Having selected a candidate core, the main questions need to be answered:

- how many turns are sufficient for acceptable InsertionVSWR at low frequencies and core loss; and
- what value of shunt capacitance best compensates the effect of leakage inductance at high frequencies?

Lets look at a simplified equivalent circuit of such a transformer, and all components are referred to the 50Ω input side of the transformer.

Above is a simplified model that will illustrate the issues. For simplicity, the model is somewhat idealised in that the components are lossless.

- L1 represents the leakage inductance;
- L2 represents the magnetising inductance; and
- C1 is a compensation capacitor.

Since the magnetising inductance is assumed lossless, this article will not address design for core loss.

So, it is obvious that the InsertionVSWR curve is pretty poor at both high and low end.

Let’s look at a Smith chart presentation of the same information, it is so much more revealing.

Above is the Smith chart plot. Remember that the points go clockwise on the arc with increasing frequency, and that InsertionVSWR is a function of the distance from the centre to the point on the locus… we want to minimise that distance. Remember also that the circles that are tangential to the left had edge are conductance circles, they are the locus of constant G.

Now lets analyse the response.

Note that from 1 to 3MHz, the shape of the response tends to a circle tangential to the left hand edge, it a constant G circle. So, G is constant but susceptance B is frequency dependent and -ve. This the the response of a constant resistance R in parallel with a constant inductance (\(B=\frac {-1} {2 \pi f L}\), \(Y= G + jB = \frac 1 R – \frac {j} {2 \pi f L}\)). A part of that susceptance (shunt inductance) is due to the magnetising inductance L2 which contributes to the poor Insertion VSWR at low frequencies.

Note that from 12 to 15Hz, the shape of the response tends to a circle tangential to the left hand edge, it a constant G circle. So, G is constant but susceptance B is frequency dependent and +ve. This the the response of a constant resistance R in parallel with a constant capacitance (\(B=2 \pi f C\), \(Y= G + jB = \frac 1 R + j 2 \pi f C\)). A part of that susceptance (shunt capacitance) is due to the compensation capacitor C1 which contributes to the poor Insertion VSWR at high frequencies.

Lets adjust L2 and C1 for a better InsertionVSWR response.

Above is the response with L2=12µH and C1=80pF. Note that the distance to the centre is improved (and therefore InsertionVSWR is improved). The kink in the response is common, that is typically the mid region where InsertionVSWR is minimum.

It is still not a good response, the InsertionVSWR at the high end is too high, and compensation with C1 does not adequately address the leakage inductance. So, as a candidate design, this one has too much leakage inductance which might be addressed by improving winding geometry and increasing core permeability.

As mentioned, real tranformers using ferrite cores have permeability that is complex (ie includes loss) and dependent on frequency (ie inductance is constant).

Above, the magenta curve is measurement of a real transformer from 1-11MHz with nominal resistance load and three compensation options:

- cyan: 0pF, too little compensation;
- magenta: 80pF, optimal compensation; and
- blue: 250pF, to much compensation.

It should be no surprise that 80pF is close to optimal. Susceptance B at the cyan X is -0.00575S, and broadly, we want to cancel that with the compensation capacitor so we come so \(C=\frac{B}{2 \pi f}=\frac{0.00575}{2 \pi 11e6}=83pF\).

With optimal compensation (80pF in this case) The insertionVSWR at 3MHz is 1.8, probably acceptable for this type of transformer but it is still quite high (4.3) at 11MHz, which hints that leakage inductance needs to be addressed by improving winding geometry and possibly increasing permeability.

Keep in mind that measurements with a nominal resistive load are a guide, measurements with the real antenna wire are very important.

]]>From the basic definition \(\mu=B/H\) we can derive the relationship that the flux density in the core with current I flowing through N turns is given by \(B=\frac{\mu_0 \mu_r N I}{2 \pi r}\).

The incremental flux at any incremental radius is proportional to the flux path length \(2 \pi r\), so the total flux due to B(r) is \(\Phi_B=\int_{a}^{b}Bc \, dr=\frac{\mu_0 \mu_r N I}{2 \pi r}c \, dr=\frac{\mu_0 \mu_r N I}{2 \pi} c \, ln \frac b a\).

