Hi Buck, Please check out my associated web page and then ask me anything that you don't understand.
Contrary to what you may have been told, you can change the 50 ohm SWR seen by your transmitter by changing the length of the 450 ohm ladder-line.
-- 73, Cecil http://www.w5dxp.com
An understanding of "mismatch loss" doesn't require SWR, reflections, power waves, "reflected power", "reflected energy", or other real or imagined complexities other than simple impedances. Here's what it means:
If you have a generator with a fixed output impedance such as a signal generator, and connect it to a conjugately matched load, the power dissipated in that load is the most you can get in any load connected to the generator. For example, if your generator produces 10 volts RMS open circuit and has a 50 ohm resistive output impedance, it can deliver 0.5 watt to a 50 ohm resistive load. If you connect any other load impedance to the generator, you'll get less power to the load. You can calculate exactly how much with simple circuit theory.
"Mismatch loss" is simply a way of expressing the reduction in power you get due to the load being mismatched, compared to how much you'd get with a matched load. For example, if you connect a 100 ohm resistor to the output of the generator, it would dissipate 0.44 watt instead of
0.5, so the mismatch loss is 10 log 0.5/0.44 = 0.51 dB(*). If you connect a 25 ohm resistor to the output, you also get 0.44 watt in the load resistor, again a "mismatch loss" of 0.51 dB. These numbers are calculated using nothing more complicated than simple lumped circuit principles.Mismatch loss is a useful concept when connecting fixed-impedance circuits together, such as in a laboratory environment. But it doesn't apply to either antennas or to VSWR. All you have to do to reduce the "mismatch loss" to zero is to insert a tuner or other matching network between the generator and the load. Presto, the generator sees 50 ohms resistive, the load dissipates 0.5 watt, and the mismatch loss is zero.
(*) For the 100 ohm example: The circuit consists of a 10 volt generator, and a 50 ohm resistance (the generator impedance) and 100 ohm resistance (the load) in series. So the current is V / R = 10 / (50 +
100) = 66.67 mA. The power dissipated in the load is I^2 * R = 0.06667^2Roy Lewallen, W7EL
billcalley wrote:
Bravo. And a great deal simpler to understand than most handwaving on these threads.
Best, Dan.
Not to mention the Noble Prize Ilya Prigogine won for "Dissipative Structures" in 1977.
Spontaneous Ordered Structures arising out of disorder. But it takes an Energy Flow to produce them.
Robert
Sorry. Forgot to include the link.
Robert
There's also the case where a perfect VSWR does no good. That happens when you connect a transmitter to an extremely lossy line and the signal dissipates before it gets to the load :>
Ok, I have seen that. It isn't that the antenna isn't tuned, it is tuned with the twin-lead instead of a conventional tuner.
I might have enough twin lead to build one of those. I'll have to consider it. It has always tweaked my interest.
Thanks
-- 73 for now Buck, N4PGW www.lumpuckeroo.com
It has a tuned feeder instead of a conventional tuner.
-- 73, Cecil http://www.w5dxp.com
I am curious to know, have you measured the power both at the antenna and the radio to see what loss there might be? I wouldn't expect very much, personally.
On a G5RV, it is taught that the twin-feedline is also part of the antenna itself, is that true in your model as well?
thanks Buck
-- 73 for now Buck, N4PGW www.lumpuckeroo.com
Cecil,
Looking more at your antenna, I am making several observations. I posted the URL at the top for other readers to know what we are talking about.
I kind of wish you would change the name of the antenna and not call it 'no-tuner' because I feel let down every time I look at it and see the feed-line tuner. I realize it isn't a conventional tuner. Don't get me wrong, I think it is a GREAT idea and I like it very much and would like to try it myself. However, I feel it is a little deceptive in the name.
I have a few questions about the system. I may have asked some already in another thread, so please bear with me. This is a better place to discuss it rather than hijacking another thread.
I see you use 450 ohm ladder line (or window-line as some call it.) I am wondering if the concept will also work with other impedance feedline such as 600, 300 or 75 ohm twin-line or even possibly with coax. ****
**** I just looked over your program. I see that you have it setup to allow 300 or 450 ohm ladder line. I wonder if you have LLWL available for 600 and 75 ohm.... I will try to put this in XL. (I don't have a basic program.) OK, I just found the compiled program.I am interested in making one, but I only have 300 ohm.
