Part I: Calculating the Magnetization Admittance of a Ferrite Toroidal Winding

Introduction

If one is to understand something about ferrite and how it behaves, some understanding of the magnetization admittance is required. The magnetization admittance will provide a means to estimate the power dissipations of ferrite transformers and inductors, their efficiencies, and their power handling capabilities.  A little bit of messy algebra is required for understanding, but it’s worth it. Ferrite is ubiquitous. It’s in many of our rigs, antenna tuners, filters, baluns, and power supplies.

Historical Information

Let’s begin with a bit of historical information about an antenna and its antenna matching device that has recently assumed the form of a ferrite transformer. This history is a bit narrow, but I hope that you will see what I am driving at later on.

Ferrite transformer-fed End-Fed Half-Wave (EFHW) antennas have grown in popularity in recent years due to their ability to operate on multiple bands in reduced spaces. Some of these antennas have been optimized for use on multiple bands by placing a small compensating inductor near the location of the first current maximum for the highest band of operation.

There is a hint to the origin of the multiband EFHW antenna on the MyAntennas website [1]. Danny Horvat, N4EXA (a.k.a. E73M), provides an interesting reference to some writing about the antenna in Karl Rothammel’s (Y21BK) Antennen Buch [2]. The cover, Figure 1, of the 12^th edition shows a drawing of harmonics on a wire antenna. The drawing shows an inductor located at the position of the first current peak of the highest order harmonic on the wire. Rothammel’s book attributes the design to Gerhard Bäz, DL7AB. Frank Dorenberg [3], N4SPP, was kind enough to forward a copy of the original journal article for this antenna that appeared in an October 1949 issue of the radio journal Funk-Technik [4] on pages 576 and 577. The compensating inductor has little effect on fundamental operation, but it has an enormous effect when the antenna is used for harmonic operation. The inductance and placement of this inductor have been analyzed in EZNEC and will be the subject of a future article. Danny, E73M, also makes reference to a transformer [5] invented by Nikola Tesla in 1897. A version of Tesla’s transformer appears in the antenna patent [6] of Joseph Fuchs, OE1JF, in 1928.

Figure 1. Cover of Rothammel’s Antennen Buch, 12^th ed. An antenna wire in harmonic operation is shown. A little inductor is located at the first current maximum for the 10m band.

Recently, the ARRL has offered a popular, inexpensive do-it-yourself kit to make an EFHW antenna [7]. Similar, ready-made versions are available from multiple sources such as MyAntennas [8], MFJ [9], PAR/LNR Precision [10], and HyEndFed [11], to name just a few. Some use the compensating inductor, while some do not. The ARRL transformer will be the subject of a worked example in Part II of this article. The ARRL antenna wire does not make use of the compensating inductor, but it may be added.

In spite of their omnipresence and popularity, there has been very little published on the subject of how the ferrite transformer version of the end-fed antenna feed actually works and how much power it can handle.

After searching the literature and the web, I came upon the blog of Owen Duffy, VK1OD. His website [12] is rich with content on the subject of ferrite. Not unlike any good teacher, he is fond of leaving calculations as a challenge to the reader. A good place to begin is with the magnetization admittance.

The Interwinding Capacitance and Magnetization Admittance

Magnetization admittances for the windings on ferrite cores are required to calculate losses in ferrite cores used in inductor and transformer applications. Owen Duffy has provided an excellent paper [13] on the subject that is somewhat short in detail. This article attempts to provide the missing steps in order to make his paper on the subject of ferrite more accessible.

In his paper, the complex impedance of a 10-turn winding on an FT140-43 ferrite core at 21 MHz has been provided as the starting point. This value could have been measured with an antenna analyzer, with a network analyzer, or could have been calculated from the manufacturer’s complex permeability data. We demonstrate how the complex impedance is derived from the complex permeability data in a separate paper.

The value of the interwinding capacitance should be included in the impedance calculation to compute the magnetization admittance value with greater accuracy. Not only is there an interwinding capacitance, but there is also capacitive coupling between primary and secondary windings. The interwinding capacitance is modeled in parallel with the winding impedance. D.W. Knight [14] has measured interwinding capacitance values for several windings on ferrite cores. The value of interwinding capacitance will have a large effect on the magnetization admittance.

