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VSWR doesn't matter?
We are all told that VSWR doesn't matter when using low loss
transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? For the RF to bounce back and forth, wouldn't the transmitter's impedance have to be very, very high (or low) when the reflected RF energy hit its output stages? I know I'm missing something vital here... Thanks! -Bill |
VSWR doesn't matter?
billcalley wrote:
We are all told that VSWR doesn't matter when using low loss transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? For the RF to bounce back and forth, wouldn't the transmitter's impedance have to be very, very high (or low) when the reflected RF energy hit its output stages? I know I'm missing something vital here... That's assuming you use an antenna tuner. The tuner will transform the transmitter's output impedance* just as it transforms the line. Were the transmitter output impedance actually at 50 ohms, on the other side of the tuner it would have the same VSWR as the line when everything was tuned up. Having said that, the VSWR _does_ matter somewhat when using low loss lines, both because the line loss is low but not zero, and the tuner loss will tend to go up as you correct for higher and higher VSWR. * I am _not_ going to start the Big Transmitter Output Impedance Debate. sed denizens -- just don't comment on what a transmitter's "actual" output impedance may be, lest you start a flame war. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html |
VSWR doesn't matter?
billcalley wrote:
... I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? For the RF to bounce back and forth, wouldn't the transmitter's impedance have to be very, very high (or low) when the reflected RF energy hit its output stages? I know I'm missing something vital here... Yep, you are missing the "total destructive interference" happening toward the source caused by a Z0-match. Here's an article that might help: http://www.w5dxp.com/energy.htm A Z0-match eliminates reflected energy from reaching the source. -- 73, Cecil http://www.w5dxp.com |
VSWR doesn't matter?
On Mar 11, 8:08�pm, "billcalley" wrote:
* * We are all told that VSWR doesn't matter when using low loss transmission lines... Well, I was never taught that -- which is a good thing because it is not true. The most intuitive way to prove this (antenna tuner or not), is to simply supply more and more power to the load. You will eventually burn it up. Now, had the VSWR been better (ie. a better network Return Loss), it would have taken even more power to achieve the same "charcoal" results. So VSWR matters. -mpm |
VSWR doesn't matter?
On Mar 12, 1:08 am, "billcalley" wrote:
We are all told that VSWR doesn't matter when using low loss transmission lines, No, we are not all told that. since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? The active part of the transmitter output isn't 50 ohm. That would cause half the power to be lost as heat in the output stage. It's only 50ohm once it becomes a moving wave in the transmission line. Bob9 |
VSWR doesn't matter?
On 11 Mar 2007 20:39:46 -0700, "Bob" wrote:
The active part of the transmitter output isn't 50 ohm. That would cause half the power to be lost as heat in the output stage. Hi Bob, Well, aside from the initial misunderstanding of how power gets to the load (much less back, and then to the load again); I will put to you a question that has NEVER been answered by those who know what the transmitter output Z ISN'T: "What Z is it?" 73's Richard Clark, KB7QHC |
VSWR doesn't matter?
Triode or pentode? ;o)
Tim -- Deep Fryer: A very philosophical monk. Website @ http://webpages.charter.net/dawill/tmoranwms "Richard Clark" wrote in message ... On 11 Mar 2007 20:39:46 -0700, "Bob" wrote: The active part of the transmitter output isn't 50 ohm. That would cause half the power to be lost as heat in the output stage. Hi Bob, Well, aside from the initial misunderstanding of how power gets to the load (much less back, and then to the load again); I will put to you a question that has NEVER been answered by those who know what the transmitter output Z ISN'T: "What Z is it?" 73's Richard Clark, KB7QHC |
VSWR doesn't matter?
