The article Baluns in antenna systems explores some different dipole and feed line configurations and the effectiveness of common mode chokes at various locations on the feed line.

Models 1, 2 and 3 particularly show the effect of a quarter wave vertical common mode conductor grounded and isolated, and a half wave vertical common mode conductor grounded.

These illustrate that those common mode conductors can be viewed to some extent as a ‘single wire’ transmission line, and the impedance presented at the dipole feed point is low or high in keeping with simple transmission line analysis of a shorted or open line of quarter or half wave length.

The question then arises with the Radcom “cable balun”, does it behave similarly, to what extent does the folding of the conductor affect its quarter wave resonance.

One way to explore this is to construct an NEC model of the structure and a reflection of itself.

Above is the serpentine structure of three quarter wavelength folded, and below it, a reflection of itself. The whole structure is fed in the middle and the impedance vs frequency charted.

Above, the impedance vs freq, both in cartesian and polar form are plotted and show a clear self resonance around 7.15MHz.

For comparison, a plain dipole of the same total wire length is modelled.

Above, the plain dipole of same total wire length.

Above, the impedance vs freq, both in cartesian and polar form are plotted and show a clear self resonance around 7.15MHz.

There is a little difference in the resistance plots as might be expected because it is mainly radiation resistance which is affected by the folding of the conductor. The reactance plots are very similar, and importantly, the resonant frequencies are almost equal.

To all intents and purposes, they are both radiating transmission lines of 3λ/4.

So there should be little surprise when the 3λ/4 “cable balun” is extended by a further λ/4 to make a total of λ and connected to ground, that a low impedance to common mode current is presented at the feed point.

Above is the current distribution diagram from the model, the dark green curve is the magnitude of current. Note the common mode current on the vertical feed line, its magnitude relative to that on the dipole elements is relevant.

It does exactly what is predicted from a very simple analysis. If you follow the thinking, you can see that it is quite naive to think the “cable balun” alone effectively reduces common mode current in general.

]]>Above is the current distribution on the half square voltage fed. It is essentially two in-phase vertical quarter waves separated a half wavelength, a broadside array.

Feed point impedance at resonance is very high 5700Ω, and being a high Q antenna, they are very sensitive to dimensions, nearby clutter etc. Note that this is calculated for an antenna in the clear, it will be different where trees or conductive mast exist nearby.

Above is a diagram of one of several feed arrangements from the ARRL.

The classic ham design would be to transform 50 to 5700 we need a turns ratio of (50/5700)^0.5=0.093, so you would design an inductor to resonate with an available capacitor and tap it at 9.3%. Of course this method cannot deal with a reactive load and pretends the transformer does not contribute significant reactance.

Well, that back of the envelope design will not work exactly for an air cored solenoid at RF because of the practical flux leakag (though sometimes it might be close), designing coupled coils (for that is what we have) is more complicated than that.

Solution 1 uses a tapped inductor and variable capacitor as in the schematic. The inductor is an air solenoid of 28t of diameter 52mm and pitch 5mm, tapped at 3.5t. These calcs assume a low loss capacitor because of the very high load impedance. The capacitor needs to be at least 200pF, and more importantly, at 1500W it needs to withstand 5kVpk… a vacuum variable is the obvious choice and has advantages for outdoor deployment.

But vacuum variables are quite expensive!

Solution 2 uses a mid Q (200) tapped inductor and fixed capacitor, and is tuned / matched by adjusting a tap on the top of the coil and one nearer the bottom of the coil. The inductor is an air solenoid of 30t of diameter 52mm and pitch 5mm, antenna tapped at about 22.3t and feed tapped at 2.8t. These calcs assume a low loss capacitor because of the very high load impedance. The capacitor used is 200pF, and more importantly, at 1500W it needs to withstand 5kVpk.. a vacuum fixed capacitor is the obvious choice and has advantages for outdoor deployment.

These can still be bought on eBay out of Russia for less than $40 inc shipping.

Above is a Smith chart model of match (from SimSmith). The three L elements are inextricably linked and cannot be individually varied as they form an equivalent circuit for the autotransformer / inductor. The inductor Q values come from distribution of the total coil resistance over the parts, and the mutual inductance components are lossless. The -1GQ is to force a near lossless inductance, and a quirk of SimSmith’s unconventional meaning of Q is that it needs to be negative in this case.

