Self-SWR is commonly known as Insertion VSWR. The article contains several errors in definition of SWR, Return Loss and Insertion Loss… but suffice to say that he uses \(InsertionLossdB=-|s21dB|\).
In the example given above, VSWR=1.2 means ReturnLoss=20.8dB (and |s11|=-20.8dB).
InsertionLoss in this case is due to input MismatchLoss (failure to capture the maximum power available from the source) and Loss (conversion of some energy to heat).
Let’s check the last statement in the quote by calculating the scenario using Calculate Loss from s11 and s21.
InsertionLoss of 0.5dB in the presence of input VSWR=1.2 (|s11|=-20.8dB) has Loss=0.4637dB. We can calculate the power lost in the switch itself
as heat as \(P=50 (1-10^{\frac{-0.4637}{10}})=5.06 \: W\), somewhat less than Gable’s 5.8W. (Since the calculator gives loss as a numeric value of 1.113, a simpler calculation is \(P=50 (1-\frac1{1.113})=5.07 \: W\).)
Note that the inferences of these measurements with a VNA apply to sources that are well represented by a Thevenin equivalent circuit with Zth=50+j0Ω. Most ham transmitters do not comply with that requirement, and the power output under mismatch is best assessed with a directional wattmeter.
It is difficult to understand why QST would publish an article that purports to drill down on these effects when it is not based on sound concepts and theory.
This article reports nanoVNA measurement of a two wire line where no common mode countermeasures were taken.
Above is a Smith chart of the complex reflection coefficient Γ (s11) looking into a length of nominally 142Ω transmission line of similar type to that in the reference article, the chart is normalised to Zref=142+j0Ω. Note the locus is a spiral, clockwise with increasing frequency, and centred on the chart prime centre Zref. More correctly it is centred on transmission line Zo, and the keen observer might note that the spirals are offset very slightly downwards, actual Zo is not exactly 142Ω, but 142-jXΩ where X is small and frequency dependent, a property of practical lines with loss.
Above is a Smith chart of the same data, but normalised to Zref=50+j0Ω, ie the prime centre is 50+j0Ω. The spirals are offset, but again they are centred on Zo. It might not look centred on Zo, but note that the spiral inwards is not symmetric on this scale, and if the line was long enough, the spirals converge on Zo.
So what does this spiral with Zref=50+j0Ω translate to in a cartesian plot of |s11| vs f?
Whilst the value of |s11| wrt actual Zo, when mapped under Zref=50+j0Ω (as in the VNA reports), it has a cyclic variation with f.
Now that we know the phenomena to expect of a normal practical (lossy) transmission line, we can review a measurement data set.
The DUT is the same test line documented at Measure transmission line Zo – nanoVNA – PVC speaker twin. It is 1m in length and that article documented nominal Zo≈142Ω and VF≈0.667.
The DUT is directly connected to the nanoVNA Port 1 jack (CH0 in nano speak), no measures have been taken to reduce potential for common mode excitation of the system, and the nanoVNA is connected to the desk computer by a USB cable… all a bit non-descript because the only detail that is important is no common mode counter measures used.
Above is a Smith chart of a scan from 1-100MHz. The form is broadly a clockwise spiral from low to high f, but notable deviations at the high and low f end. Beware, it might look like 80% of the locus is a spiral, but it represents more like just 30% of the frequency range.
Let’s look at |s11| for a better perspective.
Above is a plot of |s11| vs f. Recall that we might expect some smooth cyclic variation in |s11| with f, and that accounts for the dip at Marker 1, but the shenanigans around 30MHz are an anomaly cause by common mode excitation. Likewise the range 66-100MHz is affected.
This dataset is seriously compromised by common mode excitation of the DUT, the data is worthless and no reliable conclusions can be drawn about the pure differential mode characteristics of the DUT.
Measure transmission line Zo – nanoVNA – PVC speaker twin reported measurements and analysis of the same line section, and it described the common mode countermeasures.
Common mode current gives rise to unquantified radiation loss and affects the input impedance of the line section.
If countermeasures are apparently needed and are not described, the experimenter may be naive and none were used, and the data is questionable, possibly worthless… certainly suspect until the common mode issue can be clarified.
]]>Well, as the article showed, it is not quite the no-brainer but with care, it can give good results. This article documents such a measurement of a 0.314mm cable.
