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VSWR doesn't matter? But how about "mismatch loss"?
What I gleaned from the excellent answers for the original "VSWR
Doesn't Matter?" thread is that high VSWR doesn't really matter in a lossless transmission line environment between a transmitter's antenna tuner and the antenna, since any reflected RF energy will simply continue to "bounce" back and forth between the tuner's output impedance and the antenna's input impedance until it is, finally, completely radiated from the antenna without loss. But then why does the concept of "mismatch loss" exist in reference to antennas? I have quickly calculated that if a transmitter outputs 100 watts, and the TX antenna has an impedance that will cause a VSWR of 10:1 -- using lossless transmission line -- that the mismatch loss in this "lossless" system would be 4.81dB! (Reflected power 66.9 watts, RL -1.74). Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. But here again, I'm probably not seeing the entire picture here. What am I missing?? Confused! -Bill |
VSWR doesn't matter? But how about "mismatch loss"?
billcalley wrote:
Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. But here again, I'm probably not seeing the entire picture here. What am I missing?? If the system is Z0-matched, e.g. antenna tuner, there is a mismatch gain at the tuner that offsets the mismatch loss at the load so, in a lossless system, nothing is lost. Wave cancellation toward the source is balanced by constructive interference toward the load. -- 73, Cecil, w5dxp.com |
VSWR doesn't matter? But how about "mismatch loss"?
On Mar 29, 1:34 pm, Cecil Moore wrote:
billcalley wrote: Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. But here again, I'm probably not seeing the entire picture here. What am I missing?? If the system is Z0-matched, e.g. antenna tuner, there is a mismatch gain at the tuner that offsets the mismatch loss at the load so, in a lossless system, nothing is lost. Wave cancellation toward the source is balanced by constructive interference toward the load. -- 73, Cecil, w5dxp.com Now my head *really* hurts! This is a VERY confusing subject, to say the least. (And I also thought antenna tuners actually had a *loss* due to their limited Q...I think I'm going to change careers now and just become a pet groomer; or perhaps simply give up completely and work at Radio Shack). -Bill |
VSWR doesn't matter? But how about "mismatch loss"?
billcalley wrote:
Now my head *really* hurts! This is a VERY confusing subject, to say the least. (And I also thought antenna tuners actually had a *loss* due to their limited Q...I think I'm going to change careers now and just become a pet groomer; or perhaps simply give up completely and work at Radio Shack). Real-world antenna tuners do have a loss but we previously specified a lossless system. Of course, real world tuners and transmission lines suffer losses but we all just live with those losses while striving to minimize them. The point is that an antenna tuner reflects most of the reflected energy back toward the load thus accomplishing a mismatch gain that offsets some, if not most, of the mismatch loss. High SWR transmission lines are indeed lossier than flat matched transmission lines of the same material. -- 73, Cecil, w5dxp.com |
VSWR doesn't matter? But how about "mismatch loss"?
"billcalley" wrote in message oups.com... What I gleaned from the excellent answers for the original "VSWR Doesn't Matter?" thread is that high VSWR doesn't really matter in a lossless transmission line environment between a transmitter's antenna tuner and the antenna, since any reflected RF energy will simply continue to "bounce" back and forth between the tuner's output impedance and the antenna's input impedance until it is, finally, completely radiated from the antenna without loss. That's basically true but ignores the stress that can be placed on the output circuit of the transmitter. Why do you think VSWR shut-down circuits are so popular? One can see rather high voltages or currents that are potentially damaging to the transistors and capacitors. Now, since "lossless" is an abstraction and all materials have voltage and current limits, just make this easy on yourself and always strive for a VSWR of 2:1 or less. It simply works better and is more reliable. |
VSWR doesn't matter? But how about "mismatch loss"?
Charles Schuler wrote:
Now, since "lossless" is an abstraction and all materials have voltage and current limits, just make this easy on yourself and always strive for a VSWR of 2:1 or less. It simply works better and is more reliable. But renders many all-HF-band dipoles useless. :-) I regularly run up to an 18:1 SWR on my 450 ohm ladder-line. Owen's transmission line calculator says I'm losing about 0.8 dB in 100' of line under those conditions on 40m. IMO, it's a small price to pay for all-HF-band operation. -- 73, Cecil, w5dxp.com |
VSWR doesn't matter? But how about "mismatch loss"?
