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#41
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Reg Edwards wrote:
Now Cecil, you can stop trying to pull my leg. ;o) I'm just wondering how 99.9% of the power can be radiated from the feedline when 80% of the power turns to heat in the feedline. Is that what is known as gain? :-) -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#42
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Reg Edwards wrote:
Now Cecil, you can stop trying to pull my leg. ;o) I'm actually trying to make a technical point, Reg. Walt may be trying to make that same technical point. The SWR is based on differential energy which doesn't radiate from the feedline. It just seems to me that you are trying to have your reflected power and radiate it too. That doesn't work for radiated power any better than it works for cake. If the SWR is 1400:1 then the reflected power is almost equal to the forward power both of which are associated with *differential* currents which don't radiate. If the SWR on the feedline is 1400:1, almost no power is being radiated! The forward power is 100 watts then the reflected power is 99.7 watts. Both of these powers are based on *differential* currents and therefore don't radiate from the feedline. There's only 0.3 watts available to radiate. If 99.9% of the power is being radiated by the outside braid of the coax, then the reflected power cannot be more than 0.1% of the power and as a result of that fact, the SWR on the feedline must necessarily be very low, i.e. close to 1:1. Seems to me you need to resolve that contradiction. You cannot radiate 99.9% your reflected power and still have it available to the SWR measuring equipment. If the reflected power is available to the SWR measuring equipment, it is composed of *differential* currents and is, by definition, not radiating. If it is radiating, then it is not available to the SWR measuring equipment and the SWR is, therefore, low. You simply cannot have a sky high SWR on the feedline with the feedline radiating 99% of the power. The SWR meter cannot tell if the antenna or coax braid is doing the radiating and will report a very low SWR. -- 73, Cecil, http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#43
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dansawyeror wrote:
If it is true it proves that a driving a "bad" load can cause the coax feedline to radiate a significant portion of the feed energy. At this point I am not sure what word to use other then 'bad', but I due intend to perform experiments to measure and verify what is happening and the model. dansawyeror wrote: Below is a link to a site that claims to model coax radiation from a dipole. http://www.smeter.net/feeding/feedpowr.php We are in a position now to say what is wrong with that program. Essentially, it violates the conservation of energy principle. Energy cannot exist as both reflected power and radiated power at the same time. If the SWR is 1400:1, the feedline cannot be radiating much because the reflected power is 99.7% of the forward power and both are based on differential currents which don't radiate. That leaves only 0.3% of the power available for radiating by the feedline. If 99.9% of the power is radiated by the coax braid, then the feedline SWR must be very close to 1.6:1 because the ratio of reflected power to forward power can be no more than 0.1/100. As with cake, you cannot have your reflected power and radiate it too. Reflected power plus radiated power cannot add up to more than the forward power. In fact, Pfor = Pref + Prad If 50% of the power is radiated by the antenna plus feedline, then 50% of the power is reflected and the SWR is 5.83:1. -- 73, Cecil, http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#44
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Cecil Moore wrote:
If 99.9% of the power is radiated by the coax braid, then the feedline SWR must be very close to 1.6:1 because the ratio of Sorry, that should be 1.06:1 instead of 1.6:1. reflected power to forward power can be no more than 0.1/100. -- 73, Cecil, http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
#45
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"Cecil Moore" wrote
... If 99.9% of the power is being radiated by the outside braid of the coax, then the reflected power cannot be more than 0.1% of the power and as a result of that fact, the SWR on the feedline must necessarily be very low, i.e. close to 1:1 ... _____________ .... and therefore incapable of melting down the inner conductor of the coax, and/or causing catastrophic failure of components in the output network of the tx PA. But, given enough tx power, these failure events are common when a tx tries to supply its full output power into a very high mismatch at the end of a run of coax. I have had to fix some of these systems after this happened to them. Please explain how this fits with your theory, Reg. RF |
#46
