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.)
The sunspot drought abates…
Posted by Owen in Propagation on February 1st, 2010
January 2010 showed further relief from the sunspot drought.
The total days with no sunspots in this cycle transition continues to climb, reaching 773 days in total at end December 2009.
More information at Spotless days around solar cycle transition.
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.
Feeding multiband dipoles with open wire feeder and balun
A common configuration for multiband use is a centre fed dipole with some length of open wire feeder, a balun and ATU to deliver a 50Ω load to the transmitter.
This article explores the effect of feedline length and balun transformation ratio on ATU efficiency.
ATU efficiency is sensitive to its load impedance. Achieving efficiency on the lowest bands is the greatest challenge, so this article explores behaviour on 3.6MHz of an ATU with 220pF variable capacitors, and inductor Q assumed to be 100. The analysis assumes the balun is ideal.
Above is a Smith chart (normalised to 50Ω) showing the approximate bounds of 80% efficiency (1dB loss) for a typical ATU on 3.6MHz in the red circle. Loads outside the red circle have efficiency lower than 80%. Note that loads with low R have low efficiency, and those with larger capacitive reactance are worse. (Click on the image and zoom in for an uncluttered full size version.)
ATUs are traditionally equipped with a 4:1 voltage balun which raises two issues.
- The loss in voltage baluns is very dependent on the voltage impressed, and so operation in the region of a voltage maximum (the far right hand part of the above chart) is less efficient, and should be avoided with voltage baluns.
- A 4:1 transformer reduces load impedance, see the blue arrows for two examples, which risks transforming some lower impedance loads into even lower impedance with reduced efficiency.
The green circle is the lossless 400Ω line SWR circle for a full wave dipole (Z=4200+j0). Note that at all line lengths it is well inside the 80% efficiency contour, but the caution for use of voltage baluns at voltage maxima still applies.
The purple circle is the lossless 400Ω line SWR circle for a G5RV (Z=10-j340). Note that is is almost always outside the 80% efficiency contour, save a small region around Z=9+j300Ω. This range of extreme impedances with reasonable ATU efficiency occurs around 70% of the distance from a current maximum and the next voltage maximum on the 400Ω line. Note that T matches are unlikely to be able to match the extreme impedance found at exactly a voltage maximum if VSWR is very high, but a 30° reduction in length or adding around 150° should bring them in range of an efficient match.
The above suggests that:
- 4:1 transformation exacerbates loss at low load impedances and has no significant benefits;
- voltage baluns should not be inserted near the voltage maximum on extreme loads;
- a current balun is more suited to the application than a voltage balun, allowing ATU insertion around the voltage maximum without significant efficiency degradation;
- insertion of the ATU at a current maximum is almost the worst case for extreme loads;
- insertion of the ATU at a voltage maximum for extreme loads may be outside of the matching range of an ATU, or the match may be inefficient;
- for extreme loads, insertion of the ATU such that the load has moderately high +ve X is likely to give marginally better efficiency.
For this application, it is hard to look past a 1:1 Guanella (current) balun with high common mode impedance and high differential voltage withstand, and the shortest open wire line that is possible. The current balun is more effective in reduction of common mode current for asymmetric loads than a voltage balun.
If the ATU cannot efficiently match some extreme load cases, increasing the length of the open wire line may resolve the problem. Extreme load cases should not occur in well designed antenna systems, matching is only one of the issues that they create and if they occur, they warrant further investigation.
An explanation of the Comet CHA250
The CHA250 is a 7m vertical conductor with a feed unit that accepts 50Ω coaxial cable and is claimed to have low VSWR feed from 80m to 6m. The manufacturer recommends that it be mounted at a height of about 11m.
Above is the CHA250BXII mount with matching unit (the lower black finned module).
At lower HF frequencies, the magic component inside the CHA250 can be analysed as very coaxial short transmission line sections with toroidal cores on the outside to increase common mode impedance.
The figure above is a simplified diagram of the conductor details in the matching device. Not shown in the diagram are three ferrite sleeves fitted over each of the two tubes.