"Cecil Moore" wrote in message ... Charles Schuler wrote: Now, since "lossless" is an abstraction and all materials have voltage and current limits, just make this easy on yourself and always strive for a VSWR of 2:1 or less. It simply works better and is more reliable. But renders many all-HF-band dipoles useless. :-) I regularly run up to an 18:1 SWR on my 450 ohm ladder-line. Owen's transmission line calculator says I'm losing about 0.8 dB in 100' of line under those conditions on 40m. IMO, it's a small price to pay for all-HF-band operation. How many Ham transmitters have a balanced output? How are you feeding a balanced line? If you are using an antenna tuner with unbalanced in (50 ohms) and balanced out (variable impedance), you should be OK in most situations. |
VSWR doesn't matter? But how about "mismatch loss"?
"Charles Schuler" wrote in message . .. "billcalley" wrote in message oups.com... What I gleaned from the excellent answers for the original "VSWR Doesn't Matter?" thread is that high VSWR doesn't really matter in a lossless transmission line environment between a transmitter's antenna tuner and the antenna, since any reflected RF energy will simply continue to "bounce" back and forth between the tuner's output impedance and the antenna's input impedance until it is, finally, completely radiated from the antenna without loss. That's basically true but ignores the stress that can be placed on the output circuit of the transmitter. Why do you think VSWR shut-down circuits are so popular? One can see rather high voltages or currents that are potentially damaging to the transistors and capacitors. Now, since "lossless" is an abstraction and all materials have voltage and current limits, just make this easy on yourself and always strive for a VSWR of 2:1 or less. It simply works better and is more reliable. If a tuner is placed directly after the TX and properly adjusted the TX will always see a 50 ohm load and the shutdown circuit will always be happpy. Again as long as the TX sees a match there is no unusual stress placed on it. Remember that before the invention of coax cable SWR was rarely considered. Instead the tx was tuned for proper established operational parametrs and all was right with the world. 1:1 SWR CAN MEAN YOUR COAX IS FULL OF WATER. Jimmie |
VSWR doesn't matter? But how about "mismatch loss"?
Charles Schuler wrote:
How are you feeding a balanced line? http://www.w5dxp.com/notuner.htm If you are using an antenna tuner with unbalanced in (50 ohms) and balanced out (variable impedance), you should be OK in most situations. No tuner! I don't like tuner losses. The feedpoint impedance is always between 35 ohms and 85 ohms resistive. My choke has an impedance in the thousands of ohms. -- 73, Cecil http://www.w5dxp.com |
VSWR doesn't matter? But how about "mismatch loss"?
billcalley wrote:
What I gleaned from the excellent answers for the original "VSWR Doesn't Matter?" thread is that high VSWR doesn't really matter in a lossless transmission line environment between a transmitter's antenna tuner and the antenna, since any reflected RF energy will simply continue to "bounce" back and forth between the tuner's output impedance and the antenna's input impedance until it is, finally, completely radiated from the antenna without loss. But then why does the concept of "mismatch loss" exist in reference to antennas? I have quickly calculated that if a transmitter outputs 100 watts, and the TX antenna has an impedance that will cause a VSWR of 10:1 -- using lossless transmission line -- that the mismatch loss in this "lossless" system would be 4.81dB! (Reflected power 66.9 watts, RL -1.74). Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. But here again, I'm probably not seeing the entire picture here. What am I missing?? Your "amount of power lost due to reflection" statement would be true if the line were connected to something resistive at the line's characteristic impedance. With a properly tuned tuner, that's not the case -- instead, the impedance looking into the tuner will also reflect power, and in a way that makes it all work out so that the power all ends up being radiated, which is what you wanted in the first place. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" gives you just what it says. See details at http://www.wescottdesign.com/actfes/actfes.html |
VSWR doesn't matter? But how about "mismatch loss"?
billcalley wrote:
On Mar 29, 1:34 pm, Cecil Moore wrote: billcalley wrote: Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. But here again, I'm probably not seeing the entire picture here. What am I missing?? If the system is Z0-matched, e.g. antenna tuner, there is a mismatch gain at the tuner that offsets the mismatch loss at the load so, in a lossless system, nothing is lost. Wave cancellation toward the source is balanced by constructive interference toward the load. -- 73, Cecil, w5dxp.com Now my head *really* hurts! This is a VERY confusing subject, to say the least. (And I also thought antenna tuners actually had a *loss* due to their limited Q...I think I'm going to change careers now and just become a pet groomer; or perhaps simply give up completely and work at Radio Shack). If you start considering loss in the tuner and the line then yes, a greater mismatch between the antenna and the line will result in more lost power (and more component heating in the tuner). You really want to leave that subject be until you understand the properties of a lossless system. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" gives you just what it says. See details at http://www.wescottdesign.com/actfes/actfes.html |
VSWR doesn't matter? But how about "mismatch loss"?