Above is the Smith chart plot of the match.

Lots of ham articles show matches of this type without the shunt capacitance (the red element on the Smith chart) but the capacitance is essential to its operation. In the case of antennas (particularly VHF) that seem to work without it, it is inherent in the mount for the radiator.

Increasing Q of the tapped inductor reduces losses. Winding the inductor with 2mm copper should raise Q for the whole inductor to around 500.

The inductor is an air solenoid of 30t of diameter 52mm and pitch 5mm, antenna tapped at about 22.3t and feed tapped at 2.8t. These calcs assume a low loss capacitor because of the very high load impedance. The capacitor used is 200pF, and more importantly, at 1500W it needs to withstand 5kVpk.. a vacuum fixed capacitor is the obvious choice and has advantages for outdoor deployment.

The model above uses the higher Q inductor and the losses are reduced by about 0.3dB, about 7% improvement in radiation efficiency. Note that the turns are space to reduce proximity effect driving higher coil loss.

For a practical deployment, I will tune this initially with an analyser using a silvered mica capacitor to confirm the design, and then based on that experience, buy a vacuum capacitor that will be compatible.

Fixed capacitance choices range from transmission line sections to doorknob capacitors. both are much lower Q than a vacuum capacitor, you need to do calcs and assess whether the loss is acceptable.

Variable capacitors can be mid range Q, but long term weather resistance may be an issue.

]]>(1) Back in post #30 I showed that with a halfwave wire fed close to its end works just like the same wire fed in the center; the only difference being the feed point impedance. I let EzNec figure this out; I didn’t have to explain it with any mysterious “displacement” currents. Shown as (1) in the attached.

Since, in the model, the source is a constant current source, that forces the current on either side of the source to be equal, and the radiation pattern predicted by EzNec reflects that, because the patterns for the end-fed and center-fed match… (go back and look at post #30)

His post #30 is of a 67′ dipole at 66′ above poor ground @ 7.18MHz, fed at one end.

Above is the current distribution of my approximate re-creation of his model in NEC-4.2. It reconciles with his published graphs.

I should note that the current graph is an interpolation from stepwise segment currents, and as such it is potentially misleading. In this case the current in the driven segment (#2) is 1A and the magnitude of current in the end segment is 0.375A.

The issue is his reasoning:

Since, in the model, the source is a constant current source, that forces the current on either side of the source to be equal, and the radiation pattern predicted by EzNec reflects that, because the patterns for the end-fed and center-fed match… (go back and look at post #30)

In this scenario, it does not matter whether a current or voltage source is used.

Above is the same scenario but using a voltage source. The current distribution is identical, the type of source is not the reason for the distribution, the current distribution is determined by the model geometry.

WA7ARK later gives a diagram where he develops the idea of some feed configurations.

Above, (2) depicts an ideal transformer coupling a coaxial feed line to the dipole and shows Icm=0. In fact, the presence of the common mode conductor (the outer surface of the coax shield) in the E and H fields caused by dipole current makes Icm likely… simple linear circuit analysis of the ideal transformer ignores that coupling.

Lets just add the common mode conductor to the NEC model. In this case, a 6mm wire is configured 0.1m from the dipole end, from 2m below ground to 0.02m below the dipole wire. (Note: it is actually 2 GW elements and NEC-4 models buried wires better than NEC-2.)

Above, it should be no surprise that there is current flowing on the common mode conductor, and in this model it is not by virtue of transformer imperfections (the capacitances shown in WA7ARK’s (3)), it is because the dipole and common mode conductors are inextricably E and H field coupled.

Additionally, practical transformers incorporate coupling elements as shown in WA7ARK’s (3), and autotransformers as shown in (4) are often described and have another coupling element.

The scenario of (4) is rather simple to model.

Above, the current on the feed line shield is higher than for scenario (2) shown above. That is not surprising, and it is the reason why that type of feed configuration is used.

WA7ARK’s diagram shows Icm=I2, and it is often argued that since I2 is very small (it is at the end of the dipole, and as everyone ‘knows’ that is almost zero), Icm is insignificant.

