The sunspot drought abates…
Posted by Owen in Propagation on March 1st, 2010
February 2010 showed further relief from the sunspot drought with sunspots on every day.
The total days with no sunspots in this cycle transition has climbed to 776 days in total at end February 2010.
More information at Spotless days around solar cycle transition.
Mast tube design
***DRAFT***
A common antenna tower configuration is a lattice tower structure with a tube mast out of the top of the tower to support antennas. The tube mast is usually rotatable, cantilevered and supported in bearings to resist the lateral and thrust forces. This article discusses briefly the design of that mast, and offers a spreadsheet design support tool. The article does not consider the implications for loading of the tower structure itself.
A range of materials are available for the purpose, but for most high performance applications, performance Aluminium or Steel alloys will be best suited. Not only are these tubes expensive, but transport costs can be even greater, especially in longer lengths.
The figure above show s a range of solutions for 50mm OD and 60mmOD tubes. The notation is the <product> <OD>x<WallThickness>.
The second bar is common 40mmNB water pipe, and shows it is a low strength option.
The VK2KRR option is his recommended sleeving of an aluminium tube with a high performance steel tube, being the first and third tubes shown above. It should be noted that unless the sleeve is effectively prevented from sliding at all under the highest safe bending moments, it should be treated as a sliding interface and the maximum bending moment of the combination is no greater than the sum of the maximum bending moment of each of the tubes. That is the main reason that the Steel 4130 50.8×4.77 option has much higher strength than the VK2KRR composite even the the latter is much thicker.
Attached is a spreadsheet design tool to assist with calculations.
Above is an upper section which describes one of the antennas installed on the mast.
Above is the summary section where all of the component bending moments are summed and the wind and tilt stresses calculated.
In this case, a Safety Factor of 1 is used for the wind calculations, so that the mast is likely to bend before tower failure, though that will not necessarily save the tower. The Safety Factor for tilt is 3 to make provision for, amongst other things, dynamic forces as the tower is raised or lowered.
Design wind speeds appropriate for location and environment need to be selected.
In this example, the required yield strength is within the strength of 6061 Aluminium alloy, and a relatively low cost tube that fits the standard 50mm opening in common towers will suffice. Grab the MastCalculations spreadsheet and try raising the antenna, or adding another antenna and explore the results.
The spreadsheet makes provision for up to three antennas, cable, mast, and rotator above the top of the mast. Obviously, if the rotator is inside the tower, it does not contribute to the bending moment of the mast tube.
No allowance is made for ice, wind forces are considered horizontal.
The spreadsheet can be downloaded, click on MastCalculations .
Warning, the spreadsheet does not replace the need for design competence. It is only a support tool, you MUST NOT depend on its results without independent verification. NO RESPONSIBILITY IS ACCEPTED NOR WARRANTY OF ANY KIND OFFERED.
| Steel 4130 |
Does SWR damage HF ham transmitters?
Posted by Owen in HF, RF Power Amplfiers, Transceivers on February 20th, 2010
The statement is often made that
VSWR will damage a HF ham transmitter, and the mechanism is that the ‘reflected power’ in a standing wave will be absorbed by the Power Amplifier (PA), increasing heat dissipation and damaging the PA.
There are two problems with this statement:
- the use of “VSWR will damage” means that damage will always occur if VSWR>1; and
- that reflected power is absorbed by the PA, necessarily increasing heat dissipation, and that causing damage.
Anyone with even a modicum of experience knows that 1 is false, they have observed transmitters operating at VSWR>1 and damage has not occurred. So at best, it is an overstatement intended to frighten rather than inform.
Lets analyse 2 in some detail.
Narrowband HF transmitters can usually be analysed with sufficient accuracy by using a Frequency Domain model. A Frequency Domain model means analysis that considers a single frequency sinusoidal excitation. Anyone who talks of reactance, impedance, and VSWR is talking in the Frequency Domain, whether they know it or not.
So, what exactly is VSWR?
A wave propagating (traveling along) an RF transmission line (eg coax) has a natural ratio of voltage to current determined by the geometry of the line. This natural ratio of V/I is known as the Characteristic Impedance or Z0. If the line is terminated in an impedance equal to Z0, then the voltage and current in the traveling wave and that in the load reconcile. If the line termination is not equal to Z0, then there must be a wave traveling opposite direction (the reflected wave), a wave such that the sum of the voltages and currents for the forward and reflected waves reconciles with the ratio of V/I in the line termination. The combination of this reflected wave and the forward wave gives rise to a standing wave on the line, the ratio V/I is not equal to Z0, but changes all the way along the line. VSWR is the ratio of the voltage maximum to the voltage minimum (assuming the line is long enough to develop a full cycle of the standing wave).
