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Is it possible to have a 1:1 SWR?
I have the good ole sling shot wire antenna configuration. I picked up a
new SWR meter, hooked it up according to the directions, and the needle didn't budge when checking SWR's. The power indicator is fine, just not even a 'wiggle' on the needle for SWR reading. Bum meter? |
"Macman" wrote in message ... I have the good ole sling shot wire antenna configuration. I picked up a new SWR meter, hooked it up according to the directions, and the needle didn't budge when checking SWR's. The power indicator is fine, just not even a 'wiggle' on the needle for SWR reading. Bum meter? probably... check on a different frequency or band to be sure... or just disconnect the antenna and see what it does. |
On Fri, 12 Dec 2003 17:30:13 -0500, "Macman"
wrote: I have the good ole sling shot wire antenna configuration. I picked up a new SWR meter, hooked it up according to the directions, and the needle didn't budge when checking SWR's. The power indicator is fine, just not even a 'wiggle' on the needle for SWR reading. Bum meter? Every HF antenna here will go 1:1 "some where" although a couple of them require a tuner to get there. OTOH, Yes, at least on my 75 and 40 meter half waves, the 20, 15, and 10 meter tribander as well as the AV640 on the shop will hit 1:1 some where in the desired band without a tuner. Roger Halstead (K8RI & ARRL life member) (N833R, S# CD-2 Worlds oldest Debonair?) www.rogerhalstead.com Return address modified due to dumb virus checkers |
1:1 is possible, I've seen it, but reverse the connections to be sure.
H. NQ5H "Macman" wrote in message ... I have the good ole sling shot wire antenna configuration. I picked up a new SWR meter, hooked it up according to the directions, and the needle didn't budge when checking SWR's. The power indicator is fine, just not even a 'wiggle' on the needle for SWR reading. Bum meter? |
On Sat, 13 Dec 2003 00:06:16 GMT, Roger Halstead
wrote: | |Every HF antenna here will go 1:1 "some where" ... Not so. |
"Wes Stewart" wrote in message
... On Sat, 13 Dec 2003 00:06:16 GMT, Roger Halstead wrote: | |Every HF antenna here will go 1:1 "some where" ... Not so. I have a dipole that is 1:1 at both it's main length design frequency and the single fan dipole design frequency underneath it. Each began life 10% longer than design (formulas), one actually had to have a few inches added - and that was soldered when the line was still brand new. The other required shortening in 2" increments. Over a period of two months, the antenna still has "1:1" on those two design frequencies. Checked with reliable equipment besides the MFJ 962D tuner, which now rests on "Bypass" except for higher frequency diversions. This was designed and built for two frequencies only, and performs flawlessly in that respect. So yes, achieving 1:1 SWR is not only possible, but nothing less was acceptable in the specific need I had for that antenna. Jack Virginia Beach |
On Fri, 12 Dec 2003 22:32:07 -0700, Wes Stewart
wrote: On Sat, 13 Dec 2003 00:06:16 GMT, Roger Halstead wrote: | |Every HF antenna here will go 1:1 "some where" ... Not so. I'd like to know how you can say that? Have you measured all my antennas? You are welcome to come over and check them out. Every one "according to my meter which is a Bird" reaches 1:1 some where in the desired band. It may not be exactly the frequency the formula predicted, but it'll be relatively close. Roger Halstead (K8RI & ARRL life member) (N833R, S# CD-2 Worlds oldest Debonair?) www.rogerhalstead.com Return address modified due to dumb virus checkers |
On Wed, 17 Dec 2003 06:44:06 GMT, Roger Halstead
wrote: |On Fri, 12 Dec 2003 22:32:07 -0700, Wes Stewart |wrote: | |On Sat, 13 Dec 2003 00:06:16 GMT, Roger Halstead wrote: | || ||Every HF antenna here will go 1:1 "some where" ... | |Not so. | |I'd like to know how you can say that? I can say it because it's true. | |Have you measured all my antennas? No, but I've measured some of mine. Why just yesterday, I was out measuring my vertical that started out as a Cushcraft AV-80. I'm looking to use it on 40 meters. At resonance, at about 5.43 MHz, the base impedance was 38.5 +j0. That's a 1.3:1 SWR. And BTW that's a datum taken using a method described in HP Application Note 77-3, "Measurement of Complex Impedance 1-1000 MHz." You mistakenly believe that because an antenna *has a resonance* at some frequency, its SWR is *1:1*. Not so, as shown above. I have shown at least *one* example where you are wrong when you say *every antenna*. | You are welcome to come over and |check them out. No thanks. The weather here is much nicer for doing antenna work. | Every one "according to my meter which is a Bird" |reaches 1:1 some where in the desired band. It may not be exactly the |frequency the formula predicted, but it'll be relatively close. | |Roger Halstead (K8RI & ARRL life member) |(N833R, S# CD-2 Worlds oldest Debonair?) |www.rogerhalstead.com |Return address modified due to dumb virus checkers | |
Roger Halstead wrote:
Wes Stewart wrote: On Sat, 13 Dec 2003 00:06:16 GMT, Roger Halstead |Every HF antenna here will go 1:1 "some where" ... Not so. I'd like to know how you can say that? Have you measured all my antennas? Basic problem was the lack of a defined boundary for "here". I figured it meant, "here in the US". -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Wed, 17 Dec 2003 09:18:30 -0600, Cecil Moore
wrote: |Roger Halstead wrote: | Wes Stewart wrote: | |On Sat, 13 Dec 2003 00:06:16 GMT, Roger Halstead ||Every HF antenna here will go 1:1 "some where" ... | |Not so. | | I'd like to know how you can say that? | Have you measured all my antennas? | |Basic problem was the lack of a defined boundary for |"here". I figured it meant, "here in the US". I don't know what "here" meant for sure, but he said, "although a couple of them require a tuner to get there", even "there" his statement is incorrect. |
