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5/8 wave 2 meter antenna
Can anyone point me to a good site for construction a 5/8 wave 2M antenna???
I tried home-brewing one already with poor results .... a conductor up a piece of PVC .... same conductor wound into a coil at the bottome connected in series to the center of the coax. 4 1/4 wave radials connected to outer conductor. Thanks, -- Rob ka2pbt |
Can anyone point me to a good site for construction a 5/8 wave 2M antenna???
I tried home-brewing one already with poor results .... a conductor up a piece of PVC .... same conductor wound into a coil at the bottome connected in series to the center of the coax. 4 1/4 wave radials connected to outer conductor. I believe there's a project in the ARRL Handbook which shows how to convert a Radio Shack CB whip antenna to a 2-meter 5/8-wave - it has the details for the matching coil that is required. http://www.arrl.org/tis/info/pdf/8009022.pdf has another QST article which shows how to construct one from scratch, using a technique that doesn't require a matching coil (the matching inductor is made from a stub). I haven't tried either of these myself. With regard to the version you made - did you account for the loading effect of the PVC when you measured and trimmed the length of your radiator? -- Dave Platt AE6EO Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
With regard to the version you made - did you account for the loading effect of the PVC when you measured and trimmed the length of your radiator? Hi Dave, I was just about to recommend the QST article, but you beat me. A few comments on why his antenna did not perform: As you stated, the PVC will change the input Z of the radiator. Just about anything that is too close to the radiator will change things. According to EZNEC, a 5/8 radiator with 1/4 radials 5 WL above ground is 80-j300. As such, with a single series inductor matching section, the best 50 ohm match is 1.6:1. This could get worse or better depending on the antenna's environment. It will also change if you slope the radials. There is an article in the ARRL Antenna Compendium #1 pp.101 that deals with 5/8 antennas. It basically explains why a 5/8, depending on the application, is NOT the best performing radiator. Gary N4AST |
In article .com,
wrote: Hi Dave, I was just about to recommend the QST article, but you beat me. A few comments on why his antenna did not perform: As you stated, the PVC will change the input Z of the radiator. Just about anything that is too close to the radiator will change things. On second reading of that QST article, I admit to being a bit curious. The K4LPQ version (with a shorted inductive stub inside the radiator, soldered to it) is clear enough. However, the W9WQ version using a longer, open-circuited inductive stub wire isn't a straightforward translation of this, because there's no soldered (or other DC) connection between the radiator and anything else! I infer that in the W9WQ version, the stub wire is performing two functions at once - it's adding a series inductance, and it's also coupling the RF out onto the radiator in a capacitive fashion. This would imply that the stub needs to provide a bit more inductive reactance than in the K4LPQ shorted-coax version, with some of this reactance cancelling out the radiator's -j300 and the rest cancelling out whatever amount of capacitive reactance exists between the stub wire and the radiator. Am I reading this right, or am I missing something? [Regretfully it seems likely that both W9WQ and K4LPQ are now silent keys, so I can't ask for advice from the horses' mouths.] According to EZNEC, a 5/8 radiator with 1/4 radials 5 WL above ground is 80-j300. As such, with a single series inductor matching section, the best 50 ohm match is 1.6:1. This could get worse or better depending on the antenna's environment. It will also change if you slope the radials. I've seen a number of 5/8-wave antenna designs which deal with this issue by using something other than a simple series coil. The commonest approach seems to be to use a coil which is connected in shunt between the radiator and the ground plane, with the "hot" side of the coax being fed to a point tapped partway up on the coil. This approach transforms the radiator impedance down to the 50 ohms needed to match the coax, and also provides the series inductance needed to cancel out the reactance. It also provides DC grounding for the radiator. http://www.fluxfm.nl/schema/5-8%20go...20radialen.PDF is one such design. It requires some amount of tooling (e.g. to lathe down the plastic parts to the specified configuration) but I suspect that a version could be homebrewed up using simpler materials and methods. The photos show the way to build the tapped matching coil assembly, and could probably be adapted to other coil construction methods. I believe the ARRL Handbook article which adapts a Radio Shack CB mobile antenna uses a similar tapped coil. Another approach might be to change the length of the radiator a bit, to change the resistive part of the feedpoint impedance from 80 ohms down closer to 50 ohms, and modify the coil to suit. I haven't run any simulations to see how much change in the radiator length would be required, and what this change would do to the antenna's gain pattern. There is an article in the ARRL Antenna Compendium #1 pp.101 that deals with 5/8 antennas. It basically explains why a 5/8, depending on the application, is NOT the best performing radiator. I'll have to look it up if I can find a copy of that edition to see what they have to say. I agree, in some cases the horizon-directed gain of a 5/8 isn't what you want. For in-city and in-the-hills mobile use, a 1/4-wave might give more reliable performance, precisely because its RF energy is more broadly directed. -- Dave Platt AE6EO Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
Dave Platt wrote: In article .com, wrote: Hi Dave, I was just about to recommend the QST article, but you beat me. A few comments on why his antenna did not perform: As you stated, the PVC will change the input Z of the radiator. Just about anything that is too close to the radiator will change things. On second reading of that QST article, I admit to being a bit curious. The K4LPQ version (with a shorted inductive stub inside the radiator, soldered to it) is clear enough. However, the W9WQ version using a longer, open-circuited inductive stub wire isn't a straightforward translation of this, because there's no soldered (or other DC) connection between the radiator and anything else! I infer that in the W9WQ version, the stub wire is performing two functions at once - it's adding a series inductance, and it's also coupling the RF out onto the radiator in a capacitive fashion. This would imply that the stub needs to provide a bit more inductive reactance than in the K4LPQ shorted-coax version, with some of this reactance cancelling out the radiator's -j300 and the rest cancelling out whatever amount of capacitive reactance exists between the stub wire and the radiator. Am I reading this right, or am I missing something? [Regretfully it seems likely that both W9WQ and K4LPQ are now silent keys, so I can't ask for advice from the horses' mouths.] According to EZNEC, a 5/8 radiator with 1/4 radials 5 WL above ground is 80-j300. As such, with a single series inductor matching section, the best 50 ohm match is 1.6:1. This could get worse or better depending on the antenna's environment. It will also change if you slope the radials. I've