Firstly, Icm=I2 is true only at a point. As can be seen, both the dipole current and the feed line common mode current are standing waves. Secondly, the connection of the feed line common mode path modifies the current distribution and the current at the dipole end is small, but not insignificant. In this particular scenario, Icm near ground is significant (and greater than I2).

Most explanations of OCF dipoles ignore the existence of the feed line common mode current path and these various couplings.

This is not to represent that these effects are only present with OCF dipoles, they exist also with symmetric CF dipoles, but the key issue is the magnitude of coupling, of common mode current, and the effects on radiation pattern on transmit, and local noise pickup on receive.

Physical symmetry of dipole AND feed line helps to reduce coupling from dipole to common mode conductor.

OCF dipole antenna systems are quite complex to analyse, very sensitive to implementation detail, and it seems that in the wider world, the hams with the strongest views for and against rely upon the simplest (inadequate?) understandings.

]]>The transformer comprises a 32t of 0.65mm enamelled copper winding on a FT240-43 ferrite core, tapped at 4t to be used as an autotransformer to step down a load impedance of around 3300Ω to around 50Ω. The winding layout is unconventional, most articles describing a similar transformer seem to have their root in a single design.

Inductance was measured using a RLC meter at 10kHz. The inductances were 15.8, 824 and 1056µH.

From that we can calculate for the two winding parts, M=-(1056-(15.8+824))/2=-108.1µH, and we can calculate flux coupling factor k=108.1*(15.8*824)=0.947 at 10kHz.

The flux leakage is quite low at 10kHz by virtue of the medium core permeability, and the winding concentrated, almost close wound.

But, things change as the frequency is increased.

Above is the complex permeability characteristic of the #43 material used.

The real part of permeability decreases and the imaginary part increases (representing core loss). We can estimate the magnetising impedance and admittance from the core dimensions and the permeability characteristic above.

At 7.1MHz, rather than look like a nearly pure inductance of 15.8µH as it did at 10kHz, the 4t winding looks like an inductance of 7.3µH in series with 224Ω of resistance (due to core loss).

A first approximation of the transformer behaviour is that it is an ideal transformer shunted with its magnetising admittance 0.00143-j0.00209S.

If the transformer and load are adjusted to input VSWR(50)=1, Zin=50+j0Ω, Yin=1/50+j0S. If a component of the real part 1/50=0.02 is due to the magnetising conductance, then we can calculate the percentage of input power lost in heating the core as 0.00143/0.02*100=7.1%.

For 3.6MHz, the core loss is 0.00166/0.02*100=8.3% of input power, and the case for 14.2MHz is left as an exercise for the reader.

The transformer was swept from 1 to 30MHz with a 3250Ω resistor in series with the VNA rx port, so the transformer load is 3300Ω. The resistor was measured with an accurate ohmmeter, and it is assumed that departure from ideal is small. It is beyond the accuracy of the VNA to measure the resistor simply.

Above, we can see the uncompensated transformer. Note R falls and X increases below 3MHz, a result of low magnetising admittance.

At the high end, X increases above about 7MHz though R is fairly good up to 20MHz, a result of leakage inductance. Although flux leakage is quite small at 10kHz, at RF with reduced permeability we can see signs of significant flux leakage.

Above is the sweep with a 100pF capacitor in shunt with the input. It improves the VSWR at 15MHz from 4.0 to 2.5, but things go worse even more quickly at higher frequencies… this is the nature of this form of compensation.

The compensation capacitor should be a good quality RF capacitor, eg a 500V silver mica should suit up to 100W input, or a Class 1 ceramic capacitor should suffice. An alternative is a 1.9m length of RG6 formed into a U shape or a small coil, and BOTH ends connected in parallel with the input winding, this should have a Q of better than 1000 at 15MHz and a very high voltage withstand.

The design presented is a low cost broadband transformer for matching an n half waves end fed antenna to 50Ω feed for 80-20m with reasonably low core loss (<10%). It uses readily available materials, and has sufficient core surface area to be good for 100W SSB input.

These type of antennas find use for low power portable operation with transmitters that are tolerant of wide variation in load impedance.

A somewhat lower turns ratio may provide a better match… more when I have tested it on some real antennas.

The design depends heavily on the #43 ferrite material characteristic, core size, and winding turns. The winding are arranged as an autotransformer and near close wound to reduce flux leakage. substituting or varying details may result in a significant departure from described behaviour.

