A correspondent wrote of his project with a Guanella 4:1 balun where each pair was wound with a pair of insulated wires, and importantly the output terminals are free to float as the load demands. A Guanella 1:1 balun wound in the same way has the same characteristic.

To preserve balun choking impedance, it is best to preserve balun symmetry, and the use of a short open circuit coaxial stub across the output terminals for InsertionVSWR compensation introduces some asymmetry.

An alternative construction with coaxial cable that is more symmetric is shown above.

It is two pieces of RG58 with the shields fanned out and twisted into pigtails which will be the two terminals of the device. The inner conductors are connected across to the other cable’s shield. The cross connects are kept separated, as are the shield ends, and the space if filled with hot melt adhesive (which is electrically similar to polyethylene). Likewise, I have made sure the cut ends are clean of whiskers and used hot melt adhesive to insulate the parts from each other.

The prototype withstood 2.2kVp and the weak point was the open ends of the RG58. RG58 is strong internally, flashover occurs across the ends… so whilst the neatest smallest pigtails are good for RF, they are not conducive to high voltage withstand. By stripping the braid back 5mm and separating the braid ends similarly, it withstands 4.5kVp. Most ham ATUs cannot withstand even 2kVp, even though the capacitors may be rated for higher voltage, they tend flash over at coax connectors.

There are three open circuit transmission line stubs in parallel.

The first is formed by the outer surfaces of the two pieces of parallel RG58 with PVC jacket insulation. This turns out to have Zo close to 50Ω and VF around 0.6-0.7.

The second and third stubs are formed by the inner of each of the two pieces of RG58.

Because the three TL sections are close to Z0-6=50 and VF=0.66, we can treat them as a single stub with Zo=16.7 and VF=0.66. It is an approximation, but a good one considering the relative magnitude of error due to the pigtails when used on very short stubs as in this case.

In my correspondent’s case, his 70mm single RG58 stub could be replaced with the triple stub of length 20-25mm, enhancing symmetry by both the symmetric structure and smaller size.

Above is a measurement of the prototype. The observed capacitive reactance reconciles with 55mm of Zo=16.7 and VF=0.66 TL.

When the stub is very short electrically (<6°), its impedance is well approximated as the line’s capacitance per unit length of 1/(Co*vf*Ro). In this case, C=1/(299792458*0.66*16.7)=302pF/m, and the 55mm length should be about 302*0.055=16.6pF which reconciles with the chart above.

]]>As pointed out in the articles, the solutions cannot be simply extended to real antenna scenarios. Nevertheless, it might provoke thinking about the performance of some types of so-called balanced ATUs, indeed the naive nonsense of an “inherently balanced ATU”.

(Witt 2003) goes to some length to calculate his IMB figure of merit based on a similar load of two not necessarily equal series resistors with the mid point grounded to the ATU chassis. Witt’s IMB is equivalent to the factor |2Ic/Id|that was calculated in earlier articles in this series, and equally useless in inferring behavior in a real antenna system.

(Duffy 2010) gives an explanation of the behavior of baluns in an antenna system, and it becomes apparent that simple linear circuit solutions of a couple of resistors does not give insight into the behavior in real antenna systems.

The bottom line though is that while NEC models might be informing, there is no substitute for direct measurement of common mode current (Duffy 2011)… and it is so easy.

- Duffy, O. Dec 2010. Baluns in antenna systems. https://owenduffy.net/balun/concept/cm/index.htm (accessed 21/02/12).
- ———. May 2011. Measuring common mode current. https://owenduffy.net/measurement/icm/index.htm (accessed 21/02/12).
- ———. Feb 2012. Balanced ATUs and common mode current. https://owenduffy.net/balun/concept/BalancedAtu.htm (accessed 18/03/2019).
- Witt, Frank. Apr 1995. How to evaluate your antenna tuner In QST May 1995. Newington: ARRL.
- ———. May 1995b. How to evaluate your antenna tuner In QST May 1995. Newington: ARRL.
- ———. Sep 2003. Evaluation of Antenna Tuners and Baluns–An Update In QEX Sep 2003. Newington: ARRL.

This article reports the same asymmetric load using the MFJ-949E internal voltage balun.

The test circuit is an MFJ-949E T match ATU jumpered to use the internal balun and resistors of 50Ω and 100Ω connected from those terminals to provide a slightly asymmetric load.

The voltage between ground and each of the output terminals was measured with a scope, and currents calculated.