seen a number of 5/8-wave antenna designs which deal with this issue by using something other than a simple series coil. The commonest approach seems to be to use a coil which is connected in shunt between the radiator and the ground plane, with the "hot" side of the coax being fed to a point tapped partway up on the coil. This approach transforms the radiator impedance down to the 50 ohms needed to match the coax, and also provides the series inductance needed to cancel out the reactance. It also provides DC grounding for the radiator. http://www.fluxfm.nl/schema/5-8%20go...20radialen.PDF is one such design. It requires some amount of tooling (e.g. to lathe down the plastic parts to the specified configuration) but I suspect that a version could be homebrewed up using simpler materials and methods. The photos show the way to build the tapped matching coil assembly, and could probably be adapted to other coil construction methods. I believe the ARRL Handbook article which adapts a Radio Shack CB mobile antenna uses a similar tapped coil. Another approach might be to change the length of the radiator a bit, to change the resistive part of the feedpoint impedance from 80 ohms down closer to 50 ohms, and modify the coil to suit. I haven't run any simulations to see how much change in the radiator length would be required, and what this change would do to the antenna's gain pattern. There is an article in the ARRL Antenna Compendium #1 pp.101 that deals with 5/8 antennas. It basically explains why a 5/8, depending on the application, is NOT the best performing radiator. I'll have to look it up if I can find a copy of that edition to see what they have to say. I agree, in some cases the horizon-directed gain of a 5/8 isn't what you want. For in-city and in-the-hills mobile use, a 1/4-wave might give more reliable performance, precisely because its RF energy is more broadly directed. -- Dave Platt AE6EO Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! Hi Dave, I'm sure you are reading it right, will look and see if I can add anything later. Gary N4AST |
No I didn't account for the PVC ...... but I made it longer than my design
and then trimmed past my design length and it never seemed to make a difference. Considering my inductor was also formed around the PVC that was probably way off too..... I modeled the whole thing with nec first. This is what my input file looked like: http://mywebpages.comcast.net/ka2pbt/2M-5-8wave-dat.txt and this is what my output looked like: http://mywebpages.comcast.net/ka2pbt/2M-5-8wave-out.txt Construction was a radiator 58.5 inches long fed along a 1/2" sched-40, inductor was 19 turns around the same sched-40 2 inches long. The were 4 radials at 90 deg each 19.38" Any comments welcome .... Thanks, -- de ka2pbt "Dave Platt" wrote in message ... Can anyone point me to a good site for construction a 5/8 wave 2M antenna??? I tried home-brewing one already with poor results .... a conductor up a piece of PVC .... same conductor wound into a coil at the bottome connected in series to the center of the coax. 4 1/4 wave radials connected to outer conductor. I believe there's a project in the ARRL Handbook which shows how to convert a Radio Shack CB whip antenna to a 2-meter 5/8-wave - it has the details for the matching coil that is required. http://www.arrl.org/tis/info/pdf/8009022.pdf has another QST article which shows how to construct one from scratch, using a technique that doesn't require a matching coil (the matching inductor is made from a stub). I haven't tried either of these myself. With regard to the version you made - did you account for the loading effect of the PVC when you measured and trimmed the length of your radiator? -- Dave Platt AE6EO Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
In article ,
john doe wrote: No I didn't account for the PVC ...... but I made it longer than my design and then trimmed past my design length and it never seemed to make a difference. Considering my inductor was also formed around the PVC that was probably way off too..... I modeled the whole thing with nec first. This is what my input file looked like: http://mywebpages.comcast.net/ka2pbt/2M-5-8wave-dat.txt and this is what my output looked like: http://mywebpages.comcast.net/ka2pbt/2M-5-8wave-out.txt Construction was a radiator 58.5 inches long fed along a 1/2" sched-40, inductor was 19 turns around the same sched-40 2 inches long. The were 4 radials at 90 deg each 19.38" Is that 58.5 inch figure a typo? Your NEC model says 51 inches, and my quickie spreadsheet calculation says 50.9 inches for a 145 MHz center of band. I also wonder about the coil - it calculates out to be just over 1 microHenry, or about j910 ohms at 145 MHz. That seems like quite a bit too much, based on jgboyles's posting earlier today indicating a feedpoint Z of about 80-j300. The PDF to which I posted a link earlier today uses a coil of only 10.5 turns, spread out over a distance of about 2.5" on a 3/4" form. That's about .6 uH or j550 ohms... and it's a shunt-fed design so the actual series inductance (above the tap point) is even lower. -- Dave Platt AE6EO Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
On Mon, 14 Mar 2005 14:26:57 GMT, "john doe"
wrote: Can anyone point me to a good site for construction a 5/8 wave 2M antenna??? I tried home-brewing one already with poor results .... a conductor up a piece of PVC .... same conductor wound into a coil at the bottome connected in series to the center of the coax. 4 1/4 wave radials connected to outer conductor. Thanks, -- Rob ka2pbt Can the 5/8 wave antenna be matched with a tuning stub such as the J-Pole uses? -- 73 for now Buck N4PGW |
Can the 5/8 wave antenna be matched with a tuning stub such as the
J-Pole uses?... Probably, but seems easier to to use a 5 turn coil at the base...But I agree...A 5/8 GP is not a very good 2m antenna. Better off building a high quality 1/4 wave GP. Seriously... By high quality, I mean instead of 3-4 sloped radials, use 6-8 radials, and even better , use a 2nd set for decoupling. You can also build "sleeve" versions. 5/8's are ok on HF, where the angles are not critical, and on 10m, will usually be the best choice. But on 2m, they are usually lame. But, don't take my word. Build a 5/8 GP, and then build a 1/4 GP, and see which is best. I bet the 1/4 wave wins. For a 5/8 to work well on VHF/UHF, it needs to be a collinear dual 5/8 design, or at the least, use sloped 5/8 or 3/4 wave radials. 5/8's with 1/4 wave radials have pretty lousy patterns for 2m use. MK |
On Tue, 15 Mar 2005 02:53:22 +0000, Dave Platt wrote:
Is that 58.5 inch figure a typo? Your NEC model says 51 inches, and my quickie spreadsheet calculation says 50.9 inches for a 145 MHz center of band. Actually, the 51 you're seeing in the NEC model is probably the number of segments. One end is at 0,0,36 and the other is at 0,0,94.5; so yes it's really 58.5 inches. I came to this number by playing with the model until the REAL component of the impedance got as close to 50 as I could get it. I also wonder about the coil - it calculates out to be just over 1 microHenry, or about j910 ohms at 145 MHz. That seems like quite a bit too much, based on jgboyles's posting earlier today indicating a feedpoint Z of about 80-j300. My model comes up with a feedpoint impedance of 5.4485E+01-j2.8560E+03 So I tried to build a coil with an inductive reactance to cancel that .. I came up with 3.13 microhenries. Is my model way off????? |
Hummm...Should be about .336 mh for a grounded coil.