The material is probably not a good choice for a 7-30MHz transformer, I would be investigating a different mix.

The next step is to test the transformer at part of a complete antenna system.

Above, taps have been added to the transformer for turns ratios of 4:24,28,32. As mentioned, lower taps may better suit the system.

adf

]]>Use this to help make up your mind. Add it to the normal coax loss. http://www.csgnetwork.com/vswrlosscalc.html

This is to suggest that the feed line loss under standing waves can be calculated with that calculator.

He then berates and demeans a participant for commenting on his recommendation, bluster is par for the course in these venues.

The calculator in question states this calculator is designed to give the efficiency loss of a given antenna, based on the input of VSWR (voltage standing wave ratio) and other subsequent factors

.

This is a bit wishy washy, efficiency loss

is not very clear. The usual meaning of efficiency is PowerOut/PowerIn, and the usual meaning of loss is PowerIn/PowerOut, both can be expresssed in dB: LossdB=10*log(Loss) and EfficiencydB=10*log(Efficiency).

A sample calculation with VSWR=5 gives Loss in %

=44.44% and Loss in dB

= 2.553dB. Note that converting 44.44% to dB gives 3.5dB so that is inconsistent, an error.

How did it calculate Loss in %

?

A little exploration of the code shows the following:

LOSSPCT=100*Math.pow(((vswr-1)/(vswr+1)),2)

At the heart of this is that loss is the square of the reflection coefficient ρ, and so the value is the ratio of PowerRef/PowerFwd, 0.4444 in the test case of VSWR=5, and this value is the inverse of the well known quantity ReturnLoss (2.25 in the test case, or 3.5dB).

So where does it get Loss in dB

?

LOSSDB=-10*log10(1-(LOSSPCT()/100))

What it calculates here is (PowerFwd-PowerRef)/PowerFwd in dB, and that is the well known quantity MismatchLoss, 2.55dB for VSWR=5.

So, on both counts, the calculator entitles quantities calculated with well known quantities, but gives them different meanings… in other words it calculates well known quantities incorrectly.

Hams are won’t to appropriate industry’s well known quantities, and apply their own meaning to suit themselves. This is work in that style.

So, the calculator calculates wrong results, but is that the end of the problem?

There are two problems:

- the calculator does not attempt to calculate the change in feed line loss under standing waves; and
- use of MismatchLoss is not relevant to the discussion scenario.

In most practical transmission lines at HF, conductor loss dominates, and so under standing waves, loss per unit length is greatest in the region of current maxima, and least the region of current minima.

Most calculators that attempt to calculate the loss under standing waves are based on a formula that has a bunch of underlying assumptions which are not usually disclosed. The unstated assumptions are relevant to accuracy, and if they are non complied with, the results are not reliable.

The subject calculator does not claim to calculate line loss under mismatch, nor does it attempt to do so, which questions the poster’s understanding of the subject.

It is a common ham practice to quantify the reduction of power output from a transmitter with mismatched load of know VSWR using MismatchLoss calculated from that VSWR.

The problem in application to a transmitter is that it may not be well represented as a Thevenin equivalent source with source impedance equal to the reference impedance used to determine VSWR, and the basis for MismatchLoss collapses.

It is widely held in ham radio that transmitters are well represented by a Thevenin equivalent source impedance equal to 50+j0Ω, thanks in no small part to the teachings of Walt Maxwell; and in ham radio, popularity is commonly taken to determine fact.

Jonathon Swift wrote:

Reasoning will never make a Man correct an ill Opinion, which by Reasoning he never acquired.\

In any event, the relevant source impedance in the discussion context is that looking back into the ATU… and it isn’t a Thevenin source of 50+j0Ω.

The recommended calculator is seriously flawed in what it purports to do, and it neither purports to, nor calculates line loss under standing waves.

Don’t take tools for recommendations on face value, there is no substitute for understanding the topic and validating the tools.

The transmission line topic is not well understood in ham radio, little wonder at the smoke and mirrors used to support individual beliefs.

To many if not most practitioners, modern ham radio is an antiscience, more about beliefs and talking the talk rather than walking the walk.

]]>The broadband transformer commonly uses a medium µ ferrite toroid core, and a turns ratio of around 8:1. Flux leakage results in less than the ideal n^2 impedance transformation, and a capacitor is often connected in parallel with the 50Ω winding to compensate the transformer response on the higher bands.