"billcalley" wrote in
oups.com: What I gleaned from the excellent answers for the original "VSWR Doesn't Matter?" thread is that high VSWR doesn't really matter in a lossless transmission line environment between a transmitter's antenna tuner and the antenna, since any reflected RF energy will simply continue to "bounce" back and forth between the tuner's output impedance and the antenna's input impedance until it is, finally, completely radiated from the antenna without loss. Well, there will be tuner losses, depending on how good the tuner's components are. But then why does the concept of "mismatch loss" exist in reference to antennas? I have quickly calculated that if a transmitter outputs 100 watts, and the TX antenna has an impedance that will cause a VSWR of 10:1 -- using lossless transmission line -- that the mismatch loss in this "lossless" system would be 4.81dB! (Reflected power 66.9 watts, RL -1.74). Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. But here again, I'm probably not seeing the entire picture here. What am I missing?? Confused! Yes, you're confused. If the lossless transmission line (obviously no such animal exists) were tuned with a lossless tuner, then VSWR would not matter at all. The loss due to mismatch in any real system will depend upon frequency, VSWR, type of feedline, length of feedline, and the quality of the tuning circuits used to match the system to the transmitter. Let's take your example. 100 watt transmitter into, let's say 100ft. of feedline at 10:1 VSWR and assume tuner losses are negligble (they often aren't). Here are the losses for some different kinds of 50 ohm coax at 10mhz: Belden 8237 2.19db Belden 9913 1.63db Belden 9258 3.19db Belden 8240 3.71db Belden 9201 3.83db So, what's obvious here is that different coaxes have different losses at high SWR. Why is that? Because as power is reflected back and forth in a transmission line, the losses accumulate. So line that is very low- loss to begin with will be less affected by high SWR than line that has moderate to high losses when flat. If 10 percent of the power in a line is lost travelling from the transmitter to the antenna, and if the antenna only radiates half that power, sending the rest back down the line, then 45 percent of the transmitter power is radiated immediately, while 45 percent is reflected. But only 40.5 percent reaches the tuner or transmitter. If ALL of that is re-reflected, then only 36.45 percent of the power is available at the second reflection to the antenna. The antenna will radiate 18.225 percent of the transmitter's power at this point, making the total 63.225 percent of the transmitter's output. Another 18.225 percent will be reflected again and, of that 16.4025 percent of the transmit power will live to be re-reflected from the tuner and 14.76225 percent will arrive at the antenna on the next bounce. Of that we can expect 7.381125 percent of the transmitter's total power to end up radiated while an equal amount starts its way back to the tuner. Anyway, it becomes a pretty simple bit of limit math to predict exactly how much will be radiated and how much lost in the coax under these conditions. -- Dave Oldridge+ ICQ 1800667 |
VSWR doesn't matter? But how about "mismatch loss"?
On 29 Mar 2007 13:10:46 -0700, "billcalley"
wrote: What I gleaned from the excellent answers for the original "VSWR Doesn't Matter?" thread is that high VSWR doesn't really matter in a lossless transmission line environment between a transmitter's antenna tuner and the antenna, since any reflected RF energy will simply continue to "bounce" back and forth between the tuner's output impedance and the antenna's input impedance until it is, finally, completely radiated from the antenna without loss. Hi Bill, As a short description, that is adequate. But then why does the concept of "mismatch loss" exist in reference to antennas? "Mismatch loss" is a system description, not an antenna description. So your reference is wrong. I have quickly calculated that if a transmitter outputs 100 watts, and the TX antenna has an impedance that will cause a VSWR of 10:1 -- using lossless transmission line -- that the mismatch loss in this "lossless" system would be 4.81dB! (Reflected power 66.9 watts, RL -1.74). Two things wrong he 1. You say nothing of a tuner inline; B. Your math is wrong either way. Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. It does matter if you lack a tuner (in more ways that one). Most discussion of "mismatch loss" omits such matters as tuners as it is a separable issue. Combining these topics raises your chance of confusion. 73's Richard Clark, KB7QHC |
VSWR doesn't matter? But how about "mismatch loss"?