Above are the measured output voltage waveforms at 14MHz.

Lets work out the current amplitudes. Above, V1 (yellow) is 5.9divpp, V2 (cyan) is 7.2divpp. I1=V1/50=5.9*0.2/50=23.6mApp. I2=V2/100=7.2*0.2/100=14.4mApp.

Expanding the timebase allows better measurement of the phase difference.

V2 lags by a half cycle less 8.25µs, so V2 phase is -180+8.25e-9*14e6*360=-180+42=-138°.

Lets calculate the common mode and differential component of current in each load resistor. We will use Python as it handles complex numbers.

>>> i1=0.0236

>>> i2=0.0144*(math.cos(-138/180*math.pi)+1j*math.sin(-138/180*math.pi))

>>> ic=(i1+i2)/2

>>> abs(2*ic)

0.016100289594275147

>>> id=(i1-i2)/2

>>> abs(id)

0.01781446515461856

>>> abs(2*ic)/abs(id)

0.903776198417105

>>> 20*math.log(abs(2*ic)/abs(id))/math.log(10)

-0.8787820061070818

So, the differential component of current is 17.8mApp, and the total common mode current is 16.1mApp, the total common mode current is 90% of the differential current or 0.9dB less than differential current.

By any standard, this is appalling balance, and demonstrates why voltage baluns are unsuited to the application.

The fact that the “inherently balanced” topology is only 1.8dB better that this voltage balun experiment speaks volumes for the failure of the “inherently balanced” topology.

The measurements reported here are for a specific scenario (components, frequency and load), and should not be simply extrapolated to other scenarios.

The calculated imbalance if you like applies to the specific test circuit, and cannot really be extended to use of this balun in an antenna system scenario.

]]>This article reports the same equipment reversed so that the common mode choke is connected to the output of the MFJ-949E.

The test circuit is an MFJ-949E T match ATU followed by A low Insertion VSWR high Zcm Guanella 1:1 balun for HF. A banana jack adapter is connected to the balun output jack, and resistors of 50Ω and 100Ω connected from those terminals to provide a slightly asymmetric load.

The voltage between ground and each of the output terminals was measured with a scope, and currents calculated.

Above are the measured output voltage waveforms at 14MHz.

Lets work out the current amplitudes. Above, V1 (yellow) is 4.0divpp, V2 (cyan) is 8.0divpp. I1=V1/50=4.0*0.2/50=16.0mApp. I2=V2/100=8.0*0.2/100=16.0mApp.

Expanding the timebase allows better measurement of the phase difference.

V2 lags by a half cycle and 1.0µs, so V2 phase is -180-1.0e-9*14e6*360=-180-5=-185°.

Lets calculate the common mode and differential component of current in each load resistor. We will use Python as it handles complex numbers.

>>> i1=0.016

>>> i2=0.016*(math.cos(-185/180*math.pi)+1j*math.sin(-185/180*math.pi))

>>> ic=(i1+i2)/2

>>> abs(2*ic)

0.0013958203956907485

>>> id=(i1-i2)/2

>>> abs(id)

0.015984771545309726

>>> abs(2*ic)/abs(id)

0.0873218858170239

>>> 20*math.log(abs(2*ic)/abs(id))/math.log(10)

-21.177537875409207

So, the differential component of current is 16.0mApp, and the total common mode current is 1.40mApp, the total common mode current is 9% of the differential current or 21.2dB less than differential current.

Calculation of the common mode component of current involves the addition of two almost equal and almost opposite phase currents and is very sensitive to uncertainty in each of the measurements using this measurement method. This balun should achieve |2Ic/Id|>35dB in this scenario, but it would take a higher accuracy measurement system to measure it.

The fact that the “inherently balanced” topology measures 18dB worse that this experiment speaks volumes for the failure of the “inherently balanced” topology.

The measurements reported here are for the specific scenario (components, frequency and load), and should not be simply extrapolated to other scenarios.

The calculated imbalance if you like applies to the specific test circuit, and cannot really be extended to use of this balun in an antenna system scenario.

Continued at Inherently balanced ATUs – part 3 .

]]>LB Cebik in 2005 in his article “10 Frequency (sic) Asked Questions about the All-Band Doublet” wrote

In recent years, interest in antennas that require parallel transmission lines has surged, spurring the development of new inherently balanced tuners.

