Above, the latest repair. A new battery socket to replace the original that crumbled apart… sub-standard plastic from all appearances. This was from a reputable supplier, so it is probably a genuine JST part rather than some cheap Chinese knock off.

The blue wire is part of a mod to invoke the bootloader on power up, R5 was also changed to something small, 1k IIRC.

]]>- 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.

]]>

Above is the Airspy Youloup-2T. Try to put the two turns thing out of your mind, it is misleading, panders to some common misunderstanding, and so does not help understanding.

This is somewhat similar to the simple loop, but now the transformer primary is connected to the loop gap terminals by two parallel sections of 50Ω transmission line, the combination being effectively a 100Ω with similar parameters to the component coax sections. Because of the series connection at the transformer and parallel connection at the loop gap, there is a 1:4 impedance transformation additional to that of the coax sections themselves.

The operation of the 1:4 impedance transformer is laid out at A transmission line 1:4 impedance transformer.

The currents shown above are differential currents in the coax (ie wholly inside the coax), the current on the outside of the shield is not shown on the diagram. In this application, the current on the outside of the loop shield flows into the terminals at the left, ie 2I1 flows on the outside of the shield.

So, knowing the characteristics of the RG402 coax used, and assuming a 1:1 ideal transformer, we can estimate the load impedance seen at the gap by the loop.

It is a little different to the simple loop with 0.5:1 transformer because the coax sections perform a little impedance transformation additional to the 1:4 due to the connection arrangement… but only a little different.

So, lets continue with the analysis using the new load impedance seen by the loop. Again in the context of a linear receive system (ie no IMD) of known Noise Figure.

The receiver has an idealised input impedance of 50+j0Ω, and known internal noise implied by its Noise Figure (NF). For the purpose of this analysis we will assume the NF is 5dB and from that we can derive an equivalent noise temperature Tr of 627K.

After mentioning receiver internal noise, lets consider external noise.

For the purpose of estimating external noise, we can look to ITU-R P.372-14 for guidance, it gives us a median ambient noise figure from which we can calculate an external equivalent noise temperature. We will use the Residential precinct figures from P.372.

We can define S/N degradation (SND) to mean the reduction in external S/N \(frac{ExternalSignal}{ExternalNoise}\) by the addition of internal noise

\(frac{frac{ExternalSignal}{ExternalNoise}}{frac{ExternalSignal}{ExternalNoise+InternalNoise}}=frac{ExternalNoise+InternalNoise}{ExternalNoise}\) which we can convert to dB

\(SND=10logfrac{ExternalNoise+InternalNoise}{ExternalNoise}\)Let’s consider the single turn untuned loop with cross over connections and RG401 sections, and an ideal 1:1 broadband transformer. The loop is 2.1m perimeter and 4mm diameter conductor situated in free space. The loop has perimeter less than λ/10 up to 15MHz, so we can regard that loop current is uniform in magnitude and phase. This simplifies analysis greatly.

A simple analysis is to consider the loop to have some fixed inductance (2.2µH in this case) and in series some resistance (radiation resistance Rr, and loss resistance which we will ignore). For simplicity, we are using an ideal transformer of some known turns ratio and a receiver.

We can consider the loop to have a Thevenin equivalent circuit of a voltage source with series equivalent impedance being Rr+j2πfL.

For this scenario, the loop is loaded with some impedance being the receiver input impedance transformed by the ideal n:1 turns ratio transformer. There is a large impedance mismatch, and the antenna system gain in this scenario is entirely due to mismatch loss, Gain=-MismatchLoss.

One metric that could be used to assess performance is to calculate the SND of a receive system in a given noise environment.

To calculate SND, we need to refer internal and external noise to the same place to perform the calculation. In this instance, we will refer them to the receiver input terminals. We already have Tr=627, and we can calculate Tamr=Tam/Gain.

Above is the calculated SND from 0.3 to 9MHz.

It is a small untuned / unmatched loop and the naked truth is that the SND is significant.

The results apply to the scenario described. The results are sensitive to most parameters so they cannot be blindly applied to another scenario.

The “two turns” in the loop name is not related to a valid explanation of how it works.