Two key properties that allow a simplified analysis:
- A property of a coaxial transmission lines at frequencies where skin effect in the outer conductor is fully effective is that there are no external fields due to the current flowing on the inner conductors, all electric and magnetic fields are contained within the space bounded by the inner surface of the outer conductor. There will be a current flowing on the inner surface of the outer conductor that is equal in magnitude and opposite in direction to the sum of the currents flowing on the inner conductors.
- The transmission line sections are very short wrt wavelenth, so it can approximated by assuming that the magnitude and phase of current at one end of an inner conductor is identical to the other end.
If the current flowing to the 7m vertical tube above the feed unit is taken to be I1, it can be seen that the sum of the currents on the inner conductors of the right hand line section is 2I1 downward, and therefore the current on the inner surface of the right hand line is 2I1 upwards.
If we take the input voltage to be Vi, a current flows upwards on the outer surface of the right hand line of Vi/Zcm where Zcm is the common mode impedance of the line section, essentially the choke impedance formed the the outer surface of the line and the toroidal cores surrounding it.
Now, applying Kirchoff’s Current Law (KCL) at the node formed at the bottom of the right hand tube where the inner surface of the outer conductor, the outer surface of the outer conductor and the centre of the coax connector join, Ii=Vi/Zcm+2I1.
Looking at the left hand line section, the sum of the currents on the inner conductors is 2I1 upwards, so there is a current of 2I1 downwards on the inner surface of the outer conductor. Considering KCL at the node formed at the bottom of the left hand outer conductor, the flowing out of the node from the winding wire is I1, and the current flowing into the node on the inner surface of the outer conductor is 2I1, so the current on the outer surface of the outer conductor must be I1 out of the node (ie upwards).
Now we can treat the nodes at the top of the tubes. For simplicity, lets merge both nodes into one, so the paths and currents are:
- inner surface of left hand outer conductor inner surface, 2I1 down;
- outer surface of left hand outer conductor outer surface, I1 up;
- inner surface of right hand outer conductor inner surface, 2I1 up;
- outer surface of right hand outer conductor outer surface, Vi/Zcm up
- outer conductor of input socket, must be I1+Vi/Zcm down.
Treating the feedline, there is Vi/Zcm+2I1 flowing upwards on the outer surface of the inner conductor, therefore Vi/Zcm+2I1 flowing downwards on the inner surface of the feedline. Given that there is I1+Vi/Zcm downwards from the feed unit to the shield, there must be I1 flowing upwards on the outer surface of the feedline / support mast.
The common mode voltage drop due to I1 in the outer surface of the left hand outer conductor is in series with the vertical radiator, so it effectively loads the radiator with a series impedance of Zcm since its construction is the same as the right hand line section.
The doubled back turn in the right hand assembly is interesting, it has no effect under this analysis. In fact, in a more detailed, more accurate analysis at higher frequencies, it is a lossy s/c stub that introduces a very small inductive reactance.
In summary then, the CHA250 would appear to be an 18m vertical, fed at 7m from the top by a non-ideal lossy 1:4 transformer.
Australian LCD provisions for the low end of the 6m band
The current LCD has specific provisions for the low end of the 6m band, specifically covering 50 MHz to 52 MHz and additional to other requirements set out in the LCD. A quote from the LCD follows.
15 Operating an amateur advanced station in the frequency band 50 MHz to 52 MHz
(1) Subsections (2) and (3) apply if the licensee operates an amateur advanced station in the frequency band 50.000 MHz to 52.000 MHz.
(2) The licensee must not operate the station if it causes interference to the reception of the transmissions of television channel 0. (3) If the licensee operates the station in New South Wales, Victoria, Queensland or the Australian Capital Territory, the licensee must operatethe station:
(a) in the frequency band 50.000 MHz to 50.300 MHz only, using:
(i) emission mode 200HA1A and a maximum transmitter power of 100 watts pY; or
(ii) emission mode 1K12F1D and a maximum transmitter power of 30 watts pY; or
(iii) emission mode 4K00J3E and a maximum transmitter power of 100 watts pX; and
(b) at a place that is:
(i) at least 120 kilometres from a television channel 0 main station; and
(ii) at least 60 kilometres from a television channel 0 translator station; and
(iii) at least 60 kilometres from a television translator station that has inputs on television channel 0.