About .182 mh for a insulated coil. In the real world will usually amount to about 5 turns of coil average on say a .5 to 1 inch form. Trim coil for best match. MK |
|
In article ,
john doe wrote: Is that 58.5 inch figure a typo? Your NEC model says 51 inches, and my quickie spreadsheet calculation says 50.9 inches for a 145 MHz center of band. Actually, the 51 you're seeing in the NEC model is probably the number of segments. One end is at 0,0,36 and the other is at 0,0,94.5; so yes it's really 58.5 inches. Whups... my bad. I came to this number by playing with the model until the REAL component of the impedance got as close to 50 as I could get it. Hmmm. What did that do to the pattern? You no longer have a 5/8-wave antenna. Adding about 8 inches has brought it very close to being a 3/4-wave radiator. As such, it's going to have a substantially lower amount of towards-the-horizon energy in its pattern, and a big lobe aiming upwards at roughly 45 degrees above the horizon. This is the classic problem with running a 2-meter J-pole on 440 - it'll load up and radiate, but a lot of its radiation is aimed at airplanes rather than repeaters :-( I also wonder about the coil - it calculates out to be just over 1 microHenry, or about j910 ohms at 145 MHz. That seems like quite a bit too much, based on jgboyles's posting earlier today indicating a feedpoint Z of about 80-j300. My model comes up with a feedpoint impedance of 5.4485E+01-j2.8560E+03 So I tried to build a coil with an inductive reactance to cancel that .. I came up with 3.13 microhenries. Is my model way off????? I think you might want to take two looks at it: - Check the radiation pattern. By lengthening it to get a 50-ohm resistive component in the feedpoint, I suspect you've given up much of the gain benefit of a true 5/8-wave radiator. You may actually have less towards-the-horizon power and sensitivity than you'd get with a 1/4-wave groundplane or a 1/2-wave J-pole. - Check the formula and actual inductance for your coil. With so much capacitive reactance from the radiator to cancel out with the coil, I suspect that you may also find that you've calculated out an antenna which is going to be rather narrow-banded. Even slight frequency shifts, or errors in the coil winding (a fraction of a turn) could leave you with a lot of residual reactance and an unacceptable SWR. -- Dave Platt AE6EO Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
Dave Platt wrote: In article .com, wrote: On second reading of that QST article, I admit to being a bit curious. I infer that in the W9WQ version, the stub wire is performing two functions at once - it's adding a series inductance, and it's also coupling the RF out onto the radiator in a capacitive fashion. This would imply that the stub needs to provide a bit more inductive reactance than in the K4LPQ shorted-coax version, with some of this reactance cancelling out the radiator's -j300 and the rest cancelling out whatever amount of capacitive reactance exists between the stub wire and the radiator. Am I reading this right, or am I missing something? I took a look at the Smith Chart, and for impedances in this region of the chart, series L and parallel C is not the way to get a match. In my version, 80-j300, you need 4.7pf shunt C and .15uH series L. I have no idea how Fig. 1C in the article managed to get a good match with the single insulated wire up the middle of the radiator. I seem to recall a tri-band beam (TA-33 jr.?) that used this type of matching. Must work, so I guess I am missing something. Gary N4AST |
Wow, I didn't even consider the pattern ....