If the interwinding capacitance is 2 pf at the design frequency of 21 MHz, the capacitive reactance calculated from,

will be,

The interwinding capacitive reactance that is in parallel with the winding impedance is 3789.4 ohms.

Digression:

The subject of complex numbers is something that we seldom see in use in everyday life. If this all seems new to you, please follow along. The expressions can be messy, but they all work out. These days, circuit modeling replaces these messy calculations, but it’s good to perform them by hand at least once for understanding.

The complex impedance is composed of two parts, a purely resistive part called the real part and a purely reactive part called the imaginary part. The letter i or j in front of the reactive part lets us know that it is the imaginary part.  The other thing that we must remember is that i or j is defined by,

and

If we multiply i or j by itself, we get -1.  We will make use of this property when we begin to multiply complex numbers together.

Regression:

Suppose that the winding impedance on our ferrite core has been determined to be 2627 + j 2044 by derivation from the manufacturer’s data. There are two methods to calculate the parallel combination of the interwinding capacitive reactance with the winding impedance. The first method makes use of the formula

for which,

Z₁ = 2627 + j2044

where +j denotes an inductive reactance. Since there is a real part, the phase angle is other than +90°.

Z₂ = 0 – j3789.4

Since there is no real part, the -j denotes a capacitive reactance whose voltage lags the current by 90°.

Method I

Entering the values into the equation, we obtain

we could multiply out all of the terms in the numerator and combine the terms and rationalize the denominator, or we could convert the expressions to polar form.

In order to accomplish this, we must first find the magnitude of Z₁,

where RE is the real part of Z₁ and IM is the imaginary part of Z₁.

Next, we must find the phase angle,

In polar form

Similarly, for Z₂, because there is no real part, the magnitude is just

A few more words about the phase angle in AC circuits for pure resistors, capacitors, and inductors – in a resistor, the voltage and current are in phase, so

In a capacitor, the voltage lags the current by 90° because it requires time for the capacitor to charge up, so

In an inductor, the voltage leads the current by 90° because the inductor opposes the flow of current, so

Thus, the phase angle for the parallel capacitor,

If the impedance possesses both real and imaginary parts, there will be a phase angle other than ±90°.

In polar form for the interwinding capacitance,

Next, we need to combine terms in the denominator before converting to polar form,

The magnitude of the denominator is given by,

The phase angle of the denominator is given by,

The denominator may be written in polar form,

Finally, the expression for Z_total may be written in polar form,

This may be converted back to rectangular form with some help from the Euler identity,

where,

Substituting

The easiest way to obtain the magnetization admittance is to work with the polar form of Z_total

In order to calculate the core loss, it is necessary to obtain the real part of the magnetization admittance, the conductance, by conversion back to rectangular form,

The real part of Y_total is the conductance, G, in units of Siemens,

G = 2.371E-04 Siemens

The imaginary part of Y_total is the susceptance, B, in units of Siemens

B = 7.948E-5 Siemens

Method II

There is a second method for obtaining the magnetization admittance. This may be calculated by adding the individual admittances of the winding and the  interwinding capacitance.

We won’t be able to add the admittances in polar form unless the phase angles are the same, in which case the phasors lie on the same line. We must first convert to rectangular form and then convert back to polar form.

The second term may be written by inspection, but it is best not to leave out any steps. That means that the denominators have to be rationalized first to free them of imaginary numbers. If you have forgotten how to do this, the method is shown below. You multiply the top and bottom of each factor by the complex conjugate of their respective denominators. To obtain the complex conjugate, you just invert the sign of the imaginary part.

which is rectangular form. This form provides the real part of the admittance, which is the conductance value, G, required to calculate the ferrite loss,

G = 2.371E-04 Siemens

as before.

In order to convert to polar form, the procedure in the first method is repeated. The magnitude of the admittance is given by,

The phase angle is given by,

The polar form may be written

Power Lost to the Ferrite Core

Now that the value of the conductance, G, has been obtained using both methods, it is possible to calculate the percentage of power lost to the ferrite core,

where

For this 10-turn inductor on FT140-43 at 21 MHz, the power lost to the core expressed as a percent is 1.18%.

The subject of heat dissipation in ferrite transformers and inductors is treated in Part II of this article.

