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An antenna question--43 ft vertical
As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat
metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. While I'm pretty happy with the antenna, I'd like to simplify the matching. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? |
An antenna question--43 ft vertical
On 6/29/2015 10:48 AM, Wayne wrote:
As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. You use a 16ft vertical as a lead-in? For what and how is that done? What are the dimensions of the metal roof? While I'm pretty happy with the antenna, I'd like to simplify the matching. To what matching do you refer? You don't want to use the tuner, or is there some other stuff you have not mentioned? Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) You wrote that you were interested in a 16ft vertical. Now it is a 43ft vertical? I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? All this depends on your answers to the above questions. |
An antenna question--43 ft vertical
"John S" wrote in message ... On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. You use a 16ft vertical as a lead-in? For what and how is that done? Grammatically, the description of the vertical is a lead in for the question, not an actual antenna lead. What are the dimensions of the metal roof? Somewhat irrelevant to my question. But it's about 20 by 35 feet. I'm not looking for an analysis of the existing antenna. While I'm pretty happy with the antenna, I'd like to simplify the matching. To what matching do you refer? You don't want to use the tuner, or is there some other stuff you have not mentioned? I want the tuner matching to be less awkward on some bands. I'm willing to live with the existing high SWRs on the upper bands. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) You wrote that you were interested in a 16ft vertical. Now it is a 43ft vertical? Please disregard all about the 16 ft vertical. I'm asking about a 43 ft vertical 1:4 unun. I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? All this depends on your answers to the above questions. So, lets begin again, with no distractions. What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical. |
An antenna question--43 ft vertical
In article ,
Wayne wrote: So, lets begin again, with no distractions. What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical. http://www.eham.net/articles/21272 has a nice analysis. It looks to me as if: - Without a 4:1 unun, the antenna provides a very nice match at three frequencies with in the HF band. At other frequencies, the SWR is up over 10:1 much of the time - high enough that a coaxial feed can be rather lossy. - With a 4:1 unun, you do lose the excellent match at those three frequencies... but the match gets better at most other frequencies. The SWR across the HF band is much more uniform, and lower on average... low enough to cut the coax losses somewhat and (I think) within the matching range of many rigs' "line flattener" built-in autotuners. |
An antenna question--43 ft vertical
"Dave Platt" wrote in message ... In article , Wayne wrote: So, lets begin again, with no distractions. What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical. # http://www.eham.net/articles/21272 has a nice analysis. # It looks to me as if: # - Without a 4:1 unun, the antenna provides a very nice match at three # frequencies with in the HF band. At other frequencies, the SWR is # up over 10:1 much of the time - high enough that a coaxial feed # can be rather lossy. # - With a 4:1 unun, you do lose the excellent match at those three # frequencies... but the match gets better at most other # frequencies. The SWR across the HF band is much more uniform, and # lower on average... low enough to cut the coax losses somewhat and # (I think) within the matching range of many rigs' "line flattener" # built-in autotuners. Thanks Dave. I'll have to spend some more time studying it, but the article is along the lines of what I was looking for. I would assume that the 1:4 causes behavior just as you say....worse SWR at nearly matched frequencies and better SWR elsewhere. I'll have to pull out some textbooks and see how the math works out for a Z seen through a 1:4 unun. In practice, I've had good results with SWRs even in the 30:1 range with short coax feeds. More research...and thanks. |
An antenna question--43 ft vertical
On Monday, June 29, 2015 at 8:46:47 PM UTC-4, Wayne wrote:
"Dave Platt" wrote in message ... In article , Wayne wrote: So, lets begin again, with no distractions. What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical. # http://www.eham.net/articles/21272 has a nice analysis. # It looks to me as if: # - Without a 4:1 unun, the antenna provides a very nice match at three # frequencies with in the HF band. At other frequencies, the SWR is # up over 10:1 much of the time - high enough that a coaxial feed # can be rather lossy. # - With a 4:1 unun, you do lose the excellent match at those three # frequencies... but the match gets better at most other # frequencies. The SWR across the HF band is much more uniform, and # lower on average... low enough to cut the coax losses somewhat and # (I think) within the matching range of many rigs' "line flattener" # built-in autotuners. Thanks Dave. I'll have to spend some more time studying it, but the article is along the lines of what I was looking for. I would assume that the 1:4 causes behavior just as you say....worse SWR at nearly matched frequencies and better SWR elsewhere. I'll have to pull out some textbooks and see how the math works out for a Z seen through a 1:4 unun. In practice, I've had good results with SWRs even in the 30:1 range with short coax feeds. More research...and thanks. I know that what I am about to say is provocative to some but I still think it is worth saying. If you look at the way that commercial and military radios are matched to antennas you will notice that most of the matching is done as close to the feed point as practical. Since only the power that actually reaches the antenna can be radiated I have a hard time seeing the point of matching the transmitter to the feed line. Matching at the feed line connection point will prevent damage to the transmitter but if that were the main objective a dummy load would accomplish that. When you couple the antenna to the load at the feed point you can have extremely low losses in the feed line. When you do the matching at the feed point you will transfer the most energy possible to the antenna and will get the highest available effective radiated power. Since the objective is the transfer of the highest practical amount of power to the antenna the place to do that is at the feed point were possible. I do realize that it is often simpler and easier to match at the feed line connection but I felt obliged to point out that is is not the most effective place to do the job. Tom |
An antenna question--43 ft vertical
On 6/30/2015 12:40 PM, Tom W3TDH wrote:
I know that what I am about to say is provocative to some but I still think it is worth saying. If you look at the way that commercial and military radios are matched to antennas you will notice that most of the matching is done as close to the feed point as practical. Since only the power that actually reaches the antenna can be radiated I have a hard time seeing the point of matching the transmitter to the feed line. Matching at the feed line connection point will prevent damage to the transmitter but if that were the main objective a dummy load would accomplish that. When you couple the antenna to the load at the feed point you can have extremely low losses in the feed line. When you do the matching at the feed point you will transfer the most energy possible to the antenna and will get the highest available effective radiated power. Since the objective is the transfer of the highest practical amount of power to the antenna the place to do that is at the feed point were possible. I do realize that it is often simpler and easier to match at the feed line connection but I felt obliged to point out that is is not the most effective place to do the job. Has it occurred to you that it might be important to match impedance both at the transmitter and at the antenna? When the feed line is not impedance matched to the transmitter output the maximum power is not transferred into the feed line. Then you have already lost power that can't be recovered by the matching at the antenna even if it is perfect. Your statements are not really provocative, they are just incomplete and/or wrong. -- Rick |
An antenna question--43 ft vertical
On 6/30/2015 12:40 PM, Tom W3TDH wrote:
On Monday, June 29, 2015 at 8:46:47 PM UTC-4, Wayne wrote: "Dave Platt" wrote in message ... In article , Wayne wrote: So, lets begin again, with no distractions. What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical. # http://www.eham.net/articles/21272 has a nice analysis. # It looks to me as if: # - Without a 4:1 unun, the antenna provides a very nice match at three # frequencies with in the HF band. At other frequencies, the SWR is # up over 10:1 much of the time - high enough that a coaxial feed # can be rather lossy. # - With a 4:1 unun, you do lose the excellent match at those three # frequencies... but the match gets better at most other # frequencies. The SWR across the HF band is much more uniform, and # lower on average... low enough to cut the coax losses somewhat and # (I think) within the matching range of many rigs' "line flattener" # built-in autotuners. Thanks Dave. I'll have to spend some more time studying it, but the article is along the lines of what I was looking for. I would assume that the 1:4 causes behavior just as you say....worse SWR at nearly matched frequencies and better SWR elsewhere. I'll have to pull out some textbooks and see how the math works out for a Z seen through a 1:4 unun. In practice, I've had good results with SWRs even in the 30:1 range with short coax feeds. More research...and thanks. I know that what I am about to say is provocative to some but I still think it is worth saying. If you look at the way that commercial and military radios are matched to antennas you will notice that most of the matching is done as close to the feed point as practical. Since only the power that actually reaches the antenna can be radiated I have a hard time seeing the point of matching the transmitter to the feed line. Matching at the feed line connection point will prevent damage to the transmitter but if that were the main objective a dummy load would accomplish that. When you couple the antenna to the load at the feed point you can have extremely low losses in the feed line. When you do the matching at the feed point you will transfer the most energy possible to the antenna and will get the highest available effective radiated power. Since the objective is the transfer of the highest practical amount of power to the antenna the place to do that is at the feed point were possible. I do realize that it is often simpler and easier to match at the feed line connection but I felt obliged to point out that is is not the most effective place to do the job. Tom Tom, very close. Yes, it's most effective to match the feedline to the antenna at the antenna connection. But it's also important to match the transmitter to the feedline. This latter piece is often ignored because people will use a feedline who's characteristic impedance matches the transmitter already (i.e. 50 ohm line for a 50 ohm transmitter). However, there are exceptions. For instance, if you're feeding a 75 ohm antenna (i.e. a dipole) with 75 ohm coax, a 1:1 balun at the antenna will provide a good match (ideally, 1:1). But there will be a 1.5:1 mismatch to a 50 ohm transmitter. In this case it would be better to have the matching network at the transmitter. You could also feed the antenna with 50 ohm feedline and place the matching network at the antenna. The effect would still be a 1:1 SWR, but the lower impedance of the coax would create higher i^2R losses; not important if you're talking a short line, but a longer one would lower output at the antenna. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
"Tom W3TDH" wrote in message ... On Monday, June 29, 2015 at 8:46:47 PM UTC-4, Wayne wrote: "Dave Platt" wrote in message ... In article , Wayne wrote: So, lets begin again, with no distractions. What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical. # http://www.eham.net/articles/21272 has a nice analysis. # It looks to me as if: # - Without a 4:1 unun, the antenna provides a very nice match at three # frequencies with in the HF band. At other frequencies, the SWR is # up over 10:1 much of the time - high enough that a coaxial feed # can be rather lossy. # - With a 4:1 unun, you do lose the excellent match at those three # frequencies... but the match gets better at most other # frequencies. The SWR across the HF band is much more uniform, and # lower on average... low enough to cut the coax losses somewhat and # (I think) within the matching range of many rigs' "line flattener" # built-in autotuners. Thanks Dave. I'll have to spend some more time studying it, but the article is along the lines of what I was looking for. I would assume that the 1:4 causes behavior just as you say....worse SWR at nearly matched frequencies and better SWR elsewhere. I'll have to pull out some textbooks and see how the math works out for a Z seen through a 1:4 unun. In practice, I've had good results with SWRs even in the 30:1 range with short coax feeds. More research...and thanks. # I know that what I am about to say is provocative to some but I still think it is worth saying. If you look at the way that commercial and military radios are matched to antennas you will notice that most of the matching is done as close to the # feed point as practical. # Since only the power that actually reaches the antenna can be radiated I have a hard time seeing the point of matching the transmitter to the feed line. Matching at the feed line connection point will prevent damage to the transmitter but if # that were the main objective a dummy load would accomplish that. # When you couple the antenna to the load at the feed point you can have extremely low losses in the feed line. When you do the matching at the feed point you will transfer the most energy possible to the antenna and will get the highest # available effective radiated power. Since the objective is the transfer of the highest practical amount of power to the antenna the place to do that is at the feed point were possible. #I do realize that it is often simpler and easier to match at the feed line connection but I felt obliged to point out that is is not the most effective place to do the job. I agree. However, the extra trouble of matching at the antenna feed point may not necessarily provide a noticeable improvement. On one of my antennas, I just provide a conjugate match for the antenna-feedline at a tuner within the shack. Then I accept whatever additional loss there is in the coax. For my 25 foot run of RG8, there is about 0.5 dB of loss with a match at 30 MHz. From the charts, there will be about an additional 2.7 dB of loss if the SWR is 20:1. For a 10:1 SWR the additional loss is around 0.9 dB. Below 30 MHz, the numbers get smaller. Of course, my automatic tuner (in the shack) fizzles out around SWR 5:1, but my manual tuner can be used instead at higher SWRs. In the case of the 43 foot vertical, it seems to me that a conjugate match in the shack would still be optimum, but perhaps not significantly beneficial. On the 43 foot vertical it could be that the 4:1 unun provides a lower amount of SWR induced additional feedline loss on high Z feedpoints. Of course, the unun might increase the SWR induced additional feedline loss for smaller Z. That's why the original question, and what I'm trying to understand. |
An antenna question--43 ft vertical
In article ,
Wayne wrote: On the 43 foot vertical it could be that the 4:1 unun provides a lower amount of SWR induced additional feedline loss on high Z feedpoints. Of course, the unun might increase the SWR induced additional feedline loss for smaller Z. From the charts, that kinda does appear to be the case. Add into consideration the fact that losses go up with the square of the current. Another issue is the other aspect of SWR - voltage. If you're trying to run "legal limit" or close to it, high SWR on the feedline coax could exceed the voltage rating of the coax dielectric, and you'd get arcing in the coax (or arcing at the connector between your feedline and shack tuner). Installing a hefty 4:1 unun right at the antenna would reduce the worse-case voltage on the feedline, and inside the shack tuner considerably. Depending on feedline length, that might be even more of a consideration than the increase in peak feedline current and the associated I^2*R losses. |
An antenna question--43 ft vertical
On Tuesday, June 30, 2015 at 2:36:51 PM UTC-4, rickman wrote:
