Some of us use a resistor as a load for testing a transmitter or other RF source. In this application they are often rated for quite high power and commonly called a dummy load. In that role, they usually do not need to be of highly accurate impedance, and commercial dummy loads will often be specified to have maximum VSWR in the range 1.1 to 1.5 (Return Loss (RL) from 26 to 14dB) over a specified frequency range.

We also use a known value resistor for measurement purposes, and often relatively low power rating but higher impedance accuracy. They are commonly caused terminations, and will often be specified to have maximum VSWR in the range 1.01 to 1.1 (RL from 46 to 26dB) over a specified frequency range.

It is more logical to discuss this subject in terms of Return Loss rather than VSWR.

Return Loss is defined as the ratio of incident to reflected power at a reference plane of a network. It is expressed in dB as 20*log(Vfwd/Vref).

Calibration of directional couplers often uses a termination of known value, and the accuracy of the termination naturally rolls into the accuracy of the calibration and the measurement results.

A simple example is that of a Return Loss Bridge (RLB) where a known reference termination is compared to an open circuit and then an unknown load to find the Return Loss (being the difference between them).

Let use look at three examples of RF load resistors at hand and consider their performance as a calibration reference. The discussion uses datasheet VSWR or RL figures which are the best one can rely upon unless high accuracy measurements of made of the device.

The MFJ-264N is a high power ‘dummy load’ with max VSWR specified as 1.3 to 650MHz, which is equivalent to RL>17.6dB. In a very good RLB, the directivity will approach the reference termination’s RL, so we can regard the RLB directivity in this case to be 18dB in round numbers.

We can calculate the uncertainty in measuring a given VSWR given the minimum directivity of the RLB.

Let’s say we wanted to measure VSWR down to 1.5, and we wish to know the uncertainty (error bounds).

Above is a calculation of the scenario. It can be seen that with a true VSWR=1.5 load, the RLB may indicate anywhere between VSWR 1.16 and 1.97.

The MFJ-264N is a high power ‘dummy load’ with max VSWR specified as 1.1 from 30 to 500MHz, which is equivalent to RL>26.4dB. In a very good RLB, the directivity will approach the reference termination’s RL, so we can regard the RLB directivity in this case to be 18dB in round numbers.

We can calculate the uncertainty in measuring a given VSWR given the minimum directivity of the RLB.

Let’s say we wanted to measure VSWR down to 1.5, and we wish to know the uncertainty (error bounds).

Above is a calculation of the scenario. It can be seen that with a true VSWR=1.5 load, the RLB may indicate anywhere between VSWR 1.35 and 1.67.

Definitely better than the MFJ-264N.

The KARN-50-18+ is a low power ‘termination’ with RL specified on the chart above. In a very good RLB, the directivity will approach the reference termination’s RL, so we can regard the RLB directivity in this case to be >46dB in round numbers up to 1000MHz.

We can calculate the uncertainty in measuring a given VSWR given the minimum directivity of the RLB.

Let’s say we wanted to measure VSWR down to 1.5, and we wish to know the uncertainty (error bounds).

Above is a calculation of the scenario. It can be seen that with a true VSWR=1.5 load, the RLB may indicate anywhere between VSWR 1.48 and 1.52.

Much better than either of the previous examples, but it is only rated for 2W so it unsuitable as a load for a high power device.

High power RF resistors tend to have poor RL, yet a high RL high power resistor is needed for checking or calibrating high power directional wattmeters.

A possible solution is to use a good RLB with good reference termination to ‘calibrate’ a high power load via an ATU, and use the latter for high power measurements. This typically is a single frequency technique, and there is unavoidable uncertainty introduce in this calibration process.

Another technique is to use an ATU + high power load on the directional coupler, adjusting the ATU for null reflection indication. Then move the cable from the directional coupler to a VNA or analyser and measure the impedance seen by the DUT. Again, being an indirect method, uncertainty flows from cascading measurements.

Resistor loads of lower RL lead to high uncertainty of measurements using them as a reference (directly or indirectly).

The uncertainty is worse as measured RL of the unknown approaches the RL of the reference used.

Depending on the accuracy needed of measurements, RL of the reference typically needs to be 10dB or more better than the intended measurement.

Watch the blog for continuing postings in the series Exploiting your antenna analyser. See also Exploiting your antenna analyser – contents.

]]>Above is a diagram of a Pawsey Balun used with a half wave dipole (ARRL).

Pawsey Balun on an asymmetric load reported model results in an asymetric dipole antenna, and showed very high common mode feed line current.

Pawsey Balun on an asymmetric load – bench load simulation showed that although the Pawsey balun is not of itself an effective voltage balun or current balun, it can be augmented to be one or the other.

So, you might ask what they do, what they are good for, and why they are used.

If you were to construct a quite symmetric half wave dipole and directly connected a coax transmission line to the centre, you would destroy the symmetry of the system as connection of the shield to one dipole leg only effectively connects the common mode conductor (the outer surface of the shield) to one leg of the dipole.