References

[1] https://myantennas.com/wp/efhw-8010-is-this-the-ultimate-magic-antenna/
[2] Rothammel, Karl, Antennen Buch, 12th ed., Deutscher Amateur Radio Club (DARC), pp. 228-229.
[3] https://www.nonstopsystems.com/radio/frank_radio_antenna_multiband_end-fed.htm
[4] https://nvhrbiblio.nl/biblio/tijdschrift/Funktechnik/1949/FT_1949_Heft_19-OCR.pdf
[5] https://teslauniverse.com/nikola-tesla/patents/us-patent-593138-electrical-transformer
[6] http://www.antentop.org/016/files/oe1jf_016.pdf
[7] https://home.arrl.org/action/Store/Product-Details/productId/133267
[8] MyAntennas, op. cit.
[9] https://mfjenterprises.com/collections/all/products/mfj-1982hp
[10] https://www.parelectronics.com/end-fedz.php
[11] https://www.hyendcompany.nl/antenna/multiband_8040302017151210m#main
[12] https://owenduffy.net/blog/
[13] https://owenduffy.net/files/EstimateZFerriteToroidInductor.pdf
[14] https://www.g3ynh.info/zdocs/magnetics/appendix/self_res/self-res.pdf

Introduction

When planning for an amateur radio installation or when purchasing such items as an antenna tuner, coax, or balun, it is useful to know what we may be up against for transmission line voltages under conditions of antenna mismatch. This article explores what the worst-case standing wave voltage might be on a transmission line when the only mismatch is due to the antenna. Furthermore,  the transmitter power is deliberately increased to compensate for the mismatch.

Calculation

When computing the magnitude of the maximum peak voltage on a transmission line under conditions of increasing VSWR, it has been stated without proof [1][2] that this voltage is proportional to the square root of the VSWR. This factor may be derived by assuming that the amount of power delivered to the load will remain constant under conditions of increasing mismatch, which should be the worst imaginable case. This operating condition should be possible if a) there is sufficient forward transmitter power to compensate for reflected power and b) a matching network such as an antenna tuner is placed between the transmitter and the load to protect the transmitter.

Define:

Numerical Example 1

A 100W transmitter feeds an antenna with a 3:1 VSWR. By how much must the transmitter power be increased to maintain 100W into the antenna?

The magnitude of the voltage reflection coefficient is 0.5. The fraction of power returned from the load is given by,

The amount of power reflected from the antenna is 25%. Next, we must calculate the amount by which we need to increase the transmitter power to maintain 100W at the antenna input,

So, the amount by which the transmitter power must increase is 33.33W. Let’s check our answer to see how much power is reaching the load after we increase the power,

where

P_antenna is the amount of power reaching the antenna for a 3:1 VSWR after compensating accordingly.

Thus, we have demonstrated that for a 3:1 VSWR with 133.33W input power, the power incident on the antenna will be 100W when the only mismatch is in the antenna.

Continuing on, from the voltage standing wave definition,

where,

The reflected voltage may be written in terms of a fraction of the forward voltage,

Solution:

The corrected, forward power generated by the transmitter is given by,

and if we set S = VSWR for convenience,

The RMS magnitude of the forward voltage wave is given by,

The peak magnitude of the forward voltage wave is given by,

and the magnitude of the maximum peak voltage on the line is given by,

Finally, replacing S with VSWR yields,

Thus, it has been shown that the magnitude of the maximum peak voltage on the transmission line for the case of constant transmitter power to the antenna load is proportional to,

Numerical Example 2

What would be the maximum peak voltage on the transmission line for the case when the transmitter power is 1 kW and the antenna VSWR is 3:1?

The maximum peak voltage on the transmission line for the case where the transmitter power is 1 kW and the VSWR is 3:1 would be 547.7 Volts.

References

[1] https://owenduffy.net/blog/?p=11773
[2] ARRL Antenna Handbook, 41st Ed., 2007, Chapter 24 Transmission Lines, Line Voltages and Currents, p. 10, Equation 17.

Introduction

In Part I of this three-part series, we discuss what is meant by differential and common modes on RF transmission lines. Part II will discuss the construction of the Joe Reisert, W1JR, 1:1 balun [1] that may also be used as a common mode choke. Part III will present some test results for the common mode rejection of two common mode chokes, one constructed with #31 ferrite material and another constructed with #43 ferrite material.