Note that the core geometry is captured in the term \( c \, ln \frac b a\).

From that we can calculate inductance \(L \equiv \frac{N \Phi_B}I=\frac{\mu_0 \mu_r N^2}{2 \pi} c \, ln \frac b a \) where \(\mu_0=4\pi 10^{-7}\), the permeability of a vacuum.

Ferrite datasheets commonly give \(\mu_r=\mu^{\prime}-j\mu^{\prime\prime}\), a complex value (note it is usually a frequency dependent parameter). The imaginary term represents the core loss.

We can calculate the impedance at frequency f by substituting the values.

\(Z=j 2 \pi f L=j 2 \pi f \frac{\mu_0 (\mu^{\prime}-j\mu^{\prime\prime}) N^2}{2 \pi}c \, ln \frac b a \\\)The model can be improved for frequencies approaching SRF by addition of a small equivalent shunt capacitance \(C_s\).

\(Z=\frac1{\frac1{j 2 \pi f \frac{\mu_0 (\mu^{\prime}-j\mu^{\prime\prime}) N^2}{2 \pi} c \,ln \frac b a}+ j 2 \pi f C_s}\)The calculator Calculate ferrite cored inductor – rectangular cross section does exactly this calculation. Note that a real FT240-43 has chamfered corners, so these calculations based on sharp corners will very slightly overestimate L, but the error is trivial in terms of the tolerance of µr.

µr comes from the datasheet, but you may find Ferrite permeability interpolations convenient.

The value Cs is best obtained by observation of SRF of a particular winding, it is sensitive to winding layout.

The calculated value \(\sum{\frac{A}{l}}=\frac{c \, ln \frac b a}{2 \pi}\) and captures the core geometry in a more general form. It or its inverse often appear in datasheets and can be used to calculate Z (Calculate ferrite cored inductor – ΣA/l or Σl/A).

]]>With no other detail than the pic, it is difficult to supply a complete answer.

Nevertheless, an analysis of what is presented follows.

There are two transmission line sections of approximately equal length formed from parallel brown and white wires. These sections are wound on the core in the same sense magnetically (ie the direction in which they pass through the inside of the toroid), and are paralleled at both ends.

Without dimensions, it is a guess that characteristic impedance Zo of each line section is probably between 100 and 150Ω if they were uniformly space just touching each other. The latter is hard to achieve, so perhaps to the high end of the range mentioned… lets assume 140Ω for discussion.

Since the two line sections appear in parallel, Zo of the combination is half that of each section, so lets assume around 70Ω. (This analysis was confirmed by measurement.)

So one view of what we have is an ordinary Guanella 1:1 balun with common mode impedance due to 12t of 70Ω transmission line.

The configuration would lead to lower InsertionVSWR wt 50Ω than using a single line section of that type. Nevertheless, it is unlikely to quite achieve the InsertionVSWR when the choke is wound with good 50Ω coax as is usually done for a low InsertionVSWR balun.

Does it have lower loss?

Most loss in these types of baluns is due to magnetising losses in the ferrite core material, and that will be unchanged. Transmission line loss might appear to be lower, but if poor insulation like PVC was used, it might well be poorer in that respect than using 50Ω coax.

The transmission line configuration might well lead to a lower self resonance frequency which can be good or bad, depending on the application.

Does it work?

I have explained how I think it works, and with a suitable core material for the application, it might well work well. I would like to have seen a plot of Zcm vs freq… but balun ‘designers’ rarely publish such, and probably don’t measure the parameter.

]]>Noelec makes a small transformer, the Balun One Nine, pictured above and they offer a set of |s11| and |s12| curves.

These curves fuel online discussions about the loss, InsertionLoss, TransmissionLoss and ReturnLoss… often given in -ve dB values (it is a hammy sammy thing), and without correct application of the terms (again).

Lets calculate the loss of the pair of transformers back to back as shown from the marker data. Firstly, loss means PowerIn/PowerOut, and can be expressed in dB as 10log(PowerIn/PowerOut). For a passive network, loss is always greater than unity or +ve in dB.

