The EFHW can be deployed in a miriad of topologies, this article goes on to explore three popular practical means of feeding such a dipole.

The models are of the antenna system over average ground, and do not include conductive support structures (eg towers / masts), other conductors (power lines, antennas, conductors on or in buildings). Note that the model results apply to the exact scenarios, and extrapolation to other scenarios may introduce significant error.

A very old end fed antenna system is the End Fed Zepp. In this example, a half wave dipole at λ/4 height is driven with a λ/4 600Ω vertical feed line driven by a balanced current source (ie an effective current balun).

Above is a plot of the current magnitudes. The currents on the feed line conductor are almost exactly antiphase, and the plot of magnitude shows that they are equal at the bottom but not so at the top. The difference between the currents is the total common mode current, and it is maximum at the top and tapers down to zero at the bottom. Icm at the top is about one third of the current at the middle of the dipole.

End fed Zepp deals in more detail with the common mode current on the EFZ.

One manufacturer of a popular EFHW antenna system that uses a 2/3 terminal matching device recommends that where the fed end of the dipole is elevated, that the match device be installed there and that the coax be grounded where it reaches ground. In this model, a 2m driven ground electrode is used to ground the coax.

Above, the plot of current magnitudes shows substantial common mode current on the feed line, with a maximum at the lower end approximately the same as the current in the middle of the dipole.

The relatively high common mode current on the feed line, and particularly at lower height is a distinct disadvantage bring risk of higher rx noise and transmitter interference to nearby electronics. The magic of End Fed Half Waves (EFHW) gives further information on the common mode current in this configuration.

This antenna is advertised as “no counterpoise needed” by at least one seller, which questions the term “counterpoise”: https://owenduffy.net/files/Counterpoise.pdf.

One popular author recommends a “0.05λ counterpoise” as he calls it. Again a 2/3 terminal matching device is used the coax is grounded where it reaches ground. In this model, a 2m driven ground electrode is used to ground the coax. This is essentially the same as the previous model but with the dipole fed 0.05λ from end.

Above, the plot of current magnitudes shows substantial common mode current on the feed line, with a maximum at the lower end approximately the same as the current in the middle of the dipole.

The relatively high common mode current on the feed line, and particularly at lower height is a distinct disadvantage bring risk of higher rx noise and transmitter interference to nearby electronics.

This is almost the same as the previous model, the so-called “counterpoise” has done little for the relatively high common mode current.

The three scenarios modeled are quite similar configurations, but the detail of the feed arrangement results in the first being significantly different to the later two.

The third model shows that the so-called “counterpoise” variation to the second model has negligible effect, and questions the credibility of sources suggesting otherwise.

Relatively high feed line common mode current is a risk, again dependent on implementation.

The concept of a “no counterpoise” EFHW as commonly used is questionable.

Because of the sensitivity to implementation detail, the term End Fed Half Wave is not very descriptive.

You might like: more articles on EFHW.

- Duffy, O. Oct 2010. Counterpoise. https://owenduffy.net/files/Counterpoise.pdf.

This article presents some NEC-4.2 model results for a 7MHz λ/2 horizontal 2mm copper wire at height of λ/4 above average ground.

The model is impractical in a sense that it does not include unavoidable by-products of a practical way to supply RF power to the antenna, but it is useful in providing insight into the basic antenna.

The NEC model has 200 segments, and varying the feed segment gives insight to what happens to feed point impedance.

Above, it can be seen that as the wire is fed closer to the end (segment 1), feed point Z includes a rapidly increase capacitive reactance.

The very large series reactance close to the end makes feeding at those points impractical as any impedance transformation network needs to deliver power to a very reactive very high impedance load.

As the feed point approaches the end, the feed point reactance approaches -∞ and the feed point voltage approaches ∞.

Let’s analyse the antenna at a feed point that is close to the end, but a compromise for more practical feed point impedance. We will look at the case at segment 10 (1m from the end) which is 5% of the dipole length or 2.5% of wavelength.

Above is a 3D magnitude and phase current plot from 4NEC2. It can be seen that there is a significant change in phase near the feed point.

Above, the magnitude of current an phase are plotted.

Many authors insist that the behavior of a half wave dipole is independent of where it is fed, that the current distribution is exactly identical for all feed points, and often give the familiar sine current distribution to support their argument.

The chart above shows that the magnitude of current is close to a sinusoidal distribution, but there is a glitch near the feed point… but more importantly there is a significant variation in phase in the last quarter wavelength, culminating in a discontinuity at the feed point.

The current distribution is not the same as a centre fed half wave dipole, though it is somewhat similar.

The gain pattern at 10° elevation shows some asymmetry, not large, but evidence again that the thing is not exactly the same as a centre fed half wave dipole, but somewhat similar.

The radiation efficiency of the modeled system is 75%, it has no feed or impedance transformation losses factored in, just conductor loss (which is very small) and ground loss.

An NEC-4.2 model of a 7MHz λ/2 horizontal 2mm copper wire at height of λ/4 above average ground reveals:

- feed point impedance vs displacement for a somewhat idealised or ‘pure’ antenna, ‘pure’ in the sense of not including a feed line or other elements that would alter its operation; and
- current distribution (magnitude and phase) that are a little different to a centre fed dipole.

1 highlights the impractical nature of feeding such a ‘pure’ antenna very close to the end.

2 questions the assumption that underlies most discussions of an EFHW that it works identically to a centre fed dipole as even in this ‘pure’ form, the current distribution is not exactly the same.

Further articles will explore the effects of popular practical feed arrangements on the system.

]]>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.

]]>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
- an 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.

]]>