Differential Mode

Two familiar balanced transmission line types that will support differential mode operation are open wire line and waveguide. This section will focus on open-wire lines. An ideal model of an open-wire line is shown in Figure 1. Current from the transmitter or other matched source enters the transmission line from the left. The transmission line may be thought of as an infinite number of distributed inductors and capacitors. Each infinitesimal length of the transmission line is made up of two tiny inductors, and each infinitesimal pair of lines forms a capacitor between them. All transmission line types, not only open wire lines, are characterized by values for inductance per unit length and capacitance per unit length.

Figure 1 Idealized Model of Open Wire Transmission Line. An open wire transmission may be modeled as an infinite number of distributed inductors and capacitors.

In reality, the conductors will have a resistance per unit length. If there is a dielectric present, as there might be in a window line, twin-lead or open wire line (the dielectric would be the spreaders and air), there will also be a leakage conductance through the dielectric between the conductors.

Consequently, all transmission lines are characterized by an impedance, Z_0, that is defined by,

where,

Z₀ is the characteristic impedance in ohms

R is the resistance of the wire per unit length

G is the leakage conductance through the dielectric per unit length

L is the inductance of the transmission line per unit length

C is the capacitance of the transmission line per unit length.

Years ago, it was quite common for roof-mounted television antennas to be fed with 300-ohm twin-lead. Twinlead is a parallel wire transmission line in which the conductors are spaced apart with plastic dielectric. The dielectric fills very little of the volume around the conductors. There is just enough plastic to cover the conductors and space them a small distance apart. Consequently, twin-lead will be treated as though it were an open-wire line. If we assume that the resistance of the wire and the leakage conductance are negligible, we can make the approximations that,

As a result, the impedance of the transmission line may be simplified to,

By making further approximations that the wire diameter, d, is much smaller than the center-to-center spacing of the conductors, D, and that the value of the dielectric constant filling the volume around the conductors is close to unity,

it is possible to approximate the values of L and C from,

where,

and,

where,

from which we get,

Furthermore,

Thus,

Substituting the numerical values, we have,

So, by making reasonable approximations, our estimate is very close to 300 ohms.

When driven by and terminated in its real, characteristic impedance, the currents and voltages anywhere along the open wire transmission line will be mostly uniform. Assuming that the wire transmission line is well made, dissipative losses in the conductors and leakage conductance will account for any nonuniformity. Since the currents in the transmission line conductors are equal and travel in opposite directions, the transmission line is said to be operating in differential mode. Simply stated, the transmission line operates in a single mode, and what you put in one end is mostly what you get out of the other end. The transmission line will not radiate signals, nor will it receive signals and noise.

Figure 2 is greatly oversimplified, but it is adequate to explain what is meant by differential mode. We note that if we take a snapshot of the currents on each half of the dipole, i_1, and i₂, they are in the same direction as are the currents, i₁ and i₂, in the open wire transmission line. From this, we may conclude that the transmission line is operating in differential mode while the antenna is operating in common mode, and that is what is causing the antenna to radiate RF in the first place. At least for this case, we have demonstrated that we may associate common mode currents with antenna radiation (and reception, too).

Figure 2. Balanced Open Wire Transmission Line Feeds a Dipole Antenna. The unbalanced transmitter or transceiver is transformed to a balanced transmission line with a balun. The transmission line operates in a single, differential mode because the currents are opposite while the antenna operates in a common mode because the currents are in the same direction.

Common Mode

For completeness, let’s begin by calculating the characteristic impedance of an unbalanced coaxial transmission line. Coaxial cable was first employed to prevent interference between transmission lines in transatlantic cables used for telegraphy prior to 1860. It was Oliver Heaviside who first described its theory of operation.

Let’s determine the characteristic impedance of RG-400/U since our common mode chokes were constructed using this type of coax. RG-400/U was chosen because of its high power handling capability and small outer diameter. If the inner diameter of the coax shield is much greater than the diameter of the center conductor,

where,

d is the diameter of the coaxial transmission line center conductor
D is the inner diameter of the coaxial transmission line shield,

the inductance per unit length and capacitance per unit length for coaxial cable are approximated by the formulas,

where,

from which we obtain,

As before,

Substituting the numerical values, we have,

Again, reasonable assumptions lead us to the expected result.