Thanks for all the advice ... I'll have another go at it and let you know how I make out. -- ka2pbt "Dave Platt" wrote in message ... In article , john doe wrote: Is that 58.5 inch figure a typo? Your NEC model says 51 inches, and my quickie spreadsheet calculation says 50.9 inches for a 145 MHz center of band. Actually, the 51 you're seeing in the NEC model is probably the number of segments. One end is at 0,0,36 and the other is at 0,0,94.5; so yes it's really 58.5 inches. Whups... my bad. I came to this number by playing with the model until the REAL component of the impedance got as close to 50 as I could get it. Hmmm. What did that do to the pattern? You no longer have a 5/8-wave antenna. Adding about 8 inches has brought it very close to being a 3/4-wave radiator. As such, it's going to have a substantially lower amount of towards-the-horizon energy in its pattern, and a big lobe aiming upwards at roughly 45 degrees above the horizon. This is the classic problem with running a 2-meter J-pole on 440 - it'll load up and radiate, but a lot of its radiation is aimed at airplanes rather than repeaters :-( I also wonder about the coil - it calculates out to be just over 1 microHenry, or about j910 ohms at 145 MHz. That seems like quite a bit too much, based on jgboyles's posting earlier today indicating a feedpoint Z of about 80-j300. My model comes up with a feedpoint impedance of 5.4485E+01-j2.8560E+03 So I tried to build a coil with an inductive reactance to cancel that .. I came up with 3.13 microhenries. Is my model way off????? I think you might want to take two looks at it: - Check the radiation pattern. By lengthening it to get a 50-ohm resistive component in the feedpoint, I suspect you've given up much of the gain benefit of a true 5/8-wave radiator. You may actually have less towards-the-horizon power and sensitivity than you'd get with a 1/4-wave groundplane or a 1/2-wave J-pole. - Check the formula and actual inductance for your coil. With so much capacitive reactance from the radiator to cancel out with the coil, I suspect that you may also find that you've calculated out an antenna which is going to be rather narrow-banded. Even slight frequency shifts, or errors in the coil winding (a fraction of a turn) could leave you with a lot of residual reactance and an unacceptable SWR. -- Dave Platt AE6EO Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
Can the 5/8 wave antenna be matched with a tuning stub such as the J-Pole uses? -- 73 for now Buck N4PGW Yes, A 5/8 radiator with an input Z of 80-j300 with series 19cm of 450 ohm line will be about 55-j0 ohms. Of course all of this is highly dependent on the antenna environment on 2m. Gary N4AST |
Am I reading this right, or am I missing something?
I took a look at the Smith Chart, and for impedances in this region of the chart, series L and parallel C is not the way to get a match. In my version, 80-j300, you need 4.7pf shunt C and .15uH series L. I have no idea how Fig. 1C in the article managed to get a good match with the single insulated wire up the middle of the radiator. It's possible he isn't creating a full match (with an L network) in this case. He might just be cancelling out the negative reactance, using a combination of (series L from the stub, and a bit of series C from the capacitive coupling between stub-feed and radiator), and not bothering with a shunt component at all. This would, perhaps, result in an 80+0j feedpoint impedance and about a 1.6 SWR at the feedpoint, which would probably end up significantly lower at the other end of the feedline due to feedline losses. Or, there might be something stranger going on, with the stub giving a bit of shunt C to ground (in the PL-259), some parallel L/C inside the radiator, and six other bits of odd voodoo. The author says that it ought to be possible to get down to below 1.5:1 on the repeater portion of the band... this suggests that the design isn't one which "tries" to achieve a true 1:1 match. The WA-2 and similar 5/8-wave antennas using a tapped coil seem to be able to get down arbitrarily close to 1:1 at their best. Beats me. Almost makes me want to try building one just to measure it out and see how well it can work. On the other hand, given the comments by Cebik and others about the somewhat illusory nature of the gain advantage of a 5/8-wave, I may just stick with J-poles and quarterwave ground planes. -- Dave Platt AE6EO Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
Ok,
I have not followed all of this thread. One of the pdf's shows a shunt tapped inductor as the base loading. This is an auto-transformer match. The entire inductor is set to cancel the capacitive reactance of the aerial, then the coax feed is tapped to the 50R point, The result is 50R J0. For 5/8 aerials you may want to look at a physical short radiator with a capacity hat to lower the angle of radiation. wrote in message ups.com... Can the 5/8 wave antenna be matched with a tuning stub such as the J-Pole uses? -- 73 for now Buck N4PGW Yes, A 5/8 radiator with an input Z of 80-j300 with series 19cm of 450 ohm line will be about 55-j0 ohms. Of course all of this is highly dependent on the antenna environment on 2m. Gary N4AST |
"Dave Platt"
... On the other hand, given the comments by Cebik and others about the somewhat illusory nature of the gain advantage of a 5/8-wave, I may just stick with J-poles and quarterwave ground planes. _______________ Just to note that for AM broadcast verticals, the FCC requires a certain antenna "efficiency" for various classes of stations, in terms of the minimum ground wave field strength produced per kilowatt of input power to the radiator. The FCC field strength minimum cannot be met by "Class A" stations (basically the 50kW-ers) using a 1/4-wave vertical radiator. At least a 1/2-wave radiator is needed in most cases. The most common radiator height for Class A non-directional AM broadcast stations operating at 50kW day and night is 195°. A radiator height of 225° (5/8 wave) maximizes ground wave field strength at a given power, but also produces a high-angle lobe that can interfere with the ground wave during night-time operation -- so rarely is used by AM broadcast stations. The ground wave field strength difference between 195° and 225° radiators is fairly small. RF Former staff engineer, WJR Detroit -- (Class A, 760kHz) |
Interesting .... I changed my number of segments in NEC from 51 to 501 and I
