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#1
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My bag is VHF, so forgive me if I've bumbled up the calculations. My
numbers show that a standard 108" whip has an input impedance of somewhere around 0.04 ohms in series with a 22 pf capacitor. These numbers are so far from what I normally deal with that I'm not sure that I'm right, and I'd appreciate somebody who actually works down in this area giving my numbers a reality check. If they ARE right, how in heaven do most people match to 0.04 ohms? The 22 pf I can resonate out with a 1 millihenry choke (or thereabouts), but how do most people match 50 ohms to fractional ohms in the homebrew arena -- that is, without just going out and buying a "magical matching box" of some sort? Jim |
#2
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On Sat, 19 Mar 2005 12:45:53 -0800, "RST Engineering"
wrote: My bag is VHF, so forgive me if I've bumbled up the calculations. My numbers show that a standard 108" whip has an input impedance of somewhere around 0.04 ohms in series with a 22 pf capacitor. Not quite. Over perfect ground the Z is ~ 0.37 -j8170 @ 1.9 MHz for a 12mm diameter radiator. So you add an inductor of +j8170 with say Q=250. That gives Z = (0.37+32.68) +j0 = 33.05 +j0. Then you add 16.95 ohm of ground loss and Z = 50 +j0. A perfect match with a gain of -17 dBi. ![]() These numbers are so far from what I normally deal with that I'm not sure that I'm right, and I'd appreciate somebody who actually works down in this area giving my numbers a reality check. If they ARE right, how in heaven do most people match to 0.04 ohms? The 22 pf I can resonate out with a 1 millihenry choke (or thereabouts), but how do most people match 50 ohms to fractional ohms in the homebrew arena -- that is, without just going out and buying a "magical matching box" of some sort? Jim |
#3
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RST Engineering wrote:
My bag is VHF, so forgive me if I've bumbled up the calculations. My numbers show that a standard 108" whip has an input impedance of somewhere around 0.04 ohms in series with a 22 pf capacitor. These numbers are so far from what I normally deal with that I'm not sure that I'm right, and I'd appreciate somebody who actually works down in this area giving my numbers a reality check. If they ARE right, how in heaven do most people match to 0.04 ohms? The 22 pf I can resonate out with a 1 millihenry choke (or thereabouts), but how do most people match 50 ohms to fractional ohms in the homebrew arena -- that is, without just going out and buying a "magical matching box" of some sort? Jim Hi Jim, Your numbers look about right for a 108" whip on 1.8mhz. To feed a small whip like this you would have to "brute force" it with a large loading coil at the base. The efficiency will be low and the bandwidth narrow. To get any efficiency at all, the 2:1 swr bandwidth will only be around 5khz. Once you tune it up, it will be a single frequency antenna. There are things like adding capacitive loading at the top of the whip, and moving the inductive loading to the center (have to cut the whip in half). This will make the feedpoint Z a little more managable. I once loaded an 8' antenna on 160, just to see if I could, but never got on the air with it. I read somewhere that in the 50's, law enforcement 2-way radios operated close to our 160m ham band. Wonder what kind of antennas they used on the police cars. Of course the distances they were interested in are different from hams. Gary N4AST |
#4
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![]() "Wes Stewart" wrote in message ... On Sat, 19 Mar 2005 12:45:53 -0800, "RST Engineering" wrote: My bag is VHF, so forgive me if I've bumbled up the calculations. My numbers show that a standard 108" whip has an input impedance of somewhere around 0.04 ohms in series with a 22 pf capacitor. Not quite. Over perfect ground the Z is ~ 0.37 -j8170 @ 1.9 MHz for a 12mm diameter radiator. So you add an inductor of +j8170 with say Q=250. That gives Z = (0.37+32.68) +j0 = 33.05 +j0. Then you add 16.95 ohm of ground loss and Z = 50 +j0. A perfect match with a gain of -17 dBi. ![]() These numbers are so far from what I normally deal with that I'm not sure that I'm right, and I'd appreciate somebody who actually works down in this area giving my numbers a reality check. If they ARE right, how in heaven do most people match to 0.04 ohms? The 22 pf I can resonate out with a 1 millihenry choke (or thereabouts), but how do most people match 50 ohms to fractional ohms in the homebrew arena -- that is, without just going out and buying a "magical matching box" of some sort? Jim The following code produces Zin = 0.117 - j2717 ohms at 1.9 MHz. What did I do wrong? Note the high segmentation to place the feed-point near the base of the antenna. Frank CM 9 ft monopole CE GW 1 108 0 0 0 0 0 108 0.25 GS 0 0 0.025400 GE 1 GN 1 EX 0 1 1 00 1 0 LD 5 1 1 108 5.8001E7 FR 0 9 0 0 1.8 0.025 RP 0 181 1 1000 -90 0 1.00000 1.00000 EN |