\(loss=\frac{PowerIn}{PowerOut}\\\)Some might also refer to this as Transmission Loss to avoid doubt, but it is the fundamental meaning of loss which might be further qualified.

So, lets find the two quantities in the right hand side using ‘powerwaves’ as used in S parameter measurement.

s11 and s21 are complex quantities, both relative to port 1 forward power, so we can use them to calculate relative PowerIn and relative PowerOut, and from that PowerIn/PowerOut.

PowerIn is port 1 forward power less the reflected power at port 1, \(PowerIn=P_{fwd} \cdot (1-|s11|^2)\).

PowerOut is port 2 forward power times less the reflected power at the load (which we take to be zero as under this test it is a good 50Ω termination), \(PowerOut=P_{fwd} \cdot |s21|^2 \).

So, we can calculate \(loss=\frac{PowerIn}{PowerOut}=\frac{\frac{PowerIn}{P_{fwd}}}{ \frac{PowerOut}{P_{fwd}}}=\frac{1-|s11|^2}{|s21|^2}\)

Let’s work an example.

From the chart marker at 0.2MHz, |s11|=-2.59dB=0.742 and |S21|=-6.00dB=0.501.

So \(\frac{PowerIn}{PowerOut}=\frac{1-0.742^2}{0.501^2}=1.486\) which we can write as \(10log(1.486)=1.719dB\).

Online experts are inclined to divide the loss equally between the two back to back transformers, but that is only an approximation, and a poor approximation if the flux level is likely to be significantly different (as in this case).

In this case for back to back transformer network, InsertionLoss is 6.00dB, ReturnLoss is 2.59dB, and loss is 1.79dB.

Another quantity often used is MismatchLoss, \(MismatchLoss=\frac{P_{fwd}}{PowerIn}=\frac1{(1-|s11|^2)}\). In this example \(MismatchLoss=2.23=3.47dB\).

It is important to understand the meaning of the various loss metrics, and the appropriate metric for the application.

]]>A recent question and the responses exposes some common misunderstanding / woolly thinking.

A beginner in search of wisdom asked if there an easy way to measure balun loss?

One of the responses was…

Measure the return loss of the balun with the balun shorted. The return loss should be about twice the balun loss. This is similar to measuring the loss of a shorted or open piece of cable.

This was expanded by others, basically supporting the concept.

Let’s put it to the test with an example transformer in Simsmith.

Above is a model of a 1:1 ferrite cored transformer where the flux leakage reactance is represented by L1 and L3, the magnetising impedance is represented by parallel equivalent components L2 and R1. L2 has a reactance of 500Ω at the model frequency.

We can see from the screenshot by observing the power in the load that InsertionLoss is 0.46dB.

Now let’s follow the advice quoted above and explore the model with a short circuit at the load L and find half the return loss which it was stated should be about equal to the InsertionLoss.

Above is the short circuit scenario as recommended. The calculated value hrl at lower right is half return loss.

In the short circuit scenario, half return loss is 0.001748dB, nothing like the correct InsertionLoss of 0.46dB shown in the original model.

So the quoted advice is clearly nonsense in this instance. There might be scenarios where the half return loss of the short circuited network looks believable, but being correct is a whole lot more than simply looking believable.

This is application of basic linear circuit theory, and it takes only one case to disprove the expert’s method.

There are variations on this method sometimes described, but they tend to fail under some scenarios which means they are not soundly based.

]]>- Small untuned loop for receiving – simple model with transformer ;
- A transmission line 1:4 impedance transformer ; and
- Towards understanding the YouLoop-2T at MF/lowHF .

The first and third articles explained the concept of signal/noise degradation (SND) statistic, and gave graphs of the behavior of the subject antennas.

This article draws together those SND plots for two antennas, and some variations to the configurations.

Above, the “simple loop” with 0.5:1 ideal transformer. It could be implemented as a shielded loop (with transformer) with similar behavior (but improved common mode suppression).