A real antenna should perform similarly, note though that a real receiver will probably depart from ideal in impedance and NF, the model does not include transformer loss (perhaps some tenths of a dB), proximity of ground (ie ground gain) and the largest source of variance may be in the actual ambient noise figure.

]]>Above is a diagram of the device. The currents shown are differential currents in the coax (ie wholly inside the coax), the current on the outside of the shield is not shown on the diagram.

At very low frequencies it may be intuitive that \(V_1\approx V_2\) and \(I_1\approx I_2\), but as frequency increases, a more exact solution is needed.

We can write an expression for input impedance of each transmission line section in terms of its terminating impedance \(Z_t=V_2/I_2\), the line length l and frequency dependent complex propagation coefficient \(\gamma=\alpha+j\beta\\\).

\(Z_{i}=Z_0 \frac{Z_t/Z_0+tanh((\alpha+j\beta)l)}{1+(Z_t/Z_0)tanh((\alpha+j\beta)l)}\)

We can see intuitively that because the TL sections are of equal length and that the voltages at the left hand side are equal (by virtue of the parallel connection) and that the currents at the right hand side are equal (by virtue of the series connection) that the voltages and currents in both TL sections are equal. That means that \(Z_t=Z_{load}/2\), and knowing \(Z_0, \: \gamma\) and \( l\) we can calculate the combined input impedance \(Z_{input}=Z_i/2\).

Above is the calculated input impedance for a pair of 1m long sections of RG402 with 50+j0Ω load. It can be seen that at low frequencies, input impedance approaches 50/4Ω but as frequency increases the impedance transformation in each line section becomes more significant.

Note that the example and discussion assume no common mode TL current, and the explanation holds valid if the application does not have significant common mode current.

An upcoming article will use this device in a shielded loop antenna where the symmetry of the loop and the coupling transformer both serve to effectively minimise common mode current.

]]>This article analyses a simple untuned / unmatched loop in the context of a linear receive system (ie no IMD) of known Noise Figure.

The receiver has an idealised input impedance of 50+j0Ω, and known internal noise implied by its Noise Figure (NF). For the purpose of this analysis we will assume the NF is 5dB and from that we can derive an equivalent noise temperature Tr of 627K.

After mentioning receiver internal noise, lets consider external noise.

For the purpose of estimating external noise, we can look to ITU-R P.372-14 for guidance, it gives us a median ambient noise figure from which we can calculate an external equivalent noise temperature. We will use the Residential precinct figures from P.372.

We can define S/N degradation (SND) to mean the reduction in external S/N \(\frac{ExternalSignal}{ExternalNoise}\) by the addition of internal noise

\(\frac{\left(\frac{ExternalSignal}{ExternalNoise}\right)}{\left(\frac{ExternalSignal}{ExternalNoise+InternalNoise}\right)}=\frac{ExternalNoise+InternalNoise}{ExternalNoise}\) which we can convert to dB

\(SND=10log\frac{ExternalNoise+InternalNoise}{ExternalNoise}\)Let’s consider a simple single turn untuned loop with an ideal broadband transformer. The example loop is 2.1m perimeter and 4mm diameter conductor situated in free space. The loop has perimeter less than λ/10 up to 15MHz, so we can regard that loop current is uniform in magnitude and phase. This simplifies analysis greatly.

Above is a schematic diagram of the example loop.

A simple analysis is to consider the loop to have some fixed inductance (2.2µH in this case) and in series some resistance (radiation resistance Rr, and loss resistance which we will ignore). For simplicity, we are using an ideal transformer of some known turns ratio and a receiver.

We can consider the loop to have a Thevenin equivalent circuit of a voltage source with series equivalent impedance being Rr+j2πfL.

For this scenario, the loop is loaded with some impedance being the receiver input impedance transformed by the ideal n:1 turns ratio transformer. There is a large impedance mismatch, and the antenna system gain in this scenario is entirely due to mismatch loss, Gain=-MismatchLoss.

One metric that could be used to assess performance is to calculate the SND of a receive system in a given noise environment.

To calculate SND, we need to refer internal and external noise to the same place to perform the calculation. In this instance, we will refer them to the receiver input terminals. We already have Tr=627, and we can calculate Tamr=Tam/Gain.