In .com, billcalley said:
We are all told that VSWR doesn't matter when using low loss transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? Two problems: 1) The transmitter may well have output impedance matching the characteristic impedance of the transmission line. RF power reflected back in this case gets converted to heat in the output stage of the transmitter, in addition to whatever heat the output stage already has to dissipate. 1a) The reflection may increase requirement of the output tubes/transistors to both drop voltage and dissipate power. This can be a problem for many transistors, especially a lot of bipolar ones. It is not necessarily sufficient to stay within power, current, voltage and thermal ratings. Many bipolar transistors have reduced capability to safely dissipate power at voltages that are higher but within their ratings - sometimes even at voltages as low as 35-50 volts. This problem tends to be worse with bipolar transistors that are faster and/or better for use with higher frequencies. The keyphrase here is "forward bias second breakdown", a problem of uneven current distribution within the die at higher voltage drop. 2) It appears to me that transmitters can have output stage output impedance differing from the intended load impedance. An analog is common practice with audio amplifiers - output impedance is often ideally as close to zero as possible, as opposed to matching the load impedance. If zero output impedance is achieved in an RF output stage, I see a possible benefit - reflections do not increase output stage heating but get reflected back towards the antenna. Then again, the impedance of the input end of the transmission line could be low or significantly reactive depending on how the load is mismatched and how many wavelengths long the transmission line is, and that can increase heating of the output stage. In a few cases transmitted power can also increase. Not only is increased output stage heating possible and maybe fairly likely, high VSWR also causes a high chance of the output stage seeing a partially reactive load. RF bipolar transistors often do not like those due to increased need to dissipate power with higher voltage drop. As I said above, RF bipolar transistors are likely to really dislike simultaneous higher voltage drop and higher power dissipation. - Don Klipstein ) |
VSWR doesn't matter?
On Mon, 12 Mar 2007 00:09:26 -0600, "Tim Williams"
wrote: Triode or pentode? ;o) For you, cascode, push-pull triodes. 73's Richard Clark, KB7QHC |
VSWR doesn't matter?
*Sigh*
The same misconceptions keep coming up, as they have countless times on this newsgroup and I'm sure they will for decades or perhaps centuries to come. After one of the many previous discussions, I wrote a little tutorial on the topic. Originally in the form of plain text files, I've combined it into a pdf file for easier viewing. You can find it at http://eznec.com/misc/Food_for_thought.pdf. On page 8 you'll find the statement "THE REVERSE POWER IS NOT DISSIPATED IN OR ABSORBED BY THE SOURCE RESISTANCE". Above it is a chart of several examples which clearly show that there's no relationship between the "reverse power" and the source dissipation. The remainder of the tutorial explains why. Any theory about "forward" and "reverse" power, what they do, and their interaction with the source, will have to explain the values in the example chart on page 8. Does yours? Roy Lewallen, W7EL Don Klipstein wrote: In .com, billcalley said: We are all told that VSWR doesn't matter when using low loss transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? Two problems: 1) The transmitter may well have output impedance matching the characteristic impedance of the transmission line. RF power reflected back in this case gets converted to heat in the output stage of the transmitter, in addition to whatever heat the output stage already has to dissipate. 1a) The reflection may increase requirement of the output tubes/transistors to both drop voltage and dissipate power. This can be a problem for many transistors, especially a lot of bipolar ones. It is not necessarily sufficient to stay within power, current, voltage and thermal ratings. Many bipolar transistors have reduced capability to safely dissipate power at voltages that are higher but within their ratings - sometimes even at voltages as low as 35-50 volts. This problem tends to be worse with bipolar transistors that are faster and/or better for use with higher frequencies. The keyphrase here is "forward bias second breakdown", a problem of uneven current distribution within the die at higher voltage drop. 2) It appears to me that transmitters can have output stage output impedance differing from the intended load impedance. An analog is common practice with audio amplifiers - output impedance is often ideally as close to zero as possible, as opposed to matching the load impedance. If zero output impedance is achieved in an RF output stage, I see a possible benefit - reflections do not increase output stage heating but get reflected back towards the antenna. Then again, the impedance of the input end of the transmission line could be low or significantly reactive depending on how the load is mismatched and how many wavelengths long the transmission line is, and that can increase heating of the output stage. In a few cases transmitted power can also increase. Not only is increased output stage heating possible and maybe fairly likely, high VSWR also causes a high chance of the output stage seeing a partially reactive load. RF bipolar transistors often do not like those due to increased need to dissipate power with higher voltage drop. As I said above, RF bipolar transistors are likely to really dislike simultaneous higher voltage drop and higher power dissipation. - Don Klipstein ) |
VSWR doesn't matter?
