Starting with some basic magnetism…

The inductance of an inductor is given by \(L=N\frac{\phi}{I}\).

For a closed magnetic circuit of high permeability such as a ferrite cored toroid, the flux is almost entirely contained in the core and the relationship is \(\mathcal{F}=\phi \mathcal{R}\) where \(\mathcal{F}\) is the magnetomotive force, \(\phi\) is the flux, and \(\mathcal{R}\) is the magnetic reluctance. (Note the similarity to Ohm’s law.)

Rearranging that we have \(\phi=\frac{\mathcal{F}}{\mathcal{R}}\).

Permeance \(\mathcal{P}=\frac1{\mathcal{R}}\) we can rewrite the above as \(\phi=\mathcal{F} \mathcal{P}\). Permeances of parallel magnetic paths add, so if we stack two cores sharing the same winding, the total permeance is the sum of that of each core \(\mathcal{P}_t=\mathcal{P}_1+\mathcal{P}_2+…\).

So, returning to the inductance of the toroidal ferrite cored inductor, we can write that \(L=N \frac{\mathcal{F} \mathcal{P}}{I}\) and since \(\mathcal{F}=N I\), \(L=N \frac{N I \mathcal{P}}{I}\) which simplifies to \(L=N^2 \mathcal{P}\).

Now for a toroid \(\mathcal{P}=\mu\frac{A}{2 \pi r}\) and so \(L=N^2\mu\frac{A}{2 \pi r}\). Since A=f(r), we must integrate A over r. (Note that \(\mu=\mu_0 \mu_r=4e-7 \pi\mu_r\).)

Inductance of a toroidal ferrite cored inductor then is given by \(L=\mu N^2 \int \frac{A}{2 \pi r}dr\) (noting that µ is a complex quantity and frequency dependent). More properly, the ‘inductor’ is a resonator and as you approach its self resonant frequency, inductance alone is not an adequate model… nevertheless consideration of the simpler inductance calculation gives valuable insight.

If we stack two cores of the same physical size side by side, then µ is not uniform across the cross section, so we must capture µ in the integral \(L=N^2 \int \frac{\mu A}{2 \pi r}dr\).

In the simple case where we stack n1 cores of µ1 and n2 cores of µ2, then the expression can be simplified to \(L=(n_1 \mu_1 + n_2 \mu_2) N^2 \int \frac{A}{2 \pi r}dr\) where \(\frac{A}{2 \pi r}\) is the geometry of the consituent cores.

Readers will see that stacking one #61 mix core with one #43 mix core of the same sizes is roughly equivalent to a core of the combined cross section area with µ characteristic an average of the two mixes.

This is not equivalent to series connection of two separate inductors with each core type and same number of turns, the effects around self resonances will differ. Since to some extent, common mode chokes rely upon self resonance (albeit low Q) for their operation, this difference in response is quite relevant. Dissipation capacity is likely to be different.

In the light of that understanding, put your thinking cap on when you see magic properties ascribed to this configuration.

Note, this analysis does not address the behavior near or above the self resonant frequency.

]]>Pictured is a dual UnUn. I made this for experimenting. It’s both a 49 and 64 to 1 UnUn.

The 49 to 1 tap uses the SS eye bolt for the feed through electrical connection and the SS machine screw on the top is the 64 to 1 connection. If I want to use the 49 to 1 ratio, there’s a jumper on the eye bolt that connects to the top machine screw where the antenna wire is attached. The jumper shorts out the last two turns of the UnUn. Disconnect the jumper from the top connection and now you have a 64 to 1 ratio.

The advice to short the section of the winding (the white wire in the pic) is really bad advice.

Tapping an air cored solenoid can be effective and with low loss… can be… but not unconditionally. Tapping a ferrite cored inductor almost always has quite high Insertion Loss, it is akin to shorted turns in an iron cored transformer, … so if you try it, measure it to see if the outcome is acceptable.

Keep in mind that flux leakage degrades broadband performance, so conductors wound loosely around the core (as in the pic) and wide spaced single layer windings (as in the core) tend to have higher flux leakage and poorer broadband performance. Measure what you make to verify that it did what you think.

An S parameter file from a two port sweep over HF would be informative.