Taking it step by step…
Under (1) the clause 15 declares that it applies to amateur advanced stations operating in the 50 MHz to 52 MHz band, and should be read as additional to any other requirements.
Under (2), clause 15 prohibits any operation that causes interference to channel 0.
Under (3), two further conditions must all be met for all stations located in the listed states:
- operation must be “in the frequency band 50.000 MHz to 50.300 MHz only“, “only” meaning to the exclusion of any other part of the 50 MHz to 52 MHz band using exclusively (still riding on the word “only” in the sentence) emission and maximum transmitter power of one of the three listed combinations;
- at a place that is at least the specified distance from the various types of channel 0 transmissions.
One could be forgiven for thinking that the provisions of 15(3)b are more honoured in the breach than the observance.
The latest thrust seems to be to encourage an interpretation that the 50.300 MHz to 52.000 MHz segment (around the critical chrominance and sound subcarriers for channel 0) is not restricted in any way by clause 15, and to encourage high power operation at 50.350 MHz (around the critical chrominance and sound subcarriers for channel 0).
The article PSK31 on the low end of 6m in VK raises another area of non compliance with the ‘deal’ for access to the low end of the 6m band whilst channel 0 was still used for television broadcast.
Elsewhere we have recently seen one east coast ham arguing that his 1500W permit on 6m allows operation at 7500W PEP SSB.
Our compliance history should give the ACMA good reason to never consider such a compromise in the future… and that might not be far away if there are other intended uses of the 50 MHz to 52 MHz band after cessation of channel 0 transmissions.
What is RF Peak Envelope Power (PEP)
Posted by Owen in Measurement, RF Power Amplfiers, Transceivers on December 16th, 2009
Definition of Power
Power is the rate of doing work or the rate of flow of energy.
Power in a DC circuit
In a DC circuit, power is given by Joules first law and Ohms law as V*I.
Power in an AC circuit
In an AC circuit with steady sinusoidal excitation, voltage and current vary with time and the instantaneous power (p(t)=v(t)*i(t)) varies with time. The power in an AC circuit is usually taken as the power averaged over a whole AC cycle.
The figure above shows v(theta) and i(theta) in an AC circuit (theta=ω*t or theta=2*π*f*t). Also plotted is the instantaneous power p(theta) and the power averaged over a whole cycle, Pav.
The notation is that V is the amplitude of the wave, and v(theta) is the instantaneous voltage as a function of theta.
In the simple case where the load is purely resistive, the current is in-phase with the voltage, and it can be shown that Pav=V2/2/R or I2/2*R.
In fact, in this case, the average power is the average (or mean) of the sum of v(theta)2/R for all values of theta over one cycle, which can be expressed as the the mean of the squares of v(theta), divided by R. An equivalent DC voltage which would develop the power in R would therefore be the square root of the the mean of the squares of v(theta), and it is known as the RMS voltage. Power would be VRMS2/R, and for a sine wave, it can be seen that VRMS=0.707*V.
Peak Envelope Power
An RF waveform under any form of amplitude modulation is not a steady sine wave, and a common measure is of the power at the crest of modulation.
The ITU Radio Regulations define the terms Peak Envelope Power as:
- Peak Envelope Power ‘pX’ (s1.157) means the average power supplied to the antenna transmission line by a transmitter during one radiofrequency cycle at the crest of the modulation envelope taken under normal operating conditions.
This is not very different to the method of determining average power in an AC circuit, except that for the purpose of determining Peak Envelope Power, the power is averaged over a cycle at the crest of the modulation envelope.
Peak Envelope Power is usually measured in a resistive load, and in that case it is simply given by PEP=V2/2/R (where V is the peak RF voltage) or PEP=VRMS2/R.
The challenge in measurement of PEP of waves like SSB telephony is to design an instrument that captures the very short crests of modulation and holds the sample for display. For more information, see Peak amplifier for an RF wattmeter.
An example
Above is a CRO display of an AM transmitter modulated by a single sine wave and connected to a 50 ohm dummy load with a 50 ohm transmission line.
- I have an oscilloscope connected to a -40dB sampling port on the dummy load.
- The oscilloscope has a 10:1 probe and the Y input sensitivity is set to 20mV/cm.