Does the feedline act as part of the antenna? I am sure it will, at least up to the 'no-tuner' if it acts like a G5RV, but do you know if the feedline radiates?
Have you measured the power at both sides of the 'no-tuner' to see what loss there might be? I doubt there would be much considering that you are using window-line.
I see a couple of the frequencies are above 1.5:1, which I am not comfortable going over with solid state rigs. Do you think that could be fine-tuned with the addition of a 1/2 foot section or maybe with that and a 1/4 foot section?
Would it be safe to assume that I can create a mono-band dipole, maybe even multi-band - if I am lucky, by fixing the length of the dipole and the feedline such that the increment gives me the imax at the balun for the desired frequency(s)?
Finally, you have a 1:1 choke at the feeder. I see that the better quality coax, the more toroids are needed. Would there be a problem with using a foot of RG-58 with the fewer toroids and then connect that to the high-quality line? I suspect that by then the common-mode currents will be gone and there would be minimum loss in such a short strand of lower-grade coax for HF. I doubt there would be a noticeable signal loss.
Does the 'no-tuner' feedline need to be spread out. I see from the photo that your 16 foot section is one large loop, I figure it must be close to a four-foot diameter loop.
Is there a similar system that would work with a vertical?
Thank you,
I appreciate your taking time to answer this. I hope many of us can learn from it.
Buck N4PGW
-- 73 for now Buck, N4PGW www.lumpuckeroo.com
Please refer to my new thread. Thanks
-- 73 for now Buck, N4PGW www.lumpuckeroo.com
That answer would be 'no'. I see that the purpose of the feedline length is to get the maximum current at the center point of the dipole.
Based on the fact that the antenna is being fed maximum current at the center, I would guess that this should work.
-- 73 for now Buck, N4PGW www.lumpuckeroo.com
Might as well go into some detail here. The feedpoint impedances encountered in a 130 foot dipole may range from ~50 ohms to ~5000 ohms. In order to limit the maximum SWR the feedline Z0 should be ~SQRT(50)(5000) = ~500 ohms. Thus the choice of 450 ohm line. Given that the feedpoint impedances may range from ~50 to ~5000 ohms, here are the maximum SWRs that may be expected for the different Z0s. Since the antenna system is fed at a current maximum point, I have included the impedance at the current maximum point which needs to be between 25 ohms and 100 ohms in order to avoid foldback.
Z0 SWRmax Imax Impedance
600 12:1 50 ohms 450 11:1 41 ohms 300 17:1 18 ohms 75 67:1 1 ohm 50 100:1 0.5 ohmsIt should be readily apparent why coax is a no-no for this antenna system. Even 300 ohm twinlead will result in a
50 ohm SWR of 50/18 = 2.8:1, high enough to cause foldback.
Let's take 40m as an example. The dipole is one wavelength and will have a 300 ohm SWR of 16:1 on 7.2 MHz according to EZNEC. 300/16 = 18.75 ohms at the current maximum point resulting in a 50 ohm SWR of 2.7:1 enough to cause foldback without an antenna tuner. Such is life.
I do use 300 ohm line on my 20m rotatable dipole and it does work 20m-10m but there are probably 10m frequencies where the 50 SWR is too high.
I monitor the current balance in my feedlines. They are so well balanced that there is no hint of common-mode current on the coax side of my current/choke/balun. Balanced currents radiate a negligible amount.
No I haven't measured the losses. I have trusted Owen's transmission line calculator for that data.
The disadvantage of this method is that it lacks one dimension of tuning necessary to achieve 50 ohms. Unless you do one more impedance transformation, no amount of fine-tuning the ladder- line length will get any closer.
One such example is at:
Just use RG-58 entirely unless you are running high power.
I have my 16 foot loop coiled in a 4 turn spiral around a piece of fiberglas fishing pole.
Verticals are not usually fed with ladder-line so probably more trouble than it is worth.
--
73, Cecil, w5dxp.comYes, and it should hurt , because we are using English and text to show something that s/b shown in pictures .
Im E.E. , KC7CC, and more..
i can simply show you with pictures. It will be instantly clear .