The Pawsey stub or balun is a narrowband device (ie tuned) that adapts the coaxial feed line to a pair of symmetric terminals for attachment to the antenna feed point.

In a perfectly symmetric system (source, feedline type and topology, antenna), current in the radiator will be symmetric and there will be negligible common mode current on the feed line.

Symmetry is easier to achieve with some types of VHF/UHF/SHF antennas than at HF.

Equivalent circuit of an antenna system gives measurements of a fairly symmetric G5RV Inverted V dipole + feed line and in that case, the Z1 and Z2 values are different on the two bands reported, more so on 80m.

On the other hand, a corner reflector with half wave dipole feed for 1296MHz can be constructed with very good symmetry, and fed from behind the reflector, a Pawsey balun should give the necessary feed symmetry to preserve system symmetry and have symmetric dipole currents and negligible common mode current on the feed assembly.

The question of why are they used is more difficult than the other questions. They do have application, but they are also used inappropriately and given that it is most unusual to seem validation of balun performance by measurement, such use highlights the bliss of ignorance.

]]>Pawsey Balun on an asymmetric load reported model results in an asymetric dipole antenna, and showed very high common mode feed line current.

This article looks at two test bench configurations modelled in NEC.

The configurations are of a horizontal Pawsey balun for 7MHz constructed 0.1m over a perfect ground plane. The ‘balanced’ terminals are attached to the ground plan by two short 0.1m vertical conductors which are loaded with 33 and 66Ω resistances. At the other end, the horizontal transmission line is extended by two different lengths and connected to the ground plane using a 0.1m vertical conductor. The two extension lengths are almost zero and a quarter wavelength.

The total horizontal length from the ‘balanced terminals’ to the grounded end of the transmission line is a quarter wavelength for the Pawsey balun and a further 20mm making approximately a quarter wavelength in total.

Above is a plot of current magnitude and phase from 4NEC2. The current on the two vertical conductors containing the 33 and 66Ω loads is quite different, and the product gives load voltages that are approximately equal in magnitude and opposite in phase.

You could be forgiven for thinking that the Pawsey stub itself is a good voltage balun, but in fact the voltage balun behaviour is due to the fact that the transmission line and Pawsey stub conductors in common mode are approximately a half wave electrical length, and being grounded at the far end, the common mode impedance looking into the ‘balanced’ terminals is very low… a hallmark of a good voltage balun.

The total horizontal length from the ‘balanced terminals’ to the grounded end of the transmission line is a quarter wavelength for the Pawsey balun and a further quarter wave making a half wavelength in total.

Above is a plot of current magnitude and phase from 4NEC2. The current on the two vertical conductors containing the 33 and 66Ω loads is approximately equal in magnitude and opposite in phase.

You could be forgiven for thinking that the Pawsey stub itself is a good current balun, but in fact the current balun behaviour is due to the fact that the transmission line and Pawsey stub conductors in common mode are approximately a quarter wave electrical length, and being grounded at the far end, the common mode impedance looking into the ‘balanced’ terminals is very high… a hallmark of a good current balun.

At intermediate lengths, the common mode impedance will range from one extreme to the other, and for the most part, it will be neither a good current balun nor a good voltage balun.

The Pawsey stub or balun is not of itself a good current balun or a good voltage balun, but can be used as part of a more complete solution to act as either a good current balun or a good voltage balun.

Creating that context may be impractical for many antenna topologies.

Without careful implementation of the context, the Pawsey balun or stub is anyone’s guess. Nevertheless they are written up this way in textbooks and find practical application, even though their performance is likely to be unpredictable and unmeasured.

]]>Above is a diagram of a Pawsey Balun used with a half wave dipole (ARRL).

Whilst these have been quite popular with VHF/UHF antennas, the question arises as to how they work, and whether they are effective in reducing common mode current IIcm) for a wide range of load scenarios.

To find an answer, and NEC model was constructed of an OCF half wave dipole a half wave above a perfect earth conductor, vertical feed line to ground, and the current magnitude and phase for all conductors evaluated.

Above is the current plot. Note that horizontal conductors are defined from low X to high X, and vertical conductors from low Z to high Z.

It can be seen that there is relatively high Icm at ground level, and when the currents in both Pawsey Balun conductors are added near the feed point, again relatively high Icm. Also of interest is that the currents in the dipole legs are flowing in opposite directions at the feed point.

This antenna acts more like a top loaded vertical monopole than a horizontal dipole.

Above, the radiation pattern reinforces the view that it is behaving as a top loaded vertical monopole.

In this scenario, the Pawsey Balun has not effectively suppressed common mode current (as would a good current balun), indeed it seems to facilitate it (as a good voltage balun would).

The Pawsey Balun as shown in the diagram is not of itself either a good current balun or a good voltage balun.

]]>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.