Above is the calculated SND from 0.3 to 15MHz.

Above, the Airspy YouLoop-2T with ideal transformer.

Above is the calculated SND from 0.3 to 9MHz.

A simpler loop is a shielded loop (without transformer).

Above is the detail of a simpler shielded loop using coaxial cable and presenting a load of 50+j0Ω to the loop gap. With 50Ω coax and 50Ω receiver, there are no standing waves on the feed line or coaxial loop section so the impedance presented at the loop gap is a broadband 50+j0Ω.

Above is the calculated SND from 0.3 to 15MHz.

An alternate way to construct a shielded loop with flexibility to choose a loop load other than 50Ω is to insert a transformer at the loop gap of a basic shielded loop.

Above, a miniature ferrite cored broadband transformer or autotransformer could be used at the gap of the Simpler loop to provide broadband impedance transformation of a 50Ω receiver without the effects of standing waves and impedance transformation in the loop coax sections.

Above is the calculated SND from 0.3 to 15MHz with a 0.5:1 turns ratio transformer.

Above is a variation on the YouLoop-2T where the top crossover unit is replace with a simple through connection of the centre conductor at the gap. This removes the 1:4 impedance transformation, but there remains some transformation on the loop coax sections which have standing waves.

Above is the calculated SND from 0.3 to 9MHz.

Above is a chart comparing the simple loop and the simpler loop. The key differences is the load impedance for the simpler loop is 50+j0Ω whereas the simple loop is 12.5Ω which gives rise to the different SND response.

Some of the variations given above have some amount of transformation due to the loop coax sections so they aren’t simply 12.5Ω or 50Ω which gives rise to some small difference in their responses.

The load impedance presented to the loop proper has a great influence on SND response. The other key parameters are:

- loop diameter; and
- loop conductor diameter.

Doubling loop diameter increases loop inductance and increases radiation resistance, and delivers a large improvement in SND (of the order of 5dB). Note that the frequency were it qualifies as a small loop is reduced. That is not to suggest it does not work where perimeter>λ/10, just that analysis based on small loop assumptions becomes invalid.

Doubling conductor diameter results in a small reduction in loop inductance and small improvement in SND (tenths of a dB).

Above is the calculated SND from 0.3 to 15MHz of a 4m perimeter shielded loop of 12mm conductor diameter with 50Ω load.

Signal/Noise degradation with the small untuned loop is influenced by several key parameters:

- loop perimeter;
- loop conductor diameter;
- loop load impedance;
- impedance transformation due to standing waves on transmission line components;
- departure of transformers from ideal.

The SND response is a compromise which can be tailored for the intended application.

]]>

I mentioned in my article WIA 4:1 current balun that the use of a single toroidal core in the above graphic compromises the balun. This article presents some simple measurements and analysis that question whether the balun works as so many users think.

The popularity of the balun derives from the work of VK2DQ and is often known as the VK2DQ 4:1 current balun (though probably not his invention).

Often, analysis of a network as frequency approaches zero or infinity can simplify the analysis whilst allowing a reasonable test of the sanity of the design.

Above is a conventional transformer schematic of the WIA 4:1 current balun on a perfectly symmetric (balanced) load. At frequencies where the electrical length of each winding is very short, we can assume negligible phase delay along or between windings, simplifying analysis greatly.

If we assume that at such a frequency, that the core characteristics are such that:

- conductor resistance is insignificant;
- core loss is insignificant;
- permeability is relatively high and so flux leakage is very low (ie almost all flux is contained in the toroidal core; and
- magnetising impedance of a single winding is relatively high.

You might consider that this approaches an ideal transformer, and it does.

Under these conditions, the input impedance (at the left hand terminals) of the transformer (even with no load, ie Z=∞), approaches zero. That is to say that as you improve this transformer towards ideal characteristics, Zin approaches a short circuit, and the output voltage would approach zero.

That sounds a warning sign, that the ‘better’ you make this device, the worse it performs.

This balun is often pictured with a red core with rounded corners, looking very much like a powdered iron core, eg T200-2 with relatively low permeability of 10.

