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An antenna question--43 ft vertical
wrote:
Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. As we all agree, under equal output impedance and load impedance conditions, onty half the RF generated reaches the load. This is sim;ly not acceptable or likely for any real-world transmitter. Do 50kW radio station output valves dissipate 50kW? I hope not! You are attempting to mix circuit theory and transmission line theory. You rather have to if you are going to connect a practical circuit to a practical transmission line! The "valves" in a transmitter are not connected to the transmission line. The "valves" in a transmitter are a voltage source connected to an impedance matching network which then connects to a transmission line. Fair enough. Do you want to dissipate 50kW in the matching circuit A 50kW radio station does not generate 50kW of power, it generates a voltage that results in 50kW being dissipted into a 50 Ohm load. There is a difference. Not much, seeing it also has to supply the in-phase current to maintain that voltage across the resistive 50 ohm load. -- Roger Hayter |
An antenna question--43 ft vertical
On Sun, 05 Jul 2015 16:32:49 -0400, Jerry Stuckle
wrote: So why don't manufacturers design transmitters with 1 ohm output impedance, Rick? Hint: Sophomore-level AC Circuits course in virtually any EE degree, and anyone claiming an EE degree should be able to tell why. Actually, there are low output impedance RF power amps. They're quite common in NMR power amps to reduce coupling between stages: Ultra-low output impedance RF power amplifier for parallel excitation http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2663597/ Ultra-low Output Impedance RF Power Amplifier Array http://cds.ismrm.org/ismrm-2007/files/00172.pdf -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
An antenna question--43 ft vertical
Roger Hayter wrote:
wrote: Brian Reay wrote: On 05/07/2015 21:17, wrote: Roger Hayter wrote: snip The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. They are designed to drive into a 50 ohm load, that doesn't mean they have a 50 ohm source impedance. Otherwise efficiency would be rather 'disappointing'. Nope. If they didn't have a 50 Ohm source impedance, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. The SWR looking into the cable from the transmitter is unaffected by the source impedance. Indeed, it is exactly the same if the transmitter is not connected (though you have to connect some kind of generator in order to measure it, it matters little what kind it is.) If the input end is not connected, SWR is meaningless. SWR bridges are calibrated to the source impedance, not the load. -- Jim Pennino |
An antenna question--43 ft vertical
Roger Hayter wrote:
wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. As we all agree, under equal output impedance and load impedance conditions, onty half the RF generated reaches the load. This is sim;ly not acceptable or likely for any real-world transmitter. Do 50kW radio station output valves dissipate 50kW? I hope not! You are attempting to mix circuit theory and transmission line theory. You rather have to if you are going to connect a practical circuit to a practical transmission line! Nope. In the real world you pick a system impedance to match the transmission line you want to use. You design the transmitter with circuit theory to match the line. You design the antenna with electromagetic theory to also match the line. The "valves" in a transmitter are not connected to the transmission line. The "valves" in a transmitter are a voltage source connected to an impedance matching network which then connects to a transmission line. Fair enough. Do you want to dissipate 50kW in the matching circuit What anyone wants is irrelevant to physics. A 50kW radio station does not generate 50kW of power, it generates a voltage that results in 50kW being dissipted into a 50 Ohm load. There is a difference. Not much, seeing it also has to supply the in-phase current to maintain that voltage across the resistive 50 ohm load. The phase of a transmitter is what it is and this is a red herring. -- Jim Pennino |
An antenna question--43 ft vertical
On Sun, 05 Jul 2015 20:22:19 +0100, Brian Reay wrote:
As for a simpler way, I'd recommend a remote auto-matcher like an SGC at the antenna base. It will minimise coax losses and should give you a good match, at least for most bands. I've used a similar set up (with radials) and achieved a good match even on 80m. If your radio has a built in tuner, then it can be used to 'tweak' the match in the event the radio isn't 'seeing' 1.5:1. Turn it off initially. Let the SGC find a match. If it isn't ideal, use the local ATU for a final tweak. I never found this was required but YMMV. Not everyone is a true believer in antenna tuners: http://www.qsl.net/g3tso/Hombrew-Mobile%20Antennas.html I've done some admittedly crude testing of various matching contrivances by measuring the resultant field strength for a given RF power level (measured at the antenna connector). Although not conclusive or spectacular, the early model automagic antenna tuner was rather lossy. Incidentally, I contrived a rather crude but effective way to measure relative overall efficiency. I measured the power consumption from the AC line with a Kill-a-Watt meter (in watts, not VA) and adjusted the CW RF output for some reference level on the field strength meter. While this would not give me a real number for the efficiency, it does produce relative numbers for comparing antenna matching devices. Unfortunately, I can't find my results, but I do recall that the winner was a simple 4:1 torroidal matching xformer. Remember, you really want a low SWR for two reasons, one because modern radios demand it but also to reduce coax (or feeder) loss. With an matcher at the antenna feed point, coax losses are minimised. An ATU at the TX end does nothing to reduce coax losses in real terms. Coax losses below 500 MHz are mostly in the I^2R losses of the copper (as limited by skin effect). Above 500MHz, the dielectric gets involved. See Fig 4: http://www.maximintegrated.com/en/app-notes/index.mvp/id/4303 Higher RF currents, caused by low impedance terminations, will cause higher I^2R losses. These higher losses are why very low impedances are not popular for RF power devices. This has NOTHING to do with matching. For a given RF current through the coax, the contribution of the coax to overall losses will be independent of the VSWR. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
An antenna question--43 ft vertical
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An antenna question--43 ft vertical
On 7/5/2015 6:48 PM, Jeff Liebermann wrote:
On Sun, 05 Jul 2015 16:32:49 -0400, Jerry Stuckle wrote: So why don't manufacturers design transmitters with 1 ohm output impedance, Rick? Hint: Sophomore-level AC Circuits course in virtually any EE degree, and anyone claiming an EE degree should be able to tell why. Actually, there are low output impedance RF power amps. They're quite common in NMR power amps to reduce coupling between stages: Ultra-low output impedance RF power amplifier for parallel excitation http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2663597/ Ultra-low Output Impedance RF Power Amplifier Array http://cds.ismrm.org/ismrm-2007/files/00172.pdf We're not talking amateur transmitters, troll. Try to stay on topic - if that's at all possible. -- ================== Remove the "x" from my email address Jerry Stuckle ================== |
An antenna question--43 ft vertical
On 7/5/2015 6:33 PM, Roger Hayter wrote:
Jerry Stuckle wrote: On 7/5/2015 5:37 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. Because it doesn't have a "50 ohm output" it has an output designed for a 50 ohm load. No, it has a 50 ohm output - as did the transmitter I designed back in my EE class days. But if what you say is correct, how is it designed for 50 ohms? If you claim it as a low output impedance, then it should work equally well on a 75 ohm antenna, or even a 1,000 ohm antenna, as long as the feedline matches. I can assure you that is NOT the case. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? See comment above. If you look at the spec. it probably will not specify the output impedance, just the load impedance. See comment above. What makes the load impedance special? Why shouldn't it work with any sufficiently high load impedance - in fact, the higher, the better? What makes the load resistance special is that the voltage output of the transmitter will drive the correct load resistance with just the right amount of current to provide the design output power without dissipating too much heat. Too high a load resistance may simply not take enough power, but also may upset the operating conditions of the PA in exact-ciruit dependent ways. Too low a load resistance will draw too much current and overheat the amplifier. The design has absolutely nothing to do with making the output impedance equal to the load resistance. So, what happens when I cut my 100W transmitter down to 10W? The voltage output is different - but the impedance doesn't change. You really need to learn transmitter design and impedance