On Wed, 17 Dec 2003 07:58:25 -0700, Wes Stewart
wrote: On Wed, 17 Dec 2003 06:44:06 GMT, Roger Halstead wrote: |On Fri, 12 Dec 2003 22:32:07 -0700, Wes Stewart |wrote: | |On Sat, 13 Dec 2003 00:06:16 GMT, Roger Halstead wrote: | || ||Every HF antenna here will go 1:1 "some where" ... | |Not so. | |I'd like to know how you can say that? I can say it because it's true. | |Have you measured all my antennas? I think an important word has been missed. :-)) No, but I've measured some of mine. Why just yesterday, I was out measuring my vertical that started out as a Cushcraft AV-80. I'm looking to use it on 40 meters. At resonance, at about 5.43 MHz, the base impedance was 38.5 +j0. That's a 1.3:1 SWR. What do your antennas have to do with mine? And BTW that's a datum taken using a method described in HP Application Note 77-3, "Measurement of Complex Impedance 1-1000 MHz." I understand that. You mistakenly believe that because an antenna *has a resonance* at some frequency, its SWR is *1:1*. Not so, as shown above. I don't think I used the word resonance any where in my post although at the end I did refer to the antennas being cut according to formula and the 1:1 point not being where the formula said it should. I have shown at least *one* example where you are wrong when you say *every antenna*. Ummm...no you haven't. Nor did I say every antenna. I said "Every HF antenna here", paraphrased, "Every HF antenna at this location does show a 1:1 some where in the band for which it's used". I am speaking solely about the HF antennas at this location and was careful to point that out. All have been checked with a Bird 43 watt meter and each one has a point where it shows no reflected power. Their heights and angles are such that they are non reactive at those points, or at least as near as I can tell on a Palomar bridge and are so close to 50 ohms I can't measure the difference. So as the question was asked and the thread titled. "Is it possible to have a 1:1 SWR", the answer is yes. If I can do it, so can someone else. I didn't give any probabilities, but the odds are relatively slim that any one throwing up a wire is going to be lucky and find it 1:1 with J=0. However if some one reads 1:1 on their bridge it doesn't mean the bridge is necessarily bad. I should also point out that the wire antennas here are relatively high at an 45 degrees to horizontal. Probably not a typical installation. | You are welcome to come over and |check them out. No thanks. The weather here is much nicer for doing antenna work. Hey, it's nicer than when I put up the tribander. I think the chill factor was 4 degrees that day. Current conditions: Temperature 29 °F / -2 °C Windchill 17 °F / -8 °C Humidity 69% Dew Point 20 °F / -7 °C Wind WNW at 17 mph / 27.4 km/h Wind Gust - Pressure 29.70 in / 1005 hPa (Steady) Conditions Overcast Visibility 10 miles / 16 kilometers Clouds (Above Ground Level) (FEW) : 2700 ft / 824 m Overcast (OVC) : 6500 ft / 1983 m It's a veritable heat wave. :-)) Although we did have nearly two inches of snow today. | Every one "according to my meter which is a Bird" |reaches 1:1 some where in the desired band. It may not be exactly the |frequency the formula predicted, but it'll be relatively close. They are cut to formula plus a tad. Shortening equal amounts did not put the 1:1 point at the same frequency on each of the 75 meter antennas. However both do reach 1:1 with j=0 or as near as I can measure. The Palomar bridge isn't exactly a precision instrument. OTOH the AV-640 Hy-Gain multi band vertical reaches 1:1 (50 ohms and J=0) at some point on every band 40 through 6. However I'm convinced it's a perfectly matched dummy load on 20. Roger Halstead (K8RI & ARRL life member) (N833R, S# CD-2 Worlds oldest Debonair?) www.rogerhalstead.com Return address modified due to dumb virus checkers | |Roger Halstead (K8RI & ARRL life member) |(N833R, S# CD-2 Worlds oldest Debonair?) |www.rogerhalstead.com |Return address modified due to dumb virus checkers | |
On Wed, 17 Dec 2003 09:18:30 -0600, Cecil Moore
wrote: Roger Halstead wrote: Wes Stewart wrote: On Sat, 13 Dec 2003 00:06:16 GMT, Roger Halstead |Every HF antenna here will go 1:1 "some where" ... Not so. I'd like to know how you can say that? Have you measured all my antennas? Basic problem was the lack of a defined boundary for "here". I figured it meant, "here in the US". Here as in http://www.rogerhalstead.com/ham_files/tower.htm which is about 5 miles west of down town Midland MI. IE, my own installation. :-)) Roger Halstead (K8RI & ARRL life member) (N833R, S# CD-2 Worlds oldest Debonair?) www.rogerhalstead.com Return address modified due to dumb virus checkers |
an antenna has an impedence at the frequency it is being used at, and an
impedence at its resonant frequency. If either of these impedences happen to be 50 ohms and the coax being used is 50 ohms, and the transciever is working at 50 ohms, then the swr is 1:1, and the swr is on the transmission line, not on the antenna. The antenna does not have to have an impedence of 50 ohms at either the frequency being used at or at its resonant frequency, and these two freqeuncies could be the same, and the transmission line does not have to be at 50 ohms, and for that matter neither does the transciever. If any one of these is mismatched, then the swr is not 1:1. An impedence transformer at the antenna-transmission line junction will transform a mismatch so there is no reflection on the transmission line, amd if this impedence is the same as that of the transmitter, then the swr is 1:1, if the impedence is not the same, then the swr is not 1:1 unless it is also transformed at the transmitter, and again the swr would be 1:1 on the transmission line, which is where the swr is, it is not on the antenna. |
