing using half a wavelength of coaxial cable. If the parasites are closely spaced, then the feedpoint resistance will drop, but for wide band antennas, such as for television, wide spacings are used. The antenna manufacturers are not too concerned with impedance mismatch as it does not reduce gain very much. If no balun is used, the pattern can be distorted a little. But if the cable proceeds from the dipole at right angles the effects are small, especially as the cable is long.

Ah, I recognise that from trying to build my own DTV Yagi, although I was just following instructions from a Yagi calculator.

What I was looking for was an equivalent circuit depicting that PCB balun as discrete components, rather like the lumped element model of transmission lines. Your reply (post #2) makes perfect sense, but I don’t yet see how it relates to fundamentals.

For example, it’s not immediately obvious that a doubling of voltage means a quadrupling of impedance, unless we’re talking transformer action, or relating impedance to $V^2$ with constant power.

If the former, are we looking at an autotransformer balun? Is the length of that wiggly track a half wavelength, to bring the two dipole terminals in phase?

 

[Averagesupernova]

For example, it’s not immediately obvious that a doubling of voltage means a quadrupling of impedance,

Look at it this way: Take a snapshot in time when the voltage at the node where the two coax cables join is at peak voltage. The opposite end of the half wave section will be 180° out of phase. So the voltage between the ends of the half wave section are double the voltage at the node. You have to quadruple the impedance. If you didn't you would be picking up that power out of nowhere. Or losing it, depending on which direction power flow is.

 

[Guineafowl]

Look at it this way: Take a snapshot in time when the voltage at the node where the two coax cables join is at peak voltage. The opposite end of the half wave section will be 180° out of phase. So the voltage between the ends of the half wave section are double the voltage at the node. You have to quadruple the impedance. If you didn't you would be picking up that power out of nowhere. Or losing it, depending on which direction power flow is.

The half-wavelength makes sense, representing the p-p voltage rather than peak above ground, thanks.

So, relating V, P and Z: $P=\frac{V^2}{Z}$ And, keeping power constant: $Z\propto V^2$ ?

So the wiggly track must be a half wavelength of the centre frequency? If so, that nicely takes care of the impedance transformation.

As for the bal -> un conversion, am I looking for a transformer?

Is the primary formed by the PCB track, and the secondary by the ground plane as in the drawing? It looks like some form of autotransformer, given the primary and secondary are common between dipole and coax on the right-hand side.

In some small way, it looks a bit like a final distribution mains transformer, where the incoming line is floating with respect to earth/ground (balanced above and below), and the secondary is tied on one side to earth/ground, forming the unbalanced (Live/Neutral) side. If that is the case, that would be nice.

 

[tech99]

This particular balun is not a transformer but it uses a half-wave transmission line. The PCB ground plane is connected to the coaxial outer and forms one conductor of the half wave transmission line. The arms of the dipole are fed with +V and -V, giving a total of 2V. For constant power this gives R proportional to V^2 as you mention, giving 300 Ohms.

 

[Guineafowl]

The arms of the dipole are fed with +V and -V, giving a total of 2V.

I see, so it’s this effect that does both the impedance and the bal -> un transformations. Very clever and deceptively simple. Does the half-wavelength distance constitute the ‘delay line’, or is that a different effect?

I take it the ground plane has to remain isolated, or the coax outer shield would become part of the dipole.

 

[Baluncore]

It’s this type, commonly fitted to UHF Yagis for TV signals:

1. Is that balun used with a simple 75Ω dipole, or with a 300Ω folded dipole?

2. Is there a ground plane under the pictured PCB?
3. How is that GP connected to the GP on the pictured side, edge, or vias, or screws?
4. Is the PCB made from glass reinforced epoxy?

5. What range of UHF frequencies, wavelengths, is the antenna supposed to cover?
6. I have seen no scale, what is the diameter of the PCB? and the coax?
7. What is the PCB line length in mm?

 

[tech99]

1. Is that balun used with a simple 75Ω dipole, or with a 300Ω folded dipole?

2. Is there a ground plane under the pictured PCB?
3. How is that GP connected to the GP on the pictured side, edge, or vias, or screws?
4. Is the PCB made from glass reinforced epoxy?

5. What range of UHF frequencies, wavelengths, is the antenna supposed to cover?
6. I have seen no scale, what is the diameter of the PCB? and the coax?
7. What is the PCB line length in mm?

Looking at the picture closely, it looks as if the PCB is single sided and the GP surrounds the line in a single plane but is not beneath it.

 

[Baluncore]

Looking at the picture closely, it looks as if the PCB is single sided and the GP surrounds the line in a single plane but is not beneath it.

What can you see that suggests that there is not a GP on the back?

There are too many unspecified parameters in this thread. We need some certainty and substance before the device can be analysed, or reverse engineered.

 

[Guineafowl]

Folded dipole, receiving DTV signals at around 640 MHz. What sparked the question was not seeing this common balun fitted to my current aerial (post #3). I’m afraid I haven’t got one on me for measurements, so this will have to remain a qualitative analysis.