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VSWR doesn't matter?
On Mon, 12 Mar 2007 11:00:04 GMT, Jan Panteltje
wrote: On a sunny day (Mon, 12 Mar 2007 07:36:25 +0000 (UTC)) it happened (Don Klipstein) wrote in : snipped good stuff Not only is increased output stage heating possible and maybe fairly likely, high VSWR also causes a high chance of the output stage seeing a partially reactive load. RF bipolar transistors often do not like those due to increased need to dissipate power with higher voltage drop. As I said above, RF bipolar transistors are likely to really dislike simultaneous higher voltage drop and higher power dissipation. - Don Klipstein ) All true. Also normally, there is a pi type filter (to prevent harmonics), between amplifier and antenna. This filter _WILL_ match the antenna to the output impedance of the transmitter, so _even_ if the transmitter output impedance is very very low (low voltage high current output stage for example), the reflected power will be nicely converted to match the transmitter, and heat up the output amp, with its possible destruction as result. Jan and Don, Both of you gentlemen really need to read Walter's book Reflections (any edition) and put that myth to rest once and for all. Danny, K6MHE |
VSWR doesn't matter?
Bill,
an exellent treatment on this question has been published in QEX December 94 under the title "Where does the power go?" 73 de Ulrich, DF6JB "billcalley" schrieb im Newsbeitrag oups.com... We are all told that VSWR doesn't matter when using low loss transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? For the RF to bounce back and forth, wouldn't the transmitter's impedance have to be very, very high (or low) when the reflected RF energy hit its output stages? I know I'm missing something vital here... Thanks! -Bill |
VSWR doesn't matter?
Richard Clark wrote:
I will put to you a question that has NEVER been answered by those who know what the transmitter output Z ISN'T: "What Z is it?" It really doesn't matter except for overall efficiency. A 10 ohm source outputting 100 volts into a local load of R +/- jX sources the same amount of power as a 100 ohm source outputting 100 volts into the same local load. -- 73, Cecil http://www.w5dxp.com |
VSWR doesn't matter?
Don Klipstein wrote:
If zero output impedance is achieved in an RF output stage, I see a possible benefit - Yep, 100% efficiency would be quite a benefit. -- 73, Cecil http://www.w5dxp.com |
VSWR doesn't matter?
Roy Lewallen wrote:
*Sigh* The same misconceptions keep coming up, as they have countless times on this newsgroup and I'm sure they will for decades or perhaps centuries to come. After one of the many previous discussions, I wrote a little tutorial on the topic. Originally in the form of plain text files, I've combined it into a pdf file for easier viewing. You can find it at http://eznec.com/misc/Food_for_thought.pdf. On page 8 you'll find the statement "THE REVERSE POWER IS NOT DISSIPATED IN OR ABSORBED BY THE SOURCE RESISTANCE". Above it is a chart of several examples which clearly show that there's no relationship between the "reverse power" and the source dissipation. The remainder of the tutorial explains why. Any theory about "forward" and "reverse" power, what they do, and their interaction with the source, will have to explain the values in the example chart on page 8. Does yours? Mine does. All of your values can be understood by looking at the destructive and constructive interference and applying the irradiance (power density) equations from the field of optics. You see, optical engineers and physicists don't have the luxury of measuring voltage and current in their EM waves. All they can measure is power density and interference and thus their entire body of knowledge of EM waves rests upon measurements of those quantities. Those power density and interference theories and equations are directly applicable and 100% compatible with RF theories and equations. Any analysis based on power density and interference will yield identical results to the ones you reported in your "food for thought" article which includes the following false statement: "While the nature of the voltage and current waves when encountering an