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Author Topic: Optimum length of ladder line  (Read 2159 times)
VK1OD
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« on: February 03, 2010, 09:54:55 PM »

A question often asked is whether there is an optimum length of ladder line, for example in feeding a dipole / ladder line / balun / ATU.

As a follow up to recent discussion, I have constructed a model of a typical ATU on 80m, 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 plotted data tells a story, see Optimum length of ladder line.

73
Owen
« Last Edit: February 04, 2010, 11:25:17 AM by Owen Duffy » Logged
WA3SKN
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« Reply #1 on: February 04, 2010, 04:04:52 AM »

I find the optimum length of ladder line to be just a little longer than the distance between the antenna and the tuner... it's nice to have a service/drip loop in case there is a breakage over the years!
73s.

-Mike.
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W5DXP
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« Reply #2 on: February 04, 2010, 05:36:19 AM »

> VK1OD wrote: I have constructed a model of a typical ATU on 80m, and its performance calculated for the load presented by a 400? line at various load end VSWRs (80, 40, 20, 10, 5) and (electrical) line lengths in 1° increments. <

An interesting analysis which leaves out the fact that one the purposes of feeding at a current maximum point is to completely eliminate the tuner along with 100% of the tuner losses. In such a scenario the tuner, if it exists at all, is in bypass mode, i.e. zero tuner losses. ZERO TUNER LOSSES IS HARD TO BEAT.

http://www.w5dxp.com/notuner.htm

Let's take my 130 foot no tuner all-HF-band dipole, which doesn't require a tuner at all, as an example. The impedance measured at the 1:1 current-choke-balun (current maximum point) for each of the eight HF bands is: 3.8 MHz; 69 ohms, 7.2 MHz; 40 ohms, 10.125 MHz; 50 ohms, 14.2 MHz; 53 ohms, 18.14 MHz; 81 ohms, 21.3 MHz; 70 ohms, 24.95 MHz; 65 ohms, 28.4 MHz; 87 ohms.

The worst case SWR is on 10m where the 50 ohm SWR is 1.7:1. The power reflection coefficient when the SWR is 1.7:1 is 0.067, i.e. 0.28 dB of 50 ohm return loss worst case. That's about equal to the losses in your tuner *under best case conditions*. The return loss on all other bands is lower, with 50 ohm SWRs averaging 1.34:1 for a 0.09 dB average 50 ohm return loss, lower than the average in-line tuner losses.

Feeding my no tuner ladder-line fed dipole system at a current maximum point results in purely resistive impedances at the 1:1 current-choke-balun ranging from 40 ohms to 87 ohms. THE MAXIMUM (system resonant) SWR ON THE 400 OHM LADDER-LINE IS 10:1 (40m). Without a tuner, the return loss is lower than the minimum losses presented on your tuner loss graph.

I came up with this system of feeding at a current maximum point when I bought my SG-500 amplifier with its fixed 50 ohm output. I didn't feel like buying or building a 500 watt tuner and I don't like tuner losses. Feeding at a current maximum point was the solution. Of course, one needs to know what one is doing and avoid such undesirable configurations as an 80:1 SWR on the ladder-line. A center-fed dipole that is at least 1/2WL at the lowest frequency of operation will satisfy that requirement.

If the antenna design limits the SWR on the 400 ohm ladder-line to between 4:1 and 16:1, the 50 ohm SWR through a 1:1 current-choke-balun located at the current maximum point will be less than 2:1 on all HF frequencies. I have designed and used three such dipoles where the lowest frequency of operation was 3.5 MHz, 7 MHz, and 14 MHz. My 33 foot dipole is rotatable and used on all HF bands above 14 MHz. Every one of these dipoles fulfilled the necessary conditions for no tuner operation on all HF frequencies higher than the lowest frequency of operation.
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73, Cecil, w5dxp.com
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73, Cecil, www.w5dxp.com
The purpose of an antenna tuner is to increase the current through the radiation resistance at the antenna to the maximum available magnitude resulting in a radiated power of I2(RRAD) from the antenna.
W5DXP
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« Reply #3 on: February 04, 2010, 07:06:10 AM »

Incidentally, a German amateur radio magazine, "QRP Report", has published the information about my no-tuner all-HF-band dipole antenna. Here's what it looks like in German.

http://www.w5dxp.com/QRPReport.JPG
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73, Cecil, w5dxp.com
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73, Cecil, www.w5dxp.com
The purpose of an antenna tuner is to increase the current through the radiation resistance at the antenna to the maximum available magnitude resulting in a radiated power of I2(RRAD) from the antenna.
VK1OD
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« Reply #4 on: February 04, 2010, 10:20:06 AM »

...
The worst case SWR is on 10m where the 50 ohm SWR is 1.7:1. The power reflection coefficient when the SWR is 1.7:1 is 0.067, i.e. 0.28 dB of 50 ohm return loss worst case. That's about equal to the losses in your tuner *under best case conditions*. The return loss on all other bands is lower, with 50 ohm SWRs averaging 1.34:1 for a 0.09 dB average 50 ohm return loss, lower than the average in-line tuner losses.