Open wire lines require current balance to minimise radiation and pick up, the balance objective is current balance at all points on the line.

Cebik goes on to give examples of his “inherently balanced tuners”.

Above, Cebik’s “inherently balanced tuners” all have a common mode choke at the input, and some type of adjustable network to the output terminals.

Cebik was not the originator of the idea, many others had written of the virtue of the configuration, but I cannot recall seeing meaningful measurement to support the claims.

Lets take the last circuit, and simulate it using A low Insertion VSWR high Zcm Guanella 1:1 balun for HF followed by an MFJ-949E T match ATU. The MFJ949E stands on its insulating feet on a large conductive sheet to serve as the ground, the balun in connected to the ATU input jack and the input jack of the balun is grounded to the aluminium sheet. A banana jack adapter is connected to the ATU Coax 1 output jack, and resistors of 50Ω and 100Ω connected from those terminals to provide a slightly asymmetric load.

The voltage between ground and each of the output terminals was measured with a scope, and currents calculated.

Above are the measured output voltage waveforms at 14MHz.

The meanings of currents used here is given at Differential and common mode components of current in a two wire transmission line.

Lets work out the current amplitudes. Above, V1 (yellow) is 4.8divpp, V2 (cyan) is 7.4divpp. I1=V1/50=4.8*0.2/50=19.2mApp. I2=V2/100=7.4*0.2/100=14.8mApp.

Expanding the timebase allows better measurement of the phase difference.

V2 lags by a half cycle and 7.5µs, so V2 phase is -180-7.5e-9*14e6*360=-180-38=-218°.

Lets calculate the common mode and differential component of current in each load resistor. We will use Python as it handles complex numbers.

>>> i1=0.0192

>>> i2=0.0148*(math.cos(-218/180*math.pi)+1j*math.sin(-218/180*math.pi))

>>> ic=(i1+i2)/2

>>> abs(2*ic)

0.011825300356025956

>>> id=(i1-i2)/2

>>> abs(id)

0.016089765935912277

>>> abs(2*ic)/abs(id)

0.7349578858466762

>>> 20*math.log(abs(2*ic)/abs(id))/math.log(10)

-2.674750918136747

So, the differential component of current is 16.1mApp, and the total common mode current is 11.8mApp, the total common mode current is more than a two thirds the differential current or 2.7dB less than differential current.

By any standard, this is appalling balance.

The measurements reported here are for a specific scenario (components, frequency and load), and should not be simply extrapolated to other scenarios.

The calculated imbalance if you like applies to the specific test circuit, and cannot really be extended to use of this balun in an antenna system scenario.

The problem starts with that it is near impossible to build such an ATU with perfect symmetry, meaning the distributed inductances and capacitances to ground are symmetric so that with a symmetric load, the entire system was symmetric and there was very low common mode load current.

Achieving that symmetry does not guarantee symmetric currents in an asymmetric load. Fig 4 and the associated text at Balanced ATUs and common mode current deal with this problem.

“Inherent balance” is a belief of the very naive… and snake oil salesmen who would relieve them of their money whilst selling the satisfaction that they have something rather special!

Superlatives like “true balanced tuner”, “fully balanced tuner”, “superb current balance” are bait for naive hams who do not test the claims, hams who do not measure the balance objective… common mode current.

Continued at Inherently balanced ATUs – part 2 .

]]>Above is Ruthroff’s equivalent circuit, Fig 3 from his paper (Ruthroff 1959). Focusing on the left hand circuit which explains the balun as a transmission line transformer (TLT), and taking the node 1 as the reference, the loaded source voltage appears at the bottom end of the combined 4R load, and transformed by the transmission line formed by the two wires of the winding, and inverted, at the top end of the combined 4R load.

It is the transformation on this transmission line that gives rise to loss of symmetry.

The complex ratio Vout/Vin is dependent on the complex reflection coefficient Gamma at both ends of the line and the line propagation constant gamma, all of which are frequency dependent complex quantities.

Vout/Vin=(1+GammaLoad)/(1+GammaSource)*e^(gamma*l)

In baluns, the real part is often close to unity, the phase is more significant.

TWLLC and family tools calculate this quantity in the long output, but I do not recall seeing it calculated in other tools. Values for Gamma may be shown in other tools, but again I have not seen gamma shown directly.

Astute readers will realise that a more correct balun could be made by including another TLT to supply the non-inverted output. By then, it looks like a Guanella 4:1 balun with the output centre tap grounded (so it behaves like a voltage balun) and has better balance… on symmetric loads.