The nanoVNA was carefully SOLT calibrated from 1 to 201MHz. Care includes that connectors are torqued to specification torque… no room here for hand tight, whether or not with some kind of handwheel adapter or surgical rubber tube etc.
Above is the Smith chart view over the frequency range from a little under λ/8 to a little over λ/8. It is as expected, a quite circular arc with no anomalies. Since the DUT is coax, and the connector is tightened to specification torque, we would expected nothing less. The situation may be different with two wire lines if great care is not taken to minimise common mode excitation. The sotware does not show Marker 2 properly, it should be between ‘c’ and ‘i’ of the word Capacitive.
Above is the R,X scan, Marker 2 is at the λ/4 resonance (X≈0) and Marker 1 is exactly half of that frequency, so λ/8 electrical length at which point |X|=|Zo|.
From the λ/8 measurement, we can calculate VF=0.314/(1.767384/4)=0.710 which reconciles with the datasheet.
]]>The transformer of interest is the one to the left, and if you follow the tracks, the multiturn winding is connected between ground and a track that routes across to the through line. The transformer primary appears in shunt with the through line.
The transformer appears to comprise a 24t primary on a 2646665802 #46 toroid. Note that #46 is a MgZn ferrite, where other designs tend to use either powdered iron or NiZn ferrite.
For the purpose of this article, flux leakage is ignored as it will have very little effect on the calculated ReturnLoss due to this component alone.
Above is the calculated ReturnLoss due to the magnetising admittance of the voltage transformer that is in shunt with the through line based on the published permeability characteristics and assuming 3.5pF of equivalent shunt capacitance to model self resonance. ReturnLoss is presented here as InsertionVSWR is very very small, VSWR=1.02 equates to ReturnLoss=40dB.
ReturnLoss at the lowest specified frequency, 1.8MHz, is 47dB.
Now this is not the only source of mismatch that would drive low ReturnLoss, but in most ham designs, this single component is responsible for ReturnLoss less than 20dB at their specified lower frequency limits.
There is little point having an instrument that indicates VSWR down to say 1.1 when it invisibly causes VSWR=1.3 looking into it (eg
Grebenkember’s original Tandem match).
Measurement of VK4MQ’s wattmeter using one of these couplers showed ReturnLoss (-|s11|) jack to jack at 1.8MHz to be 46dB, and it remains above 30dB to 30MHz, 25dB at 50MHz.
I purchased one of the couplers for use with a DIY digital display, and although I have had it longer, it isn’t yet realised!
A common failing of almost all hammy Sammy designs is appalling InsertionVSWR at the lower end of the specified frequency range. This coupler is specified for 1.8-54MHz, and differently to most, has meaningful published characteristics.
In this implementation, 60mm lengths of solder soaked braid coax similar to Succoform 141 were used between the PCB and box N connectors. The expected matched line loss of both of these is about 0.01dB @ 50MHz.
The measurements here were made by VK4MQ using an Agilent E5061A ENA, data analysis by myself.
Above are the raw s parameter measurements plotted. It is a full 2 port measurement, and it can be observed that the device is not perfectly symmetric, quite adequate, and quite good compared to other ham designs that I have measured.
ReturnLoss is -|s11|, it is simply stunning, a huge departure from most ham designs. Some that I have measured were less than 20dB at their specified lower frequency limit.
An interesting perspective is to disaggregate InsertionLoss (-|s21|) into Mismatch Loss and (Transmission) Loss.
As we expect from the |s11| plot, MismatchLoss is very low, InsertionLoss is dominated by Loss. Loss tells us how much of the actual input power is converted to heat.
Above is a plot of the expected dissipation due to Loss. Most of it is likely to be in the ferrite cores, roughly equally with a 50Ω load.
Difficulties were encountered in adapting Kiciak’s design to the unmodified coupler. Kiciak didn’t deal with intercept and gain calibration entirely in the coupler or the display electronics, it is spread between them.
The VK3AMP coupler would have been much easier to use if it incorporated pots for intercept and gain calibration in each channel. This will not be an issue in my digital meter project as each channel (fwd, rev) will have slope and intercept calibration in the firmware.
Custom meter scales were designed to suit two available taut band meters.