#5
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Frank wrote:
The following code produces Zin = 0.117 - j2717 ohms at 1.9 MHz. What did I do wrong? Note the high segmentation to place the feed-point near the base of the antenna. Frank CM 9 ft monopole CE GW 1 108 0 0 0 0 0 108 0.25 GS 0 0 0.025400 GE 1 GN 1 EX 0 1 1 00 1 0 LD 5 1 1 108 5.8001E7 FR 0 9 0 0 1.8 0.025 RP 0 181 1 1000 -90 0 1.00000 1.00000 EN If you follow NEC guidelines on the minimum recommended segment length/wire radius ratio, you should have no more than 17 segments; the model has 127. In this case, though, it doesn't seem to be disturbing NEC-2 very much. Also, the half inch diameter is somewhat greater than most CB whips. If you drop the diameter to a quarter inch, you'll get a substantially greater reactance which would require a larger loading inductor and hence result in lower efficiency. Otherwise it looks ok to me. As Wes pointed out, the ground system and loading inductor losses will be so large as to make the feedpoint resistance (and wire loss) insignificant. The zero-loss feedpoint resistance is, however, useful in determining efficiency. If you set the wire loss to zero by removing the LD "card", you get the more useful value of feedpoint resistance when losses are zero, 0.098 ohms. Using Wes' value of about 50 ohm feedpoint resistance when losses are included, you can calculate the efficiency as 0.098/50 = 0.2%, or 2 watts radiated for each kW applied. In American mobile terms that's 1.5 watts per horsepower. I'm always glad to see more QRP signals on the band. Roy Lewallen, W7EL |
#6
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To obtain obtain some idea of the numerical values involved with
mobile, short, antennas on 160 meter and other low frequency bands, download programs - HELICAL3 VERTLOAD LOADCOIL from website below. Takes only a few seconds to download. Not zipped up. Run immediately. With a very short antenna, to obtain a useful efficiency, a physically large loading coil is essential. The only space available is up the antenna. So the best thing to do is extend the coil up the antenna from near the base, in the form of a long slender helical winding on a pvc plastic pipe. The whole thing behaves as a short 1/4-wave resonant vertical. There is a short rod at the top which is pruned to resonate the antenna in the required band. It is inevitably a single frequency job. The bandwidth on 160m is only a few kHz. With a 100 watt transmitter, when both vehicles are in the low-noise countryside, it is possible to work 100 miles or more, on groundwave, in daylight. Much further via skywave in darkness. G3YXM has worked transatlantic from a car in Scotland to a base station in Canada on occasions. Maximum overall antenna heights are about 9 feet above roof of car with the coil about 2" in diameter, the coil extending to a height of 6 or 7 feet. The most tedious procedure is pruning all antenas in a group of mobiles to the same frequency on the 160 meter band. But it has been done. Mobile to a base station is easy. For many years the standard UK frequency for mobiles was 1930 kHz. Matching from antenna base to a 50-ohm transmiier can be accomplished by a single or bank of mica capacitors of several hundred pF. For quite a number of years, mobile working on 160m was quite popular in the UK using helically-wound antennas. ---- .................................................. .......... Regards from Reg, G4FGQ For Free Radio Design Software go to http://www.btinternet.com/~g4fgq.regp .................................................. .......... |
#7
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![]() Frank wrote: "Wes Stewart" wrote in message ... On Sat, 19 Mar 2005 12:45:53 -0800, "RST Engineering" wrote: My bag is VHF, so forgive me if I've bumbled up the calculations. My numbers show that a standard 108" whip has an input impedance of somewhere around 0.04 ohms in series with a 22 pf capacitor. Not quite. Over perfect ground the Z is ~ 0.37 -j8170 @ 1.9 MHz for a 12mm diameter radiator. So you add an inductor of +j8170 with say Q=250. That gives Z = (0.37+32.68) +j0 = 33.05 +j0. Then you add 16.95 ohm of ground loss and Z = 50 +j0. A perfect match with a gain of -17 dBi. ![]() These numbers are so far from what I normally deal with that I'm not sure that I'm right, and I'd appreciate somebody who actually works down in this area giving my numbers a reality check. If they ARE right, how in heaven do most people match to 0.04 ohms? The 22 pf I can resonate out with a 1 millihenry choke (or thereabouts), but how do most people match 50 ohms to fractional ohms in the homebrew arena -- that is, without just going out and buying a "magical matching box" of some sort? Jim The following code produces Zin = 0.117 - j2717 ohms at 1.9 MHz. What did I do wrong? Note the high segmentation to place the feed-point near the base of the antenna. Frank CM 9 ft monopole CE GW 1 108 0 0 0 0 0 108 0.25 GS 0 0 0.025400 GE 1 GN 1 EX 0 1 1 00 1 0 LD 5 1 1 108 5.8001E7 FR 0 9 0 0 1.8 0.025 RP 0 181 1 1000 -90 0 1.00000 1.00000 EN Hi Frank, Well, I got different numbers than both of you. Must depend on the ground type of the models, as well as the # of segments. If you are trying to match an antenna with fractions of an ohm resistive and thousands reactive, the differences we are all observing will not make a whole lot of difference in the matching network you decide to use. It will be big and bulky, and very inefficient. Gary N4AST |