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

It is a small untuned / unmatched loop and the naked truth is that the SND is significant.

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

Above is the calculated SND from 0.3 to 15MHz using a 0.2:1 ratio transformer.

The results apply to the scenario described. The results are sensitive to most parameters so they cannot be blindly applied to another scenario.

]]>

Above, the 10dB attenuator is semi permanently attached to Port 2 principally to improve the Return Loss (or impedance match) of Port 2, a parameter that becomes quite important when testing some types of networks than depend on proper termination (eg many filters). I should remind readers that the improvement in Port 2 Return Loss comes at a cost, the dynamic range of Port 2 is reduced by 10dB.

Several correspondents have suggested the there is also benefit in permanently attaching such an attenuator to Port 1. These suggestions were traced to the assertions of one online expert.

The nanoVNA-H v3.3 samples the signal supplied to the measurement bridge, albeit prior to a voltage divider and we might assume that enters into calculation of the value of s11. I have not examined the source code to see if it includes the effect of that voltage divider and its variable load, but I suspect not as the voltage divider action is dispensed with in v3.4 and they both use the same firmware. I suspect normal VNA correction algorithms are relied upon to deal with non ideal hardware design.

While adding an external attenuator to Port 1 does nothing to address shortcomings in the implementation of the measurement bridge, it does degrade the dynamic range for s11 and s21 measurement.

I think it wrong to think that a network such as a filter will not measure properly if the source is not Thevenin impedance matched. The VNA is using a measurement bridge and its calibration / correction processes to come to a steady state value that depends entirely on the (complex) ratio V/I at the measurement plane, and that ratio is determined by things downstream from the measurement plane and NOT by things upstream of the measurement plane. This might not be accepted by disciples of Walt Maxwell’s re-re-reflection proposition.

So, whilst a network such as a filter might not exhibit a correct response in an ordinary sweep generator – filter – detector configurations (such as a spectrum analyser with tracking generator) unless both generator and detector are matched to the design impedances of the filter, in the case of a VNA measurement it is important that the filter is terminated in a ‘matched’ load, the measurement / calculation of s11 and s21 is corrected by the calibration and measurement processes of the VNA.

Above is an extract from the schematic for v3.3. A first approximation for the source impedance seen by the measurement bridge is the Si5351A output impedance (50Ω) in parallel with with R16+R17 (497Ω) for a combined 45.43Ω, in series with R13 (150Ω) for 195.43Ω, in parallel with R14 (56Ω) for combined 43.53Ω.

Measuring the output power @ 10MHz and applying a second termination results in a drop of 3.3dB on the power meter indication. We would expect 3.52dB for a 50Ω source and 3.32dB for a 45.53Ω source, validating the approximate circuit analysis.

Above is an extract from the schematic for v3.4. A first approximation for the source impedance seen by the measurement bridge is the Si5351A output impedance (50Ω) in series with R13 (0) in parallel with with R16+R17 (497Ω) for a combined 45.43Ω, in parallel with R14 (56Ω) for combined 25.08Ω.

The ‘improvements’ to v3.4 move further from an idealised 50Ω Zs for the measurement bridge…. but the instrument works with the same firmware as v3.3 hardware.

I don’t have a v3.4 to test, but for a double termination test, we would expect 3.52dB for a 50Ω source and 2.5dB for a 25.08Ω source. Happy to quote a valid experiment here if data is offered.

I wrote at Return Loss Bridge – some important details of the dependence of an ordinary Return Loss Bridge and its calibration / measurement process on near ideal Thevenin source impedance equaling the calibration or reference impedance. In that case, failure to supply a correct Thevenin source cannot be solved by putting an attenuator on the ‘unknown’ port.

]]>Above is a schematic for discussion. It is somewhat simplified, but it is complete and will work.

To analyse the circuit, we can use the mesh currents method. Mesh currents i1, i2 and i3 are annotated on the schematic.