impedance discontinuity is well understood, we're lacking a model of what happens to this "reverse power" we've calculated." We are not lacking a model of what happens to this "reverse power" we've calculated. The model is explained fully in "Optics", by Hecht. When one has standing waves of light in free space, it is hard to hide the details under the transmission line rug. In general, it is just as easy, and sometimes easier, to deal with the energy values and then calculate voltage and current as it is to start with voltage and current and then calculate the power. All this is explained in my WorldRadio article at: http://www.w5dxp.com/energy.htm The great majority of amateur antenna systems are Z0-matched. For such systems, an energy analysis is definitely easier to perform than a voltage analysis. Here's an example: 100W------50 ohm---+---Z050 ohms-----load Pfor1=100W-- Pfor2-- --Pref1=0W --Pref2 The power reflection coefficient is 0.5 at point '+'. The power reflection coefficient is 0.5 at the load. What are the values of Pfor2 and Pref2? What is the physics equation governing what happens to Pref2 at point '+'? -- 73, Cecil http://www.w5dxp.com |
VSWR doesn't matter?
Jan Panteltje wrote:
Also normally, there is a pi type filter (to prevent harmonics), between amplifier and antenna. This filter _WILL_ match the antenna to the output impedance of the transmitter, so _even_ if the transmitter output impedance is very very low (low voltage high current output stage for example), the reflected power will be nicely converted to match the transmitter, and heat up the output amp, with its possible destruction as result. Some gurus will say that it's the voltage and/or current that is destroying the final, not the reflected energy. They have yet to explain how those dangerous voltages and/or currents can exist without assistance from the ExH joules/second in the reflected energy wave. Depending upon phase, the E in the ExH reflected wave is what causes the overvoltage due to SWR. The H in the ExH reflected wave is what causes the overcurrent due to SWR. The impedance seen by the source is Z = (Vfor+Vref)/(Ifor+Iref) Where '+' indicates phasor (vector) addition. The above equation also gives the impedance anywhere along the transmission line and anywhere along a standing-wave antenna. -- 73, Cecil http://www.w5dxp.com |
VSWR doesn't matter?
Ulrich Bangert wrote:
an exellent treatment on this question has been published in QEX December 94 under the title "Where does the power go?" Unfortunately, that article doesn't explain where the power does go. A much better treatment of the subject is in "Optics", by Hecht. To understand where the power does go, one must understand destructive and constructive interference. Please see my transmission line example in another posting. The energy content of a transmission line during steady-state is always exactly enough to support the forward traveling wave and the reverse traveling wave without which there would be no standing wave. -- 73, Cecil http://www.w5dxp.com |
VSWR doesn't matter?
In , Cecil Moore wrote:
Don Klipstein wrote: If zero output impedance is achieved in an RF output stage, I see a possible benefit - Yep, 100% efficiency would be quite a benefit. There are audio amplifiers with output impedance around .1 ohm, driving 8 ohm speakers, and having efficiency nowhere near 80/81. The theoretical limit for efficiency of a class B amp driving a resistive load with a sinewave is 78.54%. - Don Klipstein ) |
VSWR doesn't matter?
Don Klipstein wrote:
Cecil Moore wrote: Yep, 100% efficiency would be quite a benefit. There are audio amplifiers with output impedance around .1 ohm, driving 8 ohm speakers, and having efficiency nowhere near 80/81. The theoretical limit for efficiency of a class B amp driving a resistive load with a sinewave is 78.54%. Of course, that was a tongue-in-cheek posting. But if you could design a Thevenin equivalent source with a 0.1 ohm source impedance, wouldn't the efficiency calculate out to be pretty high? -- 73, Cecil http://www.w5dxp.com |
VSWR doesn't matter?
Thanks guys -- some really great posts here -- it will take me quite a
while to digest all this!! Thanks Again, -Bill |
VSWR doesn't matter?