Above, the forward power meter scaled in dBm (Bruce’s preference). The sharp mind will recognise 56dBm is 400W.
Above, the return loss meter scaled in dB.
Above is the interior of VK4MQ’s build.
Above is the inside of the front panel. The electronics is built dead bug style on the PCB that is visible, the two meter movements are in the lower part.
Above is the front panel with the rescaled meters recovered from some e-waste. The meter does not overhang the edge, it is a perspective issue with the pic.
This article shows the use of the Smith chart to look for departures from pure transmission line behavior in that test, or any other that depends on measuring purely Zin of a length of line in purely differential mode with short circuit or open circuit termination.
Above is a Smith chart plot of what we should see looking into a line of similar characteristic swept from 1 to 20MHz. There is no magic there, this is basic transmission lines and Smith chart.
Above is the Smith chart plot of a scan of the NEC model with the USB cable included. A significant anomaly can be seen and the measurement frequency is in the seriously disturbed area, little wonder the test gave wrong results. The whole curve is wrong, and will give erroneous results.
Any significant common mode current gives rise to power lost to radiation, and the measured impedance is not that of the transmission line alone. The Smith chart might not have strong hints as in the above, but it is a first place to look for departure from expected pure transmission line behavior.
You would not see this huge hint of an invalid experimental setup if you focus on the narrow band of frequencies where the electrical length is λ/8.
Above is the NEC Smith chart plot with the USB cable removed, quite as expected.
]]>Apparent gross failures are often wrongly attributed to factors like manufacturing tolerances, polluted line surface, other esoteric factors etc that might imply a knowledgeable author… but that is social media, an unreliable source of information.
Let’s explore an estimate using measurements with a nanoVNA using the popular eighth wavelength (λ/8) method.
The λ/8 method relies upon the property of a lossless line terminated in an open circuit that differential impedance \(Z_d=\jmath X=- \jmath \left| Z_0 \right| cot \left(\pi/4\right)=- \jmath\left| Z_0 \right|\). So, if you measure the reactance looking into the λ/8 (\(\frac{\piᶜ}{4} \:or\: 45°\)), you can estimate Zo as equal to the magnitude of the reactance.
A similar expression can be written for the case of a short circuit termination and it leads to the same result that you can estimate Zo as equal to the magnitude of the reactance (an exercise for the reader).
The fact that the two cases lead to the same result can be used to verify that the line length is in fact λ/8 (they will not be equal if the length is a little different to λ/8)… though writeups rarely mention this, or perform the test.
So, the method depends critically on:
Most online articles do not include details of the measurement setup, perhaps thinking that it not all that relevant. Of course, one of the greatest failings in experiments is to ignore some factor that is in fact relevant.
The nanoVNA is such a limited device without a computer attached, lets model a scenario that might well be used by the naive.
Above is a graphic of the scenario. It comprises a vertical section of the modelled transmission line from height 2 to 7m, so it is 5m in length and comprises 1mm diameter copper conductors spaced 20mm, dielectric is a vacuum. Impedance is measured in the horizontal segment bonding both sides of the transmission line.
We might expect the quarter wave resonance of this part of the scenario alone would be approximately 15MHz, but the NEC model gives a slightly lower frequency of 14.97MHz and therefore λ/8 is 7.485MHz.
Also included is a connection from one side of the transmission line to real ground to represent the ground connection of the nanoVNA via its USB cable to a computer used to capture the measurement results.
Above is the impedance plot from the model, it looks well behaved and we might suggest that we would expect that the nanoVNA would measure Zin=Rin+jXin=92.5-j195Ω or |Z|=215Ω.
If you not into the ‘j value’ stuff, you might then say that Zo=215Ω. If you were a little more savvy, you might say that Zo=|X|=195Ω. Not a whole lot of difference… but the difference should be concerning. In fact, the relatively large value of R in Z should sound a warning that the naive might overlook.
Let’s look at a simple model of the transmission line using TWLLC.