I offer this analysis without measurement evidence to prove the case, but sometimes an understanding of basic circuit analysis allows one to avoid wasting time on poor designs.

]]>Above is the model topology. D1 is a daemon block which essentially, calculates key values for the other blocks based on exposed parameters and the named ferrite material complex permeability data file. The prototype used a Fair-rite 2643625002 (#43) core.

D1 code:

//Misc //Updates Tfmr, CoreLoss, and Cse. $data=file[]; // core mu aol; np; ratio; cse; cores; k; $u1=$data.R; $u2=$data.X; //u1=$u1; //u2=$u2; $ns=np*ratio; Ym=(2*Pi*G.MHz*1e6*(4*Pi*1e-7*$u1*aol*cores*1e9)*np^2*1e-9*(1j+$u2/$u1))^-1; Cse.F=cse; CoreLoss.ohms=1/Ym.R; $l1=1/(2*Pi*G.MHz*1e6*-Ym.I); $l2=$l1*(($ns-np)/np)^2; $lm=k*($l1*$l2)^0.5; Tfmr.L1_=$l1+$lm; Tfmr.L2_=$l2+$lm; Tfmr.L3_=-$lm;

L is the load block.

Cse models the self resonance of the transformer at lowish frequencies, Cse is an equivalent shunt capacitance.

Tfmr is a coupled coils model (above) of a (nearly) lossless autotransformer (core loss will come later). Wire loss is usually insignificant in these type of transformers and is ignored. (It seems that Simsmith will not simulate a pure inductance, in this cases the loss of the ‘lossless transformer is of the order of 5e-6dB, so satisfactory.) It is possible to simplify declaration of this component by using Simsmith’s coupled coils in a RUSE block, I have chosen to take full control and solve the mutual inductance effects explicitly.

(The model assumes that k is independent of frequency which is not strictly correct, but for medium to high µ cores, measurement suggests it is a fairly good assumption.)

Coreloss brings the ferrite core loss to book.

Ccomp models a compensation capacitor used to improve broadband InsertionVSWR.

The G block provides the source and plot definitions.

Plots code:

//Plots //check lossless Tfmr behavior //tfmrloss_dB=10*Log10(Tfmr.P/(Tfmr.P-Tfmr.p)); //Plot("TfmrLoss",tfmrloss_dB,"LossdB",y1); coreloss_dB=10*Log10(CoreLoss.P/(CoreLoss.P-CoreLoss.p)); loss_dB=10*Log10(G.P/L.P); mismatchloss_dB=10*Log10(1/G.P); Plot("CoreLoss",coreloss_dB,"LossdB",y1); Plot("Loss",loss_dB,"LossdB",y1);;

The model was calibrated to measurement of the prototype, and the fit is quite good given tolerances on components.

The model allows convenient interactive sensitivity analysis where parameters can be dialed up and down with the mouse wheel and the response changes observed.

- FT82-43 matching transformer for an EFHW
- Find |Z|,R,|X| from VSWR,|Z|,R,Ro
- A new impedance calculator for RF inductors on ferrite cores
- Calculate ferrite cored inductor (from Al)
- Calculate VSWR and Return Loss from Zload (or Yload) and Zo
- Duffy, O. 2015. A method for estimating the impedance of a ferrite cored toroidal inductor at RF. https://owenduffy.net/files/EstimateZFerriteToroidInductor.pdf.
- ———. 2006. A method for estimating the impedance of a ferrite cored toroidal inductor at RF. VK1OD.net (offline).

The design is for a system power output of about 80W on a 24V supply, it is a combination that should work with practical system components with good efficiency.

Above is a first step, an estimate of an initial load line for the PA. The calculator is written in valve terms, but is quite applicable to this scenario.

Let’s talk in terms of MOSFET devices. So, we expect that the drain to drain load needs to be around 12.6Ω, that is close to 50/4=12.5Ω and would suit a 1:2 turns ratio broadband transformer, so we will base the Smith chart on Zo=12.5Ω.

The article assumes the reader has a good understanding of these type of amplifiers.

This article assumes familiarity with Simsmith, including its right to left signal flow, and concepts of complex numbers, impedance, admittance, circuit analysis, transformer equivalent circuits etc.