got an impedance more in line with 80-j300 ..... I'll go re-work my coil now :) de ka2pbt "Dave Platt" wrote in message ... In article , john doe wrote: Is that 58.5 inch figure a typo? Your NEC model says 51 inches, and my quickie spreadsheet calculation says 50.9 inches for a 145 MHz center of band. Actually, the 51 you're seeing in the NEC model is probably the number of segments. One end is at 0,0,36 and the other is at 0,0,94.5; so yes it's really 58.5 inches. Whups... my bad. I came to this number by playing with the model until the REAL component of the impedance got as close to 50 as I could get it. Hmmm. What did that do to the pattern? You no longer have a 5/8-wave antenna. Adding about 8 inches has brought it very close to being a 3/4-wave radiator. As such, it's going to have a substantially lower amount of towards-the-horizon energy in its pattern, and a big lobe aiming upwards at roughly 45 degrees above the horizon. This is the classic problem with running a 2-meter J-pole on 440 - it'll load up and radiate, but a lot of its radiation is aimed at airplanes rather than repeaters :-( I also wonder about the coil - it calculates out to be just over 1 microHenry, or about j910 ohms at 145 MHz. That seems like quite a bit too much, based on jgboyles's posting earlier today indicating a feedpoint Z of about 80-j300. My model comes up with a feedpoint impedance of 5.4485E+01-j2.8560E+03 So I tried to build a coil with an inductive reactance to cancel that .. I came up with 3.13 microhenries. Is my model way off????? I think you might want to take two looks at it: - Check the radiation pattern. By lengthening it to get a 50-ohm resistive component in the feedpoint, I suspect you've given up much of the gain benefit of a true 5/8-wave radiator. You may actually have less towards-the-horizon power and sensitivity than you'd get with a 1/4-wave groundplane or a 1/2-wave J-pole. - Check the formula and actual inductance for your coil. With so much capacitive reactance from the radiator to cancel out with the coil, I suspect that you may also find that you've calculated out an antenna which is going to be rather narrow-banded. Even slight frequency shifts, or errors in the coil winding (a fraction of a turn) could leave you with a lot of residual reactance and an unacceptable SWR. -- Dave Platt AE6EO Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
Richard Fry wrote:
"The most common radiator height for Class A non-directional AM broadcast stations operating at 50kW day and night is 195-degrees." I won`t challenge that as I have conducted no survey. WJR Detroit is shown on the broadcast allocations map book as an unlimited (day and night) 50 kilowatt Class 1 station. Class 1 stations operate in a clear channel (this does not mean alone on the channel) with an assigned power between 10kW and 50kW A Class 2 station operates in a clear channel with an assigned power between 250W and 50kW. They must operate so as to not cause interference to the Class 1 stations. There are 29 clear channels which permit class 2 station operation. Class 3 stations share regional channels and operate with assigned powers between 500W and 5kW. Thyere are 41 regional channels and more than 2000 Class 3 stations. These numbers were taken before expansion of the AM broadcast band which has grown the totals. Class 4 stations operate in assigned local channels with no more than 1kW day and 250W night assignments. There are 6 local channels with 150 or more Class 4 stations on each channel. Primary service area is the statiob`s groundwave coverage. Secondary service is uninterfered skywave coverage. Intermittent service lies between the primary and secondary service areas. A clear channel has one or more high-powered stations which serve wide areas. All primary service areas and a substantial portion of their secondary service areas are cleared of objectional interference. A regional channel has stations not exceeding 5kW which have coverage contours which limit the primary service interference between these stations. A local channel has stations not exceeding 1 kW daytime and 250 watts at night. Primary coverage is limited by interference. Assignments are made to limit interference. Radio waves are radiated into a hemisphere as space below the antenna is hidden by the surface of the earth. This results in a formula for the field power at one mile from a perfect radiator emitting 1 kilowatt: P = 1000 / 16266419 = 0.00006 watts/sq mtr E = sq rt (PR) and R=377 ohms Volts/mtr=152 at 1 mile from a perfect infinitely short uniform hemispherical radiator. From a 1/4-wave vertical, it`s about 195 millivolts per mtr at 1 mile. From a 1/2-wave vertical, it`s about 236 millivolts per mtr at 1 mile. From a 5/8-wave vertical, it`s about 267 millivolts per mtr at 1 mile. Volts vary as the square root of the power. So, for 50 killowatts, multiply the 1 kilowatt values by 7.07. The field strengths are the inverse distance or lossless values. Real earth has losses. Best regards, Richard Harrison, KB5WZI |
I found that increasing the number of segments had a significant change
in the input Z. The material I read on 5/8 antennas indicated the real part of the Z was near 50 ohms. I could not get that result until I increased the number of segments. Guess it is cause 146 MHZ antennas are a good bit shorter than 3.5 MHZ antennas, and any small deviation such as lenght, or # of segments will change the end results. Gary N4AST |
A source is spread out over an entire segment. So when you change the
number of segments, you change both the length and the effective position of the source. When the source is at the bottom of a quarter wavelength radiator, small changes in source position don't make much difference in the impedance seen by the source. However, when the antenna approaches a half wavelength, the source impedance changes quite dramatically with source position. Consequently, you'll see substantial changes in reported source impedance with segmentation in that case. This might or might not be the cause of what you're seeing. As an experiment, you might try moving the source up one segment and see how big a difference it makes. Whenever the result is very sensitive to small changes in the model, you shouldn't expect a real antenna to come out exactly like the model predicts, since small differences between the model and real antenna will likewise cause significant differences. The absolute length doesn't matter -- a 146 MHz antenna will be no more or less sensitive to the same amount of change (in terms of percentage of the antenna size or of the wavelength) than a 3.5 MHz antenna if both are proportioned the same. In fact, 146 MHz antennas are typically considerably fatter in terms of wavelength than 3.5 MHz antennas, and this makes them less sensitive to small changes. Roy Lewallen, W7EL wrote: I found that increasing the number of segments had a significant change in the input Z. The material I read on 5/8 antennas indicated the real part of the Z was near 50 ohms. I could not get that result until I increased the number of segments. Guess it is cause 146 MHZ antennas are a good bit shorter than 3.5 MHZ antennas, and any small deviation such as lenght, or # of segments will change the end results. Gary N4AST |