"billcalley" wrote in message oups.com... We are all told that VSWR doesn't matter when using low loss transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? For the RF to bounce back and forth, wouldn't the transmitter's impedance have to be very, very high (or low) when the reflected RF energy hit its output stages? I know I'm missing something vital here... Thanks! -Bill Your mistake is that you assume the output of the tx is 50 ohms, in the case you stated the transmitter must be matched to the impedance it sees looking into the transmission line. |
VSWR doesn't matter?
On Mon, 12 Mar 2007 11:00:04 GMT, Jan Panteltje
wrote: Also normally, there is a pi type filter (to prevent harmonics), between amplifier and antenna. This filter _WILL_ match the antenna to the output impedance of the transmitter, so _even_ if the transmitter output impedance is very very low (low voltage high current output stage for example), the reflected power will be nicely converted to match the transmitter, and heat up the output amp, with its possible destruction as result. Hi Jan, Actually, there is a transformer there in the typical Ham transmitter (and probably in every general purpose power source) that typically transforms the native Z to the output Z. This is a step up for solid state, and step down for tubes. In the solid state rigs, it is a literal transformer feeding the 1-2 Ohms through a 5:1 winding ratio to a switched bank of low pass filters. This stuff is mud ordinary. As for the reflected energy, depending upon the phase it will either combine destructively (heat) or constructively (cool) in the extremes. There are, of course, 179 degrees of variation between these extremes before they repeat themselves again. Cooling, of course, is something of a misnomer as nothing useful is happening (poor power transfer) so perhaps the terms should be destructive through uselessly benign. 73's Richard Clark, KB7QHC |
VSWR doesn't matter?
I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? In my opinion the simplest way to answer your question is that you are assuming that the transmitter is equivalent to a 50 ohm load, which is not true because the transmitter is instead equivalent to the series of a 50 ohm load and a voltage generator. A simple DC example grossly clarifies thre issue: connect a 12V battery to the series of a 50-ohm load and another 12V battery. How much current flows through the load? Naught (assuming the correct polarity). So no power is dissipated in it. 73 Tony I0JX |
VSWR doesn't matter?
In message .com,
billcalley writes We are all told that VSWR doesn't matter when using low loss transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? For the RF to bounce back and forth, wouldn't the transmitter's impedance have to be very, very high (or low) when the reflected RF energy hit its output stages? I know I'm missing something vital here... It matters when it changes suddenly, like mine did recently on my 70MHz beam, when one of the elements came off in a gale. Brian GM4DIJ -- Brian Howie |
VSWR doesn't matter?
billcalley wrote:
We are all told that VSWR doesn't matter when using low loss transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. As pointed out, VSWR does matter. A lot of bouncing means you heat the transmission line with the power instead of radiating the power. 'Doesn't matter', really means it can be tolerated if need be. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? For the RF to bounce back and forth, wouldn't the transmitter's impedance have to be very, very high (or low) when the reflected RF energy hit its output stages? I know I'm missing something vital here... Here is what you are missing. In the case of the output, (real/resistive component of the transmitter), seeing the reflected wave, it is _not_ reflecting that power back up the transmission line as you think it is. It would go back to that real impedance and heat the transmitter. Here is what is done with a miss match in the real world. trans-output - match - line - antenna The 'match' is where the magic happens. All the energy coming down the line that got reflected from the antenna 'sees' the 'trans-output - match' as a perfect reflector and gets bounced back[*]. On the other side of the match is the trans-output. There the trans-output sees a perfect impedance, (technically, the conjugate of the trans-output), so that all the power travels through the match toward the antenna. The magic is that when the match is tuned, both of the above conditions are satisfied. *The reflected wave sees a purely reactive reflector not just because of the network but also because of the output power of the transmitter. Without transmitter power the impedance as seen from the load will dramatically change. Best, Dan. |
VSWR doesn't matter?