The mesh equations are easy to write:

\(vs=(zs+zref+zref) \cdot i1-zref \cdot i2-zref \cdot i3\\

0=-zref \cdot i1+(zref+zref+zd) \cdot i2-zd \cdot i3\\0=-zref \cdot i1-zd \cdot i2+(zref+zd+zu) \cdot i3\\

\)

This is a system of 3 linear simultaneous equations in three unknowns. In matrix notation:

\(\begin{vmatrix}i1\\i2\\i3 \end{vmatrix}=

\begin{vmatrix}

zs+zref+zref & -zref & -zref\\

-zref & zref+zref+zd & -zd\\

-zref & -zd & zref+zd+zu

\end{vmatrix}^{-1} \times

\begin{vmatrix}1\\0\\0\end{vmatrix}\\

\)

which we can write in Octave as

\(\begin{vmatrix}i1\\i2\\i3 \end{vmatrix}=

\begin{vmatrix}

zs+zref+zref & -zref & -zref\\

-zref & zref+zref+zd & -zd\\

-zref & -zd & zref+zd+zu

\end{vmatrix} \setminus

\begin{vmatrix}1\\0\\0\end{vmatrix}\\

\)

Lets solve it in GNU Octave (since we are going to solve for some different input values) for vs=1V.

First pass: zs, zref, zd are 50Ω, and zu is short circuit (1e-300 to avoid division by zero) for calibration and 25Ω for measurement.

zs=50 zref=50 zd=50 #s/c cal zu=1e-300 A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu] x=A\[1;0;0] vcal=(x(2)-x(3))*zd #zu=25 zu=25 A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu] x=A\[1;0;0] vm=(x(2)-x(3))*zd rl=-20*log10(abs(vm)/abs(vcal))

This gives calculated rl=9.5424dB which is correct.

Now lets change zs to 40Ω.

zs=40 zref=50 zd=50 #s/c cal zu=1e-300 A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu] x=A\[1;0;0] vcal=(x(2)-x(3))*zd #zu=25 zu=25 A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu] x=A\[1;0;0] vm=(x(2)-x(3))*zd rl=-20*log10(abs(vm)/abs(vcal))

This gives calculated rl=9.0852dB which is wrong.

Now lets change zs back to 50Ω and zd to 40Ω.

zs=40 zref=50 zd=40 #s/c cal zu=1e-300 A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu] x=A\[1;0;0] vcal=(x(2)-x(3))*zd #zu=25 zu=25 A=[zs+zref+zref,-zref,-zref;-zref,zref+zref+zd,-zd;-zref,-zd,zref+zd+zu] x=A\[1;0;0] vm=(x(2)-x(3))*zd rl=-20*log10(abs(vm)/abs(vcal))

This gives calculated rl=9.0852dB which is wrong.

So these examples show that accuracy depends not only on the three bridge components being zref, but the Thevenin equivalent source impedance must be zref, and so must the detector zd.

Common failings are:

- the source is not well represented as a Thevenin equivalent circuit;
- the Thevenin equivalent source impedance is not zref;
- the detector does not present a load of zref at the bridge (eg possibly due to imperfections in a common mode choke); and
- an unbalanced common mode choke.

The Return Loss Bridge is a deceptively simple thing… it does take careful attention to all details to obtain accurate results.

]]>I make the index finger nail width exactly the same as the round part of the SMA nut which is 7.6mm. That is a very tiny hand… or the image is a composite fraudulently not to scale.

The small screen size is one of the most criticized ‘features’ of the nanoVNA.

My nanoVNA-H v3.3 screen escutcheon width is 60mm (2.4″), diagonal 74.7mm (2.94″), the active area of the screen is a little less. There are models with bigger screens, but this pic is not one of them, it is a fraud.

My hand is not particularly large, but it does make the nanoVNA-H v3.3 look a lot smaller than the first pic. (The second pic is not a mash up, it is not retouched or edited.)

So Chinese!

]]>Eager owners are trying to apply them to solve lots of problems, often without sufficient knowledge or experience to properly inform the measurements.

An example that has a appeared a few times on online forums in the last weeks is measuring the matched line loss (MLL) of a section of RG6 coax… to inform a decision to discard it or keep it.

The common approach is to use a measurement of |s11| and to calculate Return Loss and infer the MLL.