matching. The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. Maximum power transfer for a given voltage generator, not maximum power transfer for a given dissipation available. You know electronics. Just do two simple circuits on the back of an envolope, or cigarette pack if available. A voltage generator in series with a 5 ohm internal resistance and a 50 ohm load. And another one with 50 ohm internal resistance and a 50ohm load. Make the output power 100 W, that is an RMS voltage of 70 volts across the load. Now calculate the voltage of the generator, and the total power produced, for both cases. So if what you say is true, I should be able to feed a 300 ohm antenna through 300 ohm feedline and a 1:1 balun with no ill effects. You may think you know electronics - but you do not understand transmission theory. I don't need to draw circuits on a cigarette pack - all I need to do is hook up my wattmeter to my transmitter to prove you wrong. The output impedance of a practical RF power amplifier has exactly zero to do with transmission line theory. (The *effect* of a practical PA output impedance on the transmission line is where we came in, but that only arises *after* we've sorted out the PA output impedance.) It has everything to do with matching the output of the transmitter to the transmission line and load. But you've already shown you don't understand that part. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
"Jeff Liebermann" wrote in message ... On Sun, 05 Jul 2015 20:22:19 +0100, Brian Reay wrote: As for a simpler way, I'd recommend a remote auto-matcher like an SGC at the antenna base. It will minimise coax losses and should give you a good match, at least for most bands. I've used a similar set up (with radials) and achieved a good match even on 80m. If your radio has a built in tuner, then it can be used to 'tweak' the match in the event the radio isn't 'seeing' 1.5:1. Turn it off initially. Let the SGC find a match. If it isn't ideal, use the local ATU for a final tweak. I never found this was required but YMMV. Not everyone is a true believer in antenna tuners: http://www.qsl.net/g3tso/Hombrew-Mobile%20Antennas.html Interesting. I'm off on a different approach. I have an RF ammeter mounted in a box. The box is in the shack between the ATU and the antenna. I simply adjust the ATU for max current on the ammeter. |
An antenna question--43 ft vertical
Jerry Stuckle wrote:
On 7/5/2015 6:48 PM, Jeff Liebermann wrote: On Sun, 05 Jul 2015 16:32:49 -0400, Jerry Stuckle wrote: So why don't manufacturers design transmitters with 1 ohm output impedance, Rick? Hint: Sophomore-level AC Circuits course in virtually any EE degree, and anyone claiming an EE degree should be able to tell why. Actually, there are low output impedance RF power amps. They're quite common in NMR power amps to reduce coupling between stages: Ultra-low output impedance RF power amplifier for parallel excitation http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2663597/ Ultra-low Output Impedance RF Power Amplifier Array http://cds.ismrm.org/ismrm-2007/files/00172.pdf We're not talking amateur transmitters, troll. Try to stay on topic - if that's at all possible. What kind of transmitters should be talk about in an amateur group? Are amateur transmitters different than any other kind of transmitter? -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. -- Jim Pennino |
An antenna question--43 ft vertical
On Sun, 05 Jul 2015 20:05:17 -0400, Jerry Stuckle
wrote: On 7/5/2015 6:48 PM, Jeff Liebermann wrote: On Sun, 05 Jul 2015 16:32:49 -0400, Jerry Stuckle wrote: So why don't manufacturers design transmitters with 1 ohm output impedance, Rick? Hint: Sophomore-level AC Circuits course in virtually any EE degree, and anyone claiming an EE degree should be able to tell why. Actually, there are low output impedance RF power amps. They're quite common in NMR power amps to reduce coupling between stages: Ultra-low output impedance RF power amplifier for parallel excitation http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2663597/ Ultra-low Output Impedance RF Power Amplifier Array http://cds.ismrm.org/ismrm-2007/files/00172.pdf We're not talking amateur transmitters, troll. Try to stay on topic - if that's at all possible. NMR is Nuclear Magnetic Resonance. To the best of my knowledge, medical imaging in not performed by amateurs. I realize that the translation from Chinese to English is somewhat lacking in the above articles. I think if you make the effort, you might learn something about low output impedance power amplifiers, which seems to be a topic drift that *YOU* started with the above "1 ohm" question. Since the original question of the purpose of the matching transformer on the 43 ft vertical has been correctly answered at least 3 times, I would guess that some topic drift might be acceptable. Do continue. Since the original question is addressed to Rick, I won't ruin your fun by providing an answer. No need to thank me. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