- The crest of the modulated envelope measures 8cm peak to peak, the valley measures 2cm peak to peak (don’t read from the graticule in the picture, use the values stated in this text).
What is the PEP (ITU pX)?
For the answer, see RF Power Terms.
Ham myths about PEP
- PEP can only be measured on SSB signals
- PEP can only be measured using a two tone signal
- A traditional CRO is the only, or at least the superior way to measure PEP on SSB telephony
- PEP is peak voltage multiplied by peak current
- PEP of an unmodulated wave is different to its average power
- Common in-line wattmeters measure average power of complex modulated waves
A method for estimating the impedance of a ferrite cored toroidal inductor at RF
Taking an example of an FT140-43 or Fairrite 5943002701 with 10 turns at 21MHz.
Inductance of an inductor wound on a high permeability core can be given as L=N2Al where Al=(µA)/(2πr) where N is the number of turns, A is the cross section area of the core, and r is the radius of the core, and µ is the permeability of the core. µ=µ0*µr where µ0=4πe-7 and µr is the relative permeability.
The impedance of an ideal inductor can be expressed as Z=jωL, or Z=j2πfL. Substituting for L, Z=j2πfN2Al.
Relative permeability can be expressed as a complex quantity to capture both the ability of the core to increase flux and the core loss. The figure above taken from the data sheet gives µ’ and µ” over a range of frequencies.
The published data for the core gives Al=885e-9, but Al is based on the permeability at low frequencies (initial permeability, µi) where it is relatively constant, and must be adjusted for higher frequencies by multiplying by (µ’-jµ”)/µi.
So, we now have Z=j2πfN2Al(µ’-jµ”)/µi.
Let us put some numbers around that.
From the graph above, at 21MHz µ’=140 and µ”=180, and from the datasheet, µi=800.
Z=j2πfN2Al(µ’-jµ”)/µi.
Z=j2π21e6*102*885e-9(140-j180)/800=2627+j2044Ω. Note this is quite a different answer to calculation based on uncorrected Al, which suggests an inductance of 88µH and an impedance therefore of 0+j11680Ω.
It is also important to understand that a small amount of stray capacitance exists and it may significantly modify the inductor impedance.
If for example the inductor calculated to have Z=2627+j2044Ω is shunted with 2pF of stray capacitance, the impedance of the combination is given as 1/(1/(2627+j2044)+j2πf*2e-12)=3792-j1270Ω. Note that the effect of the stray capacitance is to make this inductor look like a high resistance with a series capacitive reactance… not what is perhaps expected of an inductor.
Some manufacturer publish graphs for specific cores of R and X for a single turn. These can be multiplied by turns squared, but the effects of stray capacitance need to then be factored in as above.
Beware of simplistic calculators and tools and techniques that take shortcuts that are not valid at radio frequencies.
Counting turns on a toroidal core
Current flowing on a conductor through the hole in a toroidal core give rise to magnetic flux in space surrounding the conductor, including the core. If the core material is of high permeability, almost all of the flux that is created is within the core material.
If the conductor is taken around the core and through the hole again in the same direction as the first, the flux in the core doubles due to the magnetising force of the current flowing in the second pass through the core. The flux due to the rest of the additional ‘turn’ is insignificant, the part of the turn outside the core does not produce significant flux in the core, nor is its flux significant compared to the flux in the high permeability core.
So, the number of ‘turns’ of a conductor wound on a high permeability toroidal core is the number of passes through the core in the same direction. There are no half turns, when you see someone describe a winding as having half turns, they don’t know what they are talking about, or are perhaps suggesting a physical layout, but the number of electrical turns is always a whole number.
Inductance of a coil on a toroidal high permeability core is proportional to turns squared, so is very sensitive to winding the correct number of turns.
Above is a current balun that is interesting. The designer thought it had three turns, but look closely, there are two passes in one direction, and one in the opposite direction, so the net number of turns is one, the inductance of this choke is about 11% of the designer’s intention.
The effective number of turns is the net passes of current carrying conductor through the hole in the same direction.
For lower permeability cores, because the flux in the core is lower, there is a significant portion of the total flux outside the core, so the situation is not as clear as with a high permeability core. However, you won’t go far wrong if you apply the case for high permeability cores more generally.