---BTW------ Im doin ARM computers , I will give a free Op System . It will NOT use English nor ASCII . No arbitrary definitions ... No C++ , No Linux , No M$ , no more "Free Lunch" .....
We use Coax for its isolation from nearby absorbers .. parallel line is much lower loss but absorbs into other objects close . We do not use caps , but stubs . But they are tuned ( freq dependent ) . The fast way to follow this , is to draw a picture , then edit it as you go . English will only get you a college degree and a free lunch ( Liberals ) .
__________________________________
How about considering the localized W5DXP system? :-)
-- 73, Cecil http://www.w5dxp.com
Dave, I'm pretty sure I know what you mean here, but it should be noted that this isn't entirely true. The line would have to have a couple of other characteristics besides being simply "lossless" for VSWR not to matter at all.
The problem, of course, is that a VSWR of other than 1:1 implies (by definition!) that the voltages and currents along the length of the line are not constant; there are cyclic variations in each, with maxima and minima located at half-wavelength intervals (that's the whole "standing wave" thing in the first place, right?). Particularly in high-power situations, it is possible for the maxima to exceed the ratings of the source or of the line itself.
Bob M. (KC0EW)
Not at all; "entropy increases" is with respect to the total entropy of a closed system. But in the case of "evolution," the only closed system which makes any sense to consider is the entire solar system, or at the very least the Sun-Earth system. It is entirely pemissible for an overall increase in entropy to occur (i.e., the Sun slowly loses energy to the rest of the universe) while at the same time this overall trend results in a very localized DECREASE in entropy (increase in order; in this case, the evolution of complex systems on Earth).
Bob M.
Bill ...
I would bet that most of the "confustion" comes from the conditions people put on their answers to the question.
Some postulate a "tuner" between generator and load.
Some postulate a specific internal impedance of the generator.
Some postulate a specific length of feeder line (either lossless or resistive, which is another parameter in and of itself).
Some postulate lots of other stuff, almost all of which is valid in the context of their answer.
What exactly do we mean when we say that we have a "100 watt transmitter"? What we are actually saying is that the transmitter will cause a resistor of a specific value to dissipate 100 watts of energy when tied to the transmitter output port and the transmitter keyed. Let's not muddy the waters up by asking if we are talking about peak power, PE power, average power, or whatnot. Let's just assume an unmodulated carrier putting out a constant power into the resistor that gets just as hot as when 100 dc watts (E*I) is pumped into it from a battery.
What value resistor? Whatever the designer/engineer/manufacturer specifies.
32 ohms? Sure. 50 ohms? No problem. 300 ohms? Certainly. Any competent engineer can give you a specified power into a specified resistive load.The crux of the question becomes, "What happens if my transmitter is specified into a 50 ohm resistor and I put a 100 ohm resistor as the load? How much "loss" do I get (or another way of asking that same question is how much power is dissipated in the 100 ohm resistor)?"
The answer is that it is impossible to tell without making the measurement. That may seem like a "wiggle" answer, but the truth of it is that the output stage design of the transmitter will dictate how it handles a "VSWR load". In some output stages, the output voltage will increase to the point of nearly driving 100 watts into the 2:1 VSWR resistor. Some will shut themselves down with a protection circuit. Some will go into parasitic oscillation. Some will fry the output devices.
Now increase the magnitude (and probably the sign also) of the problem to toss in a complex impedance instead of a resistive load and the confusion factor goes up rapidly. What DOES that output stage do when the load has an inductive component? Or a capacitive component? And both a resistive component and an imaginary component that varies with frequency?
The simple answer to your question outside the "laboratory environment" where everything is nicely matched and the internal impedances are set "just so" is that there IS NO SINGLE RIGHT ANSWER to this simple question.
And that is most probably the cause of your confusion.
Jim
But here again, I'm probably not seeing the
"Bob Myers" wrote in news:MmdPh.1809 $ snipped-for-privacy@news.cpqcorp.net:
were
Not if there's a lossless tuner.
In which case it's not truly "lossless" (and after breakdown becomes VERY lossy).
-- Dave Oldridge+ ICQ 1800667
Duh, life always violates it locally, but makes the sum total of entropy higer than it would have been. You aren't the idiot you appear, I hope.
tom K0TAR
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