David, VK3IL posted EFHW matching unit in which he describes a ferrite cored transformer matching unit that is of a common / popular style.

Above is David’s pic of his implementation. It is a FT140-43 toroid with 3 and 24t windings and note the 150pF capacitor in shunt with the coax connector.

The article End fed matching – analysis of VK3IL’s measurements gives the following graph showing the effects of compensation for various resistive loads.

An online expert opined in the context of a 100W antenna system use a 3Kv or better cap across the input to help with SWR on the higher bands

.

Since the advice was offered with no explanation, it is questionable. Whilst a 3kV rated capacitor is likely to be adequate, they are quite hard to obtain and it is worth considering a rational design rather than a figure picked out of the air with no supporting justification.

It would be naive to use a capacitor rated for no more than the nominal working voltage, safety factors are appropriate. Lets declare two safety factors to be applied:

- VSWR up to 3:1 which could result in voltage of 3^0.5 times nominal; and
- sn overall 100% increase in the voltage rating for conservative reserve.

The peak working voltage for 100W in 50Ω is given as (2*P*R)^0.5=(2*100*50)^0.5=100V. Applying a safety factor of 2*3^0.5, we get 346V.

So, for adequate voltage breakdown we would select an off the shelf component with a rating of greater then 346V, a common value for these caps is 600V.

Clearly the online expert’s recommendation of 3kV is over the top, 500% of what is prudent, a further reserve of 14dB which isn’t necessarily a bad thing… except that a 3kV low loss component is much harder to find.

For lower power, lower voltage withstand is adequate, eg on the same basis, 77V or better is sufficient for a nominal 5W.

I mentioned loss!

Lots of published articles show the use of inexpensive ‘moulded mud’ ceramic capacitors in this application at low power levels.

Not only have these low grade capacitors low voltage ratings (commonly 50V), but the dielectric is lossy, capacitor Q might not be better than 50.

Lets consider the loss in a 150pF shunt capacitor with Q=50 at 30MHz with 100W input to the matching network. Xc=1/(2*π*f*c)=35Ω, and by virtue of Q=50, it has an equivalent shunt loss resistance of Q*Xc=50*35=1750Ω. The voltage impressed on that capacitor at 100W in 50 is (P*R)^0.5=(100*50)^0.5=70Vrms. The power dissipated in the capacitor is E^2/R=70^2/1750=2.8W… way in excess of what is safe for a tiny capacitor.

Voltage rating of the compensation capacitor is easy to calculate, and worth the effort as low loss capacitors suited to the application may be hard to find, especially with higher voltage rating.

These capacitors should be of a type that has low loss at the frequencies in use.

]]>The magic is that it is supposed to give closer to ideal behaviour of the transformers by way of minimising flux leakage.

The transformer above is styled on the common design, and it consists of a 2t primary and 16t secondary where the primary is wound bifilar, and a third 2t winding wound over the primary end of the transformer between the other turns.

Inductance of the two 2t windings each was measured at around 700kHz (µ is close to µi), and the pair in opposition (L’). The inductances were 4.02, 4.01 and 0.02µH respectively. We can calculate mutual inductance M=(L1+L2-L’)/2=4.005µH. We can calculate flux coupling factor k=M/(L1*L2)^0.5=0.997.

There is negligible flux leakage between the 2t windings with µ in the region of 800.

A further test was made of inductance from the non-shared ends of each of the 2t windings to the 16t winding, and the inductances were both 184.4µH and the 16t winding was 237µH. If we follow the same procedure, we will find that the flux coupling factor between either 2t and 16t winding is 0.919, there is very little flux leakage, and the results between the two 2t windings are indistinguishable.

Note that the secondary winding has been spread out to occupy about half of the core, but common designs utilise all of the core, with or without the W1JR crossover, and using more of the core tends to increase flux leakage though the effect for this medium µ core will be small.

Cores with lower permeability have higher flux leakage for the same winding geometry, and there are other changes.

The permeability of core material may vary with frequency, particularly ferrite.

Though these transformers are claimed to be broadband and have near ideal transformation, that is hardly the case. They are often frequency compensated with a capacitor in shunt with the primary, and the transformation is not simply Zin=Zload/n^2. That doesn’t make them unsuitable, it just makes the simplistic explanations inadequate.