- Duffy, O. 2007. A model of a practical Ruthroff 1:4 balun.
- Ruthroff C. 1959. Some broad band transformers. Proceedings of the IRE.

A low Insertion VSWR high Zcm Guanella 1:1 balun for HF – more detail #2 discussed the imperfection caused by the quite short pigtails, and although small, it is measurable.

Chris, NX0E, related experience with Dr E M T Jones at TCI where they made, among other things, TCI’s HF baluns. These baluns were compensated using capacitors, and we see that very occasionally in ham grade baluns.

The pigtails can be seen as a short transmission line of higher Zo, and although not uniform, it provides a model for understanding their effect.

Above is a Simsmith model that treats the pigtails as short sections of 300Ω line, the lengths adjusted to calibrate the model to the observed impedance at 30MHz.

The excursion on the Smith chart shows that in each case, the section of high Zo line increases X with little effect on R, and could be viewed as an equivalent series inductance.

A possible compensation scheme may be to introduce an equivalent shunt capacitance to restore the L/C ratio of 50Ω line.

The model above uses two fixed capacitors attached to the coax connectors terminals for convenience to approximately compensate the pigtails. The model’s Insertion VSWR is extremely good, 1.004 @ 30MHz.

Note that these capacitors must withstand high voltage, and need to be low loss capacitors (eg silvered mica).

Now we should note that while this compensation is extremely good at 30Mhz, it is not so good at frequencies either side of that, and as frequency increases, Insertion VSWR soon increases more rapidly than the uncompensated balun.

Let’s explore a transmission line alternative.

Above is a model of compensation using short open circuit stubs of RG58A/U, again attached to the coax connectors terminals for convenience. These stub sections withstand high voltage, and have a Q of around 500 at 30MHz. An alternative would be stubs of RG213 for a Q around 1500.

The advantage of coax stubs is that they are low cost, medium to good Q (when very short), can be cut to requirement.

Compensation at a single frequency is appropriate to a single frequency application, but lets rework the last scheme for a broad band application.

Above, sacrificing near perfect performance at 30MHz for a wider band limit, shortening the compensation stubs gives InsertionVSWR<1.07 up to 100MHz, which is a more practical result for a ham broadband balun.

Note that RG213 stubs would be a better choice at 100MHz, but for up to 60MHz RG58A/U should be adequate.

Of course, this model has oversimplified the stubs as it has not captured the effects of the stub pigtails. The stubs (with pigtails) will need to be longer, and it is sensitive to the length and form of the pigtails (all of them), so in practice, one would cut the stubs longer (rather than rely on a model for dimensions), and trim them to length observing the response on a VNA. When finally adjusted, the stubs need to be reworked to prevent flashover at the cut end, and the simplest method is to remake them with the center conductor 5mm longer at the open circuit end, and some heatshrink to cover the end (see the references for an example).

- Duffy, O. May 2012. High frequency compensation of T/R relay. https://owenduffy.net/transmissionline/concept/RelayComp/index.htm.
- Duffy, O. Dec 2012. Another Morse beacon keyer – A/B RF switching. https://owenduffy.net/module/smbk/AB.htm.
- Exploiting your antenna analyser #23

]]>

Insertion VSWR is the VSWR looking into the balun with a matched load (termination) on its output, it is a measure of imperfection of the balun. It ought to be a specification item for low Insertion VSWR baluns, but it rarely given.

What is not mentioned in the above definition is the symmetry or balance of the load.

Above is a Smith chart plot of input Z of the balun with an isolated load of 50+j0Ω. Isolated to mean that there is no direct path from either load terminal to ground, it could be seen as a symmetric load with extremely high common mode impedance. All of the external connections use N type connectors with Zo=50Ω.

Let’s remind ourselves of the internal layout.

Note the pigtails at each coax connector, they are a departure from Zo of the coax and connectors. They can be seen as short sections of transmission line with Zo perhaps 200Ω or more. The effect of these is to transform impedance and so cause the input VSWR to depart from ideal.

The departure is seen on the Smith chart above, almost no effect at 1MHz and Insertion VSWR rises to 1.15 at 29.5MHz.

To improve that, one really needs to use coax connectors that terminate the coax shield circumferential and preserve Zo from coax through the connector. That could have been done, but before deciding that it would be better in practice, let’s look at the effect of load asymmetry.