Parameters | |
Conductivity | 5.800e+7 S/m |
Rel permeability | 1.000 |
Diameter | 0.001000 m |
Spacing | 0.020000 m |
Velocity factor | 1.000 |
Loss tangent | 0.000e+0 |
Frequency | 7.485 MHz |
Twist rate | 0 t/m |
Length | 0.125 wl |
Zload | 1.000e+100+j0.000e+0 Ω |
Yload | 0.000000+j0.000000 S |
Results | |
Zo | 444.04-j1.48 Ω |
Velocity Factor | 1.0000 |
Length | 45.000 °, 0.125000 λ, 5.006554 m, 1.670e+4 ps |
Line Loss (matched) | 2.28e-2 dB |
Line Loss | >100 dB |
Efficiency | ~0 % |
Zin | 8.486e-1-j4.418e+2 Ω |
Above is an extract of the output.
Key calculated results are:
Note that:
By comparison with the TLLC prediction, a key point of reconciliation failure is that in the measurement Rin is relatively quite large and quite inconsistent with even a low loss line, and so the premise for applying the λ/8 method vanishes.
Let’s model a better DUT,
Above is a graphic of the scenario. It comprises a vertical section of the modelled transmission line from height 2 to 7m, so it is 5m in length and comprises 1mm diameter copper conductors spaced 20mm, dielectric is a vacuum. Impedance is measured in the horizontal segment bonding both sides of the transmission line.
We might expect the quarter wave resonance of this part of the scenario alone would be approximately 15MHz, but the NEC model gives a slightly lower frequency of 14.97MHz and therefore λ/8 is 7.485MHz.
There is no connection to ground, the instrument and the transmission line section are relatively isolated from ground.
Above is the impedance plot from the model, it looks well behaved and we might suggest that we would expect that the nanoVNA would measure Zin=Zoc=Rin+jXin=0.879-j458Ω.
As a simple check, Rin is relatively small, and we might accept that Zo=|Xin|=458Ω… but this value is sensitive to the electrical length.
Running the same model with a short circuit termination, Zsc=3.98234+j463.85Ω.
We can calculate Zo=Ro+jXo from Zoc and Zsc.
As you see, Ro lies between the |Xoc| and |Xsc|.
This calculation if of higher accuracy than basing Zo on |Xoc| or |Xsc| alone of an assumed λ/8 section.
A simpler approximation for low loss line is \(Z_o\approx \sqrt{-X_{oc} X_{sc}}=460.8\) which reduces sensitivity to actual length.
Calculated Zo based on the second NEC model does not reconcile exactly with the TWLLC calculation as they use different methods of modelling, but the results are reasonably close.
What’s the problem was asked. The problem is that the λ/8 method depends on a valid differential mode impedance measurement, and that did not happen in the first example, the common mode current path prevented pure differential mode excitation of the DUT.
Instantly something looks wrong.
The usual shape of a Smith chart plot is that in any set of nearby points, the locus of the points forms an arc and with increasing frequency, the points are located clockwise about the centre of that arc. Nothing in that statement constrains the location of the centre of that arc, it is not necessarily the middle of the chart.
Above is an example that obeys that rule. The highest frequency is the point marked X, and the lowest is marked O. In any set of nearby points, the locus forms an arc, and points of increasing frequency are located clockwise about the centre of the arc.
Experienced analysts use this property to navigate around the curve when there is no frequency axis to the chart.
So what is phoney about the original chart?
The problem is that implied under that rule:
Here is another example of an implausible Smith chart.
Again the locus of the plot breaks the ‘arcs’ rule of real systems, this time as a result of poor guesses by software of the sign of reactance from unsigned reactance data from a low end analyser.
Try this one…
Is the low frequency end the X or the O?
]]>It means that the line geometry imposes a natural constraint on a wave travelling in the line that V/I=50… but remember that TEM waves are free to travel in (only) two directions. This natural ratio of V/I is called the characteristic impedance Zo.
Let’s say that at some point on a 50Ω line (it could be start, finish, or anywhere in between), V=100V and I=1A, therefore Z=V/I=100/1=100Ω. This situation could be created at the end of a transmission line by simply attaching a 100Ω load.
Zo is a complex value (having real and imaginary parts), but for simplicity we will assume that Zo is a purely real value, Zo=50+j0Ω. It makes the maths easier, it becomes arithmetic you can do in your head, and in fact practical transmission lines at 100MHz have Zo that is very very close to a purely real number.
So how can V/I at a point on a Zo=50Ω line be 100/1=100Ω?