Richard Fry wrote:
"The most common radiator height for Class A non-directional AM broadcast stations operating at 50kW day and night is 195-degrees." "Richard Harrison" responded (in part) I won`t challenge that as I have conducted no survey. WJR Detroit is shown on the broadcast allocations map book as an unlimited (day and night) 50 kilowatt Class 1 station. __________________ Your text is referenced to out-of-date versions of applicable FCC Rules. The FCC adopted the metric standard over 15 years ago, and the classification of AM broadcast stations no longer is defined as Class 1 to 4 but Classes A, B, C & D. The current versions of the applicable Rules are contained in 47CFR Part 73, and for this topic are dated October 1, 2004. Minimum radiator heights in meters for Class A, B and C AM broadcast stations are shown in Figure 7 of 47CFR73.190. Radiator efficiency for Class A stations (other than in Alaska) must be such as to produce a ground wave of least 362 mV/m at 1km for 1kW of antenna input power. A 90° omni radiator cannot do that. RF |
Richard Fry wrote:
"Radiator efficiency for Class A stations (other than in Alaska) must be such as to produce a groundwave of at least 352 mV/m at 1 km for 1kW of antenna input power. A 90-degree omni radiator cannot do that." Most such antennas have long been in place. Surely they are grandfathered until the station is modified. Best regards, Richard Harrison, KB5WZI |
Richard Harrison"
Richard Fry wrote: "Radiator efficiency for Class A stations (other than in Alaska) must be such as to produce a groundwave of at least 352 mV/m at 1 km for 1kW of antenna input power. A 90-degree omni radiator cannot do that." Most such antennas have long been in place. Surely they are grandfathered until the station is modified. ________________ The minimum efficiency for Class A is 362mV/m as I posted, not 352mV/m as you "quoted me" above. As far as the 50kW / non-D / 24-hr operations are concerned, their facilities have avoided the use of 90° radiators nearly from their start, due to the commercial/competitive need to generate a very strong groundwave while avoiding self interference to that groundwave at night from their own high-angle radiation. The advantages of using greater fractional wavelength radiators to accomplish this has long been known and confirmed by field measurements. RF |
I'm trying to understand the shunt tapped inductor as a circuit.
How does it work??? As long as the inductor cancels out the reactance of the radiator you just need to find the 50 ohm point on the coil ... or for that matter any feedline impedance??? Pardon my lack of knowledge. de ka2pbt |
"Richard Fry" wrote
As far as the 50kW / non-D / 24-hr operations are concerned, their facilities have avoided the use of 90° radiators... ___________________ The list below illustrates my statement above. This data for most of the 50kW, 24-hr stations with the same radiation pattern day and night was taken from a secondary source linked to the FCC database, current as of October 2004 (typos, if any, excepted). The average height of these radiators is 188.8°, not including WFAN and KDKA. Data Format = Freq, Call Letters, Tower Height in Degrees, Notation 640 KFI 175.7 650 WSM 190.3 660 WFAN 155.3 (top loaded) 670 WSCR 181.5 700 WLW 189.3 710 WOR 177 720 WGN 195 750 WSB 179.3 760 WJR 194.7 770 WABC 180.3 780 WBBM 194.1 810 WGY 182.9 820 WBAP 192.1 830 WCCO 194.4 840 WHAS 201.1 850 KOA 207.8 870 WWL 182.1 880 WCBS 207.1 890 WLS 189.3 1020 KDKA 280.28 (sectionalized) 1030 WBZ 188.5 1050 WEPN 186 1060 KYW 180 1070 KNX 193.5 1100 WTAM 185.2 1120 KMOX 192.7 1140 WRVA 185 1160 KSL 193.2 1180 WHAM 177.1 1200 WOAI 193.2 1210 WPHT 186 RF |
Richard Fry wrote:
"The list below illustrates my statement above." Yes, and it proves your point that non-directional 50 KW broadcast stations in the U.S. use tall towers in terms of wavelength. The r-f is precious and it makes sense to make the most of it. Thank you for the list of high-powered stations which can be heard far and wide. I still regret the end of WLW`s 500 KW days. Best regards, Richard Harrison, KB5WZI |
john doe wrote: I'm trying to understand the shunt tapped inductor as a circuit. How does it work??? As long as the inductor cancels out the reactance of the radiator you just need to find the 50 ohm point on the coil ... or for that matter any feedline impedance??? Pardon my lack of knowledge. de ka2pbt Hi John, Is the shunt tapped inductor you are refering to is a coil with one end connected to the 5/8 radiator, the other end to the radials, the coax shield to the radials, and the center conductor tapped up the coil from the radials? If this is the case, according to the Smith Chart, you can only get a perfect 50 ohm match if the real part of the impedance you are trying to match less than 50 ohms. Our models indicate this is not the case with a 5/8 radiator. The shunt tapped inductor is actually a 2 element matching system, with both elements being inductors. As you stated, the inductor in series with the radiator cancels the reactance, and the tapped shunt inductor provides a 50+j0 point. Works only if the real part of the impedance is 50 ohms. Gary N4AST |
Let's suppose you have a vertical antenna with a base feedpoint
impedance of, say, 75 - j300 ohms (75 ohms resistance in series with 300 ohms of capacitive reactance), typical of a thin 5/8 wave vertical. A network consisting of 1275 ohms in *parallel* with 319 ohms of capacitive reactance has the same impedance. (Note how the Xc isn't much different from the Xc of the series circuit in this case.) If we put an inductor with 319 ohms of inductive reactance in parallel with the antenna (that is, from the base feedpoint to ground), the reactance of the inductor cancels out the capacitive reactance of the antenna, and we're left with 1275 ohms of pure resistance from the antenna base to ground (that is, across the inductor). We can use the inductor as an autotransformer. If we tap up on the inductor some fraction k of the whole way up, the impedance we'll see at that tap will be very nearly 1275 * k^2, and it'll be purely resistive (no reactance) because the impedance across the whole coil is purely resistive. For example, half way up the coil we'll see 1275 * (.5)^2 = 1275 / 4 = 319 ohms. So to get 50 ohms, we tap up sqrt(50/1275) = 20% of the way up the coil. Roy Lewallen, W7EL john doe wrote: I'm trying to understand the shunt tapped inductor as a circuit. How does it work??? As long as the inductor cancels out the reactance of the radiator you just need to find the 50 ohm point on the coil ... or for that matter any feedline impedance??? Pardon my lack of knowledge. de ka2pbt |
Thanks Roy!!!