On 6/30/2015 12:40 PM, Tom W3TDH wrote: I know that what I am about to say is provocative to some but I still think it is worth saying. If you look at the way that commercial and military radios are matched to antennas you will notice that most of the matching is done as close to the feed point as practical. Since only the power that actually reaches the antenna can be radiated I have a hard time seeing the point of matching the transmitter to the feed line. Matching at the feed line connection point will prevent damage to the transmitter but if that were the main objective a dummy load would accomplish that. When you couple the antenna to the load at the feed point you can have extremely low losses in the feed line. When you do the matching at the feed point you will transfer the most energy possible to the antenna and will get the highest available effective radiated power. Since the objective is the transfer of the highest practical amount of power to the antenna the place to do that is at the feed point were possible. I do realize that it is often simpler and easier to match at the feed line connection but I felt obliged to point out that is is not the most effective place to do the job. Has it occurred to you that it might be important to match impedance both at the transmitter and at the antenna? When the feed line is not impedance matched to the transmitter output the maximum power is not transferred into the feed line. Then you have already lost power that can't be recovered by the matching at the antenna even if it is perfect. Your statements are not really provocative, they are just incomplete and/or wrong. -- Rick Rick OK I'll buy incomplete and therefore wrong. Now given a Fifty Ohm feed line connected to a transmitter that is designed for that impedance at the antenna connector does not the actual mismatch occur at the antenna feed point? Certainly that can be compensated for at the transmitter but isn't there a likelihood or at least a risk that you will loose significant effective radiated power in spite of adjusting the apparent feed line impedance to the transmitter? If I do the matching at the feed point will I not maximize the effective radiated power of the antenna by installing the tuner at the feed point. I have already conceded that it is not as convenient to do the matching at the feed point. I do not allege that doing the matching at the transmitter end of the feed line is inherently ineffective only that there is a greater likelihood of loosing ERP needlessly and invisibly if the matching is done at transmitter end of the feed line. By this I mean to ask if I may well deceive the power meeter into showing more power out then I am actually getting. If any power lost is very likely to be insignificant at a practical level than help me to understand why that would be true and I will sell off my Icon AH-4, together with the control converter that allows my Yaesu FT-857D to control it, and my SGC SG-235 and go back to using the Yaesu FC-30 tuner with my FT-857D and the built in tuner on my Yaesu FT-1000. This is especially important for me to get right with my FT-857D since it is the transceiver that I use for my personal go kit. If putting the Icon AH-4 on the mast and running the control line in addition to the coaxial cable is a waste of time I would really appreciate knowing that. Thank you for helping with my education on this issue. -- Tom Horne W3TDH |
An antenna question--43 ft vertical
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An antenna question--43 ft vertical
On 6/29/2015 3:47 PM, Wayne wrote:
"John S" wrote in message ... On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. You use a 16ft vertical as a lead-in? For what and how is that done? Grammatically, the description of the vertical is a lead in for the question, not an actual antenna lead. What are the dimensions of the metal roof? Somewhat irrelevant to my question. But it's about 20 by 35 feet. I'm not looking for an analysis of the existing antenna. While I'm pretty happy with the antenna, I'd like to simplify the matching. To what matching do you refer? You don't want to use the tuner, or is there some other stuff you have not mentioned? I want the tuner matching to be less awkward on some bands. I'm willing to live with the existing high SWRs on the upper bands. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) You wrote that you were interested in a 16ft vertical. Now it is a 43ft vertical? Please disregard all about the 16 ft vertical. I'm asking about a 43 ft vertical 1:4 unun. I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? All this depends on your answers to the above questions. So, lets begin again, with no distractions. What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical. Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? |
An antenna question--43 ft vertical
In message , John S
writes On 6/29/2015 3:47 PM, Wayne wrote: "John S" wrote in message ... On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. You use a 16ft vertical as a lead-in? For what and how is that done? Grammatically, the description of the vertical is a lead in for the question, not an actual antenna lead. What are the dimensions of the metal roof? Somewhat irrelevant to my question. But it's about 20 by 35 feet. I'm not looking for an analysis of the existing antenna. While I'm pretty happy with the antenna, I'd like to simplify the matching. To what matching do you refer? You don't want to use the tuner, or is there some other stuff you have not mentioned? I want the tuner matching to be less awkward on some bands. I'm willing to live with the existing high SWRs on the upper bands. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) You wrote that you were interested in a 16ft vertical. Now it is a 43ft vertical? Please disregard all about the 16 ft vertical. I'm asking about a 43 ft vertical 1:4 unun. I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? All this depends on your answers to the above questions. So, lets begin again, with no distractions. What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical. Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? It was been pointed out to me that the figures for feeder loss with an imperfect SWR are only correct when the length is fairly long (at least an electrical wavelength?). How much loss does 25' of RG-8 really have at 12MHz, when there's a halfwave hanging on the far end? -- Ian |
An antenna question--43 ft vertical
On Tue, 30 Jun 2015 15:13:55 -0400, Jerry Stuckle
wrote: Yes, it's most effective to match the feedline to the antenna at the antenna connection. But it's also important to match the transmitter to the feedline. This latter piece is often ignored because people will use a feedline who's characteristic impedance matches the transmitter already (i.e. 50 ohm line for a 50 ohm transmitter). However, there are exceptions. For instance, if you're feeding a 75 ohm antenna (i.e. a dipole) with 75 ohm coax, a 1:1 balun at the antenna will provide a good match (ideally, 1:1). But there will be a 1.5:1 mismatch to a 50 ohm transmitter. In this case it would be better to have the matching network at the transmitter. We may have had this discussion before. Matching a 75 ohm load to a 50 ohm source might be academically interesting, but the actual loss is almost negligible. for a VSWR of 1.5, the return loss is 14dB and the load mismatch attenuation is 0.177dB. That's about what I would expect to lose in two coax connector pairs. You could also feed the antenna with 50 ohm feedline and place the matching network at the antenna. The effect would still be a 1:1 SWR, but the lower impedance of the coax would create higher i^2R losses; not important if you're talking a short line, but a longer one would lower output at the antenna. True, but for roughly equivalent sizes of coax cables, the 75 ohm cable has less loss and the equivalent 50 ohm cable. If you want to handle high power, use 50 ohms. If you want low loss, use 75 ohms: http://www.belden.com/blog/broadcastav/50-Ohms-The-Forgotten-Impedance.cfm Note that these are for air dielectric cables. Things are not so neat if we consider the dielectric. See the bottom paragraph and graphs: http://www.microwaves101.com/encyclopedias/why-fifty-ohms Dielectric Dielectric const Minimum loss impedance solid PTFE 2.2 50 ohms foam PTFE 1.43 60 air 1.0 75 RG-6/u CATV 75 ohm foam coax still has slightly less loss than the equivalent 50 ohm cable, but not as much as I've previously claimed. This is cute: http://cablesondemandblog.com/wordpress1/2014/03/06/whats-the-difference-between-50-ohm-and-75-ohm-coaxial-cable/ "A good rule of thumb is that if the device being connected via coaxial cable is a receiver of some kind, 75 Ohm Coax is ideal." -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
An antenna question--43 ft vertical
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An antenna question--43 ft vertical
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An antenna question--43 ft vertical
"Dave Platt" wrote in message ... In article , Wayne wrote: On the 43 foot vertical it could be that the 4:1 unun provides a lower amount of SWR induced additional feedline loss on high Z feedpoints. Of course, the unun might increase the SWR induced additional feedline loss for smaller Z. From the charts, that kinda does appear to be the case. Add into consideration the fact that losses go up with the square of the current. Another issue is the other aspect of SWR - voltage. If you're trying to run "legal limit" or close to it, high SWR on the feedline coax could exceed the voltage rating of the coax dielectric, and you'd get arcing in the coax (or arcing at the connector between your feedline and shack tuner). Installing a hefty 4:1 unun right at the antenna would reduce the worse-case voltage on the feedline, and inside the shack tuner considerably. Depending on feedline length, that might be even more of a consideration than the increase in peak feedline current and the associated I^2*R losses. Good points. I'm running low powers at the moment, but there was a time where I burned up a lot of stuff running a KW :) |