billcalley wrote:
What I gleaned from the excellent answers for the original "VSWR Doesn't Matter?" thread is that high VSWR doesn't really matter in a lossless transmission line environment between a transmitter's antenna tuner and the antenna, since any reflected RF energy will simply continue to "bounce" back and forth between the tuner's output impedance and the antenna's input impedance until it is, finally, completely radiated from the antenna without loss. But then why does the concept of "mismatch loss" exist in reference to antennas? I have quickly calculated that if a transmitter outputs 100 watts, and the TX antenna has an impedance that will cause a VSWR of 10:1 -- using lossless transmission line -- that the mismatch loss in this "lossless" system would be 4.81dB! (Reflected power 66.9 watts, RL -1.74). Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. But here again, I'm probably not seeing the entire picture here. What am I missing?? Confused! -Bill I think the confusion (which *always* seems to arise) comes from the mix of the concept with the real world. The concept claims that the system is lossless, so the power bounces around until it eventually exits the "system" via the antenna. Real world, the system is lossy, so with all the bouncing around some of the power fails to leave the system via RF radiation and instead leaves the system via IR radiation. More heat, less RF. Tastes great, less filling. :-) Ed |
VSWR doesn't matter? But how about "mismatch loss"?
Tim Wescott wrote:
billcalley wrote: What I gleaned from the excellent answers for the original "VSWR Doesn't Matter?" thread is that high VSWR doesn't really matter in a lossless transmission line environment between a transmitter's antenna tuner and the antenna, since any reflected RF energy will simply continue to "bounce" back and forth between the tuner's output impedance and the antenna's input impedance until it is, finally, completely radiated from the antenna without loss. But then why does the concept of "mismatch loss" exist in reference to antennas? I have quickly calculated that if a transmitter outputs 100 watts, and the TX antenna has an impedance that will cause a VSWR of 10:1 -- using lossless transmission line -- that the mismatch loss in this "lossless" system would be 4.81dB! (Reflected power 66.9 watts, RL -1.74). Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. But here again, I'm probably not seeing the entire picture here. What am I missing?? Your "amount of power lost due to reflection" statement would be true if the line were connected to something resistive at the line's characteristic impedance. With a properly tuned tuner, that's not the case -- instead, the impedance looking into the tuner will also reflect power, and in a way that makes it all work out so that the power all ends up being radiated, which is what you wanted in the first place. Unless your coax decides that it had enough of all this current. I had that happen once. The SWR had gone up and I pressed on. Then a muffled boom outside and it became 100% reflective. Now I had to get on a ladder and painstakingly scrape all the molten gunk off the stucco. -- Regards, Joerg http://www.analogconsultants.com |
VSWR doesn't matter? But how about "mismatch loss"?
On Thu, 29 Mar 2007 13:10:46 -0700, billcalley wrote:
What I gleaned from the excellent answers for the original "VSWR Doesn't Matter?" thread is that high VSWR doesn't really matter in a lossless transmission line environment between a transmitter's antenna tuner and the antenna, since any reflected RF energy will simply continue to "bounce" back and forth between the tuner's output impedance and the antenna's input impedance until it is, finally, completely radiated from the antenna without loss. But then why does the concept of "mismatch loss" exist in reference to antennas? I have quickly calculated that if a transmitter outputs 100 watts, and the TX antenna has an impedance that will cause a VSWR of 10:1 -- using lossless transmission line -- that the mismatch loss in this "lossless" system would be 4.81dB! (Reflected power 66.9 watts, RL -1.74). Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. But here again, I'm probably not seeing the entire picture here. What am I missing?? Confused! It truly wouldn't matter if there were no such thing as resistance and so on. The whole circuit could be tuned, with the transmission line a part of it, and all of the power would go out the antenna. Unfortunately, that's not the way reality works, more's the pity. Every time those "standing waves" bounce back and forth, they warm up the transmission line, the connectors, the transmitter tank, the transmitter itself, etc, etc, and Entropy is conserved. ;-) Hope This Helps! Rich |
VSWR doesn't matter? But how about "mismatch loss"?