The compensation capacitor depends on the core type, turns, frequency range and acceptable VSWR limit. I see designs where the core type was changed yet the compensation capacitor value was retained, resulting in an overcompensated transformer with worse behaviour than if the capacitor was omitted.

It is not unusual to see that a constructor has taken a design and substituted their own favourite ferrite mix in the belief that it is and independent variable of the design, and that their superficial analysis is meaningful.

Essentially, if the design was good in the first place (and that is often questionable), copy it exactly. If the design data does not give useful measurement data (including loss performance, thermals etc) to validate the design, perhaps keep looking.

]]>

The purpose of the balun is to minimise common mode feed line current which may contribute to EMC problems when transmitting, and contribute to increased ambient noise when receiving. Reduction of feed line common mode current also helps in achievement of expected load impedance characteristic, radiation pattern and gain. This article gives measured Zcm, but the definitive test of the effectiveness of such a balun is direct measurement of common mode current Icm… and it is so easy.

Example applications are half wave centre fed dipoles, fan dipoles, trapped dipoles, G5RV with hybrid feed, ZS6BKW, trapped verticals, monopoles, ground planes.

To obtain low Insertion VSWR, the choke will be wound with 50Ω coax, to demonstrate the practicality of the design budget (but good quality) regular (ie solid PE dielectric) RG58C/U will be used. Foam dielectric is NOT recommended. Solid PTFE coax could be used, but avoid coax with steel cored inner conductor, it may be lossier than you think at low frequencies with the silver cladding is relatively thin.

The candidate core is the readily available FT240-43 (Fair-rite 2643803802, 5943003801), it is a low cost NiZn ferrite with medium µr, and its µr and loss characteristic contributes to a broad high impedance choke well suited to this application.

Above is a model of the expected Zcm with 11 turns of RG58C/U coax and an equivalent shunt capacitance of 4.6pF.

Hams wax on about the capacity of RG58 to withstand the voltages experienced with severely mismatched antenna, and some online experts glean voltage withstand figures of 400-600V. Lets test a sample of the coax used.

Above, a sample of the coax withstands 6.17kV RMS, 8.7kVpk. This is more than the coax connectors and pigtails of the connections will withstand, the coax is quite up to this task.

The pic above shows the basic winding of 11 turns in Reisert cross over configuration so that the input and output are conveniently on opposite sides of the box. (The bend radius here is well less than datasheet, that may become an issue if the cable becomes quite hot.)

The choke could be employed as-is in a feed line, or it could be encapsulated in a non-conductive enclosure. I recommend against using a conductive enclosure.

Above, the choke is house in a non-conductive box to avoid compromising Zcm with unnecessary shunt capacitance to ground.

The prototype uses N connectors because they are good electrically, but also that they are waterproof when mated (assuming connector bodies that seal to the cable jacket).

Above is a plot of the measured Zcm. Self Resonant Frequency is about 8.5MHz. The measurement is made with a Rigexpert AA-600 and SOL calibrated fixture, and plot is produced with a back level version of Antscope (the latest version’s ‘enhancements’ prevent making this plot). |Z|>1500Ω over all of HF and exceeds 2000Ω from 2.4 to 24MHz.

Above is a plot of the Insertion VSWR. Not brilliant, but good enough for this task.

Lets look at a Smith chart plot of a matched load.

Above, the Smith chart plot shows a spiral centred on R=54Ω hinting that the Zo of the coax might be closer to 54Ω than 50Ω, a consequence possibly of the very tight radius bends around the core (about half of the specified min bend radius). It might also be an artifact of the budget coax, but other uses of it have tested ok. I would expect foam dielectric coax to be even less tolerant of the tight bends and hence my reason to recommend against it.

Balun sellers tend not to publish this information, but to their credit, Balun Designs do. By way of comparison, the Balun Designs 1115i which uses a smaller teflon coax shows Insertion VSWR=1.07 (ReturnLoss=29.4dB) at 31MHz whereas this balun had Insertion VSWR=1.15 (ReturnLoss=23.4dB) at 31MHz.

Cost of the whole thing was less than $25, more like half that if it is not enclosed in a box.