An antenna question--43 ft vertical
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An antenna question--43 ft vertical
"Jeff" wrote in message ... If there is a mismatch at the antenna (and there is no matching at the antenna), then maximum power transfer will occur when the conjugate match is applied at the transmitter end of the feedline. Surely a conjugate match will only match the load if the coax length is 1/2 wavelength or multiple thereof, and the feeder is also lossless. Any other coax length will introduce a phase shift that will require a different match. Yes, I'm assuming that the antenna tuner conjugate match is for the end of the feedline, not for the antenna itself. Wayne W5GIE exiled to W6 :) |
An antenna question--43 ft vertical
Jeff wrote:
If there is a mismatch at the antenna (and there is no matching at the antenna), then maximum power transfer will occur when the conjugate match is applied at the transmitter end of the feedline. Surely a conjugate match will only match the load if the coax length is 1/2 wavelength or multiple thereof, and the feeder is also lossless. Any other coax length will introduce a phase shift that will require a different match. Jeff You just have to match whatever impedance the aerial impedance has been transformed to at the transmitter end. Then you will get maximum power into the radiation resistance of the aerial (less the second order losses in the feeder). A remaining reactive mismatch between the feeder and the aerial will result in increased voltages and currents and increased feeder loss (a second order effect at HF) but will not prevent substantially full power transfer. We had this discussion about very short aerials quite recently, You have to have a very extreme radiation resistance for this not to work. Choosing a length of aerial with no extreme values on the bands you are using is where we came in. -- Roger Hayter |
An antenna question--43 ft vertical
On 7/1/2015 12:26 PM, Jeff Liebermann wrote:
On Tue, 30 Jun 2015 15:13:55 -0400, Jerry Stuckle wrote: Yes, it's most effective to match the feedline to the antenna at the antenna connection. But it's also important to match the transmitter to the feedline. This latter piece is often ignored because people will use a feedline who's characteristic impedance matches the transmitter already (i.e. 50 ohm line for a 50 ohm transmitter). However, there are exceptions. For instance, if you're feeding a 75 ohm antenna (i.e. a dipole) with 75 ohm coax, a 1:1 balun at the antenna will provide a good match (ideally, 1:1). But there will be a 1.5:1 mismatch to a 50 ohm transmitter. In this case it would be better to have the matching network at the transmitter. We may have had this discussion before. Matching a 75 ohm load to a 50 ohm source might be academically interesting, but the actual loss is almost negligible. for a VSWR of 1.5, the return loss is 14dB and the load mismatch attenuation is 0.177dB. That's about what I would expect to lose in two coax connector pairs. You could also feed the antenna with 50 ohm feedline and place the matching network at the antenna. The effect would still be a 1:1 SWR, but the lower impedance of the coax would create higher i^2R losses; not important if you're talking a short line, but a longer one would lower output at the antenna. True, but for roughly equivalent sizes of coax cables, the 75 ohm cable has less loss and the equivalent 50 ohm cable. If you want to handle high power, use 50 ohms. If you want low loss, use 75 ohms: http://www.belden.com/blog/broadcastav/50-Ohms-The-Forgotten-Impedance.cfm Note that these are for air dielectric cables. Things are not so neat if we consider the dielectric. See the bottom paragraph and graphs: http://www.microwaves101.com/encyclopedias/why-fifty-ohms Dielectric Dielectric const Minimum loss impedance solid PTFE 2.2 50 ohms foam PTFE 1.43 60 air 1.0 75 RG-6/u CATV 75 ohm foam coax still has slightly less loss than the equivalent 50 ohm cable, but not as much as I've previously claimed. This is cute: http://cablesondemandblog.com/wordpress1/2014/03/06/whats-the-difference-between-50-ohm-and-75-ohm-coaxial-cable/ "A good rule of thumb is that if the device being connected via coaxial cable is a receiver of some kind, 75 Ohm Coax is ideal." Jeff, do you always miss the forest for the trees? That was an EXAMPLE. The same would be true if you were feeding a 300 ohm yagi with 300 ohm twinlead and a transmitter with a 10 ohm output impedance. And BTW - when calculating, you forgot about the transmitters which cut back power to protect the finals. Many will do so even with a 1.5:1 SWR. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On 7/1/2015 4:23 PM, Jerry Stuckle wrote:
On 7/1/2015 12:26 PM, Jeff Liebermann wrote: On Tue, 30 Jun 2015 15:13:55 -0400, Jerry Stuckle wrote: Yes, it's most effective to match the feedline to the antenna at the antenna connection. But it's also important to match the transmitter to the feedline. This latter piece is often ignored because people will use a feedline who's characteristic impedance matches the transmitter already (i.e. 50 ohm line for a 50 ohm transmitter). However, there are exceptions. For instance, if you're feeding a 75 ohm antenna (i.e. a dipole) with 75 ohm coax, a 1:1 balun at the antenna will provide a good match (ideally, 1:1). But there will be a 1.5:1 mismatch to a 50 ohm transmitter. In this case it would be better to have the matching network at the transmitter. We may have had this discussion before. Matching a 75 ohm load to a 50 ohm source might be academically interesting, but the actual loss is almost negligible. for a VSWR of 1.5, the return loss is 14dB and the load mismatch attenuation is 0.177dB. That's about what I would expect to lose in two coax connector pairs. You could also feed the antenna with 50 ohm feedline and place the matching network at the antenna. The effect would still be a 1:1 SWR, but the lower impedance of the coax would create higher i^2R losses; not important if you're talking a short line, but a longer one would lower output at the antenna. True, but for roughly equivalent sizes of coax cables, the 75 ohm cable has less loss and the equivalent 50 ohm cable. If you want to handle high power, use 50 ohms. If you want low loss, use 75 ohms: http://www.belden.com/blog/broadcastav/50-Ohms-The-Forgotten-Impedance.cfm Note that these are for air dielectric cables. Things are not so neat if we consider the dielectric. See the bottom paragraph and graphs: http://www.microwaves101.com/encyclopedias/why-fifty-ohms Dielectric Dielectric const Minimum loss impedance solid PTFE 2.2 50 ohms foam PTFE 1.43 60 air 1.0 75 RG-6/u CATV 75 ohm foam coax still has slightly less loss than the equivalent 50 ohm cable, but not as much as I've previously claimed. This is cute: http://cablesondemandblog.com/wordpress1/2014/03/06/whats-the-difference-between-50-ohm-and-75-ohm-coaxial-cable/ "A good rule of thumb is that if the device being connected via coaxial cable is a receiver of some kind, 75 Ohm Coax is ideal." Jeff, do you always miss the forest for the trees? That was an EXAMPLE. The same would be true if you were feeding a 300 ohm yagi with 300 ohm twinlead and a transmitter with a 10 ohm output impedance. Transmitter output impedance does not determine SWR. And BTW - when calculating, you forgot about the transmitters which cut back power to protect the finals. Many will do so even with a 1.5:1 SWR. |
An antenna question--43 ft vertical
On Wed, 01 Jul 2015 17:23:54 -0400, Jerry Stuckle