An antenna question--43 ft vertical
On 7/5/2015 7:21 PM, wrote:
John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. So, it can't be used in a 50 ohm environment? What does that have to do with anything? The chart has a SWR graph and nowhere does it need source impedance. If you disagree, please link to one. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. Please show the EZNEC statement that "assumes the transmission line matches the transmitter". Look in the help section if you have EZNEC and can cut and paste or just refer me to the chapter and verse. Also, if you have EZNEC, you can insert a transmission line with arbitrary characteristic impedance, put a load on the far end matching the line, and look at the SWR. It will still be 1:1 because the LOAD matches the LINE. Not because EZNEC assumes a source impedance. Try it with and report back here. There is no way that a source initiates reflections. That is a property of the line and load only. It may re-reflect a wave reflected from the load, but that is all. You can also verify this in LTSPICE if you wish. |
An antenna question--43 ft vertical
On 7/5/2015 8:23 PM, Jeff Liebermann wrote:
On Sun, 05 Jul 2015 20:05:17 -0400, Jerry Stuckle wrote: On 7/5/2015 6:48 PM, Jeff Liebermann wrote: On Sun, 05 Jul 2015 16:32:49 -0400, Jerry Stuckle wrote: So why don't manufacturers design transmitters with 1 ohm output impedance, Rick? Hint: Sophomore-level AC Circuits course in virtually any EE degree, and anyone claiming an EE degree should be able to tell why. Actually, there are low output impedance RF power amps. They're quite common in NMR power amps to reduce coupling between stages: Ultra-low output impedance RF power amplifier for parallel excitation http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2663597/ Ultra-low Output Impedance RF Power Amplifier Array http://cds.ismrm.org/ismrm-2007/files/00172.pdf We're not talking amateur transmitters, troll. Try to stay on topic - if that's at all possible. NMR is Nuclear Magnetic Resonance. To the best of my knowledge, medical imaging in not performed by amateurs. I realize that the translation from Chinese to English is somewhat lacking in the above articles. I think if you make the effort, you might learn something about low output impedance power amplifiers, which seems to be a topic drift that *YOU* started with the above "1 ohm" question. Since the original question of the purpose of the matching transformer on the 43 ft vertical has been correctly answered at least 3 times, I would guess that some topic drift might be acceptable. Do continue. Since the original question is addressed to Rick, I won't ruin your fun by providing an answer. No need to thank me. We're not talking about NMR. We're discussing amateur ratio. Try to stay on target, troll. And no, I didn't start ANY topic on output impedance drift. But you're too stoopid to understand the topic at hand, so I can see how you can come to that conclusion. Fortunately, intelligent people understand the thread and the fact I didn't start the topic on impedance drift. The fact you think I did proves your stoopidity. As well as the fact that amateurs have anything to do with NMR. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On Sun, 5 Jul 2015 17:08:56 -0700, "Wayne"
wrote: I'm off on a different approach. I have an RF ammeter mounted in a box. The box is in the shack between the ATU and the antenna. I simply adjust the ATU for max current on the ammeter. Rewind. I just noticed that you're planning to put the RF ammeter between the ATU and the antenna. That will work, but with approximately 1200 ohms antenna impedance, you are going to see 50/1200 = 0.04 times the antenna current that you would see on the 50 ohm line between the xmitter and the ATU. If you're running lots of power, that might work, but offhand, methinks not. Also, you can't adjust the ATU for maximum current. It adjusts itself based on it's own internal VSWR sensor. All you can do is watch the light show and listen to the relays clatter. You might be able to have some control if it were a motorized antenna tuner. Certainly a manual antenna tuner would work. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
An antenna question--43 ft vertical