The model is a simple approximation of the behavior of the PA and makes some assumptions to make a very complex system into a simpler approximation that leads to an understanding of the main influences on system power out.

The model assumes ideal Class B operation of an ideally linear active device with conduction angle 180°. The conduction angle in Class AB is typically so close to 180° to make little difference to the calculated result, insignificant in terms of other uncertainty.

The model assumes an ideal linear transfer characteristic. The error of this assumption is small in estimating power output at the fundamental, but relevant to IMD / harmonic content… which is outside the scope of this model.

That said, the real test of linearity is an Intermodulation Distortion (IMD) test, statistics like 1dB compression power (P1dB) are pretty weak. The analysis presented hear avoids voltage or current saturation, slightly higher power might be obtainable at acceptable IMD figures.

The nature of the active device, whether transistor or FET is that with a given level of drive, it tends to produce a collector or drain current independent of load impedance. Constraints on output power in linear operation are:

- the available voltage swing (approximately supply voltage V+ less saturation voltage (Vsat) of the active device); and
- maximum peak current available.

With adjustable drive available (eg using ALC), we can consider that the output device is a nearly ideal voltage source Vmax=Vdd-Vsat able to supply current up to some Imax.

In the Simsmith model, that has been specified in the Zo sub element as

Vdd; Vsat; Imax; absV(Min(Vmax*2^-0.5,Mag(Zin)*Imax*2^-0.5));

where the first three items are declarations of variables so they appear first in the list below. The important part is that it sets the source G to a RMS voltage that is the lesser of \(\frac{V_{max}}{\sqrt2}\) and \(\frac{|Z|I_{max}}{\sqrt2}\). Don’t be confused by **absV**, it does not set the source absolute voltage or amplitude, but the RMS voltage.

If you have notions of an idealised Thevenin or Norton source, they are not a good fit to this design problem.

The Plt property of G specifies some calcs and plots.

//Plots Vmax=2*(Vdd-Vsat); Idc=2*Mag(I)*2^0.5*2/Pi; efficiency=G.P/(Vdd*Idc)*100; Plot("CoreLoss (Wc)",Core.R1.p,"Wc",y1); Plot("Dissipation (W)",Vdd*Idc-G.P,"PWR"); Plot("Efficiency %",efficiency,"Eff",y2);

Vmax is calculated from Vdd and Vsat, and Idc is calculated from G.I using the factor for ideal push-pull Class B operation. Two plots are specified, one of core loss and one of dissipation of the active devices being the difference between DC input power and RF output power. Active device efficiency is also calculated in %.

Above is an example set of plots for a contrived scenario to show departures from ideal PA behavior. There is a distinct roll-off below about 3MHz and a sharp change around 1MHz, and a broad efficiency roll-off above about 10MHz another sharp change in power out around 26MHz. We will discuss the cause of these later.

C2 models the shunt capacitance at the drain or collector of the active devices. This is an average approximation of what is actually a capacitance that varies instantaneously with voltage.

The Tfmr element models a transformer with a lossless ferrite core, it assumes very high flux coupling factor (which you can tweak). It is used with the Core element which has precedence setting some of the values.

(The model assumes that k is independent of frequency which is not strictly correct, but for medium to high µ cores, measurement suggests it is a fairly good assumption.)

The core element models the ferrite transformer core, and feeds the Tfmr with key values.

Above is the schematic of the Core element.

The core element models the ferrite core referred to the secondary side as a shunt admittance Ym. The susceptance and turns ratio are used to calculate the primary and secondary inductances of the Tfmr element, and the conductance creates a shunt element to approximate the core loss of the transformer referred to the secondary (LHS). It precedes Tfmr so that it sets Tfmr values prior to execution of Tfmr.

Below is the content of the Core element code.