William F. Hagen wrote:
an antenna has an impedence at the frequency it is being used at, and an impedence at its resonant frequency. If either of these impedences happen to be 50 ohms and the coax being used is 50 ohms, and the transciever is working at 50 ohms, then the swr is 1:1, and the swr is on the transmission line, not on the antenna. One source of confusion is, on systems with both coax and ladder-line, the SWR on the coax Vs the SWR on the ladder-line. A 12:1 SWR on 600 ohm ladder-line can result in a 50 ohm SWR of 1:1 without a tuner. The ladder-line can be used as an impedance transformer. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil Moore wrote:
William F. Hagen wrote: an antenna has an impedence at the frequency it is being used at, and an impedence at its resonant frequency. If either of these impedences happen to be 50 ohms and the coax being used is 50 ohms, and the transciever is working at 50 ohms, then the swr is 1:1, and the swr is on the transmission line, not on the antenna. One source of confusion is, on systems with both coax and ladder-line, the SWR on the coax Vs the SWR on the ladder-line. A 12:1 SWR on 600 ohm ladder-line can result in a 50 ohm SWR of 1:1 without a tuner. The ladder-line can be used as an impedance transformer. That lily really didn't need the gold paint job, Cecil. But thanks for providing a source of confusion. How could we have a good argument without one. :-) 73, Jim AC6XG |
Jim Kelley wrote:
Cecil Moore wrote: William F. Hagen wrote: an antenna has an impedence at the frequency it is being used at, and an impedance at its resonant frequency. If either of these impedences happen to be 50 ohms and the coax being used is 50 ohms, and the transceiver is working at 50 ohms, then the swr is 1:1, and the swr is on the transmission line, not on the antenna. One source of confusion is, on systems with both coax and ladder-line, the SWR on the coax Vs the SWR on the ladder-line. A 12:1 SWR on 600 ohm ladder-line can result in a 50 ohm SWR of 1:1 without a tuner. The ladder-line can be used as an impedance transformer. That lily really didn't need the gold paint job, Cecil. But thanks for providing a source of confusion. How could we have a good argument without one. :-) 73, Jim AC6XG For additional confusion, *IF* your transmission line (coax or ladder line) is low loss, and if your rig can load into it, SWR doesn't much matter. Reflected power will "bounce" off the rig and go back to the antenna. Our rigs actually present a very low impedance to the antenna and transmission line. This is by design; we want all of the RF we manufactured to go to the antenna and none wasted as heat in the rig. Our rigs don't really look like the Thevinian equivalent (voltage source with internal resistor of 50 ohms) feeding a 50 ohm load. For a fixed voltage and fixed internal resistance, using a 50 ohm load gets you max power *into the load*, ignoring the wasted power in the source resistance. Your electric power company doesn't do that, efficiency would suck. They want all the energy used to be in paying customer's loads. They do that by keeping their source impedance very low. Actually, our rigs have a source impedance of only a few ohms, and are designed to pump power into a 50 ohm load. There is a delay involved with the reflected power getting to the antenna, but for the narrow bandwidth signals we transmit (SSB voice or code) this is not significant. It will matter for amateur television up on UHF, though. You can get a lower SWR reading than what your antenna is doing if you have lossy feedline. The lossy feedline is absorbing some of the reflected power. So don't be suprised at the worse SWR if you upgrade your coax. As long as you can load up into it, it's not a real problem. ============================================= "What did Santa say at the house of ill repute?" "Ho ho ho!" ============================================ Keep Santa in Xmas |
Standing wave ratios on ordinary 1/2-wave dipoles can soar as high as
10-to-1. With all the fuss made about excessive SWR everywhere else, why is it the guru's never mention it, let alone show anxiety about it? What are they trying to cover up? |
Reg Edwards wrote:
Standing wave ratios on ordinary 1/2-wave dipoles can soar as high as 10-to-1. With all the fuss made about excessive SWR everywhere else, why is it the guru's never mention it, let alone show anxiety about it? What are they trying to cover up? It's a communist plot! ;-) Actually, if your feedline is low loss, and is able to handle the higher voltages you'd get with high SWRs, and you're using a narrowband mode like voice SSB, and your rig can tune load into it, high SWR is not a problem. BTW, if you're using coax to feed a dipole, be sure to make a coil of the coax feedline of several turns at the antenna feedpoint. The object of this is to keep RF currents from traveling down the *OUTSIDE* of the coax shield. Otherwise your antenna's radiation pattern will be goofed up, and also the SWR will get even worse. This also keeps stray RF out of the shack. |
On Wed, 24 Dec 2003 06:11:10 GMT, Robert Casey