On 7/5/2015 4:45 PM, Jerry Stuckle wrote:
On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. -- Rick |
An antenna question--43 ft vertical
On 7/6/2015 12:39 AM, rickman wrote:
On 7/5/2015 4:45 PM, Jerry Stuckle wrote: Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. In case anyone wants to duplicate the simulation... Version 4 SHEET 1 880 680 WIRE 48 48 0 48 WIRE 144 48 48 48 WIRE 304 48 224 48 WIRE 336 48 304 48 WIRE 0 80 0 48 WIRE 336 80 336 48 WIRE 0 192 0 160 WIRE 336 192 336 160 WIRE 48 304 0 304 WIRE 144 304 48 304 WIRE 304 304 224 304 WIRE 336 304 304 304 WIRE 0 336 0 304 WIRE 336 336 336 304 WIRE 0 448 0 416 WIRE 336 448 336 416 FLAG 0 192 0 FLAG 336 192 0 FLAG 48 48 V1 FLAG 304 48 Vload1 FLAG 0 448 0 FLAG 336 448 0 FLAG 48 304 V2 FLAG 304 304 Vload2 SYMBOL voltage 0 64 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V1 SYMATTR Value SINE(0 200 100K) SYMBOL res 320 64 R0 SYMATTR InstName RL1 SYMATTR Value 50 SYMBOL res 128 64 R270 WINDOW 0 32 56 VTop 2 WINDOW 3 0 56 VBottom 2 SYMATTR InstName RS1 SYMATTR Value 50 SYMBOL voltage 0 320 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V2 SYMATTR Value SINE(0 200 100K) SYMBOL res 320 320 R0 SYMATTR InstName RL2 SYMATTR Value 50 SYMBOL res 128 320 R270 WINDOW 0 32 56 VTop 2 WINDOW 3 0 56 VBottom 2 SYMATTR InstName RS2 SYMATTR Value 1 TEXT -34 472 Left 2 !.tran 10us -- Rick |
An antenna question--43 ft vertical
On 7/5/2015 5:19 PM, Brian Reay wrote:
On 05/07/2015 21:17, wrote: Roger Hayter wrote: The impedance of a transmitter output will be nothing like 50 ohms resistive, as this would result in an efficiency well below 50%, with all the normal amplfier losses plus the actual RF power produced being 50% dissipated in the PA. This is why matching in the forward direction coexists with a mjaor mismatch in the reverse direction. This is good because if there is any reflected wave we don't want it to add yet more to the PA dissipation. But it does explain what is happening, and why there are increased losses in the feeder as well as the matching networks. The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. They are designed to drive into a 50 ohm load, that doesn't mean they have a 50 ohm source impedance. Otherwise efficiency would be rather 'disappointing'. The PA stages are designed to operate safely with a load equivalent to a SWR of (typically) 1.5:1 . Any higher, and it means the load is out of spec, and the PA leaves its safe area of operation (assuming there is no mechanism to reduce the power). This is were the myth of RF 'entering' the PA came from - people thinking that a high SWR meant the reflected RF was getting into the PA and causing damage. In fact, it 'sees' a mismatch and therefore can't enter the PA. I designed a line driver circuit once that used "synthetic impedance" to simulate a 50 ohm output impedance with just a 12.5 ohm resistor. Basically it uses positive feedback to sense the load current through the output resistor and manage the drive to appear to be a higher voltage source with a higher output impedance while only dissipating the power of the smaller resistor as well as utilizing a lower power supply voltage. As to your mismatch theory, a mismatch doesn't mean *no* power enters the amp from the feed line. It just means not all the power will be transferred into the amp. -- Rick |
An antenna question--43 ft vertical
John S wrote:
On 7/5/2015 7:21 PM, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. So, it can't be used in a 50 ohm environment? What does that have to do with anything? The chart has a SWR graph and nowhere does it need source impedance. If you disagree, please link to one. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. Please show the EZNEC statement that "assumes the transmission line matches the transmitter". Look in the help section if you have EZNEC and can cut and paste or just refer me to the chapter and verse. Also, if you have EZNEC, you can insert a transmission line with arbitrary characteristic impedance, put a load on the far end matching the line, and look at the SWR. It will still be 1:1 because the LOAD matches the LINE. Not because EZNEC assumes a source impedance. Try it with and report back here. There is no way that a source initiates reflections. That is a property of the line and load only. It may re-reflect a wave reflected from the load, but that is all. You can also verify this in LTSPICE if you wish. What happens if you take any off the shelf commercial amateur radio transmitter that does not have a built in tuner and: Attach a 10 Ohm load. Attach a 200 Ohm load. Attach a 1,000 Ohm load. Attach a 1 Ohm load. Attach a 50 Ohm load. -- Jim Pennino |