Is it possible to cancel the capacitive reactance of the antenna by placing an inductor in series with the radiator?? i.e center of coax connected to one end of a coil with the other end of the coil connected to the radiator. Coax shield connected to ground plane radials. de ka2pbt "Roy Lewallen" wrote in message ... Let's suppose you have a vertical antenna with a base feedpoint impedance of, say, 75 - j300 ohms (75 ohms resistance in series with 300 ohms of capacitive reactance), typical of a thin 5/8 wave vertical. A network consisting of 1275 ohms in *parallel* with 319 ohms of capacitive reactance has the same impedance. (Note how the Xc isn't much different from the Xc of the series circuit in this case.) If we put an inductor with 319 ohms of inductive reactance in parallel with the antenna (that is, from the base feedpoint to ground), the reactance of the inductor cancels out the capacitive reactance of the antenna, and we're left with 1275 ohms of pure resistance from the antenna base to ground (that is, across the inductor). We can use the inductor as an autotransformer. If we tap up on the inductor some fraction k of the whole way up, the impedance we'll see at that tap will be very nearly 1275 * k^2, and it'll be purely resistive (no reactance) because the impedance across the whole coil is purely resistive. For example, half way up the coil we'll see 1275 * (.5)^2 = 1275 / 4 = 319 ohms. So to get 50 ohms, we tap up sqrt(50/1275) = 20% of the way up the coil. Roy Lewallen, W7EL john doe wrote: I'm trying to understand the shunt tapped inductor as a circuit. How does it work??? As long as the inductor cancels out the reactance of the radiator you just need to find the 50 ohm point on the coil ... or for that matter any feedline impedance??? Pardon my lack of knowledge. de ka2pbt |
"john doe" wrote in message ... Thanks Roy!!! Is it possible to cancel the capacitive reactance of the antenna by placing an inductor in series with the radiator?? i.e center of coax connected to one end of a coil with the other end of the coil connected to the radiator. Coax shield connected to ground plane radials. de ka2pbt John DK2 PBT I'm no Roy. But, I have thought about Smith Charts alot. I wonder if you'd consider trying to identify where the antenna's input impedance must be located so that it might be matched with a series inductor. I'd be way out of line to assume that you want to look at a Smith Chart for knowledge about the effectiveness of that series inductance. But, the Smith Chart can be a real big aid in estimating 'what it takes' to match any antenna's input impedance. Jerry "Roy Lewallen" wrote in message ... Let's suppose you have a vertical antenna with a base feedpoint impedance of, say, 75 - j300 ohms (75 ohms resistance in series with 300 ohms of capacitive reactance), typical of a thin 5/8 wave vertical. A network consisting of 1275 ohms in *parallel* with 319 ohms of capacitive reactance has the same impedance. (Note how the Xc isn't much different from the Xc of the series circuit in this case.) If we put an inductor with 319 ohms of inductive reactance in parallel with the antenna (that is, from the base feedpoint to ground), the reactance of the inductor cancels out the capacitive reactance of the antenna, and we're left with 1275 ohms of pure resistance from the antenna base to ground (that is, across the inductor). We can use the inductor as an autotransformer. If we tap up on the inductor some fraction k of the whole way up, the impedance we'll see at that tap will be very nearly 1275 * k^2, and it'll be purely resistive (no reactance) because the impedance across the whole coil is purely resistive. For example, half way up the coil we'll see 1275 * (.5)^2 = 1275 / 4 = 319 ohms. So to get 50 ohms, we tap up sqrt(50/1275) = 20% of the way up the coil. Roy Lewallen, W7EL john doe wrote: I'm trying to understand the shunt tapped inductor as a circuit. How does it work??? As long as the inductor cancels out the reactance of the radiator you just need to find the 50 ohm point on the coil ... or for that matter any feedline impedance??? Pardon my lack of knowledge. de ka2pbt |
john doe wrote:
Thanks Roy!!! Is it possible to cancel the capacitive reactance of the antenna by placing an inductor in series with the radiator?? i.e center of coax connected to one end of a coil with the other end of the coil connected to the radiator. Coax shield connected to ground plane radials. de ka2pbt Sure. In the example I gave, where the antenna has a 75 - j300 ohm feedpoint impedance, you'd just need an inductor with 300 ohms of reactance. Then you'd have a feedpoint impedance of 75 ohms resistive. Of course, this would be useful in feeding an antenna only if the resistive part of the feedpoint resistance is close to 50 ohms or some other value your system can conveniently feed. Unlike the tapped coil method, it doesn't give you any way of transforming the resistance. If you need to match any arbitrary impedance, say 250 - j85 or something, to 50 ohms or some other impedance, you need two things you can adjust or choose, since you have two things (R and X or magnitude and phase) which you need to transform. We could get 50 ohms resistive using the tapped inductor scheme because we could choose the capacitor and the tap position. Another very common solution is an L network, which of course has two components. In theory, we can match (or transform) anything to anything with an L network of some sort. In some special cases, one of the elements of the L network is zero or infinite, such as if we're transforming 50 - j200 to 50 + j0, where all we need is a single inductor. But this isn't generally so. A 5/8 wave radiator conveniently has a feedpoint resistance in the neighborhood of 50 ohms, so it can often be matched well enough by simply adding a series inductor as you suggest. Roy Lewallen, W7EL |
Roy Lewallen wrote:
We can use the inductor as an autotransformer. Someone should tell that to Kurt N. Sterba. This month he says: "But for a half-wave or 5/8 wave vertical you need a parallel tuned circuit at the base if you want to feed it there with coaxial cable." He also says: "A reader is irate because Kurt has stated that coaxial cable does not radiate. Actually, Kurt is just quoting from textbooks." He apparently forgot what he wrote previously: "The Resonant Feedline Dipole takes advantage of the fact that coaxial cable acts like a three conductor cable for RF. There is the inner conductor, the inside of the shield, and the outside of the shield. RF signals travel down the inside of the cable with equal currents on the inner conductor and the shield. The current on the shield does not penetrate the shield." -- 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 =--- |
john doe wrote:
Is it possible to cancel the capacitive reactance of the antenna by placing an inductor in series with the radiator?? That's exactly what happens with the 5/8WL vertical and base coil. coil 5/8 WL GND---/////////-------------------------------------- ^ | 50 ohm tap The coil and 5/8 WL whip form a series-resonant circuit. -- 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 =--- |
Cecil Moore wrote:
john doe wrote: Is it possible to cancel the capacitive reactance of the antenna by placing an inductor in series with the radiator?? That's exactly what happens with the 5/8WL vertical and base coil. coil 5/8 WL GND---/////////-------------------------------------- ^ | 50 ohm tap The coil and 5/8 WL whip form a series-resonant circuit. -- 73, Cecil http://www.qsl.net/w5dxp Not quite. If you look at the analysis I posted regarding the tap feed, you'll see that the inductive reactance required for feeding the antenna this way isn't equal to the series capacitive reactance of the antenna. They're often close (as in the example, where the feedpoint series X = -300 and the required Xl = 319 ohms), but how close they are depends on the Q (Xs/Rs) of the feedpoint Z. When you feed the antenna this way, what you have is not a series resonant circuit but a parallel resonant one, and the required inductance is different for the two. The requirement for this type of feed is not series resonance, but that the Z at the top of the coil be completely resistive, which series resonance doesn't produce. As I pointed out in my most recent posting, you can make a series resonant circuit consisting of the antenna and an inductor, and put the source in series with that. But you need a different size inductor. Roy Lewallen, W7EL |
Roy Lewallen wrote: We can use the inductor as an autotransformer. If we tap up on the inductor some fraction k of the whole way up, the impedance we'll see at that tap will be very nearly 1275 * k^2, and it'll be purely resistive (no reactance) because the impedance across the whole coil is purely resistive. For example, half way up the coil we'll see 1275 * (.5)^2 = 1275 / 4 = 319 ohms. So to get 50 ohms, we tap up sqrt(50/1275) = 20% of the way up the coil. Roy Lewallen, W7EL Hi Roy, I'm OK with 319+j0 half way up the coil, but at 20% of the way up I keep getting 50+j50. I did it on the Smith Chart, and on a Spice based circuit analysis program. Too old to do this stuff by hand. That's not a real good match for 50 ohm coax. The are other taps that will provide a better match, but no where did I find 50+j0. In an earlier post I stated that it looked like the real part of the antenna impedance needed to be less than 50 ohms to get a perfect match, using this method. The impedance across the whole coil is not purely resistive, it is at the 50% point. Apparently I am modeling incorrectly, or missing something. Gary N4AST |
"Roy Lewallen" wrote in message
... Let's suppose you have a vertical antenna with a base feedpoint impedance of, say, 75 - j300 ohms (75 ohms resistance in series with 300 ohms of capacitive reactance), typical of a thin 5/8 wave vertical. A network consisting of 1275 ohms in *parallel* with 319 ohms of capacitive reactance has the same impedance. Roy, how did you come up with "1275 ohms in *parallel* with 319 ohms " equatating to "75 ohms resistance in series with 300 ohms of capacitive reactance" ??? Just wondering. de ka2pbt |
Rob Roschewsk wrote:
Roy, how did you come up with "1275 ohms in *parallel* with 319 ohms " equatating to "75 ohms resistance in series with 300 ohms of capacitive reactance" ??? Just wondering. I got it from a routine I keep on my HP48GX calculator, which comes from the following series-parallel transformations: Let: Rs = resistance of series equivalent circuit Xs = reactance of series equivalent circuit Rp = resistance of parallel equivalent circuit Xp = reactance of parallel equivalent circuit To convert a series circuit to a parallel circuit which has an identical impedance: Rp = (Rs^2 + Xs^2) / Rs Xp = (Rs^2 + Xs^2) / Xs To convert a parallel circuit to a series circuit which has an identical impedance: Rs = (Rp * Xp^2) / (Rp^2 + Xp^2) Xs = (Rp^2 * Xp) / (Rp^2 + Xp^2) These aren't very difficult to derive if you're comfortable with complex arithmetic. They should be in the toolkit of everyone who works with electrical circuits. Important things to keep in mind when using these transformations: 1. Although frequency isn't explicitly involved in the conversions, when you make an equivalent circuit from a resistor and inductor or capacitor, Xp and Xs will change with frequency. Therefore a transformed circuit will have the same impedance as the original only at one frequency. If the frequency changes, new values of resistance and capacitance or inductance have to be calculated for the equivalent circuit. 2. Because point 1, one circuit or the other will usually be better for modeling a real circuit over a range of frequencies, because the impedance of the real circuit will change with frequency more like one or the other of the two equivalent circuits. In the example, where the feedpoint Z = 75 - j300: Rs = 75 Xs = 300 so Rp = (75^2 + (-300)^2) / 75 = 1275 Xp = (75^2 + (-300)^2) / (-300) = -318.75 You can check this if you'd like. You'll find that the parallel combination of 1275 and -j318.75 ohms is 75 - j300. Roy Lewallen, W7EL |
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