When we speak of common mode for transmission lines, we are discussing signals that may enter or leave the conductors in the same direction. The most common cause of common mode current is an unbalanced transmission line. Imagine, if you would, a dipole antenna being fed by the coaxial transmission line of Figure 3. The currents inside the coax will be opposite. Now, suppose that the currents in the coax reach the antenna. If we take a snapshot of the currents on each half of the dipole, i_4, and i₂, they correspond to the directions of the currents, i_3, and i₁, on the inside of the transmission line (but not their amplitudes). From this, we may conclude from the currents on the antenna that the antenna operates in common mode, as before.

However, we also notice that there are currents, i₅, on the outside of the coax shield and the current, i₁, on the center conductor that are in the same direction. These currents operate in common mode. The outside of the coaxial cable shield operates as a single conductor transmission line, a distinct mode. This mode operates separately from the mode represented by currents i₃ and i₁, which operate in, essentially, differential mode. Thus, we have a transmission line system that operates in two distinct modes. The outer cable shield will radiate upon transmit and will be susceptible to receiving signals and noise upon receive.

Figure 3. Common Mode Currents on Unbalanced Coaxial Transmission Line. Because of its construction, there is no way to keep the current i₃ from dividing into currents i₄ and i₅. Since currents i₁ and i₅ are in the same direction, they operate in common mode. Since the currents i₁ and i₃ are in opposite directions, they operate in differential mode. Thus, we have a transmission line system that operates in two distinct modes. The common mode conductor will radiate and also be susceptible to receiving signals and noise.

To reiterate, since one side of our dipole antenna is connected to the shield, any current that is traveling inside the shield may split between the antenna and the outside of the shield. In this configuration, there is nothing to stop this from happening. Now, we have a center conductor and the outside of the shield acting like a pair of conductors with currents traveling in the same direction. This is very much like a single wire transmission line, and the outer shield will radiate and receive power quite nicely in common mode. Another observation is that the currents on the antenna halves are asymmetric, and this asymmetry will corrupt the antenna pattern. Notice that the current on the outer shield may be returned to the chassis of the transmitter. This can become very unpleasant for the operator.

A remedy for this is to convert the unbalanced coaxial line to a balanced line where it feeds the antenna and provides a means to suppress current i₅. This is done with a device called a choke balun (balanced-to-unbalanced). The choke balun effectively disconnects the inner shield from the outer shield so that most of the current will no longer flow on the outer shield. Can there still be common mode currents on the coax? The answer is yes. The shield can still couple some of the antenna’s radiated emission back to the shack, or signals and noise on the shield may originate from elsewhere. Either or both may occur because it’s not unusual to place a choke balun at the feedpoint of the antenna. RF can still couple to the coax beyond where the choke is located. For this reason, it is not unusual to place another choke at another current maximum on the coax outer shield where it may be effective and at a location that is close to the entrance to the shack. This point on the transmission line may be found with a clip-on antenna current probe like the MFJ-854 [2], or by modeling the antenna in something like EZNEC [3]. It is incorrect to assume that the placement of a common mode choke is arbitrary. If we want the common mode choke to work, it should be located near a voltage null on the outside coax shield. Another effective way to reduce the common mode signals and noise on the outer shield from reaching the shack is to bury at least some of the coax.

Baluns may be constructed from sections of coax or from wire or coax wrapped on ferrite cores. Baluns constructed from coax alone, rely upon the electrical length of the coax to work, so they tend to be narrow band. Baluns constructed from coax are more practical for UHF and VHF because of the short length of transmission line required. Coax baluns do not possess the same choking properties that ferrite baluns have.

Part II of this three-part series will discuss the construction of a Joe Reisert, W1JR, 1:1 balun that may also be used as a common mode choke.

References

1. Reisert, Joe, Simple and Efficient Broadband Balun, Ham Radio, September 1978, pp. 12-15. https://worldradiohistory.com/Archive-DX/Ham Radio/70s/Ham-Radio-197809.pdf
2. https://mfjenterprises.com/products/mfj-854
3. https://www.eznec.com/