Nevertheless the text further down recommends a FT140-61 ferrite core (which will usually be unpainted and have sharper chamfered corners.

Above is a balun wound to the directions given in the article on a real FT140-61.

Earlier, we discussed that the magnetising impedance would approach zero for an ideal transformer of this configuration. The core used is low to medium permeability, so there will be significant flux leakage… so lets measure it.

Above is measured magnetising impedance. The graph is windowed at ±500Ω as we would want that the magnitude magnetising impedance was OUTSIDE this range in a 50Ω system. It is outside ±500Ω from about 9-18MHz.

So the low magnetising impedance (caused by the series opposed windings) will compromise InsertionVSWR (and therefore nominal impedance transformation) on 80, 40, 15, 12 and 10m.

Above is the InsertionVSWR plot for the transformer terminated in an asymmetric 200+j0Ω load. The response is never good, particularly poor at the lower frequencies, but supporters will hold that it works real good

.

Zcm gives a good indication of whether a current balun is likely to give good common mode current reduction in most scenarios.

Above is the Zcm response, it shows a quite sharp self resonance around 22MHz. The sharp response is quite expected with #61 material which is a relatively low loss material at these frequencies.

The graph is windowed at ±2000Ω as we would want that the magnitude of Zcm was OUTSIDE this range for good common mode current reduction. It is outside ±2000Ω from about 16-30MHz.

So the low Zcm (caused by the series opposed windings) will compromise common mode current reduction on 80, 40, 30, and 20m.

Make up your own mind, it is clearly not a good HF 4:1 current balun, but hey, you might work some DX and prove that it works real good

.

If you use the balun with an OCF dipole, your antenna asymmetry and poor current balun performance working against you, so expect higher local noise pickup and increased risk of electromagnetic compatibility. The cry of mediocrity often heard is any antenna is better than none at all

.

My own view: I do not recommend this balun for any scenario.

- WIA 4:1 current balun
- WIA 4:1 current balun – further explanation.
- 4:1 current balun – identifying bad ones
- Bertrand, R. 2005. Understanding and building the OCF dipole.
- Duffy, O. 2008. A review of the Guanella 4:1 balun on a shared magnetic circuit.

Baluns are commonly employed to obtain nearly balanced feed line currents (ie negligible common mode current) in two wire lines or negligible common mode current on coaxial feed lines. This article discusses baluns for that application.

A very common 4:1 current balun is Guanella’s 4:1 current balun, but there are others including pretenders.

(Guanella 1944) described a 4:1 current balun in his 1944 article, he did not show the winding pairs coupled by a magnetic core as shown above.

Above is Guanella’s circuit, and he does not show coupling between the two winding pairs.

Properly implemented, this device is known to work very well.

Let us look at Sevick’s device because it underlies so many failures.

If you look at it very carefully, you will see that the two output wires emerge from opposite sides of the core, the left hand wire exiting under the core was wound from front to back of the core and the right hand wire exiting on top of the core was wound from back to front of the core.

(Sevick 2001) describes a balun that he states works only on Isolated Loads, but Isolated Loads MUST have perfect current balance, they do not require a balun.

You might ask whether such a balun which works only on an Isolated Load which itself MUST be current balanced, works due only to the load isolation.

Most HF wire antennas are not well represented as a Isolated Load. Whilst you might at first thing there are only two terminals, there is a third terminal in most cases, ground.

Nevertheless the ‘net abounds with instructions on how to build this and there are plenty of commercial sellers who apparently endorse the maxim the customer is always right

… but it does speak to their technical expertise.

The following diagram is from WIA 4:1 current balun, it is a design made very popular by the work of VK2DQ (though he may not have originated it).

Note that the output wires at the top both emerge from the top side of the core, both were wound from back to front.

To be brief, the common issues are:

- Sevick’s single core 4:1 current balun on a single core will not work well for generalised three terminal loads;
- VK2DQ’s balun as has series opposed windings which cause the device to almost short circuit the input;
- poor choice of ferrite core (size, material);
- wrong number of turns for application;
- poor transmission line characteristics for the application;
- mis-wiring which may prevent the device working as intended.