Feeding my no tuner ladder-line fed dipole system at a current maximum point results in purely resistive impedances at the 1:1 current-choke-balun ranging from 40 ohms to 87 ohms. THE MAXIMUM (system resonant) SWR ON THE 400 OHM LADDER-LINE IS 10:1 (40m). Without a tuner, the return loss is lower than the minimum losses presented on your tuner loss graph.

Cecil, this all sounds pretty technical, but you need to review the meaning of the term "Return Loss". In the event, and allowing for your your incorrect use of "Return Loss", you have presented no evidence that your transmitter, or a transmitter in general will deliver any specific percentage of its rated power into the W5DXP no-tuner antenna.

Tuned feeder solutions such as yours are one feasible option for an antenna design, and your "no-tuner / variable feedline length" design is a variant, but they are outside the scope of my original posting which was "A question often asked is whether there is an optimum length of ladder line, for example in feeding a dipole / ladder line / balun / ATU."

I wouldn't opine that one or other option is better, rather my analysis might throw some light on the sensitivity of ATU losses to line length for high VSWR operation of ladder line, perhaps more insight that given by some other traditional advice or tools based on such. It might inspire some further thinking about "the theory behind the 'good' and 'bad' length designations" as you put it elsewhere.

Owen
« Last Edit: February 07, 2010, 09:21:21 PM by Owen Duffy » Logged
VK1OD
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« Reply #5 on: February 04, 2010, 10:33:33 AM »

I find the optimum length of ladder line to be just a little longer than the distance between the antenna and the tuner... it's nice to have a service/drip loop in case there is a breakage over the years!

Thanks Mike, that is one of the popular traditional solutions and I have added it to the article.

Owen
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W5DXP
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« Reply #6 on: February 04, 2010, 02:09:18 PM »

>  VK1OD wrote: ... you need to review the meaning of the term "Return Loss". ... you haven't presented any evidence that your transmitter, or a transmitter in general will deliver any specific percentage of its rated power. <

If you want to disprove what I am saying, here's an example. The feedpoint impedance of my 130 foot ladder-line fed dipole at 40 feet is 116-j510 on 30m according to EZNEC. The ladder-line length is 100'. The SWR at the current maximum point at the transmitter is measured at 1:1. Please explain how you can possibly improve on that by adding antenna tuner losses and additional transmission line losses?

From "The IEEE Dictionary", "return loss - the ratio of incident to reflected power at a reference plane of a network." I don't see any reason why that ratio cannot be expressed in dB. I am even assuming that, in the absence of a Z0-match provided by an antenna tuner, all the reflected power is dissipated by the source. Even with that worst case assumption (which is only correct for special worse cases of interference) my no-tuner antenna system usually beats an antenna tuner as far as losses go (for the same length feedline).

Remember, your objection was to feeding at a current maximum point. Eliminating the tuner losses by directly feeding at a current maximum point in my *properly designed antenna system* seems to shoot down your blanket objection to such a feedpoint. Hint: A flat coax fed 1/2WL resonant 50 ohm dipole is fed at a current maximum point. So why not feed at a current maximum point if the impedance is 50+j0 ohms???

Anyone who would throw up a random length antenna at a random height fed with a random length of a randomly selected feedline through a randomly selected balun has performed a foolish act. Hopefully, their selection of frequency is not nearly that random. :-)

As far as delivered power is concerned, I am in agreement with the conservation of energy principle. Power to the load = Source power minus system losses.
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73, Cecil, w5dxp.com
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73, Cecil, www.w5dxp.com
The purpose of an antenna tuner is to increase the current through the radiation resistance at the antenna to the maximum available magnitude resulting in a radiated power of I2(RRAD) from the antenna.
VK1OD
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« Reply #7 on: February 04, 2010, 02:21:32 PM »

The worst case SWR is on 10m where the 50 ohm SWR is 1.7:1. The power reflection coefficient when the SWR is 1.7:1 is 0.067, i.e. 0.28 dB of 50 ohm return loss worst case. That's about equal to the losses in your tuner *under best case conditions*. The return loss on all other bands is lower, with 50 ohm SWRs averaging 1.34:1 for a 0.09 dB average 50 ohm return loss, lower than the average in-line tuner losses.

If you apply the definition of Return Loss that you quoted from your IEEE dictionary, you will find that the Return Loss for a VSWR of 1.7 is 11.7dB, not the 0.28dB you give above. Although you have quoted the IEEE definition, the use above suggests a lack of understanding of the meaning of Return Loss.

The power delivered to a non 50 ohm load by a specific ham transmitter cannot simply be calculated in the way you have attempted for a number of reasons (including for example that the equivalent series output impedance is not necessarily 50+j0, saturation effects in the PA, effects of PA VSWR protection, PA power control).