On 29 Mar 2007 13:10:46 -0700, "billcalley"
wrote: I have quickly calculated that if a transmitter outputs 100 watts, and the TX antenna has an impedance that will cause a VSWR of 10:1 -- using lossless transmission line -- that the mismatch loss in this "lossless" system would be 4.81dB! (Reflected power 66.9 watts, RL -1.74). Taking this at face value, yes, the "mismatch loss" is 4.8dB. 73's Richard Clark, KB7QHC |
VSWR doesn't matter? But how about "mismatch loss"?
On Mar 29, 8:15 pm, Rich Grise wrote:
Every time those "standing waves" bounce back and forth, they warm up the transmission line, the connectors, the transmitter tank, the transmitter itself, etc, etc, and Entropy is conserved. ;-) Humor noted, but Entropy is not conserved. Entropy increases |
18:1 all band dipole?
On Thu, 29 Mar 2007 22:44:24 GMT, Cecil Moore
wrote: No tuner! I don't like tuner losses. The feedpoint impedance is always between 35 ohms and 85 ohms resistive. My choke has an impedance in the thousands of ohms. Cecil, Please explain your antenna and radio. I am assuming you have a solid state rig with an so-239 connector on it for the basic and then you do what? Thanks -- 73 for now Buck, N4PGW www.lumpuckeroo.com |
VSWR doesn't matter? But how about "mismatch loss"?
bw wrote:
Entropy increases Seems to me that evolution violates that principle. -- 73, Cecil http://www.w5dxp.com |
18:1 all band dipole?
Buck wrote:
Cecil Moore wrote: Please explain your antenna and radio. I am assuming you have a solid state rig with an so-239 connector on it for the basic and then you do what? Hi Buck, Please check out my associated web page and then ask me anything that you don't understand. http://www.w5dxp.com/notuner.htm Contrary to what you may have been told, you can change the 50 ohm SWR seen by your transmitter by changing the length of the 450 ohm ladder-line. -- 73, Cecil http://www.w5dxp.com |
VSWR doesn't matter? But how about "mismatch loss"?
An understanding of "mismatch loss" doesn't require SWR, reflections,
power waves, "reflected power", "reflected energy", or other real or imagined complexities other than simple impedances. Here's what it means: If you have a generator with a fixed output impedance such as a signal generator, and connect it to a conjugately matched load, the power dissipated in that load is the most you can get in any load connected to the generator. For example, if your generator produces 10 volts RMS open circuit and has a 50 ohm resistive output impedance, it can deliver 0.5 watt to a 50 ohm resistive load. If you connect any other load impedance to the generator, you'll get less power to the load. You can calculate exactly how much with simple circuit theory. "Mismatch loss" is simply a way of expressing the reduction in power you get due to the load being mismatched, compared to how much you'd get with a matched load. For example, if you connect a 100 ohm resistor to the output of the generator, it would dissipate 0.44 watt instead of 0.5, so the mismatch loss is 10 log 0.5/0.44 = 0.51 dB(*). If you connect a 25 ohm resistor to the output, you also get 0.44 watt in the load resistor, again a "mismatch loss" of 0.51 dB. These numbers are calculated using nothing more complicated than simple lumped circuit principles. Mismatch loss is a useful concept when connecting fixed-impedance circuits together, such as in a laboratory environment. But it doesn't apply to either antennas or to VSWR. All you have to do to reduce the "mismatch loss" to zero is to insert a tuner or other matching network between the generator and the load. Presto, the generator sees 50 ohms resistive, the load dissipates 0.5 watt, and the mismatch loss is zero. (*) For the 100 ohm example: The circuit consists of a 10 volt generator, and a 50 ohm resistance (the generator impedance) and 100 ohm resistance (the load) in series. So the current is V / R = 10 / (50 + 100) = 66.67 mA. The power dissipated in the load is I^2 * R = 0.06667^2 * 100 ~ 0.44 watt. No reflections, VSWR, transmission lines, or bouncing power waves required. Roy Lewallen, W7EL billcalley wrote: What I gleaned from the excellent answers for the original "VSWR Doesn't Matter?" thread is that high VSWR doesn't really matter in a lossless transmission line environment between a transmitter's antenna tuner and the antenna, since any reflected RF energy will simply continue to "bounce" back and forth between the tuner's output impedance and the antenna's input impedance until it is, finally, completely radiated from the antenna without loss. But then why does the concept of "mismatch loss" exist in reference to antennas? I have quickly calculated that if a transmitter outputs 100 watts, and the TX antenna has an impedance that will cause a VSWR of 10:1 -- using lossless transmission line -- that the mismatch loss in this "lossless" system would be 4.81dB! (Reflected power 66.9 watts, RL -1.74). Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. But here again, I'm probably not seeing the entire picture here. What am I missing?? Confused! -Bill |
VSWR doesn't matter? But how about "mismatch loss"?