//Core model //Calculates magnetising admittance from mu data. //Updates Tfmr Hl, Hr. $data=file[]; // core mu aol; Ns; cores; Cse; $u1=$data.R; $u2=$data.X; //u1=$u1; //u2=$u2; Ym=(2*Pi*G.MHz*1e6*(4*Pi*1e-7*$u1*aol*cores*1e9)*Ns^2*1e-9*(1j+$u2/$u1))^-1; Tfmr.Hl=1/(2*Pi*G.MHz*1e6*-Core.Ym.I); Tfmr.Hr=Tfmr.Hl*(1/Core.Ns)^2;

The input values are:

- file: a csv file of complex permeability vs frequency;
- aol: core geometry as \(\int \frac{area}{length}\);
- Ns: number of turns on transformer secondary (primary is 1t);
- cores: number of cores in cascade;
- Cse: equivalent shunt capacitance.

The C1 element models the Tfmr self resonance as an equivalent shunt capacitance.

The L element is the nominal load, for most cases this will be 50+j0Ω. You can change it to observe the sensitivity of the design to variations in load impedance.

The example scenario is contrived to exhibit some departures from ideal for the purpose of discussion.

Above is a Smith chart of the impedance seen by source G normalised to Zo=12.5Ω. At O, freq=5MHz, it is pretty close to the target 12.5Ω. Power out is 84.5W (remember this should be followed by a LPF), efficiency is 75%, close to the theoretical efficiency of an ideal Class B stage.

Core loss is around 15W which is quite high. The design needs lower core loss.

Efficiency rolls off below 3MHz due to increasing core loss, magnetising admittance is too high. The design needs lower core loss, first point would be to address the core selection, cross section, number of cascaded cores, and material characteristics. Change the **cores** value and observe the result.

The sharp change in power output near 1MHz is due to current saturation, the active device cannot produce the current required to drive the load with the growing magnetising admittance. Increasing MOSFET size and drive can solve this, but that does not solve the attendant efficiency problem.

Efficiency roll-off above 10MHz id due mainly to the increasing effect of C2, but flux leakage in Tfmr contributes also. Fix this by choosing MOSFETs with lower capacitance, and making sure the transformer has low flux leakage (high permeability, structure).

The sharp change in power out near 26MHz is again due to current saturation.

Core loss varies with the magnetising current and depends on core characteristics including the frequency dependent complex permeability. Select material and geometry to improve core loss.

Efficiency is degraded whenever the MOSFETs are not operating at fully voltage swing, so presentation of low load impedance might allow acceptable output power at acceptable IMD, but at degraded efficiency. It is worth measuring efficiency as an indicator of correct setup of the stage.

If you design for rated power into a load impedance with some tolerance, eg 60+j0Ω, that will result in reduced efficiency with a 55+j0Ω load. That is just part of the cost of a tolerant design so that half the units coming off the production line don’t fail to make rated power. In the same vein, the MOSFETs need to be able to drive a 45+j0Ω load. The Simsmith model allows exploring these variations.

Above is an overview of the model applied to a commonly eBay listed nominal 70W Chinese PA. Yes, it falls short of the advertised 70W… but the fine print says that is on 16V… so try it.

But overall, the expected performance is quite good from 1.5-30MHz, and core loss is very low. (Note this core loss is based on VK1EA’s measurement of a single core, there is no assurance the cores used in products sold are of the same type.)

The model will often predict lower power than some products claim. It is likely that they depend on driving the PA into saturation to achieve their stated power, causing degraded IMD performance (rarely stated).

The model provides a framework for exploring the second stage of a design process. At the end of the day, testing some prototypes is necessary to complete a design.

RfBPaSs.zip contains the two Simsmith v18.1g model files and their dependent csv files. The component files are available on github where updates will be posted. They are offered for study without any warranty.

]]>Others opine that saturation is a practical design limit, and for example that Bs/2 is a safe / appropriate design target.

Let us consider a ferrite cored inductor at 7MHz. The inductor comprises 11t on a 11t on Fair-rite 5943003801 (FT240-43) toroid. This is a medium to high permeability ferrite material, and for that reason, has significant loss at HF. Higher and lower permeability materials are fashionable at different times, the higher permeability #31 mix is fashionable at this time.

I will work in MKS units.

Above is Fair-rite’s B-H curves for #43 material. Let’s take saturation flux density Bs to be 1500gauss or 0.15T.

Most inductance calculators I see are based on Al, here is popular one, toroids.info.