For discussion, lets consider an example of 30′ of Belden 1694A RG6 solved in Simsmith. We should note that unlike most RG6 in the market today, this uses a solid copper centre conductor.

Some authors insist that the half return loss method is to be performed using a short circuit test section. Bird does this in their Bird 43 manual.

Above is a plot of calculated |s11| (-ReturnLoss) from 1 to 20MHz for the test section. The three plots are of |s11| wrt 50Ω, 75Ω and frequency dependent actual Zo (as calculated for the model). The cursor shows that the actual |s11| is -0.37474dB (ReturnLoss=0.37474dB). Using the half return loss method MLL=ReturnLoss/2=0.37474=0.187dB/m.

Now to the other two traces.

|s11|(50)=-0.4241dB, quite different from the correct value 0.37474dB (the red curve).

|s11|(75)=-0.4521dB, quite different from the correct value 0.37474dB (the red curve).

Note that at other frequencies the error is different.

Above is a plot of calculated |s11| (-ReturnLoss) from 1 to 20MHz for the test section. The three plots are of |s11| wrt 50Ω, 75Ω and frequency dependent actual Zo (as calculated for the model). The cursor shows that the actual |s11| is -0.37474dB (ReturnLoss=0.37474dB). Using the half return loss method MLL=ReturnLoss/2=0.37474=0.187dB/m.

Now to the other two traces.

|s11|(50)=-0.2636dB, quite different from the correct value 0.37474dB (the red curve).

|s11|(75)=-0.2922dB, quite different from the correct value 0.37474dB (the red curve).

Note that at other frequencies the error is different.

If we average the |s11|(50) measurements for short and open short we get (-0.4241+-0.2636)/2=-0.3439 for MLL=0.1719dB, well below the correct value of 0.187dB.

If we average the |s11|(75) measurements for short and open short we get (-0.4521+-0.2922)/2=-0.3722 for MLL=0.186dB, very close to the correct value of 0.187dB.

The method of inferring Matched Line Loss from measured Return Loss of an open circuit or short section line section is soundly based but often fails in the execution. The Return Loss (or -|s11|) used as the basis MUST be with reference to the actual Zo of the line, though a good approximation can be obtained by averaging the Return Loss measurements for S/C and O/C section when Zref is close to Zo.

The knowledge and experience important to exploitation of the nanoVNA does not come in the box.

]]>Most RG6 type cable sold these days at low cost uses a copper clad steel centre conductor, and much of it has insufficient copper cladding thickness for copper like performance at HF.

Above is a pic N0TZU gave of the centre conductor cross section. It is possible to measure the cladding thickness from the pic knowing that the overall diameter is 1.024mm. The copper thickness measured 13.7µm, lets round it to 14µm.

There is a rule of thumb that where a copper tube (the cladding) thickness exceeds three skin depths (δ), effective RF resistance will be only very slightly higher than solid copper, and can be considered as good as solid copper.

Skin depth in copper at 10MHz is 20.6µm, 3δ is 61.8µm, and this conductor comes nowhere near it… so we should not expect copper like performance.

On the 3δ criteria, a 14µm cladding delivers copper like performance above 200MHz. That is not to say it isn’t usable below 200MHz, just that its MLL will be higher than an equivalent cable with solid copper centre conductor. Indeed, you might be hard put to measure the difference above 85MHz where the cladding is 2δ.

This analysis has focussed on the centre conductor cladding thickness, there may be other issues with the actual cable.

Of course in some applications, the higher MLL might be of little consequence and the cable may represent good value for money.

Beyond rules of thumb, here is a plot of the current distribution in the conductor at 10MHz. The current density magnitude falls exponentially as explained by skin effect, but the cladding is too thin to allow the full exponential decay and the current distribution is essentially truncated at 14µm depth… leading to the higher effective 10MHz RF resistance of about 1.6 times that of a solid copper conductor.

RG6 CCS cables are commonly sold without meaningful specifications, but occasionally they have useful specs. Some cables site ASTM B-869 for the centre conductor, basically 21% IACS conductivity, and for a #18 centre conductor, DC resistance will be around 100Ω/km. The cladding thickness in this case is 30.7µm which is 3δ at 40MHz, 2δ at 18MHz.

]]>