So the voltage at the point of interest point is 100V and I=1A, Z=100Ω. (For simplicity, I am assuming I in phase with V, so the actual power is easily calculated P=V*I=100*1=100W. P is the energy flow past the point averaged over a full RF cycle.
This 100W is the power that the transmitter ‘makes’.
Because of the Zo constraint that a travelling wave MUST have V/I=50, then this 100V/1A can only be due to the sum of two waves travelling in opposite directions, there MUST be a forward wave and a non-zero reflected wave such that their sums deliver the observed voltage and current.
A Bird 43 50Ω wattmeter ‘explains’ V and I at a point as being due to two waves travelling in opposite directions (taking into account their relative phase). The Bird wattmeter is calibrated for a purely real Zo, 50Ω in this example.
Now because there is 100W flowing, then the difference between the indicated forward wave Pf and reflected wave Pr MUST be 100W, ie \(P=P_f-P_r=100\). This does not tell us the value of Pf or Pr.
To find Pr and Pf, we need to calculate the complex reflection coefficient: \(\Gamma=\frac{Z-Z_0}{Z+Z_0}=\frac{100-50}{100+50}=0.33\).
Now \(V=V_f(1+\Gamma)\), which we can rearrange to \(V_f=\frac{V}{(1+\Gamma)}=\frac{100}{1+0.33}=75\) and therefore \(V_r=V-V_f=100-75=25\).
So, we know Vf and Vr, we can calculate \(P_f=\frac{{V_f}^2}{Z_0}=\frac{{75}^2}{50}=112.5\) and \(P_r=\frac{{V_r}^2}{Z_0}=\frac{{25}^2}{50}=12.5\) and \(P=P_f-P_r=100\).
Similarly we can say \(I=I_f(1-\Gamma)\) (-ve due to the wave travelling in the opposite direction), which we can rearrange to \(I_f=\frac{I}{(1-\Gamma)}=\frac{1}{1-0.33}=1.5\) and therefore \(I_r=I-I_f=1-1.5=-0.5\).
Forward power can be greater than the transmitter ‘makes’, you must deduct the reflected power to find the ‘net’ power, and that ‘net’ power cannot be greater than the actual transmitter output power under those load conditions (though it may be greater than the transmitter’s notional ‘rated’ power).
The transmission line supports TEM waves travelling in two possible directions, and you must consider both waves to calculate the power at a point on the line. Forward power is a notional quantity associated with one of two possible travelling waves.
The law of conservation of energy is not violated by any of the discussion above.
]]>Relationship between angle of reflection coefficient and angle of impedance
It was stated above that the angle (or phase) of s11 or Γ is not the same as the angle (or phase) of Z.
Given Zo and Γ, we can find θ, the angle of Z.
\(
Z=Z_0\frac{1+\Gamma}{1-\Gamma}\)Zo and Γ are complex values, so we will separate them into the modulus and angle.
\(
\left | Z \right | \angle \theta =\left | Z_0 \right | \angle \psi \frac{1+\left| \Gamma \right | \angle \phi}{1-\left| \Gamma \right | \angle \phi} \\
\theta =arg \left ( \left | Z_0 \right | \angle \psi \frac{1+\left| \Gamma \right | \angle \phi}{1-\left| \Gamma \right | \angle \phi} \right )\)We can see that the θ, the angle of Z, is not simply equal to φ, the angle of Γ, but is a function of four variables: \(\left | Z_0 \right |, \psi , \left| \Gamma \right |, \& \: \phi\) .
It is true that if ψ=0 and φ=0 that θ=0, but that does not imply a wider simple equality. This particular combination is sometimes convenient, particularly when ψ=0 as if often the case with a VNA.
This article offers a simulation of a load similar to a 7MHz half wave dipole.
The load comprises L, L1, and C1 and the phase of s11 (or Γ) and phase of Z (seen at the source G) are plotted, along with VSWR.
Firstly, note that the two phase plots are very different, but in this case they cross over at phase=0 at 7.077MHz.
Secondly, note that even thought both phases are zero at 7.077MHz, the VSWR is 1.20. Neither phase demonstrates the best conditions for least feed line loss, minimum VSWR is slightly lower at 7.099MHz.
Maximum power in the load coincides with minimum VSWR at 7.099MHz.
Beware of claims that phase (of something) is the optimisation target, the author probably doesn’t really understand this stuff.
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