It calculates an inductance of 130µH and impedance of 0+j5710Ω, both are nonsense. You might be forgiven to thinking the calculated inductance is at 7MHz, it is actually for 10kHz.

We can calculate the expected impedance of the inductor with a better calculator.

The inductor has an impedance of 5490+j1320Ω. Also calculated is the equivalent inductance of the series reactance, 30.1µH.

This size core can safely dissipate around 10W continuous, lets use that figure.

The effective area of the core from the datasheet is 1.58e-4m^2.

The current to cause 10W of dissipation is given by \(I=\sqrt\frac{P}{R}=\sqrt\frac{10}{5490}=42.7 \; mA\).

Recalling that \(L=n\frac{\phi}{i}\) we can calculate peak flux as \(\phi=\frac{L \sqrt{2} I}{n}=\frac{3e-5 \cdot \sqrt2 \cdot 0.0427}{11}=1.65e-7 \; Wb\) and \(B=\frac{\phi}{A}=\frac{1.65e-7}{1.58 e-4} \frac{3e-5 \cdot 0.0427}{11}=1.04 \; mT\)

So, the core is dissipating 10W at 0.7% of Bs… it is nowhere near saturation.

Unsurprisingly, measurements of real devices reconciles will with small signal based models.

To illustrate how absurd the Bs/2 target is for this scenario, if the material was sufficiently linear to Bs/2, the impressed voltage would be 17kV and it would dissipate 52kW. Of course the core would quickly reach Curie temperature at which point losses become dramatically lower.

None of this is to say that saturation isn’t relevant to design, but it will turn out that for many if not most HF RF applications, dissipation is a more constraining factor than saturation.

Let’s use VK1SV’s flux density calculator (B), one of many.

A common mistake is to apply a formula that the user doesn’t really understand.

Above is a calculation that overestimates B because it is for a lossless inductor.

When the inductor’s loss resistance if factored into the calculation, the voltage across the pure inductance is substantially lower, and B is lower at 10.4 gauss or 1.04 mT which reconciles with the earlier calculation.

This worked example shows that naive / mindless use of online calculators can give significantly wrong results.

Note that if there is a DC component to current through an inductor, that must be considered in flux and saturation calculations.

]]>This article offers an explanation of how the the alternative output circuit at Fig 5 of EB104 works.

Let’s look at the schematic diagram of the PA.

Above is the schematic from EB104, of interest for this article is the output circuit comprising T2 and T3 which are intended ideally to provide a drain to drain load of 50/9=5.55Ω.

Let’s run a simple initial load line model of the amplifier (albeit in tube terminology).

Above, we find that if we assume a saturation voltage for the MOSFETs of 0V, a 40.823V supply delivers 600w into exactly 5.555Ω load for ideal Class B. As noted, 5.555Ω is 50/9, so an ideal transformer with 1:3 turns ratio should provide that.

The drain current in an ideal Class B stage is a half wave of sine wave. If we were to perform a Fourier analysis of such a wave of say 1A peak amplitude, we would find that the fundamental RF component is 0.5A. (There are higher order harmonics, but they are not of interest to us in our frequency domain analysis at the fundamental frequency.)

In this case, the calculator results tell us that the drain current amplitude is 29.40A and the amplitude of the fundamental RF current is 14.70A or 10.39Arms. Both FETs are driving such a current but of opposite phase.

Above is a diagram of the ideal Class B current wave shape, fundamental RF component and DC component (albeit in tube terms).

Let’s deal with T2, the magnetising impedance is sufficient that we will ignore the current flowing in its windings for an approximate solution.

Now to the alternative arrangment shown in Fig 5.

Granberg offers a simplification of the alternative to T3 as a cascade of a transformer and 1:1 Guanella balun.

Performing a frequency domain analysis of that simplified circuit, we have two current sources supplying a sinusoidal current to the coupling network.

Let’s annotate the diagram with currents relative to the current into the 50Ω load.

We can simplify the analysis by making some assumptions and depending on coax with well developed skin effect in TEM mode:

- the purpose of the ferrite sleeves is to increase the common mode impedance of the coax section and make its common mode current (the current that flows on the outer surface of the shield) insignificant, we will assume zero;
- in the interior of the coax, at any point along the coax, the current on the inner surface of the outer conductor is exactly equal to that on the outer surface of the inner conductor but of the opposite phase (ie flowing in the opposite direction); and
- the coax sections are so short the the current phase is uniform along the section.