wrote: Jeff, do you always miss the forest for the trees? That was an EXAMPLE. Perhaps you missed my point. I don't care about VSWR as long as the system is reasonably efficient, does not protect itself, and produces adequate TX power. In my world, that means I'm concerned about losses, not VSWR. The losses involved in using 75 ohm feedline and coax in a 50 ohm system are negligible. The same would be true if you were feeding a 300 ohm yagi with 300 ohm twinlead and a transmitter with a 10 ohm output impedance. One of my favorite methods of argumentation is to provide a ridiculous and extreme example, and then use it as the basis for discussion. I guess it's a form of "straw man" argument, where I'm now expected to defend my point of view against your ridiculous and extreme example. Please forgive me for not following your lead and continuing to be more concerned with the 50/75 ohm problem: https://en.wikipedia.org/wiki/Straw_man And BTW - when calculating, you forgot about the transmitters which cut back power to protect the finals. Many will do so even with a 1.5:1 SWR. Could you provide me with one example of such a transmitter? I've seen such radios on the bench, but they're usually mistuned or misadjusted. I did some Googling and found that the typical threshold for both AM and FM broadcast xmitters is 1.5:1. https://www.google.com/#q=transmitter+high+vswr+threshold+1.5:1 Same with most HF radios that I could find. The stuff I designed for marine use (150 watts PEP) was set to operate up to 2.0:1 because at the time, ATU's were just appearing and the typical vessel HF antenna was problematic (23ft vertical with a dubious ground system). Today, 1.5:1 threshold would probably work. Note that such a threshold does NOT mean that at 1.4999:1, the radio would work normally, and at 1.5001:1 would shut down. For (marine) radios that are expected to work with random antennas in emergencies, shutting down at 1.5:1 is absurd. What is normally done is to slowly reduce the drive starting at 1.5:1 until it gets to some point below where either the final current is too high to maintain safe dissipation, or the voltage across the final xsistor or FET is too high to prevent breakdown. My guess(tm) is that's about 5:1 or more with todays radios but I'll admit that I'm guessing and haven't actually tried it. In the past, I used a test load, that someone else built, that would provide a resistive 2:1 VSWR at 25/37.5/50/75/100 ohms, and an adjustable phase angle with a big variable capacitor and roller inductor. I had to be VERY careful not to accidentally tune the inductor and capacitor to resonance, or I would end up with a short or open load. The dials had an accompanying chart that followed varioius constant VSWR circles around the Smith chart. The tricky part was not making the power amp work over a 4:1 impedance range. The tricky part was making a VSWR sensor that would be fairly flat over the entire 2 to 30 MHz range. Another headache was when a mismatch caused the PA to draw more current. It wasn't final heating that caused instabilities and odd behavior. It was my worthless bench power supply that would detect the overcurrent and protect itself by dropping the output voltage. So, which radio shuts down at less than 1.5:1 VSWR? I know of a few possible candidates, but I would like to see what you've observed first. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
An antenna question--43 ft vertical
On Tue, 30 Jun 2015 17:21:08 -0700, "Wayne"
wrote: # Since only the power that actually reaches the antenna can be radiated I have a hard time seeing the point of matching the transmitter to the feed line. Matching at the feed line connection point will prevent damage to the transmitter but if # that were the main objective a dummy load would accomplish that. Yep. Note that the electric utility companies do not bother to match the transmitter (generators) with the impedance of the transmission lines and the load. That was one of the reasons that Edison and Westinghouse has so much trouble with the experts when they proposed electric power transmission. The experts assumed that the source had to be matched to the load, which would cause the generators to dissipate as much power as is dissipated in the load. Incidentally, one reason Tesla/Westinghouse eventually went with 60 Hz instead of 133Hz, 400 Hz, or higher frequencies (which use less iron in the xformers) was the danger of creating standing waves on the transmission lines because of the mismatch. When the wavelength of 60 Hz (3100 miles or 5000 km) is longer than the width of the country, it's a safe bet that there aren't going to be any standing waves. So, why don't we run transmitters with lower than 50 ohm output impedances? Well... 1. The gain of the PA stage would be reduced possibly requiring an additional gain stage. 2. The current in the PA stage would increase, possibly causing the power supply to complain. 3. The low pass harmonic filter will require physically larger parts. 4. The coax cable between the PA stage and the RF connector will need to have a very thin dielectric to work at low impedances. Same with the RF output connector. Instead of dealing with these aforementioned hassles, it's probably better to run the transmitter at some impedance that provides a benefit and let everything else conform to that standard. That's where the maximum power at 50 ohms for transmitters, and lowest loss for 75 ohms (air dielectric) for CATV were derived. The rest of the connected devices (PA, filter, antenna) simply conformed to these standards. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
An antenna question--43 ft vertical
On 7/2/2015 6:31 AM, John S wrote:
On 7/1/2015 4:23 PM, Jerry Stuckle wrote: On 7/1/2015 12:26 PM, Jeff Liebermann wrote: On Tue, 30 Jun 2015 15:13:55 -0400, Jerry Stuckle wrote: Yes, it's most effective to match the feedline to the antenna at the antenna connection. But it's also important to match the transmitter to the feedline. This latter piece is often ignored because people will use a feedline who's characteristic impedance matches the transmitter already (i.e. 50 ohm line for a 50 ohm transmitter). However, there are exceptions. For instance, if you're feeding a 75 ohm antenna (i.e. a dipole) with 75 ohm coax, a 1:1 balun at the antenna will provide a good match (ideally, 1:1). But there will be a 1.5:1 mismatch to a 50 ohm transmitter. In this case it would be better to have the matching network at the transmitter. We may have had this discussion before. Matching a 75 ohm load to a 50 ohm source might be academically interesting, but the actual loss is almost negligible. for a VSWR of 1.5, the return loss is 14dB and the load mismatch attenuation is 0.177dB. That's about what I would expect to lose in two coax connector pairs. You could also feed the antenna with 50 ohm feedline and place the matching network at the antenna. The effect would still be a 1:1 SWR, but the lower impedance of the coax would create higher i^2R losses; not important if you're talking a short line, but a longer one would lower output at the antenna. True, but for roughly equivalent sizes of coax cables, the 75 ohm cable has less loss and the equivalent 50 ohm cable. If you want to handle high power, use 50 ohms. If you want low loss, use 75 ohms: http://www.belden.com/blog/broadcastav/50-Ohms-The-Forgotten-Impedance.cfm Note that these are for air dielectric cables. Things are not so neat if we consider the dielectric. See the bottom paragraph and graphs: http://www.microwaves101.com/encyclopedias/why-fifty-ohms Dielectric Dielectric const Minimum loss impedance solid PTFE 2.2 50 ohms foam PTFE 1.43 60 air 1.0 75 RG-6/u CATV 75 ohm foam coax still has slightly less loss than the equivalent 50 ohm cable, but not as much as I've previously claimed. This is cute: http://cablesondemandblog.com/wordpress1/2014/03/06/whats-the-difference-between-50-ohm-and-75-ohm-coaxial-cable/ "A good rule of thumb is that if the device being connected via coaxial cable is a receiver of some kind, 75 Ohm Coax is ideal." Jeff, do you always miss the forest for the trees? That was an EXAMPLE. The same would be true if you were feeding a 300 ohm yagi with 300 ohm twinlead and a transmitter with a 10 ohm output impedance. Transmitter output impedance does not determine SWR. Transmitter output impedance vs. feedline impedance does determine SWR at one end of the system. If you have a mismatch, you will have a non-1:1 SWR. And BTW - when calculating, you forgot about the transmitters which cut back power to protect the finals. Many will do so even with a 1.5:1 SWR. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On 7/2/2015 8:43 AM, Jeff Liebermann wrote:
On Wed, 01 Jul 2015 17:23:54 -0400, Jerry Stuckle wrote: Jeff, do you always miss the forest for the trees? That was an EXAMPLE. Perhaps you missed my point. I don't care about VSWR as long as the system is reasonably efficient, does not protect itself, and produces adequate TX power. In my world, that means I'm concerned about losses, not VSWR. The losses involved in using 75 ohm feedline and coax in a 50 ohm system are negligible. I didn't miss your point. But you can't see the forest for the trees. The same would be true if you were feeding a 300 ohm yagi with 300 ohm twinlead and a transmitter with a 10 ohm output impedance. One of my favorite methods of argumentation is to provide a ridiculous and extreme example, and then use it as the basis for discussion. I guess it's a form of "straw man" argument, where I'm now expected to defend my point of view against your ridiculous and extreme example. Please forgive me for not following your lead and continuing to be more concerned with the 50/75 ohm problem: https://en.wikipedia.org/wiki/Straw_man I prefer to use realistic examples to show a point. But you have to nitpick with off-topic comments. And BTW - when calculating, you forgot about the transmitters which cut back power to protect the finals. Many will do so even with a 1.5:1 SWR. Could you provide me with one example of such a transmitter? I've seen such radios on the bench, but they're usually mistuned or misadjusted. Many of the solid state finals amateur transmitters will start cutting back well before 2:1 SWR. Even my early 80's era IC-720A would start dropping power before then. I did some Googling and found that the typical threshold for both AM and FM broadcast xmitters is 1.5:1. https://www.google.com/#q=transmitter+high+vswr+threshold+1.5:1 Same with most HF radios that I could find. The stuff I designed for marine use (150 watts PEP) was set to operate up to 2.0:1 because at the time, ATU's were just appearing and the typical vessel HF antenna was problematic (23ft vertical with a dubious ground system). Today, 1.5:1 threshold would probably work. We're not talking AM and FM broadcast transmitters (which are immaterial because the antenna system is tuned to get as close to a 1:1 match as possible - much easier with one frequency). Pretty much the same with marine use - a very limited band of frequencies. Additionally, the limited power on marine radios allow you to use higher power finals so they can dissipate the additional heat caused by a mismatch. Note that such a threshold does NOT mean that at 1.4999:1, the radio would work normally, and at 1.5001:1 would shut down. For (marine) radios that are expected to work with random antennas in emergencies, shutting down at 1.5:1 is absurd. What is normally done is to slowly reduce the drive starting at 1.5:1 until it gets to some point below where either the final current is too high to maintain safe dissipation, or the voltage across the final xsistor or FET is too high to prevent breakdown. My guess(tm) is that's about 5:1 or more with todays radios but I'll admit that I'm guessing and haven't actually tried it. In the past, I used a test load, that someone else built, that would provide a resistive 2:1 VSWR at 25/37.5/50/75/100 ohms, and an adjustable phase angle with a big variable capacitor and roller inductor. I had to be VERY careful not to accidentally tune the inductor and capacitor to resonance, or I would end up with a short or open load. The dials had an accompanying chart that followed varioius constant VSWR circles around the Smith chart. I never said shut down. I said cut back. But you can't read very well, either, can you? The tricky part was not making the power amp work over a 4:1 impedance range. The tricky part was making a VSWR sensor that would be fairly flat over the entire 2 to 30 MHz range. Another headache was when a mismatch caused the PA to draw more current. It wasn't final heating that caused instabilities and odd behavior. It was my worthless bench power supply that would detect the overcurrent and protect itself by dropping the output voltage. So, which radio shuts down at less than 1.5:1 VSWR? I know of a few possible candidates, but I would like to see what you've observed first. Once again, you can't read. But I know you're just foaming at the mouth to contradict me, as you always do. But that's OK. I know what you are, and I'm not going to bite. Now go away, troll, and play with your CB radios. They're your speed. Leave this discussion to adults - who can read and understand the points being made. And BTW - you can also go running to your mommy and tell her the mean old man called you a troll. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On Thu, 02 Jul 2015 06:12:33 -0700, Jeff Liebermann
wrote: (...) where the maximum power at 50 ohms for transmitters, and lowest loss for 75 ohms (air dielectric) for CATV were derived. The rest of the connected devices (PA, filter, antenna) simply conformed to these standards. Continuing from where I accidentally hit the "send" button... It might be interesting to measure the output impedance of your HF xmitter. All you need is a dummy load, and an RF voltmeter, RF probe and voltmeter, or oscilloscope. 1. Turn down the xmitter RF output to some level where you won't blow up your test equipment and so that it doesn't go into high VSWR protect mode. My guess is about 10 watts is about right. 2. Measure the RF voltage across the output connector both with a load (Vload) and without a load (Vno_load). 3. If measuring peak voltage, convert RMS by multiplying by 0.707. If measuring peak-to-peak, divide by 2 and then multiply by 0.707. Output_Impedance = 50 ohms (Vno_load - Vload) / Vload It's been many years since I've done this, so I can't recall the range of values that I obtained. I do recall that it was surprisingly large and precipitated a few heated discussions in the lab. Also, the output impedance will change with output power level but I don't recall how much. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
An antenna question--43 ft vertical
On 7/1/2015 10:56 AM, Ian Jackson wrote:
In message , John S writes On 6/29/2015 3:47 PM, Wayne wrote: "John S" wrote in message ... On 6/29/2015 10:48 AM, Wayne wrote: As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat metal roof. The antenna is fed with about 25 feet of RG-8, and there is a tuner at the transmit end. You use a 16ft vertical as a lead-in? For what and how is that done? Grammatically, the description of the vertical is a lead in for the question, not an actual antenna lead. What are the dimensions of the metal roof? Somewhat irrelevant to my question. But it's about 20 by 35 feet. I'm not looking for an analysis of the existing antenna. While I'm pretty happy with the antenna, I'd like to simplify the matching. To what matching do you refer? You don't want to use the tuner, or is there some other stuff you have not mentioned? I want the tuner matching to be less awkward on some bands. I'm willing to live with the existing high SWRs on the upper bands. Thus, the question: what is the purpose of a 1:4 unun on a 43 foot vertical? ( I assume the "4" side is on the antenna side.) You wrote that you were interested in a 16ft vertical. Now it is a 43ft vertical? Please disregard all about the 16 ft vertical. I'm asking about a 43 ft vertical 1:4 unun. I'd expect a better coax to antenna match when the antenna feedpoint is a high Z (example, at 30 meters), but I'd also expect a worse coax to antenna match when the feedpoint is a low Z (example, at 10 meters). Is that the way it works, or is there other magic involved? All this depends on your answers to the above questions. So, lets begin again, with no distractions. What is the purpose (or benefit) of using a 1:4 unun on a 43 ft vertical. Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? It was been pointed out to me that the figures for feeder loss with an imperfect SWR are only correct when the length is fairly long (at least an electrical wavelength?). How much loss does 25' of RG-8 really have at 12MHz, when there's a halfwave hanging on the far end? A *resonant* half wave at 12MHz is about 36.7 feet long and it presents an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The current at the antenna end is 0.0245A while one watt is applied at the source end. This means that the power applied to the antenna is about 0.687W. So, about 68% of the applied power reaches the antenna. So, about 32% of the power is lost in the RG-8 for this example. Does this help? |
An antenna question--43 ft vertical
On 7/2/2015 8:14 AM, Jerry Stuckle wrote:
On 7/2/2015 6:31 AM, John S wrote: On 7/1/2015 4:23 PM, Jerry Stuckle wrote: On 7/1/2015 12:26 PM, Jeff Liebermann wrote: On Tue, 30 Jun 2015 15:13:55 -0400, Jerry Stuckle wrote: Yes, it's most effective to match the feedline to the antenna at the antenna connection. But it's also important to match the transmitter to the feedline. This latter piece is often ignored because people will use a feedline who's characteristic impedance matches the transmitter already (i.e. 50 ohm line for a 50 ohm transmitter). However, there are exceptions. For instance, if you're feeding a 75 ohm antenna (i.e. a dipole) with 75 ohm coax, a 1:1 balun at the antenna will provide a good match (ideally, 1:1). But there will be a 1.5:1 mismatch to a 50 ohm transmitter. In this case it would be better to have the matching network at the transmitter. We may have had this discussion before. Matching a 75 ohm load to a 50 ohm source might be academically interesting, but the actual loss is almost negligible. for a VSWR of 1.5, the return loss is 14dB and the load mismatch attenuation is 0.177dB. That's about what I would expect to lose in two coax connector pairs. You could also feed the antenna with 50 ohm feedline and place the matching network at the antenna. The effect would still be a 1:1 SWR, but the lower impedance of the coax would create higher i^2R losses; not important if you're talking a short line, but a longer one would lower output at the antenna. True, but for