On Mon, 12 Mar 2007 13:37:00 -0400, "Jimmie D"
wrote: Your mistake is that you assume the output of the tx is 50 ohms, Hi Jimmie, At the risk of yet another, non-quantitative reply I will repeat: a question that has NEVER been answered by those who know what the transmitter output Z ISN'T: "What Z is it?" in the case you stated the transmitter must be matched to the impedance it sees looking into the transmission line. THAT is true, and it brings us to the point of all this energy sloshing around until the antenna finally dissipates it out into the Æther. It is the reflection off the mismatch of the tuner (the mismatch seen by the antenna as source to the line going back) that prevents energy from presenting any destructive results to the source - the whole point of using a tuner in the first place. 73's Richard Clark, KB7QHC |
VSWR doesn't matter?
"Richard Clark" wrote in message ... On Mon, 12 Mar 2007 13:37:00 -0400, "Jimmie D" wrote: Your mistake is that you assume the output of the tx is 50 ohms, Hi Jimmie, At the risk of yet another, non-quantitative reply I will repeat: a question that has NEVER been answered by those who know what the transmitter output Z ISN'T: "What Z is it?" in the case you stated the transmitter must be matched to the impedance it sees looking into the transmission line. THAT is true, and it brings us to the point of all this energy sloshing around until the antenna finally dissipates it out into the Æther. It is the reflection off the mismatch of the tuner (the mismatch seen by the antenna as source to the line going back) that prevents energy from presenting any destructive results to the source - the whole point of using a tuner in the first place. 73's Richard Clark, KB7QHC Correct but I just want to remember that the purpose of the tuner is to match the impedance of the transmitter to the impedance of the antenna/ transmission line.The standing waves can be viewed as a reflect voltage, a reflect current or as a reflected impedance. Besides I thought there had been enough quanitative analysis of the question and was hoping a simple answer may be enough to turn on the light bulb for the OP. If he still wanted to know more I figure he would ask. |
VSWR doesn't matter?
On Mon, 12 Mar 2007 19:30:59 +0100, "Antonio Vernucci"
wrote: A simple DC example grossly clarifies thre issue: connect a 12V battery to the series of a 50-ohm load and another 12V battery. How much current flows through the load? Naught (assuming the correct polarity). So no power is dissipated in it. Hi Tony, Turn the second battery over. Double the power is dissipated in it. Phase, you can't live with it, you can't live without it. 73's Richard Clark, KB7QHC |
VSWR doesn't matter?
A simple DC example grossly clarifies thre issue: connect a 12V battery to the series of a 50-ohm load and another 12V battery. How much current flows through the load? Naught (assuming the correct polarity). So no power is dissipated in it. Hi Tony, Turn the second battery over. Double the power is dissipated in it. Phase, you can't live with it, you can't live without it. 73's Richard Clark, KB7QHC Of course. Mine was just a DC example to illustrate things in a simple manner. When the transmitter is properly tuned, the phase relationship is such that the reflected wave does not get dissipated at all into the 50 ohm output of the transmitter, and is then reflected back to the antenna Tony I0JX |
VSWR doesn't matter?
Bob wrote:
On Mar 12, 1:08 am, "billcalley" wrote: We are all told that VSWR doesn't matter when using low loss transmission lines, No, we are not all told that. since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? The active part of the transmitter output isn't 50 ohm. That would cause half the power to be lost as heat in the output stage. It's only 50ohm once it becomes a moving wave in the transmission line. Bob9 Kewl, then I'll just run a tap directly to the 5000 ohm plates and start a long chat up ... what the heck is all those pi matching components in the way of the rf? Probably some loss there! ROFLOL!!! JS -- http://assemblywizard.tekcities.com |
VSWR doesn't matter?
Danny Richardson wrote in
: (any edition) and put that myth to rest once and for all. Wishful thinking Danny, myths are the stuff of ham radio, aren't they? Owen |
VSWR doesn't matter?