Above is from Fig 5. Starting at the right we can annotate current into the upper load terminal (the open arrow) and then observing the rules above, we can annotate currents through the whole network, arriving at the conclusion that the current flowing from the current sources representing the MOSFETs is 3I or three times the 50Ω load current, and therefore the impedance transformation is 1:3^2 from source to load, and the drain to drain load impedance is \(Z_{dd}=\frac{50}{3^2}=5.555 Ω\).

]]>The PA uses a MRF9180 dual MOSFET operating on 26V supply.

Above is the prototype PA. The text states very clearly that the output transformer uses a secondary of two turns of PTFE insulated wire, and the pic above does not provide evidence to the contrary.

Hmmm, experience suggests that may be too few turns.

Here is a little table that shows the maximum power obtainable with a transformer of this type for various supply voltages and secondary turns.

You would usually design for rated power to be a little less than shown in the power cell for a particular combination. For example, if you wanted 150W out on 26V supply, you might choose a three turn secondary (which is capable of somewhat higher power). The danger in undersizing (low Nsec) is risk of saturation and resulting non-linearity / intermodulation distortion.

Let’s look at an initial load line prediction for M0DGQ’s specification.

Above is a simple prediction of the initial load line of an amplifier of this type with 26V supply and 150W output. Though the calculator is written in terms of valves, the technique is valid in this case, and Rlaa, the anode to anode load resistance is equivalent to the drain to drain load resistance, and we might expect it in the region of 8Ω.

A back of the envelope calculation is that if we have a 26V supply, and we allow for FET saturation voltage drop and other sagging of the supply, we might expect a peak drain to drain RF voltage of 50V. If we want 150W, then \(R_l=\frac{V_p^2}{2P_o}=8.33Ω\). Now if you have a 1t primary and 2t secondary, the secondary load \(R=8.33 n^2=33.33Ω\), quite a long way from 50Ω and the implication is that the PA seems unlikely to develop 150W into 50Ω if the transformer is close to ideal (as needs to be for a broadband PA).

Above is a Simsmith model for the two turn secondary, and it includes a model for the ferrite core, and some allowance for flux leakage in the transformer… so a not quite ideal transformer but probably a good estimate of a good transformer.

(The model assumes that k is independent of frequency which is not strictly correct, but for medium to high µ cores, measurement suggests it is a fairly good assumption.)

Power output falls way short of 150W. Note that this is analysis of an ideal Class B amplifier, though with the stated limits on voltage swing (including an allowance for transistor saturation), and it should not be interpreted to mean that higher power could not be obtained by driving it into non-linearity.

The problem is that we have delivered a higher load impedance to the drain that needed. The theoretical impedance transformation needed is 50/8.333=6 which implies a turns ratio of 2.45:1… so that would require a 2.45t secondary which is not practical (for balance reasons). We can use 2t which results in lower power, or 3t which will result in higher power.

Let’s examine a three turn option for a drain to drain load around 5.5Ω.

Above is the initial load line prediction.

Above is the Simsmith model adjusted for the three turn secondary.

Reducing supply voltage brings total power down if dissipation is a problem.

]]>One of the first questions to mind is whether it is likely to deliver the rated power, so let’s review the MOSFET output circuit design from that perspective.

Sellers mostly seem to need to obscure the MOSFET type in their pics, so essentially you buy this with no assurance as to what is supplied, no comeback if the supplied MOSFET is not up to the task. Online experts suggest the MOSFET is probably a MRF9120 (or 2x IRF640 in a 70W build). The amplifier claims 100W from 12-16V DC supply.

Note that this module does not include the necessary output filter which will lose 5-10% of the power from this module.

In this case Carlos, VK1EA, connected a sample output transformer (T2) core from a recently purchased MiniPa100 kit to a EU1KY antenna analyser. The fixture is critically important, it is at my specification.

We also need to know the geometry parameter ΣA/l.

Above, from the measured dimensions of a sample core, ΣA/l=0.003415/m.