wrote: Our rigs actually present a very low impedance to the antenna and transmission line. This is by design; we want all of the RF we manufactured to go to the antenna and none wasted as heat in the rig. Our rigs don't really look like the Thevinian equivalent (voltage source with internal resistor of 50 ohms) feeding a 50 ohm load. Hi Robert, Actually none of this is true. It is the glib explanation that is bandied about commonly in this forum, but it contains its own internal inconsistency of logic. This illogic is present in the single statement: Reflected power will "bounce" off the rig and go back to the antenna. The presumption is that the point of bouncing back, the transistion point of the so-called low Z transmitter to the high Z line, performs this action. It contains to howlers: 1. if it were true, no one would ever need a tuner whose sole purpose is to do exactly that (the bouncing back); 2. if it were true, the original power coming from the transmitter would see the same reflection and bounce right back in to turn to heat (which is a fairly true representation of the problem of SWR). As for the reality of the situation, answer me this: 1. How much power does your rig transmit? 2. How much power does your rig draw? Correct me if the operation of dividing the first by the second does not reveal an efficiency of roughly 40% and a power loss to heat of roughly greater than that transmitted. Your rig has a massive heat sink with a fan, n'est pas? Too many of the pundits want to force a literal carbon composition resistor into the mix so that they can point to its absence proving Thevenin's Theorem does not apply. The same pundits ignore the fact that Thevenin did not specify a resistor, he specified an impedance to satisfy his theorem. It was Edison's pervsion of logic in trying to persuade the investors that AC distribution was for the birds when it came time to match loads. He inserted the false claim of resistance forcing inefficiency. This perversion has been with us ever since and qualifies such believers only as possible investors in the Edison DC Power distribution company (which folded immediately due to inefficiency in the market place). 73's Richard Clark, KB7QHC |
On Wed, 24 Dec 2003 08:09:55 +0000 (UTC), "Reg Edwards"
wrote: Standing wave ratios on ordinary 1/2-wave dipoles can soar as high as 10-to-1. With all the fuss made about excessive SWR everywhere else, why is it the guru's never mention it, let alone show anxiety about it? What are they trying to cover up? Ah Punchinello, You have had your crown returned to you by those same gurus who have abandoned the field, and who quixotically made just those protestations you now so miss. We will leave it in your capable hands to argue both sides of the coin on this one, in your fulsome, best oratorio. 73's and the best of the season to ye, Richard Clark, KB7QHC |
As for the reality of the situation, answer me this: 1. How much power does your rig transmit? 2. How much power does your rig draw? Correct me if the operation of dividing the first by the second does not reveal an efficiency of roughly 40% and a power loss to heat of roughly greater than that transmitted. Your rig has a massive heat sink with a fan, n'est pas? n'est pas? ? Anyway, the power lost inside the transmitter I thought was due to the inneficiencies of a class A or B amp configuration. That the amplifying device (tube or transistor) is metering out varying amounts of current from a constant DC supply. And the current has to pass thru the tube or transistor before it gets to the load. Assuming a sine wave at RF, zero crossing would be low wasted power, peak also low wasted power, but at 0.707 peak max wasted power (class B). Class C is more efficient except the power in the harmonics you get have to be filtered out, and you get a heated filter. Now if the Thevenin impedance of the transmitter was 50 ohms, then half of the remaining power you successfully converted to RF would heat the transmitter some more. That 40% would become 20%. But if you design the output right, making the Thevenin impedance very low, little RF power is wasted in the transmitter (say 1%) and 39% is pumped out into space (on the air) by a matched antenna. Now that low Thevenin impedance will reflect most of the reflected from the mismatched antenna power back to the antenna. If the transmitter Thevenin impedance was 50 ohms, it would absorb the reflections and get even hotter. Oh, this has its uses, like in a signal coming from a TTL line driver that passes thru a "source termination" resistor of 100 ohms, then thru a 100 ohm impedance transmission line (like carefully designed traces on circuit boards in that GHz computer) and then to a single destination receiving TTL gate. which looks like a high impedance load. The reflected signal gets absorbed by the source termination. A reason for not destination terminations is that this makes the source work harder pumping DC current when the signal is a "high", vs essentially idling when just source terminated. No daisy chaining allowed, the signal looks like crap except at the very end. |
On Thu, 25 Dec 2003 05:47:12 GMT, Robert Casey
wrote: As for the reality of the situation, answer me this: 1. How much power does your rig transmit? 2. How much power does your rig draw? Correct me if the operation of dividing the first by the second does not reveal an efficiency of roughly 40% and a power loss to heat of roughly greater than that transmitted. Your rig has a massive heat sink with a fan, n'est pas? n'est pas? ? Anyway, the power lost inside the transmitter I thought was due to the inneficiencies of a class A or B amp configuration. Hi Robert, Given you offer no evidence to the contrary, 40% efficiency seems to be the confirmed rule. Giving it a name does nothing to reduce it or enhance it, the calories expended rob us of RF output for the power draw to generate that output. That power loss is confirmed through everyone's experience as heat. Heat sinks and fans testify to our acceptance of its loss. The ONLY advance we can claim in the last half century, is that no power is lost to lighting up filaments in transistors. That the amplifying device (tube or transistor) is metering out varying amounts of current from a constant DC supply. And the current has to pass thru the tube or transistor before it gets to the load. All of this goes without saying and changes nothing about efficiency, or Thevenin issues that isn't already obvious. If it illustrates anything is that it CONFIRMS