So, if VSWR>1, and it almost always is, the ratio of V/I is not equal to Z0, and varies all the way along the line, so that the ratio of V/I at the transmitter terminals is not equal to Z0. So, in a practical situation using a transmitter intended for a nominal 50Ω load, a length of nominal 50Ω coax, and a less than perfect antenna, the load impedance (V/I) at the transmitter terminals is not exactly 50+j0Ω (meaning 50Ω of resistance and no reactance). But that is not necessarily a problem.
Practical transmitters are designed to work with a range of load impedances. That range is often specified as a nominal load impedance, and a tolerance in terms of VSWR, eg 50+j0Ω with VSWR<1.5. Such a specification encompasses a range of impedances, including purely resistive loads and combination of resistance and reactance.
A transmitter that does not have a reasonable tolerance of load impedance is not very practical. What does “reasonable tolerance of load impedance” mean? It means that:
- the transmitter will deliver its rated power, or close to it, into a less than perfect load;
- the transmitter is unlikely to be damaged by an imperfect load.
Imagine how impractical it would be if for instance a mobile transmitter was damaged because VSWR increased as a result of a truck traveling alongside changing the antenna characteristics a little, or the antenna touching an overhead light fitting or other metal in a garage. There are countless applications where ham transmitters are required to operate into less than perfect loads.
In practice, a transmitter may deliver more or less than its rated power into a perfect load, and it may deliver more or less than that into a less than perfect load. Importantly also, the DC input to the transmitter may vary, so the efficiency may vary, and the amount of heat dissipated in the PA may vary… up or down. This is easily demonstrated by experiment, and the cases where dissipation is lower on some transmitters with some less than perfect loads questions whether the explanation at 2 is valid. If you delve into the design of a PA and explore its output, input and dissipation on different load impedances, you will find that the variation in power experienced is fully explained by conventional circuit theory considering the ration of V/I (or load impedance) at te transmitter terminals, and there is no conversion of reflected waves to heat in the PA. A steady state analysis of the PA fully explains the change in output power and dissipation in terms of the changed load impedance at the transmitter terminals.
Analysis of the explanation at 2 that shows that it is not an informed position.
So, if the explanation is bunk, the question remains, does VSWR damage ham transmitters?
If a transmitter is sensibly designed, it is designed to work with a range of load impedances. Operation outside of that range may present voltages or currents in parts of the circuit that are beyond safe operating limits for those parts, and may cause damage. For that reason, good designs incorporate protection schemes that limit operation on extreme load impedances to safe power levels. The most common scheme that is used in HF solid state transmitters that may be designed for a load of say 50Ω with VSWR<1.5, is that they automatically reduce power when the load impedance is outside that range. Operation outside the recommended load range is not usually guaranteed by the manufacturer, but experience shows that the protection schemes employed are usually very effective and the risk of damage is very low. So reliable and effective are these schemes that many mobile users tune screwdriver antennas just watching output power, relying entirely on the protection scheme to reduce output at high VSWR.
This article is not to say that high VSWR loads cannot result in transmitter damage, but to explain the effects of high VSWR loads in practical transmitters of good design, and to say that high VSWR loads do not necessarily cause damage, that damage is not caused by the mechanism explained at 2, and that damage to practical transmitters of good design due to high VSWR loads is unlikely.
Notwithstanding that, it makes good sense to supply the transmitter with a reliable antenna that is within manufacturer’s specifications, and to depend on the protection schemes for dealing with unusual circumstances (like keying up without an antenna connected, or keying up on an antenna that has developed a fault).
Collins 30L-1 and AM
Posted by Owen in HF, RF Power Amplfiers, Uncategorized on February 16th, 2010
The Collins 30L-1 is a heritage linear RF amplifier using four 811A triodes.
The manual gives no guidance on the use of the 30L-1 on AM, and that leads some users to think that it isn’t suited. This article explores the 30L-1 using the model described at RF Power Amplifier Tube Performance Computer.
The 30L-1 is specified for 1000W peak DC input in SSB telephony. This imposes one constraint on its use for AM, the other constraint is anode dissipation.