On Mon, 12 Mar 2007 01:25:00 UTC, Tim Wescott
wrote: * I am _not_ going to start the Big Transmitter Output Impedance Debate. sed denizens -- just don't comment on what a transmitter's "actual" output impedance may be, lest you start a flame war. OK ;-)) -- Jim Backus running OS/2 Warp 3 & 4, Debian Linux and Win98SE bona fide replies to j dot backus the circle thingy jita dot demon dot co dot uk |
VSWR doesn't matter?
"Dan Bloomquist" wrote in message ... billcalley wrote: We are all told that VSWR doesn't matter when using low loss transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. As pointed out, VSWR does matter. A lot of bouncing means you heat the transmission line with the power instead of radiating the power. 'Doesn't matter', really means it can be tolerated if need be. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? For the RF to bounce back and forth, wouldn't the transmitter's impedance have to be very, very high (or low) when the reflected RF energy hit its output stages? I know I'm missing something vital here... Here is what you are missing. In the case of the output, (real/resistive component of the transmitter), seeing the reflected wave, it is _not_ reflecting that power back up the transmission line as you think it is. It would go back to that real impedance and heat the transmitter. Here is what is done with a miss match in the real world. trans-output - match - line - antenna The 'match' is where the magic happens. All the energy coming down the line that got reflected from the antenna 'sees' the 'trans-output - match' as a perfect reflector and gets bounced back[*]. On the other side of the match is the trans-output. There the trans-output sees a perfect impedance, (technically, the conjugate of the trans-output), so that all the power travels through the match toward the antenna. The magic is that when the match is tuned, both of the above conditions are satisfied. *The reflected wave sees a purely reactive reflector not just because of the network but also because of the output power of the transmitter. Without transmitter power the impedance as seen from the load will dramatically change. Best, Dan. Saying that SWR doesnt matter is a rather broad statement(like saying never or always) but I have know of antenna systems having an SWR of 30:1 and his was normal. The feedline was balanced line made of 1 inch copper. Of course an SWR lie this on coax could be fatal to coax and equipment. A more common example of this is the 1/4 wl matching section on a J-pole antenna. It matches 50 ohms to a few Kohms so an SWR of 60: 1 or so would not be unusal here.S oas long as the feedline can handle the current and voltage peaks without much los it doesnt matter much as long as the source impedance is matched to the impedance at the input to the transmission line.Im sure there is a practical limit though. Jimmie |
VSWR doesn't matter?
Tim Wescott wrote:
billcalley wrote: We are all told that VSWR doesn't matter when using low loss transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? For the RF to bounce back and forth, wouldn't the transmitter's impedance have to be very, very high (or low) when the reflected RF energy hit its output stages? I know I'm missing something vital here... That's assuming you use an antenna tuner. The tuner will transform the transmitter's output impedance* just as it transforms the line. Were the transmitter output impedance actually at 50 ohms, on the other side of the tuner it would have the same VSWR as the line when everything was tuned up. Having said that, the VSWR _does_ matter somewhat when using low loss lines, both because the line loss is low but not zero, and the tuner loss will tend to go up as you correct for higher and higher VSWR. * I am _not_ going to start the Big Transmitter Output Impedance Debate. sed denizens -- just don't comment on what a transmitter's "actual" output impedance may be, lest you start a flame war. If you want a quick lesson in high vswr find a ham with an old tube transmitter and see if he will hook it up to a mismatched load. The cherry red plates are the reflected energy being absorbed. Transistors will just turn to smoke under the same conditions. Dave WD9BDZ |
VSWR doesn't matter?