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![]() Cecil Moore wrote: dansawyeror wrote: If it is true it proves that a driving a "bad" load can cause the coax feedline to radiate a significant portion of the feed energy. At this point I am not sure what word to use other then 'bad', but I due intend to perform experiments to measure and verify what is happening and the model. dansawyeror wrote: Below is a link to a site that claims to model coax radiation from a dipole. http://www.smeter.net/feeding/feedpowr.php We are in a position now to say what is wrong with that program. Essentially, it violates the conservation of energy principle. Energy cannot exist as both reflected power and radiated power at the same time. If the SWR is 1400:1, the feedline cannot be radiating much because the reflected power is 99.7% of the forward power and both are based on differential currents which don't radiate. That leaves only 0.3% of the power available for radiating by the feedline. If 99.9% of the power is radiated by the coax braid, then the feedline SWR must be very close to 1.6:1 because the ratio of reflected power to forward power can be no more than 0.1/100. As with cake, you cannot have your reflected power and radiate it too. Reflected power plus radiated power cannot add up to more than the forward power. In fact, Pfor = Pref + Prad If 50% of the power is radiated by the antenna plus feedline, then 50% of the power is reflected and the SWR is 5.83:1. Bear in mind that current flowing in a conductor will always generate a field. Alternating current flowing in a (straight) conductor will always generate a radiating electromagnetic field. That radiating field will convey energy if it is not nulled by another equal and opposite field. You need to demonstrate that all the fields resulting from all the currents are nulled in this scenario. Back into the woodwork. ac6xg |
#47
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I'm not sure where in this convoluted thread to put this, and please
excuse me if the point has been made before but I just wanted to reiterate... When you put up an antenna system, pretty much _everything_ in the vicinity of the wires you think of as the "antenna" is actually part of the antenna system. Of special concern are all conductors, as well as big pieces of dielectric material and lossy material. Certainly ground has a profound effect on the radiation pattern, for example. I'm really not so concerned with "how much power is radiated from my feedline" as I am with "what is the radiation pattern (and in some cases, efficiency) of my entire antenna system." To the end of controlling the radiation pattern, I may wish to suppress antenna currents on things like support wires and feedlines. Or, I may model the system and find that antenna currents on the feedline are really not a problem. In the case where I do care, I can add "current baluns" or "choke baluns" or other structures as needed, or change the configuration to break up the unwanted currents. In some cases, a balun can be as simple as a hunk of ferrite clamped over the feedline. I've also used self-resonant coils of coaxial feedline to very effectively suppress current at a particular frequency. Breaking up support wires with insulators can be very useful. If you think that antenna current on the feedline is always a BAD thing, consider the coaxial collinear, where the sections of feedline that compose the antenna are INTENDED to radiate. On the other hand, with that antenna, it's very important to suppress antenna current on the line feeding the antenna part, because it doesn't take much antenna current on that line to mess up the radiation pattern. But with an 80 meter coax-fed dipole, the pattern may actually be better for some purposes if you don't suppress the antenna current on the feedline. Again, the question I care about is, "What is the pattern for this antenna," not "How much power does the feedline radiate." Is this really so different from caring more about properly loading a transmitter and getting power efficiently to the antenna, instead of caring specifically about transmission line SWR? Focus on what gets you the results you want, not on red herrings or old husband's tales. Cheers, Tom |
#48
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Jim Kelley wrote:
Bear in mind that current flowing in a conductor will always generate a field. Alternating current flowing in a (straight) conductor will always generate a radiating electromagnetic field. That radiating field will convey energy if it is not nulled by another equal and opposite field. You need to demonstrate that all the fields resulting from all the currents are nulled in this scenario. It is true by definition for ideal coaxial feedlines in which the radiating fields are 100% nulled by the basic shielded design, i.e. solid copper outer tubing much thicker than the skin effect depth. The problem with the program is that the conclusions violate the conservation of energy principle. It is simply impossible to have a coax SWR of 1400:1 while radiating 99.9% of the power from the feedline. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
#49
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K7ITM wrote:
Again, the question I care about is, "What is the pattern for this antenna," not "How much power does the feedline radiate." The present question has nothing to do with reality. The present question is: what is wrong with the simulation software? -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
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