Above is a Smith chart plot of input Z of the balun with a load of 50+j0Ω, and a strip of aluminium foil (low inductance) bonding the shield of the input connector to the centre conductor of the output connector. This is one extreme of an asymmetric 50+j0Ω load.

If Zo was preserved through the system, the response would be better, but not perfect because this connection makes the common mode impedance Zcm appear across one side of the load. This is the reason why Insertion VSWR is now worse than unity at the lowest frequencies, and would be even worse below 1MHz where Zcm is even less adequate.

Above is a Smith chart plot of input Z of the balun with a load of 50+j0Ω, and a strip of aluminium foil bonding the shield of the input connector to the shield of the output connector. This is the other extreme of an asymmetric 50+j0Ω load.

As a result of the asymmetric internal implementation, this response is almost identical to the first (the isolated load).

The Insertion VSWR of this balun is quite low, not perfect but sufficiently low for almost any practical antenna system implementation.

It is not instrument quality, for that one might make the balun internally symmetric. Chinese no-name 1-1000MHz Return Loss Bridge shows the internal detail of such a balun which has a second inductor to make the internal structure symmetric.

]]>Whether your antenna is a perfect 1:1 or a 10:1, a 50 foot length of coax will have HALF the loss of the exact same coax on the exact same antenna system as measured with the 100 foot piece.

Is it true? Can we learn from it?

Let’s take a worked example of Belden 8259 (RG58A/U) with a load of 5+j0Ω at 146MHz. VSWR is approximately 10.

The loss of 100′ under that mismatch calculated with TLLC is 11.367dB, for 50′ it is 7.484dB, or 66% of that for 100′.

It is not “HALF” as shouted by the expert.

The results should be no surprise to anyone who understands transmission lines.

For a line of this type most of the loss even at 146MHz is conductor loss, and so more power is lost in regions of higher current than adjacent regions of lower current. The line is quite lossy, so although the VSWR at the load is 10, it reduces towards the source due to that loss, and as the VSWR reduces, so the loss per foot becomes lower and more uniform tending towards the matched line loss (MLL). In fact if you look at the loss in the first 50′ of the 100′ case, it is 11.367-7.484=3.883dB which is barely above the MLL for 50′ (3.348dB).

Be wary of what you read online.

The online expert expressed a flawed but very common opinion on transmission line loss under standing waves. You might ask where that comes from if it is so common.

The very expression of line loss in dB/length is the start of it. The manufacturer specified loss figures are more fully known as Matched Line Loss… and “Matched” is a hint to their limitation… they apply to a line that has a matched load, ie there are no standing waves and the ratio of V/I=Zo at all places along the line.

For most good practical transmission lines at HF through VHF and most of UHF, most of the loss is conductor loss, and so more power is lost in regions of higher current than adjacent regions of lower current. Whilst loss along a matched line is uniform, it is not so under standing waves.

]]>The Ruthroff 1:1 voltage balun can be seen as two back to back Ruthroff 4:1 voltage baluns with the redundant winding removed… and that prompts the thinking that the cascade of two baluns back to front might cancel the phase delay.

Let’s measure a popular Ruthroff 1:1 voltage balun.

Above, the RAK BL50-A was a quite popular balun, and probably the balun of choice for half wave dipoles… well until the message about current baluns escaped.

Above is a scope capture of the terminal voltages with a 100Ω centre tapped load, the centre tap is bonded to the coax shield terminal.

The red trace is CH1+CH2.

Lets put those numbers into a calculator as unscaled divisions from the scope.

Above, the calculated phase difference is 10.9° which reconciles well with the estimate from the scope trace delay above.

The ratio 2Ic/Id is 20%, this is pretty awful current balance on a symmetric load.

Now this balun is worse at higher frequencies because the attenuation in the balun transmission line increases, and the phase delay increases almost proportional to frequency.

So, whilst ideal voltage baluns deliver equal but opposite currents into a perfectly symmetric load, practical Ruthroff 1:1 voltage baluns might not be all that good.

Voltage symmetry is poor, predictable, and it is load dependent, and it gives rise to significant common mode current, even in a symmetric antenna system.

If you think your balun is working well… it is probably because you have not measured common mode current.

- Duffy, O. 2007. A model of a practical Ruthroff 1:4 balun.
- Ruthroff C. 1959. Some broad band transformers. Proceedings of the IRE.