An antenna question--43 ft vertical
In message , rickman
writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). -- Ian |
An antenna question--43 ft vertical
On 7/6/2015 12:19 AM, wrote:
John S wrote: On 7/5/2015 7:21 PM, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. So, it can't be used in a 50 ohm environment? What does that have to do with anything? The chart has a SWR graph and nowhere does it need source impedance. If you disagree, please link to one. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. Please show the EZNEC statement that "assumes the transmission line matches the transmitter". Look in the help section if you have EZNEC and can cut and paste or just refer me to the chapter and verse. Also, if you have EZNEC, you can insert a transmission line with arbitrary characteristic impedance, put a load on the far end matching the line, and look at the SWR. It will still be 1:1 because the LOAD matches the LINE. Not because EZNEC assumes a source impedance. Try it with and report back here. There is no way that a source initiates reflections. That is a property of the line and load only. It may re-reflect a wave reflected from the load, but that is all. You can also verify this in LTSPICE if you wish. What happens if you take any off the shelf commercial amateur radio transmitter that does not have a built in tuner and: Attach a 10 Ohm load. Attach a 200 Ohm load. Attach a 1,000 Ohm load. Attach a 1 Ohm load. Attach a 50 Ohm load. Please address my questions first before setting up another strawman. |
An antenna question--43 ft vertical
"Jeff Liebermann" wrote in message ... On Sun, 5 Jul 2015 17:08:56 -0700, "Wayne" wrote: I'm off on a different approach. I have an RF ammeter mounted in a box. The box is in the shack between the ATU and the antenna. I simply adjust the ATU for max current on the ammeter. Rewind. I just noticed that you're planning to put the RF ammeter between the ATU and the antenna. That will work, but with approximately 1200 ohms antenna impedance, you are going to see 50/1200 = 0.04 times the antenna current that you would see on the 50 ohm line between the xmitter and the ATU. If you're running lots of power, that might work, but offhand, methinks not. Also, you can't adjust the ATU for maximum current. It adjusts itself based on it's own internal VSWR sensor. All you can do is watch the light show and listen to the relays clatter. You might be able to have some control if it were a motorized antenna tuner. Certainly a manual antenna tuner would work. This isn't something planned. I have been using the ammeter between the tuner and antenna for many years. With an automatic tuner, there is no feedback from the ammeter to the tuner. In that case it is simply an indicator that current is present. Some of my antennas have SWR outside the capabilities of my automatic tuner. A manual tuner is used in that case. With a manual tuner, I don't look for a specific reading, just a peak in the current from the tuner to the antenna. |
An antenna question--43 ft vertical
On 7/6/2015 12:39 AM, rickman wrote:
On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Your simplified response shows you never took an EE course in your life. I suggest you take some EE courses and learn how things work. It's very obvious you don't have any more knowledge than you get with ohm's law. But if what you say is correct, then I should be able to get a lot of power out of my 100 watt transmitter feeding a 1 ohm antenna. Never mind the 50:1 SWR. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On 7/6/2015 4:20 AM, Ian Jackson wrote:
In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. If what you say is correct, then it wouldn't matter what antenna impedance I had, as long as it matches the transmission line. VSWR would be immaterial. That is demonstrably false. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On 7/6/2015 11:59 AM, Jerry Stuckle wrote:
On 7/6/2015 12:39 AM, rickman wrote: On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Your simplified response shows you never took an EE course in your life. I suggest you take some EE courses and learn how things work. It's very obvious you don't have any more knowledge than you get with ohm's law. But if what you say is correct, then I should be able to get a lot of power out of my 100 watt transmitter feeding a 1 ohm antenna. Never mind the 50:1 SWR. Would anyone else like to explain to Jerry the fallacy of his argument? I get tired of explaining the obvious sometimes. -- Rick |
An antenna question--43 ft vertical
On 7/5/2015 7:08 PM, Wayne wrote:
"Jeff Liebermann" wrote in message ... On Sun, 05 Jul 2015 20:22:19 +0100, Brian Reay wrote: As for a simpler way, I'd recommend a remote auto-matcher like an SGC at the antenna base. It will minimise coax losses and should give you a good match, at least for most bands. I've used a similar set up (with radials) and achieved a good match even on 80m. If your radio has a built in tuner, then it can be used to 'tweak' the match in the event the radio isn't 'seeing' 1.5:1. Turn it off initially. Let the SGC find a match. If it isn't ideal, use the local ATU for a final tweak. I never found this was required but YMMV. Not everyone is a true believer in antenna tuners: http://www.qsl.net/g3tso/Hombrew-Mobile%20Antennas.html Interesting. I'm off on a different approach. I have an RF ammeter mounted in a box. The box is in the shack between the ATU and the antenna. I simply adjust the ATU for max current on the ammeter. Hey, Wayne - As a matter of curiosity on my part, can you find a way to measure the ammeter's resistance and let me know the full-scale value? Many thanks, John |
An antenna question--43 ft vertical
On 7/6/2015 11:01 AM, Jerry Stuckle wrote:
On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? If what you say is correct, then it wouldn't matter what antenna impedance I had, as long as it matches the transmission line. VSWR would be immaterial. There is no VSWR nor ISWR if the load matches the line. That is demonstrably false. Please demonstrate this for us as we wish to learn. |
An antenna question--43 ft vertical
Ian Jackson wrote:
In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? Because RF transmitters deliver high frequency AC to a transmission line. -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/6/2015 12:19 AM, wrote: John S wrote: On 7/5/2015 7:21 PM, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. So, it can't be used in a 50 ohm environment? What does that have to do with anything? The chart has a SWR graph and nowhere does it need source impedance. If you disagree, please link to one. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. Please show the EZNEC statement that "assumes the transmission line matches the transmitter". Look in the help section if you have EZNEC and can cut and paste or just refer me to the chapter and verse. Also, if you have EZNEC, you can insert a transmission line with arbitrary characteristic impedance, put a load on the far end matching the line, and look at the SWR. It will still be 1:1 because the LOAD matches the LINE. Not because EZNEC assumes a source impedance. Try it with and report back here. There is no way that a source initiates reflections. That is a property of the line and load only. It may re-reflect a wave reflected from the load, but that is all. You can also verify this in LTSPICE if you wish. What happens if you take any off the shelf commercial amateur radio transmitter that does not have a built in tuner and: Attach a 10 Ohm load. Attach a 200 Ohm load. Attach a 1,000 Ohm load. Attach a 1 Ohm load. Attach a 50 Ohm load. Please address my questions first before setting up another strawman. Start with Electromagnetics by Kraus and Carver, Chapter 13. -- Jim Pennino |
An antenna question--43 ft vertical
On 7/5/2015 11:39 PM, rickman wrote:
On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. There can be a lossless resistive part of source impedance according to the IEEE (and most every other well educated EEs). After all, a transmission line has a resistance but it's loss resistance is much lower. |
An antenna question--43 ft vertical
On 7/6/2015 11:44 AM, wrote:
Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? Because RF transmitters deliver high frequency AC to a transmission line. So, the laws of physics are different with AC (high frequency or not)? Please show the equations, literature, links, or any other suitable authority advancing that theory. |
An antenna question--43 ft vertical
On 7/6/2015 12:50 PM, John S wrote:
On 7/5/2015 11:39 PM, rickman wrote: On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. There can be a lossless resistive part of source impedance according to the IEEE (and most every other well educated EEs). After all, a transmission line has a resistance but it's loss resistance is much lower. Can you provide a reference to any of this? -- Rick |
An antenna question--43 ft vertical
John S wrote:
On 7/6/2015 11:01 AM, Jerry Stuckle wrote: On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? Yes, quite a lot, you get a whole new set of laws. Capacitors are an open circuit at DC and have a frequency dependant impedance at AC. Inductors are a short circuit at DC and have a frequency dependant impedance at AC. There is no such thing as a transmission line at DC. Current at DC is constant and does not cause propagation while current at AC causes a varying electromagnetic field that can propagate. There is no such thing as a phase angle at DC. A wire carrying DC current will not induce a voltage into another nearby wire but a wire carrying AC current will. More? -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/6/2015 11:44 AM, wrote: Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? Because RF transmitters deliver high frequency AC to a transmission line. So, the laws of physics are different with AC (high frequency or not)? Please show the equations, literature, links, or any other suitable authority advancing that theory. Yes, for AC you get a whole new set of laws such as: Capacitors no longer have an infinite impedance. Inductors no longer have zero impedance. Transmission lines work. Resonant circuits work. Resonant cavities work. Inductive coupling works. Generation of an electromagnetic field that can propagate works. There are many, many more differences between AC and DC. -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
There can be a lossless resistive part of source impedance according to the IEEE (and most every other well educated EEs). After all, a transmission line has a resistance but it's loss resistance is much lower. Nope, a transmission line has a characterisic, or surge, impedance, not a resistance. There is a big difference between a resistance and an impedance. The characterisic impedance of a transmission line is the square root of the series impedance per unit length divided by the shunt admittance per unit length. See any electromagnetics text. -- Jim Pennino |
An antenna question--43 ft vertical
In message , Jeff Liebermann
writes One problem with the thermocouple ammeter. It's slow. The other problem with the thermocouple ammeter is that it's easy to burn out the thermocouple. If you come across one in a sale, before you part with your money first twist the meter enthusiastically from side to side, and note how the needle swings. It should be reluctant to move - ie it's well damped by the almost short circuit of the thermocouple. If it swings freely, it almost certainly means that the thermocouple is open-circuit. But if you've been unfortunate to buy a duffer, don't totally despair. With a simple internal rewire, at least you'll have a usable* 500uA or 1mA FSD moving coil meter. *Convert to diode type? . -- Ian |
An antenna question--43 ft vertical
In message , rickman
writes The only case I am aware of that will give total reflection is when the terminal is open circuit with infinite impedance absorbing *no* signal. Also when it is a zero impedance (short circuit). -- Ian |
An antenna question--43 ft vertical
On 7/6/2015 12:41 PM, John S wrote:
On 7/6/2015 11:01 AM, Jerry Stuckle wrote: On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? Yup, AC has reactance. DC does not. Big difference. If what you say is correct, then it wouldn't matter what antenna impedance I had, as long as it matches the transmission line. VSWR would be immaterial. There is no VSWR nor ISWR if the load matches the line. Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. That is demonstrably false. Please demonstrate this for us as we wish to learn. OK, take your amateur transmitter. Connect it through a 1:1 balun to 300 ohm feedline. Connect that to a 300 ohm antenna. According to you, you should get full power output at the antenna. In reality, you will get a 6:1 SWR and about 49% of the power at the antenna, minus transmission line loss (assuming, of course, your transmitter hasn't cut it's power back). -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On 7/6/2015 12:20 PM, rickman wrote:
On 7/6/2015 11:59 AM, Jerry Stuckle wrote: On 7/6/2015 12:39 AM, rickman wrote: On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Your simplified response shows you never took an EE course in your life. I suggest you take some EE courses and learn how things work. It's very obvious you don't have any more knowledge than you get with ohm's law. But if what you say is correct, then I should be able to get a lot of power out of my 100 watt transmitter feeding a 1 ohm antenna. Never mind the 50:1 SWR. Would anyone else like to explain to Jerry the fallacy of his argument? I get tired of explaining the obvious sometimes. Of course you can't explain it, because you are clearly wrong. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
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