Tuned / loaded for rated SSB telephony
If the PA is tuned and loaded for 1000W peak DC input in SSB telephony, one would expect an output of about 680W PEP.
The carrier power for normal 100% modulated DSB AM is 25% of the PEP, so the carrier power for 680W PEP is 160W. However, at this level, the anode dissipation would be 294W, in excess of the valve ICAS ratings of 260W for four.
If drive was reduce until carrier power was just 70W, anode dissipation would be just under 260W, and PEP would reach 280W. This clearly does not exploit the 30L-1 very well.
Tuned / loaded for AM
Reloading the 30L-1 for a higher resonant load impedance at the anodes will allow operation at reduced power for AM (due to the dissipation constraint) at higher efficiency and within the dissipation limit.
The RF Power Amplifier Tube Performance Computer gives us a tool for exploring different operating conditions without risking damage to the valves.
If the PA is tuned / loaded for a little over 500W PEP (say 510W) with the least drive, the PI coupler is adjusted for the optimal resonant load for AM within the dissipation limit. AM drive is then adjusted for 122W carrier output, and modulation will take the signal to 488W PEP.
Output power is 74% higher than under the “Tuned / loaded for rated SSB telephony” heading above.
Under these conditions, anode current should be about 228mA DC, and assuming the DC input voltage is 1700V+30V drive, DC input power should be about 1730*0.228=394W, carrier out should be about 136W at the anodes, 122W at the output terminals, leaving anode dissipation at about 394-136=258W.
Summary
The above is a procedure to adjusting the 30-L1 for optimal AM telephony operation within the dissipation limits of four 811As. Under those conditions, it is capable of 490W PEP output, which is about 72% of the PEP capability in SSB telephony.
A test of Zs on an IC7000
Posted by Owen in Measurement on February 11th, 2010
A measurement of my IC7000 on 1.8MHz as per the article Is Zs of a HF ham tx typically 50+j0?.
Using a digital multimeter, I have plotted the DC meter voltage against RF power at 8 points from 10W to 100W, and created a cubic spline interpolation to be able to convert meter voltage with 0.1mV resolution to RF power.
With VSWR(50)=1, the measured forward power is 109W, and of course reflected power is 0W, so the power delivered to the load is 109W.
Now, two tests each with load VSWR(50)=1.5:
- Adjusting the load impedance by adjusting the input capacitor of the ATU until VSWR(50)=1.5, forward power is 85.9W and reflected power is 3.4W, so the power delivered to the load is 82.5W, or a loss of 1.2dB due to load impedance. Measurement of the IC7000 on this load with VSWR(50)=1.5 yields a power reduction of 1.2dB, whereas the classic MismatchLoss formula would yield 0.18dB, less than one sixth the actual loss. The considerable error in the MismatchLoss formula’s prediction is proof that the equivalent source impedance of the transmitter is not 50+j0Ω.
- But wait a minute, swinging that capacitor the other way for VSWR(50)=1.5, forward power is 109W and reflected power is 4.4W, so the power delivered to the load is 104.6W, or a loss of 0.18dB due to load impedance. Measurement of the IC7000 on this load with VSWR(50)=1.5 yields a power reduction of 0.18dB, where the classic MismatchLoss formula would yield 0.18dB, exactly the same. The low error in the MismatchLoss formula’s prediction is proof that the equivalent source impedance of the transmitter is 50+j0Ω.
So, here are two tests that indicate that equivalent source impedance is, and is not 50+j0Ω. The transmitter clearly exhibits non-linear behaviour (ie cannot be accurately represented by a Thevenin equivalent circuit) and power in the load depends on something else than just ρ(50). Calculations of MismatchLoss based on ρ(50) (the magnitude of the complex reflection coefficient Γ) are unsafe.
The tell tale sign that equivalent source impedance is not constant at 50+j0Ω is that forward power varies with change in load impedance.
Is Zs of a HF ham tx typically 50+j0?
Posted by Owen in Measurement, Transceivers, Transmission lines on February 11th, 2010
It is often assumed that a ham transmitters designed for a nominal 50+j0Ω load can be accurately represented generally by a Thevenin equivalent circuit with a source impedance of 50+j0Ω. This assumption may then be used for instance to infer the reduction in power delivered to an antenna of certain VSWR using the concept of Mismatch Loss.
Thevenin’s theorem for linear networks states that any combination of voltage sources, current sources and resistors with two terminals can be represented by an electrically equivalent circuit of a single voltage source Vs and a single series resistance Rs. In the frequency domain, the theorem can also be applied to impedance, not just resistance. A Thevenin equivalent circuit can be used to determine how much power would be developed in a load of any given impedance.