To make these measurements, you need an instrument that can measure impedance at the balun coax socket. The shorter the connections the better, a single coax adapter from balun to instrument will give acceptable results.

The instrument could be a VNA (including a one port analyser such as a MFJ-259B, FG-01 etc), a calibrated noise bridge, a high end RF impedance or admittance bridge.

Real world measurement of baluns has uncertainty that will introduce error into the measurements, so don’t expect extreme accuracy. Nevertheless, defects will usually be easily recognised by a large departure from ideal.

You will also need a load of 200+j0Ω. This could be made with two 100Ω carbon composition resistors in series, or similar SMD resistors (100 x 100Ω 1% SMD resistors can be bought for about $3). Low power / small metalfilm resistors with pigtails trimmed may measure up just fine at 3.6MHz.

Connecting wires MUST be short and direct.

If your instrument does not directly measure the sign of X but measures the magnitude of X, label your measurements |X|.

These are all tests than any good 200Ω:50Ω current balun should pass, not just the balun types discussed above.

Perform these tests @ 3.6MHz , writing down the results at each step in a table:

- measure the 200Ω resistor with your instrument to verify both;
- measure input impedance (R and X) with nothing connected to the output terminals; and
- measure input impedance (R and X) with the 200+j0Ω load connected to the output terminals;
- measure input impedance (R and X) with the 200+j0Ω load connected to the output terminals and the shield of the coax connector bonded to one output terminal; and
- measure input impedance (R and X) with the 200+j0Ω load connected to the output terminals and the shield of the coax connector bonded to the other output terminal;

Think about what you should expect:

- very close to 200+j0, VSWR≈4;
- a quite high impedance, |Z|>1000 ohms;
- very close to 50+j0, VSWR<1.2;
- very close to 50+j0, VSWR<1.2; and
- very close to 50+j0, VSWR<1.2.

A good Guanella 4:1 current balun with appropriate core material and turns will pass all of the tests.

Sevick’s single core 4:1 current balun will fail 4 and 5.

VK2DQ’s single core 4:1 current balun will fail 2, 4, 5 and probably 3.

Below are the results of performing the above tests on a 4:1 balun at hand.

The yellow cells are my actual measurements (using a nanoVNA), the other values are calculated but they may be displayed directly on some instruments.

In this case, all tests are passed easily.

It is not a current balun you say? Sure behaves like a good current balun at 3.6MHz.

FT140-43 core with 9t for each winding (per WIA instructions).

Above are the test results. The DUT fails tests 2,3,4, and 5. This does not work on the Isolated Load (Test 3), and even worse on the asymmetric loads (Tests 4 and 5).

Why? Look at the Test 2 result, to a first approximation, that low ‘magnetising impedance’ impedance shunts the transformation spoiling Test 3, a result of the series opposing sense of the two windings. Further, the common mode impedance is low which results in significant change between Test3 and Tests 4 and 5.

Some analysers may not display R and X, but they may display magnitude of X (|X|), use that and label your result column |X|.

Some analysers may display may not display R or X but may display VSWR and |Z|, or other combinations from which the others can be derived. Try the calculator at Find |Z|,|X| from VSWR,|Z|,R,Ro to find the needed data.

- Duffy, O. 2008. A review of the Guanella 4:1 balun on a shared magnetic circuit.
- ———. May 2011. Measuring common mode current. https://owenduffy.net/module/icm/index.htm (accessed 31/05/19).
- Guanella, G. Sep 1944. New methods of impedance matching in radio frequency circuits. The Brown Boveri Review.
- Sevick, J. 2001. Transmission line transformers 4th Ed. Noble Publishing Co.

The MDF is located where the underground cable enters the building. From here it rises vertically and travels some 25m across the ceiling to the VDSL modem.

The choke can be seen in the pic, it is 7 turns of CAT5 data pair wound around a LF1260 ferrite sleeve.

Above is the measured common mode impedance R,X of the choke. It is designed to peak between 3.5 and 7MHz to afford some moderately high impedance at both frequencies.

