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#31
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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 |
#32
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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 |
#33
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"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 |
#34
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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 |
#35
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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 =--- |
#36
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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 =--- |
#37
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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 |
#38
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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 |
#39
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"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 |
#40
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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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