Building this yourself won’t save you a lot of money, but it will expose you to the opportunity to learn more than buying off-the-shelf. I would not consider any off-the-shelf product that does not publish measured Zcm R,X plots (|Zcm| is less useful), otherwise you are buying on looks alone.

More at A low Insertion VSWR high Zcm Guanella 1:1 balun for HF – more detail and A low Insertion VSWR high Zcm Guanella 1:1 balun for HF – more detail #2.

]]>Steel MIG wire is often coated with copper and is claimed by some online experts to “work real good”, particularly as a stealth antenna.

But is it the makings of a reasonably efficient antenna?

This article applies the model developed at A model of current distribution in copper clad steel conductors at RF to estimate the effective RF resistance of the wire at 3.5MHz.

In fact copper is an undesirable and restricted contaminant of steel welding wire, high grade MIG wire is not copper coated.

Copper content is held to less than 0.05% in the core, and less than 0.05% in the coating… which on my calcs says the coating of common 0.91mm MIG wire is less than 0.125µm…. basically it is a small diameter wire with low conductivity and high permeability.

The permeability contributes to higher inductance per metre of wire, and very small skin depth which contributes to higher resistance per metre of wire.

Analysis of the current distribution in the wire shows that almost all the current flows in the core, and that the RF resistance with and without the coating is almost the same. The copper plating can be ignored as significant, in fact the copper plating will usually fall off with a few weeks of exposure to weather as rust develops under the copper.

Lets model a 40m long centre fed dipole in free space, this would be a half wave dipole on 80m if using a copper conductor but we will use the MIG wire equivalent, a steel wire (σ=2.1e6, µ=800). We will use NEC-4.2 (NEC2 cannot natively model ferromagnetic wires.)

The current distribution is nothing like a half wave dipole, it is more like a three half waves dipole due to the magnetic effect of the wire.

Conductor loss is a staggering 91% whereas you might expect more like 3% for a copper dipole.

Ferromagnetic wire changes the tuning of the antenna, radically.

If the objective is a stealth antenna, this is doubly stealthy, not only can no one see it, no one can hear it!

]]>A recent article by PD7MAA describes such a transformer using a BN43-202 balun core for up to 20W PEP from 7-29MHz.

Above is PD7MAA’s graphic for his transformer. It is a little confusing as an 11t wind will start and finish with ends as the blue wind, so the red winding must have and odd number of half turns which suggests the windings are actually 1t and 5.5t (pity he did not show a picture of the real transformer). Let’s proceed under that assumption.

The transformer is intended to be used with a load such that the input impedance Zin is approximately 50+j0Ω, Gin=0.02S.

Let us calculate the magnetising admittance of the 1t primary at 7MHz.

Gcore is the real part of Y, 0.0115S.

If Yin of the loaded transformer is 0.02S, we can calculate the core efficiency as 1-Gcore/Gin=1-0.0115/0.02=43%, core loss is 3.7dB.

The average power of uncompressed SSB telephony is about 3%, so the average power of a 20W PEP SSB transmitter is about 600mW, and 57% dissipation is only 340mW which the core should safely withstand.

Losses in the matching transformer are only part of the total system loss, and overall system efficiency will be lower than estimated here for the transformer alone.

Low efficiency is not uncommon in QRP systems as even when more than half the transmitter power is lost in the core, the core survives by virtue of the low input power whereas a 1kW transmitter would destroy the core in short time.

Some hams might question the wisdom of “chucking away” transmitter power when you don’t have very much to start with, but the logic might be that the purest form of QRP is to start with a low power transmitter and then waste more than necessary in an inefficient antenna system.

PD7MAA does give a measurement of Zin with a 3k3Ω resistor load as VSWR=1.48 and |Z|=60.3Ω.

Above is a calculation of Gin and |Bin|, and if the 3k3Ω load (G=0.000303S) is transformed approximately by the turns ratio (though that is a gross approximation), it would present G approximately 0.0092S in parallel with the magnetising admittance estimated at 0.0115+j0.0165S above for total Yin of 0.021+j0.016S which is in the ball park of the admittance implied by PD7MAA’s measurements (given the tolerances of ferrite and the approximations mentioned). PD7MAA’s measurements support the assumption of a 1t primary and 5.5t secondary.

- PD7MAA EFHW antenna for 40-10m qrp
- 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).

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