#8
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On Sat, 19 Mar 2005 23:23:30 GMT, "Frank"
wrote: [snip] The following code produces Zin = 0.117 - j2717 ohms at 1.9 MHz. What did I do wrong? Note the high segmentation to place the feed-point near the base of the antenna. Frank CM 9 ft monopole CE GW 1 108 0 0 0 0 0 108 0.25 GS 0 0 0.025400 GE 1 GN 1 EX 0 1 1 00 1 0 LD 5 1 1 108 5.8001E7 FR 0 9 0 0 1.8 0.025 RP 0 181 1 1000 -90 0 1.00000 1.00000 EN My bad. I hate it when that happens. A typo on the length on my part. 101 instead of 108. It's more like 0.12 -j2890 in EZNEC. I'm not real fluent in NEC decks but I think you are using too many segments. |
#9
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![]() "Roy Lewallen" wrote in message ... Frank wrote: The following code produces Zin = 0.117 - j2717 ohms at 1.9 MHz. What did I do wrong? Note the high segmentation to place the feed-point near the base of the antenna. Frank CM 9 ft monopole CE GW 1 108 0 0 0 0 0 108 0.25 GS 0 0 0.025400 GE 1 GN 1 EX 0 1 1 00 1 0 LD 5 1 1 108 5.8001E7 FR 0 9 0 0 1.8 0.025 RP 0 181 1 1000 -90 0 1.00000 1.00000 EN If you follow NEC guidelines on the minimum recommended segment length/wire radius ratio, you should have no more than 17 segments; the model has 127. In this case, though, it doesn't seem to be disturbing NEC-2 very much. Also, the half inch diameter is somewhat greater than most CB whips. If you drop the diameter to a quarter inch, you'll get a substantially greater reactance which would require a larger loading inductor and hence result in lower efficiency. Otherwise it looks ok to me. As Wes pointed out, the ground system and loading inductor losses will be so large as to make the feedpoint resistance (and wire loss) insignificant. The zero-loss feedpoint resistance is, however, useful in determining efficiency. If you set the wire loss to zero by removing the LD "card", you get the more useful value of feedpoint resistance when losses are zero, 0.098 ohms. Using Wes' value of about 50 ohm feedpoint resistance when losses are included, you can calculate the efficiency as 0.098/50 = 0.2%, or 2 watts radiated for each kW applied. In American mobile terms that's 1.5 watts per horsepower. I'm always glad to see more QRP signals on the band. Roy Lewallen, W7EL I should have read the manual before modeling, but I was so concerned that the feed-point be near the bottom. Changing the segmentation to 17, removing the LD card, and reducing the radius to 0.125 ", calculates the feed-point impedance as: 0.123 - j3286. I considered Jim's calculation of 0.37 - j8170 to be far too reactive. The fact is these points are so close, when observed on the Smith chart, as to be virtually identical. Analyzing a shunt L/series C matching network, with inductor Q at 300, indicates a loss of 21.5dB with 30 kV RMS at the base of the antenna (1.5 kW into the matching network). (In the case of 0.37 - j8170 the loss is 21.8 dB, with 42 kV at the base). These voltages are so large, I am beginning to wonder if I goofed again! The losses, at least, are the same order of magnitude as Roy's calculation. 73, Frank |
#10
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If you had made this model with EZNEC, you would have seen a warning
that the segment length is shorter than recommended. And you could have entered the dimensions directly in inches (or immediately converted between meters and inches). The free demo program is adequate for this model, providing you don't want more than 20 segments. Roy Lewallen, W7EL Frank wrote: I should have read the manual before modeling, but I was so concerned that the feed-point be near the bottom. Changing the segmentation to 17, removing the LD card, and reducing the radius to 0.125 ", calculates the feed-point impedance as: 0.123 - j3286. I considered Jim's calculation of 0.37 - j8170 to be far too reactive. The fact is these points are so close, when observed on the Smith chart, as to be virtually identical. Analyzing a shunt L/series C matching network, with inductor Q at 300, indicates a loss of 21.5dB with 30 kV RMS at the base of the antenna (1.5 kW into the matching network). (In the case of 0.37 - j8170 the loss is 21.8 dB, with 42 kV at the base). These voltages are so large, I am beginning to wonder if I goofed again! The losses, at least, are the same order of magnitude as Roy's calculation. 73, Frank |
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