The saved S parameter file was processed as described at nanoVNA-H – measure ferrite core permeability described a method for characterising an unknown ferrite material and a complex permeability curve produced.

Above, the results are fairly good and fairly much as expected, but let’s remove the noise by digitising the plots.

Above, the points sampled for the digitised output. Though there is a lot less data in the result, when points are obtained by interpolation, noise is greatly reduced.

The above pic from an eBay advertisement of the 2020 version of the PA would suggest very strongly that there are three turns on the secondary of the output transformer, and a half turn on each drain. Interestingly the 70W versions also appear to use three turns, alarm bells ring!

From all this, we can produce an approximation in Simsmith that captures most of the expected behavior of a practical transformer, including core loss.

Above is the RUSE block schematic used for Core which models the frequency dependent magnetising admittance of the transformer and sets the frequency dependent inductances of the Tfmr element.

More to come…

]]>Essentially, my analysis was that it comprises two 12t winds of two wire transmission line in parallel on the ferrite ring. The potential benefit was that the characteristic impedance Zo of each transmission line is probably close to 100Ω, and the parallel combination is probably close to 50Ω.

Online experts following fashion are opining that a low Insertion VSWR balun is better made with two wire line(s) than winding a single 50Ω coax line. They make these claims without evidence, I am not convinced.

In that vein, here is a variation on the TrxBench balun above.

The designer describes it:

Wound with 18 AWG PTFE, Solid Silver Plated Copper wire. By using that specific gauge wire with PTFE insulation tightly coupled in pairs results in a 100-Ohm transmission line. Two in parallel = 50-Ohms. The advantage using the wire over coax is it flattens and widens the bandwidth. I stress it is extremely important to pay attention to the small details. The spacing, twisting, orientation, neatness, and symmetry are extremely important.

He is quite correct regarding his last point.

His claim that the each pair of wires wound on the core has Zo of around 100Ω without supporting evidence, but it is believable based on my own experience of making and measuring similar line sections

The ‘tails’ are not just two continuations of the 100Ω transmission lines (as wrapped on the core) paralleled at the connector and load resistor, they are two line sections of some other geometry and again without evidence of their Zo. I cannot call upon experience to inform me about likely Zo, I suspect it may be significantly higher than 100Ω.

Whilst the designer explained that the picture was of a test setup for measurement with a VNA, he did not give the results of such a measurement, an InsertionVSWR plot would be informative.

I wrote a series of articles that showed how a very small length of pigtails impacted InsertionVSWR of a balun:

- A low Insertion VSWR high Zcm Guanella 1:1 balun for HF;
- A low Insertion VSWR high Zcm Guanella 1:1 balun for HF – more detail #2; and
- A low Insertion VSWR high Zcm Guanella 1:1 balun for HF – more detail #3.

So, we return to claims along the lines of If I recall correctly, that was also K9YC’s conclusion. His updated ‘cookbook’ moved away from coax to wire

which of course is fallacious if it attributes someone’s personal opinion to a well known author.

Where is the evidence of the InsertionVSWR of this ‘superior’ design? It is great to see thinking and experimentation, but a bit of scientific method would make it so much more valuable… if the claims are supported.

]]>A correspondent suggested that with a ferrite core, flux leakage is insignificant. This article calculates the coupled coils scenario.

Above is the ‘schematic’ of the balun. Note the entire path from rig to dipole.

Let’s use the impedance measurement with short circuit termination to find the inductance of the two coupled windings in series opposed.

Above is a plot of the impedance, R+jX. X at 1MHz implies L=8.6µH. Remember that this is the inductance of two series opposed coils, so it includes the effect of mutual inductance.

We can estimate reasonably by calculation that the inductance of one coil L1 @ 1MHz is 114µH.

Measurement of a SC termination gave \(L=(L1-M)+(L2-M)=8.6µH \) and since L1=L2 we can calculate \(M=114e-6-\frac{8.6e-6}{2}=109.7\;µH\) and from that the flux coupling factor \(k=\frac{M}{\sqrt {L1L2}}=\frac{109.7}{114}=0.9623\).

So, k is very high, there is very little flux leakage, but not enough to ignore… it has a huge bearing on the outcome.

]]>