An interesting property of a Thevenin source (constant Vs and with Zs=50+j0Ω) is that the ‘forward power’ indicated by a quality directional wattmeter calibrated for 50+j0Ω will be independent of load impedance. ‘Forward power’ is Vf2/Z0 The following shows that Vf=Vs/2, ie Vf is independent of Zl, and therefore ‘forward power’ is independent of Zl.
A simple and easily performed test for practical hams therefore is to connect a transmitter via a directional wattmeter to an ATU and dummy load. If one can vary the ATU settings to vary the load impedance seen by the transmitter within VSWR(50) limits of say 1.5:1, the indicated ‘forward power’ should not vary. (If you swing the load VSWR too far, you may trigger some further non-linear effects like VSWR protection, power control, voltage or current saturation of the PA, which are further reasons why the assumed Zs may be invalid.)
Try the experiment at different frequencies, different power levels, you are likely to find that the results are frequency dependent and power level dependent.
Of course, nothing is perfect so an acceptable tolerance for depending on the assumed Zs=50+j0Ω for Mismatch Loss calculations might be that ‘forward power’ doesn’t vary by more than 10% of the ‘reflected power’ at any point. The worst case ‘reflected power’ for the experiment as described is 4%, so a variation of ‘forward power’ by more than 0.4% (ie 0.4W in 100W) at any point would indicate significant error in any Mismatch calculations based on an assumed Zs=50+j0Ω.
Now, it is very hard to discern such a small variation in ‘forward power’, but for most transmitters that I have ever tested, the variation is more like 5% and readily seen on a meter.
If you perform this test on a range of transmitters, at a range of frequencies, you will probably be convinced that whilst it is possible that Zs=50+j0Ω, it is not always, or even often the case, and that calculations such as Mismatch Loss that depend on that assumption are in error.
Power in an RF transmission line
Posted by Owen in Measurement on February 10th, 2010
This article presents a derivation of the power at a point in a transmission line in terms of ρ (the magnitude of the complex reflection coefficient Γ) and Forward Power and Reflected Power as might be indicated by a Directional Wattmeter. Mismatch Loss is also explained.
We start by deriving the apparent power Pa in terms of transmission line parameters.
Pa has both reactive and real components. We are only interested in the real component of power.
Mismatch Loss is a measure of the reduction of power in a load due to mismatch. From the above, it can be seen that MismatchLoss=-10log(1-ρ2). The definition of Mismatch Loss says nothing of any change in loss or dissipation inside the source, just the reduced power available in the mismatched load.
The quantity |Vf|2/Z0 is commonly known as Pfwd, and |Vr|2/Z0 is commonly known as Pref, but Pr=Pfwd-Pref is true only when Z0 is real.
Likewise, the calculation of power and Mismatch Loss from ρ as derived above is valid only if the underlying Z0 is real and the source impedance is exactly Z0. Mismatch Loss is widely misused, if the underlying criteria are not satisfied, the basis for the calculation does not exist and the figure obtained is in error.
Is a 4:1 balun a good choice for use with an ATU on HF?
Traditional HF ATU designs have often incorporated a 4:1 voltage balun. As more users become aware that voltage baluns are quite lossy for high VSWR loads in the region of differential voltage maxima, the choice of an appropriate transformation ratio is being revisited.
In the article Feeding multiband dipoles with open wire feeder and balun I discuss the effects of a 4:1 balun on the performance of an ATU.
The reasons often proposed for use of a 4:1 transformation are:
- increased range of matching; and
- lower ATU loss.
In the article Optimum length of ladder line , I present an analysis of typical ATU losses when used with and ideal 1:1 current balun and 400Ω transmission line operating at a range of high VSWRs. The graph above is from that article, and the model is detailed in the article.
Above is a graph of the same scenario, but using an ideal 4:1 balun between the transmission line and ATU. A practical 4:1 current balun should exhibit similar behaviour to the ideal balun used in the model. A 4:1 voltage balun will depart from significantly from the ideal where the differential voltage is high (away from the current minimum at 90° in the graph), balun losses alone could be several dB near voltage maximum, and this model would grossly understate losses.
There has been some extension in the ability to match the 400Ω transmission line operating with VSWR=80 at any line length, and some small reduction of ATU loss in the region of voltage maximum (0° and 180°), but at the expense of a large increase in loss near current minimum (90°) for high VSWR cases.