roughly equivalent sizes of coax cables, the 75 ohm cable has less loss and the equivalent 50 ohm cable. If you want to handle high power, use 50 ohms. If you want low loss, use 75 ohms: http://www.belden.com/blog/broadcastav/50-Ohms-The-Forgotten-Impedance.cfm Note that these are for air dielectric cables. Things are not so neat if we consider the dielectric. See the bottom paragraph and graphs: http://www.microwaves101.com/encyclopedias/why-fifty-ohms Dielectric Dielectric const Minimum loss impedance solid PTFE 2.2 50 ohms foam PTFE 1.43 60 air 1.0 75 RG-6/u CATV 75 ohm foam coax still has slightly less loss than the equivalent 50 ohm cable, but not as much as I've previously claimed. This is cute: http://cablesondemandblog.com/wordpress1/2014/03/06/whats-the-difference-between-50-ohm-and-75-ohm-coaxial-cable/ "A good rule of thumb is that if the device being connected via coaxial cable is a receiver of some kind, 75 Ohm Coax is ideal." Jeff, do you always miss the forest for the trees? That was an EXAMPLE. The same would be true if you were feeding a 300 ohm yagi with 300 ohm twinlead and a transmitter with a 10 ohm output impedance. Transmitter output impedance does not determine SWR. Transmitter output impedance vs. feedline impedance does determine SWR at one end of the system. If you have a mismatch, you will have a non-1:1 SWR. And BTW - when calculating, you forgot about the transmitters which cut back power to protect the finals. Many will do so even with a 1.5:1 SWR. Not true. Post some links to support your position, please. |
An antenna question--43 ft vertical
On 7/2/2015 11:07 AM, John S wrote:
On 7/2/2015 8:14 AM, Jerry Stuckle wrote: On 7/2/2015 6:31 AM, John S wrote: On 7/1/2015 4:23 PM, Jerry Stuckle wrote: On 7/1/2015 12:26 PM, Jeff Liebermann wrote: On Tue, 30 Jun 2015 15:13:55 -0400, Jerry Stuckle wrote: Yes, it's most effective to match the feedline to the antenna at the antenna connection. But it's also important to match the transmitter to the feedline. This latter piece is often ignored because people will use a feedline who's characteristic impedance matches the transmitter already (i.e. 50 ohm line for a 50 ohm transmitter). However, there are exceptions. For instance, if you're feeding a 75 ohm antenna (i.e. a dipole) with 75 ohm coax, a 1:1 balun at the antenna will provide a good match (ideally, 1:1). But there will be a 1.5:1 mismatch to a 50 ohm transmitter. In this case it would be better to have the matching network at the transmitter. We may have had this discussion before. Matching a 75 ohm load to a 50 ohm source might be academically interesting, but the actual loss is almost negligible. for a VSWR of 1.5, the return loss is 14dB and the load mismatch attenuation is 0.177dB. That's about what I would expect to lose in two coax connector pairs. You could also feed the antenna with 50 ohm feedline and place the matching network at the antenna. The effect would still be a 1:1 SWR, but the lower impedance of the coax would create higher i^2R losses; not important if you're talking a short line, but a longer one would lower output at the antenna. True, but for roughly equivalent sizes of coax cables, the 75 ohm cable has less loss and the equivalent 50 ohm cable. If you want to handle high power, use 50 ohms. If you want low loss, use 75 ohms: http://www.belden.com/blog/broadcastav/50-Ohms-The-Forgotten-Impedance.cfm Note that these are for air dielectric cables. Things are not so neat if we consider the dielectric. See the bottom paragraph and graphs: http://www.microwaves101.com/encyclopedias/why-fifty-ohms Dielectric Dielectric const Minimum loss impedance solid PTFE 2.2 50 ohms foam PTFE 1.43 60 air 1.0 75 RG-6/u CATV 75 ohm foam coax still has slightly less loss than the equivalent 50 ohm cable, but not as much as I've previously claimed. This is cute: http://cablesondemandblog.com/wordpress1/2014/03/06/whats-the-difference-between-50-ohm-and-75-ohm-coaxial-cable/ "A good rule of thumb is that if the device being connected via coaxial cable is a receiver of some kind, 75 Ohm Coax is ideal." Jeff, do you always miss the forest for the trees? That was an EXAMPLE. The same would be true if you were feeding a 300 ohm yagi with 300 ohm twinlead and a transmitter with a 10 ohm output impedance. Transmitter output impedance does not determine SWR. Transmitter output impedance vs. feedline impedance does determine SWR at one end of the system. If you have a mismatch, you will have a non-1:1 SWR. And BTW - when calculating, you forgot about the transmitters which cut back power to protect the finals. Many will do so even with a 1.5:1 SWR. Not true. Post some links to support your position, please. Basic physics that anyone with even an inkling of AC theory should understand. Any time you have an impedance mismatch in a system - in this case, whether at the antenna or the transmitter end of a feedline - you will not have perfect power transfer. What is not transferred will be reflected. This causes an SWR greater than 1:1. Try this - connect the output of an HF transmitter to an SWR bridge. Now connect a piece of 75 ohm coax such as RG-59 to the output of the SWR meter, and connect that to a 75 ohm resistive load. Do you think the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
"John S" wrote in message ... On 7/1/2015 10:56 AM, Ian Jackson wrote: In message , John S writes On 6/29/2015 3:47 PM, Wayne wrote: snipped to shorten Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? It was been pointed out to me that the figures for feeder loss with an imperfect SWR are only correct when the length is fairly long (at least an electrical wavelength?). How much loss does 25' of RG-8 really have at 12MHz, when there's a halfwave hanging on the far end? # A *resonant* half wave at 12MHz is about 36.7 feet long and it presents # an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The # current at the antenna end is 0.0245A while one watt is applied at the # source end. This means that the power applied to the antenna is about # 0.687W. So, about 68% of the applied power reaches the antenna. # So, about 32% of the power is lost in the RG-8 for this example. I'm just trying to understand this, so let me ask a question about your example. Isn't the 32% lost a function of not having a conjugate match maximum power transfer? If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline, wouldn't maximum power be transferred? (Even with a SWR of about 21:1) |
An antenna question--43 ft vertical
On 7/2/2015 12:18 PM, Wayne wrote:
"John S" wrote in message ... On 7/1/2015 10:56 AM, Ian Jackson wrote: In message , John S writes On 6/29/2015 3:47 PM, Wayne wrote: snipped to shorten Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? It was been pointed out to me that the figures for feeder loss with an imperfect SWR are only correct when the length is fairly long (at least an electrical wavelength?). How much loss does 25' of RG-8 really have at 12MHz, when there's a halfwave hanging on the far end? # A *resonant* half wave at 12MHz is about 36.7 feet long and it presents # an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The # current at the antenna end is 0.0245A while one watt is applied at the # source end. This means that the power applied to the antenna is about # 0.687W. So, about 68% of the applied power reaches the antenna. # So, about 32% of the power is lost in the RG-8 for this example. I'm just trying to understand this, so let me ask a question about your example. Isn't the 32% lost a function of not having a conjugate match maximum power transfer? If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline, wouldn't maximum power be transferred? (Even with a SWR of about 21:1) Transferred where? The match at the transmitter output only matches the output to the line. There are still reflections from the mismatch at the antenna. These reflections result in extra losses in the line as well as power delivered back into the transmitter output stage (especially with a perfect impedance match). But I don't see anyone taking wavelength vs. feed line length into account. If the wavelength is long compared to the feed line I believe a lot of the "bad" stuff goes away. But then I am used to the digital transmission line where we aren't really concerned with delivering power, rather keeping a clean waveform of our (relatively) square waves. So I guess a short feed line doesn't solve the SWR problems... or does it? -- Rick |
An antenna question--43 ft vertical
"Jerry Stuckle" wrote in message ... Try this - connect the output of an HF transmitter to an SWR bridge. Now connect a piece of 75 ohm coax such as RG-59 to the output of the SWR meter, and connect that to a 75 ohm resistive load. Do you think the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1. What you have described is a case of using the wrong swr bridge. You are trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm bridge is used it will show a 1:1 SWR. The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no reflected power. |