I think I have it: [-V ->earth<- +V] at the dipole becomes [2V -> earth] at the coax centre due to the half-wave delay line. This is the bal -> un transition. The doubled voltage steps up the impedance by 4 as per $Z \propto V^2$ with constant power.

 

[Averagesupernova]

The doubled voltage steps up the impedance by 4 as per Z∝V2 with constant power.

You didn't get that from this?

You have to quadruple the impedance. If you didn't you would be picking up that power out of nowhere. Or losing it, depending on which direction power flow is.

 

[Guineafowl]

You didn't get that from this?

I did, yes - up there ^^ I was just checking $P=\frac {V^2}{Z}$ was the relevant equation. 25 years since I did A-Level (high school) physics.

 

[Averagesupernova]

I did, yes - up there ^^ I was just checking $P=\frac {V^2}{Z}$ was the relevant equation. 25 years since I did A-Level (high school) physics.

Ok. $P=\frac {V^2}{Z}$ is always relevant. We can transform voltages and impedances around but you can't transform power. You get out what you put in minus losses.

 

[tech99]

It should be R not Z, as the latter can have a reactive part which does not consume power.

 

[Guineafowl]

Of course, yes. That makes sense.

 

[Baluncore]

To make that a 4:1 balun the PCB etched transmission line will need to have an impedance equal to the coaxial line, and be electrically half a wavelength long.

Since the folded PCB line should be physically short, it needs the lowest velocity factor, that suggests it needs a high capacitance, since, vf = 1/√LC. I would expect the thinnest available PCB, with a ground plane to minimise vf, by maximising C.

A broadband balun could pass the entire UHF band, but here we have a narrowband half-wave line balun, said to be for about f=640 MHz, λ=470 mm ; λ/2 = 235 mm. Even with a low vf, that line does not look long enough to me. Maybe it is designed to operate at higher UHF, shorter wavelengths?

 

[tech99]

To make that a 4:1 balun the PCB etched transmission line will need to have an impedance equal to the coaxial line, and be electrically half a wavelength long.

For a half wave balun it seems to me that the characteristic impedance will not be important, because the same impedance appears at both ends of any half wave line. It looks to me also that it is very short, so maybe it is for higher channel numbers as you mention.

 

[Baluncore]

For a half wave balun it seems to me that the characteristic impedance will not be important, because the same impedance appears at both ends of any half wave line.

That assumes the line is lossless, and the length is an exact half-wavelength. The half-wave line is a resonator, and you are assuming an arbitrarily high Q.

I think you may be considering commercial digital TV, so things can be sloppy, and nothing really matters that much. You get what you pay for.

On the other hand, the engineering analysis of a balun requires a more rigorous approach, one that considers the entire signal bandwidth, and the sensitivity of the accumulation of detrimental gain and phase shift, at the edges of the band or channel.

 

[Guineafowl]

If I come across one I’ll post details here. To replace the aerial in post #3 I’ve dug out a log periodic type that I’d forgotten about. If it doesn’t work well, I’ll get a decent Yagi type.

There’s one remaining niggle, about the direction of the impedance transformation - it seems to be backwards.

Imagine the aerial is transmitting. Unbalanced 2V travels in the coax, and meets the balun, where the voltage is halved to a balanced V. According to $R \propto V^2$ , halving the voltage should 1/4 the impedance, but we need it x4.

 

[Averagesupernova]

Unbalanced 2V travels in the coax, and meets the balun, where the voltage is halved to a balanced V.

Why do you think it's halved?

 

[Guineafowl]

Why do you think it's halved?

So the voltage between the ends of the half wave section are double the voltage at the node.

The arms of the dipole are fed with +V and -V, giving a total of 2V.

Just re-reading the above two quotes, I see the voltage doubles as it reaches the dipole, which sets the impedance the right way round. I had the voltage doubling due to the peak at the node adding to the peak, half a wavelength late, coming down the delay line.Thanks.

 

[Swamp Thing]

In case you want to dive deeper into exactly how the voltage is doubled / halved in this family of baluns (called transmission line transformers)
https://www.allaboutcircuits.com/te...-transformers-understanding-the-bifilar-coil/

The thin wiggly track (along with its twin on the other side) is the equivalent of a bifilar coil as described in the linked page.

 

[tech99]

It does not have a twin on the other side.

 

[Swamp Thing]

It does not have a twin on the other side.

Sorry, I didn't look carefully enough at the coax version you posted. So no, although it does contain a transmission line, it's not what is referred to as a transmission line transformer.

 

[Baluncore]

It does not have a twin on the other side.

Since the PCB transmission line is not alone in space, there must be other conductive surfaces nearby that will reflect images.

If, as I suspect, the line has a ground plane on the back of the PCB, then the line will have a virtual image of the front, behind the ground plane, with the opposite current and polarity.

The engineering analysis cannot escape the image, but you can control it by providing the ground plane.