Measurements were made with a nanoVNA-H and the graphic made in Python scikit-rf from a saved .s1p file.

On test, the choke is effective on 40m.

]]>Above is the test inductor, enamelled wire on an acrylic tube.

An online expert’s advice make this task look like a no-brainer:

For a 100 nH inductor you are probably using an air wound coil so you won’t see that much change in inductance with frequency. However, inductors made with toroids will because the permeability of the core goes down with frequency.

So, this is an air cored inductor, permeability is approximately that of free space, a constant 4πe-7 independent of frequency. Nevertheless we will see that apparent inductance can change with frequency.

The inductor was designed to be around 20µH using Hamwaves Inductance calculator.

Above the physical dimensions.

Above, the equivalent circuit at 5MHz. Our measured inductor will be a little different due to some additional distributed capacitance in the text fixture.

The test jig is very simple, the inductor is connected between the centre pin of Port 1 and the shield of Port 2 connectors. (I should note that my nanoVNA-H has a shorting bridge between connector shields: Strength of reinforcement of nanoVNA-H connectors). The pigtails are about 40mm long and they contribute to the measurements to some extent, but they may be representative of how such an inductor may be used in circuit.

The fixture could be represented as a two wire transmission line of Zo almost 600Ω and 40mm in length. Later a measurement of 3.5+j250Ω @ 2MHz could be transformed to 3.48+j248.82Ω at the coil ends due to the 1° length of 600Ω line. Understanding fixtures is important, and this fixture has very little impact at the lower frequencies, those that we will find are most relevant for discovering the inductance of the coil.

Firstly, a measurement direct on the device using the Hammy Sammy Smith chart display directly showing equivalent series inductance.

Above, at 20MHz, L=37.2µH. A good deal higher than the design value of 20µH.

Above, at 5MHz, L=20.3µH, right on the design target, but half of that just measured at four times the frequency, 20MHz.

Isn’t inductance largely independent of frequency… the online expert quoted earlier stated so.

Let’s learn from that…

Above is a sweep from 0.1-30MHz. Though the chart format is stated “Series RLC” it actually shows (series) R, X and |Z|(the latter for the hams who do not truly understand impedance… but here is obscures X which I do want to talk about).

So, understanding that up to about 20MHz, the X trace lies under the green trace, we ask what should we expect the X trace to look like.

In an IDEAL inductor, the inductance IS independent of frequency and \(X=2 \cdot \pi \cdot f\) so it should be a straight line. In this case, a PRACTICAL inductor, it might be almost a straight line up to 5MHz, but it certainly is not above that.

So, the equivalent series inductance of a practical inductor is not independent of frequency… though it might approximately so at sufficiently low frequencies.

For a simple air solenoid like the DUT, at low frequencies, the equivalent resistance Req is approximately that of the conductor given skin effect and proximity effect. Broadly, increasing frequency by a factor of four should approximately double Req

Above is a scan expanded to show R at low frequencies. At the cursor @2MHz, R=3.5Ω, and at four times that frequency, 8MHz, the value is noisy but around 32Ω, not double that of 2MHz, but nearly 10 times.

The DUT might be considered simply an inductor, but more completely it is a resonator and it has a self resonant frequency in the fixture in this case of about 30MHz. The departure from ideal is not solely due to the fixture, most of it is due to the inductor itself. The inductance calculator used above gives a self resonant frequency (SRF) of 40MHz for the coil alone.

At frequencies approaching SRF, the inductor behavior is not simply that of an IDEAL inductor, and this will affect measurement and behavior in the intended application circuit.

Magnetic cores have a magnetic circuit where part or substantially all of the magnetic circuit uses a material with relative permeability not equal to 1.

For some materials the relative permeability has a significant imaginary component (ie, it is a complex value), and may be frequency dependent at some frequencies.

Differently to the online expert’s advice, frequency dependent magnetic cores are not a result of the toroidal winding, but the core material, and ferrite cores may behave quite differently to powdered iron cores or air cores.

These effects will flow into measurement.

There are many articles on this site about measuring ferrite cored inductors.

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