An antenna question--43 ft vertical
Wayne wrote:
snip In my own particular case, an automatic remotely tuned ATU would be a pain to install/maintain. This part I do not understand at all. At the antenna end is a box with a connector for the feed line and a connector for the antenna. There is nothing to maintain there. If you get an ATU that gets it's power through the coax, you put the power injector in line with the feed line in the shack. There is nothing to maintain there either and you do not need to run any extra wires out to the antenna. -- Jim Pennino |
An antenna question--43 ft vertical
On 7/2/2015 1:56 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message ... Try this - connect the output of an HF transmitter to an SWR bridge. Now connect a piece of 75 ohm coax such as RG-59 to the output of the SWR meter, and connect that to a 75 ohm resistive load. Do you think the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1. What you have described is a case of using the wrong swr bridge. You are trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm bridge is used it will show a 1:1 SWR. The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no reflected power. My knowledge of antenna systems is limited, but I do know that this is correct, there will be no reflection from the antenna. If the transmitter output is 50 ohms there will be a loss in this matching that will result in less power being delivered to the feed line, but that will not result in reflections in the feed line. I calculate the loss to be -0.177 dB or 4%. How much loss would be expected in the feed line itself if it is a moderate length? -- Rick |
An antenna question--43 ft vertical
On 7/2/2015 1:56 PM, Ralph Mowery wrote:
"Jerry Stuckle" wrote in message ... Try this - connect the output of an HF transmitter to an SWR bridge. Now connect a piece of 75 ohm coax such as RG-59 to the output of the SWR meter, and connect that to a 75 ohm resistive load. Do you think the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1. What you have described is a case of using the wrong swr bridge. You are trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm bridge is used it will show a 1:1 SWR. The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no reflected power. No, the SWR bridge is correct. The output of the transmitter is 50 ohms. You are correct in that if a 75 ohm bridge is used, the indicated SWR would be 1:1, because everything from that point on is 75 ohms. However, the mismatch (and reflection) occurs on the transmitter side of the bridge, not the antenna side. So the bridge will never see it. But an accurate bridge will show lower power output due to the mismatch. A mismatch is a mismatch, no matter where in the system it occurs. And any mismatch will cause less than 100% power to be transferred. The rest is reflected. Just look at the specs of any amateur transceiver. They show an impedance of 50 ohms. So a load of 50 ohms provides for maximum power transfer; any other impedance causes a mismatch. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
wrote in message ... Wayne wrote: snip In my own particular case, an automatic remotely tuned ATU would be a pain to install/maintain. This part I do not understand at all. At the antenna end is a box with a connector for the feed line and a connector for the antenna. There is nothing to maintain there. If you get an ATU that gets it's power through the coax, you put the power injector in line with the feed line in the shack. There is nothing to maintain there either and you do not need to run any extra wires out to the antenna. The problem is with my own particular case. The antenna is a whip mounted in the middle of a metal roof. At my age, I shouldn't be wandering around on or climbing such a roof. Once installed, any failure would require a trip to the roof. The ATU would be exposed to extreme temperature and sunlight that might eventually induce failures. |
An antenna question--43 ft vertical
"Jerry Stuckle" wrote in message ... On 7/2/2015 1:56 PM, Ralph Mowery wrote: "Jerry Stuckle" wrote in message ... You are correct in that if a 75 ohm bridge is used, the indicated SWR would be 1:1, because everything from that point on is 75 ohms. However, the mismatch (and reflection) occurs on the transmitter side of the bridge, not the antenna side. So the bridge will never see it. But an accurate bridge will show lower power output due to the mismatch. A mismatch is a mismatch, no matter where in the system it occurs. And any mismatch will cause less than 100% power to be transferred. The rest is reflected. Just look at the specs of any amateur transceiver. They show an impedance of 50 ohms. So a load of 50 ohms provides for maximum power transfer; any other impedance causes a mismatch. -- The real impedance of the transmitter is not 50 ohms. It is whatever the device is used in the final stage and the poewr level. For a 100 watt transmitter it is in the thousand ohm range and for solid state devices it is very low. The matching circuit is often fixed to be 50 ohms,but could be made for most any impedance. The older tube circuits were adjustable by the user for a range of somewhat bleow 50 ohms to around 200 ohms. Could be more or less depending on the design. The mismatch you are counting on for a 50 ohm transmitter and a 75 ohm feedline and 75 ohm antenna is in the tuned circuits/matching circuit in the transmitter. Whatever power comes out of the transmitter will make it to the antenna minus the loss of the coax, but not additional loss due to swr. The power comming out of a 50 ohm transmitter will be less due to mismatch, but not because of swr of the antenna system which is 1:1. |
An antenna question--43 ft vertical
"rickman" wrote in message ... On 7/2/2015 12:18 PM, Wayne wrote: "John S" wrote in message ... On 7/1/2015 10:56 AM, Ian Jackson wrote: In message , John S writes On 6/29/2015 3:47 PM, Wayne wrote: snipped to shorten Ok. Well, 43ft is a half wavelength at about 12MHz. The vertical will be very high impedance at that frequency and a 1:4 unun will theoretically bring that impedance down closer to the feed line impedance. Does this help? It was been pointed out to me that the figures for feeder loss with an imperfect SWR are only correct when the length is fairly long (at least an electrical wavelength?). How much loss does 25' of RG-8 really have at 12MHz, when there's a halfwave hanging on the far end? # A *resonant* half wave at 12MHz is about 36.7 feet long and it presents # an impedance of about 1063 + j0 ohms to the RG-8 at the antenna end. The # current at the antenna end is 0.0245A while one watt is applied at the # source end. This means that the power applied to the antenna is about # 0.687W. So, about 68% of the applied power reaches the antenna. # So, about 32% of the power is lost in the RG-8 for this example. I'm just trying to understand this, so let me ask a question about your example. Isn't the 32% lost a function of not having a conjugate match maximum power transfer? If the transmitter had a Z of 1063 -j0, and a lossless RG8 feedline, wouldn't maximum power be transferred? (Even with a SWR of about 21:1) # Transferred where? The match at the transmitter output only matches the # output to the line. There are still reflections from the mismatch at # the antenna. These reflections result in extra losses in the line as # well as power delivered back into the transmitter output stage # (especially with a perfect impedance match). Well, I put a few (unrealistic) qualifiers into my question: a transmitter with a a 1063 ohm output (not 50), and a lossless RG-8. Thus, the back and forth reflections would not have attenuation. And the transmitter and load are conjugately matched for maximum power transfer. # But I don't see anyone taking wavelength vs. feed line length into # account. If the wavelength is long compared to the feed line I believe # a lot of the "bad" stuff goes away. But then I am used to the digital # transmission line where we aren't really concerned with delivering # power, rather keeping a clean waveform of our (relatively) square waves. # So I guess a short feed line doesn't solve the SWR problems... or does # it? The attenuation at a given high SWR depends upon the the matched feedline loss, as reflections encounter that loss with every forward or backward trip. Thus feedline length/attenuation should be considered. As a young man I was given a problem of solving poor antenna performance on an aircraft band fixed station antenna. The SWR at the transmitter was close to 1:1, but the antenna didn't work well. I climbed up on the tower and found that the coax had never been connected to the antenna. That was with about 400 feet of coax at 120 MHz. |
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