the Thevenin mechanism (once you look under the hood, an argument no-no if you first posit the Thevenin argument). Assuming a sine wave at RF, zero crossing would be low wasted power, peak also low wasted power, but at 0.707 peak max wasted power (class B). What makes the angle an indicator of waste? The RF wave consists of 360 degrees of variation all of which is propagated. If you have your head under the hood, playing with the innards of the source, there is still nothing inherently lossy about the angle of conduction. It is correlatable to the magnitude of voltage across a source (the tube/transistor, not the load), or current through the source (same provisos), but in that respect the least loss occurs at the EXTREME of the cycle that draws down the voltage differential against the supply or the current magnitude; certainly not at the zero cross (there is no such thing at the plate or collector of a powered amplifier) and even more remotely the 0.707. Again, such discussion simply focuses on the Thevenin validation for the source Z specification. Class C is more efficient except the power in the harmonics you get have to be filtered out, and you get a heated filter. Your logic, as faulty as the presentation may be, simply validates the Thevenin specification for the Source Z. You exhibit any number of calorie sinks to satisfy some aspect of the Real R, but I see by the flawed argument below that you persist in Edison's folly of forcing an R only into the Z specification of Thevenin. However, we will next proceed to illustrate that Edison's folly is in fact demonstrably valid for the present discussion. :-) Now if the Thevenin impedance of the transmitter was 50 ohms, then half of the remaining power you successfully converted to RF would heat the transmitter some more. With a 40% efficiency, that has been demonstrated without further analysis. That 40% would become 20%. This is ENRON math. But if you design the output right, making the Thevenin impedance very low, little RF power is wasted in the transmitter (say 1%) and 39% is pumped out into space (on the air) by a matched antenna. "IF?" Are we to assume that every transmitter that has come down the pike has never been designed correctly nor optimized? Forcing such an argument without showing how to accomplish it presumes you have a method that transcends all of the current generation of Engineers capacity to bring it to the market. The force of Capitalism and the Profit Motive most sincerely invalidates that concept in a heartbeat. What about the remaining 61%? You have in fact lowered the emitted power in comparison to the drawn power in your argument above. How this goes to prove that Thevenin's source is Resistorless is an argument of the man standing behind the curtains. Now that low Thevenin impedance will reflect most of the reflected from the mismatched antenna power back to the antenna. And how does the low source Z face the prospect of the High load Z reflecting the power in turn? Thevenin explains the absurdity of this forced expectation, but you do not. Your argument contains the usual blindside, to be described below. If the transmitter Thevenin impedance was 50 ohms, it would absorb the reflections and get even hotter. It does get hot, that is demonstrably true to everyone's experience. That heat has its genesis in the Thevenin model and confirms it. Oh, this has its uses, like in a signal coming from a TTL line driver that passes thru a "source termination" resistor of 100 ohms, then thru a 100 ohm impedance transmission line (like carefully designed traces on circuit boards in that GHz computer) and then to a single destination receiving TTL gate. which looks like a high impedance load. The reflected signal gets absorbed by the source termination. A reason for not destination terminations is that this makes the source work harder pumping DC current when the signal is a "high", vs essentially idling when just source terminated. No daisy chaining allowed, the signal looks like crap except at the very end. Your final elaborations above, again, go to the proof of Thevenin, how you expect it to be otherwise has not been demonstrated. Eventually, these arguments devolve to the myopic observation that there is no "smoking gun" to be exhibited in the form of a carbon composition resistor that embodies the forced R argument of Thevenin. Such posters un/willingly ignore the obvious location found within the Emitter-Collector junction. When my push comes to shove, like right now, they generally scrabble around the spec sheets (giving every evidence of having never built from one) to point out that their transistor exhibits 2 to 3 Ohms NOT 50! They thump their thin chest with pride and then proclaim vindication. Again the myopic arguments collapse in turn when it is pointed out that the Transistor's Emitter-Collector junction is not the Source Z, but rather the abstracted or untransformed value that has yet to be observed (hence the failure of their myopic testimony). If you are going to look under the hood, you have to look at everything under the hood. For EVERY finals deck supporting the usual tandem transistor output, they ALL drive a transformer exhibiting a 1:3 winding ratio. The canonical instruction of Z transformation reveals that this same series combination of Emitter-Collector Z presented through the transformer finds at that output their abstract Source Z transformed to, guess what?, roughly 50 Ohms. However, that is not ALL of the components under the hood to be examined. We also have in EVERY finals deck, an output low pass filter with a characteristic Z of, guess what?, roughly 50 Ohms. Hence, in the deep interior where we abstract the Thevenin model to the circuit level, the equivalent R that is demanded, is found in the several Ohms described above. THAT abstracted Thevenin Source