Owen
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W5DXP
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« Reply #8 on: February 05, 2010, 06:56:21 AM »

> VK1OD wrote: Although you have quoted the IEEE definition, the use above suggests a lack of understanding of the meaning of Return Loss. <

Sorry, I should have said "The power reflection coefficient when the SWR is 1.7:1 is 0.067, i.e. 0.28 dB of loss assuming all the reflected power is lost".

Semantic details aside, the *only* point I was trying to make was: A current maximum point on a transmission line is often a good place to feed some antenna systems. Example: If I was not feeding at a current maximum point on my no-tuner dipole, my transmitter would reduce its power output.

> VK1OD wrote: The power delivered to a non 50 ohm load by a specific ham transmitter cannot simply be calculated in the way you have attempted ... <

True for the general case. The *specific* hypothetical boundary condition I had in mind was when zero interference exists within the linear source thus ensuring that all the reflected power is dissipated in the linear source resistor - again, irrelevant to the point I was trying to make.
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73, Cecil, w5dxp.com
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73, Cecil, www.w5dxp.com
The purpose of an antenna tuner is to increase the current through the radiation resistance at the antenna to the maximum available magnitude resulting in a radiated power of I2(RRAD) from the antenna.
VK1OD
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Posts: 1697




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« Reply #9 on: February 05, 2010, 12:56:42 PM »

... assuming all the reflected power is lost ...

... linear source thus ensuring that all the reflected power is dissipated in the linear source resistor ...

Both invalid assumptions when talking about a typical HF ham transmitter, and so any analysis based on those assumptions is also invalid.

Your use of the concept of Mismatch Loss (though you haven't called it that) in this situation is flawed.

Cecil, your notuner antenna might well be a good solution for some situations, but the reasons you set forth in this thread based on Mismatch Loss are nonsense.

If you simply made measurements of the power delivered to the line (meaning for example 'forward' power less 'reflected' power using a suitable directional wattmeter) using a particular transmitter, and tabulated that, calculated line losses (under the appropriate mismatch conditions), and resultant power at the feedpoint, you would have some relevant data for your own experiment.


Owen

PS: I might add that your tabulated data for your notuner antenna leaves some questions.

For example, a line from the table for 40m:

Freq-MHz..   ..T-line length = Matching Section + 1/2WL's..   ..Impedance at XMTR..   ..50 ohm SWR..   ..Impedance at Antenna..   ..450 ohm SWR..
7.2   92.0' = 30.5' + 1x61.5'   40 ohms   1.2:1   4939-j716   11.2:1

The impedance at the tx, impedance at antenna, and Zo=450 all reconcile for lossless line... but of course, the line cannot be lossless, we should expect one way loss of around 0.1xdB. The pictured line is ladder line which is nominally 450 ohms, but N7WS's measurements of Wireman line has shown Zo to be quite different to 450. I am a loss to know which (if any) any of the data is measured, and which is calculated and on what basis.

« Last Edit: February 05, 2010, 01:33:21 PM by Owen Duffy » Logged
W5DXP
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« Reply #10 on: February 05, 2010, 05:51:46 PM »

> VK1OD wrote: Both invalid assumptions when talking about a typical HF ham transmitter, and so any analysis based on those assumptions is also invalid. <

Owen, many great physicists, scientists, and engineers have presented conceptual math models (e.g., lossless transmission lines) that are invalid in reality and which date back hundreds of years. It's too bad that you reject all such historical conceptual examples, including Ramo's, Maxwell's, Balanis', Kraus', and mine.

Hint: No math model (so far) corresponds exactly to reality and is therefore invalid according to you. Good luck on your metaphysical crusade for perfection. I'm willing to bet that you don't even know why one plus one equals two and that you cannot prove the validity of such an assertion.
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73, Cecil, w5dxp.com
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73, Cecil, www.w5dxp.com
The purpose of an antenna tuner is to increase the current through the radiation resistance at the antenna to the maximum available magnitude resulting in a radiated power of I2(RRAD) from the antenna.
W5DXP
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« Reply #11 on: February 06, 2010, 07:02:25 AM »

However, because of side diversions, my original point remains unaddressed. Given my 130 foot dipole at 40 feet fed with 100 feet of 400 ohm ladder-line, the impedance on 30m looking into the 1:1 balun is 50+j0 ohms at a *current maximum point*. So why isn't the current maximum point a good place to feed that antenna?
--
73, Cecil, w5dxp
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73, Cecil, www.w5dxp.com
The purpose of an antenna tuner is to increase the current through the radiation resistance at the antenna to the maximum available magnitude resulting in a radiated power of I2(RRAD) from the antenna.
N5YPJ
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« Reply #12 on: February 07, 2010, 06:08:28 AM »

Owen thanks for the research and the nice report on it, as usual it's very informative and helpful.
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