One is not always better than the other, but my choice for general purpose use would be a 1:1 current balun for better ATU loss near current maximum.
Optimum length of ladder line
Is there an optimum length of ladder line for use with a dipole and ATU?
Ladder line is often operated at high VSWR, even extreme VSWR, resulting is the load impedance seen by the ATU being very sensitive to line length, and in turn the capacity of the ATU to match the load efficiently is affected, especially on the lower bands.
A very common configuration for manual ATUs is the high pass T-match. This discussion will focus on performance of a typical T-match ATU on 3.5MHz to explain some of the effects.
To support the discussion, a model of a typical T-match ATU was constructed, and its performance calculated for the load presented by a 400Ω line at various source end VSWRs (80, 40, 20, 10, 5) and (electrical) line lengths in 1° increments. The model ATU has stepless variable capacitors of 10-220pF (Q=2000) and a switched inductor of 0.5µH to 32µH in 12 geometric steps of 1.4x (Q=100), and an ideal 1:1 current balun.
Above is a plot of the model data. If a data point is not plotted, it is because it was outside the range of the ATU. The steps are a result of using a switched inductor.
Traditional advice is to adjust the feedline length so that the ATU is at variously, a (differential) voltage maximum, a current maximum, either, an odd number of eighth waves, and any convenient length. None of these lead to the lowest losses at high VSWR.
Placing the ATU at a voltage maximum:
- exposes the ATU to the highest operating voltage;
- causes the greatest loss in a voltage balun if located at the same point;
- excluding the previous point, may be a relatively moderate loss operating point for the ATU;
- loss is quite sensitive to small changes in length; and
- may be out of range of the ATU for loads that result from high line VSWR.
So, there is a mix of good and bad. Best to avoid if you are using a voltage balun, and could be difficult to match for extreme VSWR.
Placing the ATU at a current maximum:
- has marginally higher ATU losses for loads that result from moderate line VSWR;
- loss is quite sensitive to small changes in length; and
- losses are exacerbated if a 4:1 balun is used.
Again, not really an optimal operating point.
Is there a better option?
At low VSWR(400), say less than 10, the length doesn’t matter very much. Losses are relatively low, and load impedance should be within an ATU’s range at any electrical length.
For 400Ω line lengths around 135° longer than a voltage maximum, the losses are low and not very sensitive to small changes in length, even for extreme VSWR, and the impedance presented should be well within the range of modest ATUs.
Note that lengthening a feedline to obtain lower ATU loss has its own loss costs, and a system perspective is necessary.
(If the extreme VSWRs plotted seem too high, keep in mind that the load end VSWR(400) of a G5RV on 80m is around 60.)
Estimating T match ATU efficiency
Users of T match ATUs are often interested in estimating the efficiency of the ATU. This article explores a simple loss model for the highpass T match. The model does not include a balun, balun losses are additional and can be quite high in the case of voltage baluns deployed in the region of a voltage maximum.
A simple loss model is to consider the capacitors to be lossless, and the only loss therefore to be in the inductor. This is not a bad approximation as the loss in practical ATUs with air spaced capacitors is almost entirely in the inductor. The model assumes Q of the inductor is constant for all values of inductance. In practical variable inductors, Q does vary with inductance, but over the narrow range of inductance used on one band, the assumption is reasonable.
With these simplifications, the efficiency of the ATU can be estimated from knowledge of C1 (the input capacitor), L and intrinsic Q of L. In this analysis, C1 is taken to have a maximum of 220pF (as used in many popular ATUs), and Qli is taken as 100. The analysis is performed at 3.6MHz, as efficiency is poorest at lower frequencies.
The graph above plots ATU efficiency against L for various values of C1. (Click on the image and zoom in for a larger image.)
After adjusting an ATU, if you consult the chart, you can get a good estimate of efficiency.
For example, MFJ’s 949E has 220pF caps, and switched inductors. Lets assume that Q of the inductors is 100. Lets say loading a 50+j0Ω load that the lowest inductor tap that works is 6µH, and C1=175 will give the match. If you look the graph up, you get an efficiency of 90%… not too bad.
Now lets try a G5RV fed with just 10m of 400Ω ladder line. Load Z is about 8+j32. Lets say the lowest inductor tap is 6µH, and C1=105pF will give the match. If you look the graph up, you get an efficiency of about 73%… not so good, especially when the line efficiency of 70% if factored in.