sees the 50 Ohm Load transformed DOWN to its expected value (the usual, and expected reverse operation of transformers built with 3:1 ratios, from the perspective of the load). Most posters argue from their blind side of wholly ignoring the MATCHing operation of the finals' transformer to force their low R argument (which is a corruption of Thevenin's stated Z). Thus, there is a chain of transformation from the Emitter-Collector junction through many levels to finally meet a load that exhibits the confirmation of Thevenin that is consistent with the evidence of heat dissipated by the physical source. If that load presents a mismatch, the power returned is delivered to a MATCHED load: what was formerly the Source. The Source, by design, reflects NOTHING. When the Source is presented with a mismatched Load we either perform additional matching operations (a Tuner) or accept the additional heat burden the mismatch forces upon the Source. This can be in the form of a long term temperature rise (current phase is additive) or the sudden spark (heat, n'est pas?) of voltage breakdown (voltage phase is additive). Contrary to the lore of Transistors being low voltage devices, failure is found far more frequently in the current density at the, guess what?, Emitter-Collector junction. The catastrophic results testify, again, to the validity of the Thevenin model. As I have repaired more than my share of Electronic equipment, professional experience has shown the majority of transistor failures reveal themselves as shorts due to their inability to shed the heat of that current density (a melt down - calories - heat - R). The compelling, and obvious proof of this last is found in very few operators deliberately driving massive mismatches to anything less than equipment failure as a consequence. Of course, when that occurs, it validates the reflective mode of the Source for as many cycles of RF stored in the transmission line returning from the mismatched load; being generous, say roughly several microseconds? 73's Richard Clark, KB7QHC |
Richard sed,
no power is lost to lighting up filaments in transistors. ======================= Wrong ! Transistors don't have filaments. |
"Reg Edwards" wrote in message ... Richard sed, no power is lost to lighting up filaments in transistors. ======================= Wrong ! Transistors don't have filaments. Are some people so ready to jump on comments they see that they fail to see the comment being jumped on was made tongue-in-cheek??? OF COURSE the guy that said "no power is lost to lighting up filaments in transistors" knew there are no filaments in transistors. Sheesh. I can't believe the number of people here just waiting to try and demonstrate how brilliant they are here by jumping on such comments where the intent should have been obvious to everyone. Jerry -- Jerry Bransford To email, remove 'me' from my email address KC6TAY, PP-ASEL See the Geezer Jeep at http://members.cox.net/jerrypb/ |
Richard Clark wrote:
On Thu, 25 Dec 2003 05:47:12 GMT, Robert Casey wrote: As for the reality of the situation, answer me this: 1. How much power does your rig transmit? 2. How much power does your rig draw? Correct me if the operation of dividing the first by the second does not reveal an efficiency of roughly 40% and a power loss to heat of roughly greater than that transmitted. Your rig has a massive heat sink with a fan, n'est pas? n'est pas? ? Anyway, the power lost inside the transmitter I thought was due to the inneficiencies of a class A or B amp configuration. Hi Robert, Given you offer no evidence to the contrary, 40% efficiency seems to be the confirmed rule. Giving it a name does nothing to reduce it or enhance it, the calories expended rob us of RF output for the power draw to generate that output. That power loss is confirmed through everyone's experience as heat. Heat sinks and fans testify to our acceptance of its loss. Let's ignore, for the moment, the losses involved in the conversion of DC power to RF power. Now lets say we have 100 watts of RF. If the antenna is a perfect load (50 ohms resistive) and if the Thevenin impedance of the transmitter is 50 ohms, then yes, you got 50 watts of extra heat in the transmitter. Now if the transmiter has a very low Thevenin impedance, then more power is delivered to the antenna and less waste in the transmitter. I'm not trying to do the "transfer the max power to the load and I don't care how much waste in the source" Thevenin thing we had in EE101. If I did that, and the amp is up to it, I could transmit even more power to the antenna, but I'd waste more in the source. In any event Merry Xmas. |
"Reg Edwards" wrote in message ... Richard sed, no power is lost to lighting up filaments in transistors. ======================= Wrong ! Transistors don't have filaments. but they are full of smoke, let out the smoke and they don't work any more. WAKE UP reg, read his statement again and don't be so fast to jump on some one. |
"Jerry Bransford" wrote in message news:8bGGb.27604$gN.16638@fed1read05... "Reg Edwards" wrote in message ... Richard sed, no power is lost to lighting up filaments in transistors. ======================= Wrong ! Transistors don't have filaments. Are some people so ready to jump on comments they see that they fail to see the comment being jumped on was made tongue-in-cheek??? OF COURSE the guy that said "no power is lost to lighting up filaments in transistors" knew there are no filaments in transistors. Sheesh. I can't believe the number of people here just waiting to try and demonstrate how brilliant they are here by jumping on such comments where the intent should have been obvious to everyone. Jerry -- Jerry Bransford Wellsir, I remember doing that in High School a couple of times and demonstrating just the opposite. I'm actually fairly bright but sometimes I'm not as bright as I thoughtI was at a particular moment. Harold Burton KD5SAK |