"David G. Nagel" wrote in message ... Tim Wescott wrote: billcalley wrote: We are all told that VSWR doesn't matter when using low loss transmission lines, since the RF energy will travel from the transmitter up to the mismatched antenna, where a certain amount of this RF energy will reflect back towards the transmitter; after which the RF will then reflect back up to the antenna -- where the energy is eventually radiated after bouncing back and forth between the transmitter and antenna. I understand the concept, but what I don't quite understand is why the reflected RF energy isn't simply absorbed by the 50 ohm output of the transmitter after the first reflection? For the RF to bounce back and forth, wouldn't the transmitter's impedance have to be very, very high (or low) when the reflected RF energy hit its output stages? I know I'm missing something vital here... That's assuming you use an antenna tuner. The tuner will transform the transmitter's output impedance* just as it transforms the line. Were the transmitter output impedance actually at 50 ohms, on the other side of the tuner it would have the same VSWR as the line when everything was tuned up. Having said that, the VSWR _does_ matter somewhat when using low loss lines, both because the line loss is low but not zero, and the tuner loss will tend to go up as you correct for higher and higher VSWR. * I am _not_ going to start the Big Transmitter Output Impedance Debate. sed denizens -- just don't comment on what a transmitter's "actual" output impedance may be, lest you start a flame war. If you want a quick lesson in high vswr find a ham with an old tube transmitter and see if he will hook it up to a mismatched load. The cherry red plates are the reflected energy being absorbed. Transistors will just turn to smoke under the same conditions. Dave WD9BDZ Hi david Wouldnt it be OK to have a high VSWR along the transmission line if the "tank ckt" can be adjusted to match the load to the transmitter output impedance? That is, the VSWR along the transmission could concievely be high, yet, with proper "tank ckt" adjustment that impedance seen by the output circuit (plate) wouldnt result in a "cherry red plate". What I am asking is ? is the transmission line VSWR directly related to "plate reddening"? I'm more asking than *telling*. Jerry |
VSWR doesn't matter?
David G. Nagel wrote:
If you want a quick lesson in high vswr find a ham with an old tube transmitter and see if he will hook it up to a mismatched load. The cherry red plates are the reflected energy being absorbed. Transistors will just turn to smoke under the same conditions. Unfortunately, you'd be learning the wrong lesson. The cherry color is due to the transmitter being loaded with an impedance it's not designed for, causing the final to run at low efficiency. You can disconnect the antenna and replace it with a lumped RC or RL impedance of the same value and get exactly the same result. Alternatively, you can attach any combination of load and transmission line which give the same impedance, resulting in a wide variation of "reflected energy", and get exactly the same result. All that counts is the impedance seen by the transmitter, not the VSWR on the line or the "reflected power". The problem is that the idea of "reflected energy" turning the plates hot is so easy to understand, that people aren't willing to abandon it simply because it isn't true. See http://eznec.com/misc/Food_for_thought.pdf for more. Roy Lewallen, W7EL |
VSWR doesn't matter?
Jerry Martes wrote:
Wouldnt it be OK to have a high VSWR along the transmission line if the "tank ckt" can be adjusted to match the load to the transmitter output impedance? That is, the VSWR along the transmission could concievely be high, yet, with proper "tank ckt" adjustment that impedance seen by the output circuit (plate) wouldnt result in a "cherry red plate". Yes! All that matters to the transmitter is the impedance it sees. It doesn't know or care that you've mathematically separated the delivered power into "forward" and "reverse" components. It doesn't know or care what the SWR is on the transmission line connected to it, or even if a transmission line is connected at all. What I am asking is ? is the transmission line VSWR directly related to "plate reddening"? Absolutely not. I'm more asking than *telling*. That's the first step in learning. Roy Lewallen, W7EL |
VSWR doesn't matter?
The active part of the transmitter output isn't 50 ohm. That would cause half the power to be lost as heat in the output stage. It's only 50ohm once it becomes a moving wave in the transmission line. Bob9 Kewl, then I'll just run a tap directly to the 5000 ohm plates and start a long chat up ... what the heck is all those pi matching components in the way of the rf? Probably some loss there! ROFLOL!!! JS It appears that you agree with that part of my post but you are drawing an invalid conculsion from it. I never suggested that the passive matching network usually found in a transmitter output is unnecessary. Bob |
VSWR doesn't matter?
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