On Thu, 25 Dec 2003 19:18:17 GMT, Robert Casey
wrote: Let's ignore, for the moment, the losses involved in the conversion of DC power to RF power. Let's ignore the R resistor, and then we can prove it doesn't exist? :-) Now lets say we have 100 watts of RF. If the antenna is a perfect load (50 ohms resistive) and if the Thevenin impedance of the transmitter is 50 ohms, then yes, you got 50 watts of extra heat in the transmitter. See? Your logic failed at the gate. You ALREADY have 100W RF as a premise. :-) How did you arrive at 100W without its measure AT the Load? To force the speculation that the internal resistance (neglected but evident by such logic) drops it? You are separating those things that are inseparable and this is the common fault of all speculations that look beneath the hood of the Thevenin Model. Now if the transmiter has a very low Thevenin impedance, then more power is delivered to the antenna and less waste in the transmitter. I'm not trying to do the "transfer the max power to the load and I don't care how much waste in the source" This is another forced argument. No where, until now, was it offered that maximum power was being delivered or even demanded. We are simply observing what IS. With the efficiency sitting at 40% it is painfully obvious from the beginning that the road of maximum transfer has not been tread upon - EVER. Thevenin thing we had in EE101. If I did that, and the amp is up to it, I could transmit even more power to the antenna, but I'd waste more in the source. Brush up on your EE101 to discover that the Thevenin Model is derived from observables. I have already described ALL observables that are consistent with opening the model to look under the hood. There have been many writers here, over the years, who have offered bench tests that prove this by observation. As of yet, none has been toppled with work equal in quality (i.e. demonstration or measurement). It is no more simple that plunking the load into a bucket of water, plunking the transmitter into a bucket of water, hitting the transmit switch and measuring temperature. The transmitter will release more heat than the load. To save yourself the issue of submerging the source, common practice (as described by Thevenin) would have you measure the voltage across the source, and the current into it. Yields identical results. I've employed the HP Caloric Wattmeter for years at the Metrologist's bench to faithfully validate this concept over and over. In any event Merry Xmas. Season's Greetings Richard Clark, KB7QHC |
Richard Clark wrote:
SNIP The ONLY advance we can claim in the last half century, is that no power is lost to lighting up filaments in transistors. Oh Boy!! 50 years of hamming, 43 years of engineering, 40 years of marriage, 15 years of ministry, and 3+ years of retirement and NOW I FINALLY FIND OUT WHY TRANSISTOR DON'T LIGHT UP!!! I better hold on to my 3-500 Amplifier so I can demonstrated a ham station to the neighborhood kids ... lots of light! |
"Dave Shrader" wrote in message
news:fTZGb.45760$VB2.83730@attbi_s51... Richard Clark wrote: SNIP The ONLY advance we can claim in the last half century, is that no power is lost to lighting up filaments in transistors. Oh Boy!! 50 years of hamming, 43 years of engineering, 40 years of marriage, 15 years of ministry, and 3+ years of retirement and NOW I FINALLY FIND OUT WHY TRANSISTOR DON'T LIGHT UP!!! I better hold on to my 3-500 Amplifier so I can demonstrated a ham station to the neighborhood kids ... lots of light! I'm dating myself but I miss filaments/vacuum tubes and the wonderful smells/aromas of the hamshacks I used to visit as a kid in the sixties before I got my ham license. Luckily I got to work with enough big vacuum tube tx/rx equipment in the USAF to know how much better what we have now really is. I bought my childhood dream receiver a few years ago, a Hammarlund HQ-180, and it brought back some great memories at mediocre performance. :) 73 Jerry -- Jerry Bransford To email, remove 'me' from my email address KC6TAY, PP-ASEL See the Geezer Jeep at http://members.cox.net/jerrypb/ |
On Fri, 26 Dec 2003 17:05:15 GMT, Dave Shrader
wrote: Richard Clark wrote: SNIP The ONLY advance we can claim in the last half century, is that no power is lost to lighting up filaments in transistors. Oh Boy!! 50 years of hamming, 43 years of engineering, 40 years of marriage, 15 years of ministry, and 3+ years of retirement and NOW I FINALLY FIND OUT WHY TRANSISTOR DON'T LIGHT UP!!! I better hold on to my 3-500 Amplifier so I can demonstrated a ham station to the neighborhood kids ... lots of light! OK Fellows, Such lackluster response to this single comment! Lots of light Dave? Really? Then obviously you were not driving hard enough! When I broke into this business/hobby back then, I worked for a Ham in his TV repair business. One of the notable experiences was watching his final's plates glowing a cheery ruby red and the surrounding envelope filled with a violet light. The Amp may have not been "optimized" nor was the output free of spurs; the line voltage sagged a bit in the effort (another tribute to the Thevenin model); but no one was analyzing the situation. We didn't need a thermometer to prove where the calories in plate resistance were. It gives me the grins when these kind of debates about Thevenin Resistors ignores the obvious. Some folks demand a carbon composition resistor to fulfill their imaginings. :-) 73's Richard Clark, KB7QHC |
Is it possible to have a 1:1 SWR?
-------------------------- Yes! 73 de Jack, K9CUN |
"JDer8745" wrote in message ... Is it possible to have a 1:1 SWR? -------------------------- Yes! 73 de Jack, K9CUN Agreed! I get it all the time with my dummy load. Ed WB6WSN |
Dear Rich, your patience is admirable.
The world of electrical engineering would have been a far more more understandable place if Thevenin and other trivial theorem inventors of his ilk had never existed. As things are, the only purpose served by such superfluous statements of the bleeding obvious is to assist university professors and Ph.D's in justifying their grants and salaries. But people must be allowed to make livings and reputations in the best way they can even if, inadvertently, they turn out to be a handicap. The human race is now so wealthy the economy can well afford them. ;o) By the way, what DID Thevenin say? I don't recall ever having knew. Have I never progressed beyond V = I*R ? ---- Reg |
Reg Edwards wrote:
By the way, what DID Thevenin say? Here's what Ramo and Whinnery said about Thevenin: "It must be emphasized, as in any Thevenin equivalent circuit, that the equivalent circuit was derived to tell what happens in the load under different load conditions, and significance cannot be automatically attached to a calculation of power loss in the internal impedance of the equivalent circuit." -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Mon, 29 Dec 2003 13:38:53 +0000 (UTC), "Reg Edwards"
wrote: The world of electrical engineering would have been a far more more understandable place if Thevenin and other trivial theorem inventors of his ilk had never existed. Hi Reg, Your observation reminds me of Arthur Dent's outrage with Ford Prefect's considered opinion of the importance of Earth within the scheme of the Universe: "Mostly harmless" 73's and remember where your towel is, Richard Clark, KB7QHC |
Someone sed:
"The world of electrical engineering would have been a far more more understandable place if Thevenin and other trivial theorem inventors of his ilk had never existed." =========================== Spoken by a person who is extremely ignorant of what he writes about. It is because of Thevenin's theorem that complicated circuits can be reduced to the equivalent circuit consisting of only two components: a voltage source and an impedance. This is a powerful tool. If I didn't know it, I'd learn it. 73 de Jack, K9CUN |
"The world of electrical engineering would have been a far more more
understandable place if Thevenin and other trivial theorem inventors of his ilk had never existed." =========================== Spoken by a person who is extremely ignorant of what he writes about. It is because of Thevenin's theorem that complicated circuits can be reduced to the equivalent circuit consisting of only two components: a voltage source and an impedance. This is a powerful tool. If I didn't know it, I'd learn it. 73 de Jack, K9CUN ============================ For heaven's sake Jack, is THAT all what it's about ? How come 95% of the contributors to this newsgroup come to blows with each other about the manner of its application? Apparently Thevenin is a severe educational handicap rather than an asset. But perhaps newsgroup Guru's at loggerheads with each other are not representative of the engineering fraternity in general. Normal sixteen year-old students, with the right teacher, grasp the idea immediately without they or their teacher ever having heard of Thevenin. It can then be forgotten. It's so bleeding obvious! As John Cleese implies - students, even at that tender age, should be endowed with a first-class honors degree, including cap and gown, in the venerable practice of the ancient Babylonians, Hittites and Egyptians in stating the obvious. In this more recent age, can we soon expect a graduate of that rat-infested campus of the Rio de Janerio sewers to announce in a blaze of glory that "A complicated circuit is reduceable to an equivalent circuit consisting merely of two components: a current source and an impedance."? Let's hope his name will be pronounceable. ---- Yours, Reg. ;o) |
Reg, G4FGQ wrote:
"A complicated circuit is reduceable to an equivalent circuit consisting merely of two components: a current source and an impedance."? "Let`s hope his name will be pronounceable." True. Does the name "Norton", as in Norton`s Theorem, seem pronounceable? Reg`s assertion is indubitable. In a linear circuit, any generator of electric power may be considered equivalent, at specified frequency, to a current generator whose current is equal to the short-circuit current in shunt with an admittance whose magnitude is equal to that when the generator is inactive and there is no load connected to it, says Henney in his "Radio Engineering Handbook". The Thevenin`s Theorem impedance is the open-circuit voltage divided by the short-circuit current. Terman says on page 95 of his 1937 2nd edition of "Radio Engineering": "According to Thevenin`s theorem, any linear network containing one or more sources of voltage and having two terminals behaves, insofar as a load impedance connected across the terminals is concerned, as though the network and its generators were equivalent to a simple generator having an internal impedance Z and a generator voltage E, where E is the voltage that appears when no load is connected and Z is the impedance that is measured between the terminals when all sources of voltage in the network are short-circuited." We have argued about the linearity required to make the Thevenin equivalent valid. Terman says that linearity is the only limitation to validity. Best regards, Richard Harrison, KB5WZI |
Reg, G4FGQ wrote:
"---students, even at that tender age, should be endowed with a first-class honors degree, including cap and gown---" Ha ratos com toda a certeza nas "favelas" de Rio, mas nao somente nos esgotos, e tambem, nao tengo certeza que os estudantes se usem beretes e capas para mostrar seus elevacoes de grau. It`s been more than 50 years since my 2 weeks of Berlitz instruction in spoken Portuguese, and I`m out of practice and don`t have a dictionary handy but this is what I tried to write above: It is certain that there are rats in the "bean shacks" of Rio, but not only in the sewers, and also, I am not certain that the students use caps and gowns to show their advancements in grade. My grammer may be fractured and my spelling may be poor, but I`ve lived in Lisbon though long ago, and I`ve been to Rio. I think a resident of Rio could decipher what I`ve written. I could more easily write it it in Spanish, which we hear in Texas every day, but it`s different from Portuguese. It`s over 3 months since I was in England, so I`m not so sure I`m doing so well in English either. Spanglish is often spoken here. Best regards, Richard Harrison, KB5WZI |
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