Roy Lewallen wrote:
An understanding of "mismatch loss" doesn't require SWR, reflections, power waves, "reflected power", "reflected energy", or other real or imagined complexities other than simple impedances. Here's what it means: If you have a generator with a fixed output impedance such as a signal generator, and connect it to a conjugately matched load, the power dissipated in that load is the most you can get in any load connected to the generator. For example, if your generator produces 10 volts RMS open circuit and has a 50 ohm resistive output impedance, it can deliver 0.5 watt to a 50 ohm resistive load. If you connect any other load impedance to the generator, you'll get less power to the load. You can calculate exactly how much with simple circuit theory. "Mismatch loss" is simply a way of expressing the reduction in power you get due to the load being mismatched, compared to how much you'd get with a matched load. For example, if you connect a 100 ohm resistor to the output of the generator, it would dissipate 0.44 watt instead of 0.5, so the mismatch loss is 10 log 0.5/0.44 = 0.51 dB(*). If you connect a 25 ohm resistor to the output, you also get 0.44 watt in the load resistor, again a "mismatch loss" of 0.51 dB. These numbers are calculated using nothing more complicated than simple lumped circuit principles. Mismatch loss is a useful concept when connecting fixed-impedance circuits together, such as in a laboratory environment. But it doesn't apply to either antennas or to VSWR. All you have to do to reduce the "mismatch loss" to zero is to insert a tuner or other matching network between the generator and the load. Presto, the generator sees 50 ohms resistive, the load dissipates 0.5 watt, and the mismatch loss is zero. Bravo. And a great deal simpler to understand than most handwaving on these threads. Roy Lewallen, W7EL Best, Dan. |
VSWR doesn't matter? But how about "mismatch loss"?
"Cecil Moore" wrote in message . net... bw wrote: Entropy increases Seems to me that evolution violates that principle. -- 73, Cecil http://www.w5dxp.com Not to mention the Noble Prize Ilya Prigogine won for "Dissipative Structures" in 1977. Spontaneous Ordered Structures arising out of disorder. But it takes an Energy Flow to produce them. Robert |
VSWR doesn't matter? But how about "mismatch loss"?
"Robert" wrote in message link.net... "Cecil Moore" wrote in message . net... bw wrote: Entropy increases Seems to me that evolution violates that principle. -- 73, Cecil http://www.w5dxp.com Not to mention the Noble Prize Ilya Prigogine won for "Dissipative Structures" in 1977. Spontaneous Ordered Structures arising out of disorder. But it takes an Energy Flow to produce them. Robert Sorry. Forgot to include the link. http://en.wikipedia.org/wiki/Ilya_Prigogine Robert |
VSWR doesn't matter? But how about "mismatch loss"?
On Mar 29, 2:10 pm, "billcalley" wrote:
What I gleaned from the excellent answers for the original "VSWR Doesn't Matter?" thread is that high VSWR doesn't really matter in a lossless transmission line environment between a transmitter's antenna tuner and the antenna, since any reflected RF energy will simply continue to "bounce" back and forth between the tuner's output impedance and the antenna's input impedance until it is, finally, completely radiated from the antenna without loss. There's also the case where a perfect VSWR does no good. That happens when you connect a transmitter to an extremely lossy line and the signal dissipates before it gets to the load : But then why does the concept of "mismatch loss" exist in reference to antennas? I have quickly calculated that if a transmitter outputs 100 watts, and the TX antenna has an impedance that will cause a VSWR of 10:1 -- using lossless transmission line -- that the mismatch loss in this "lossless" system would be 4.81dB! (Reflected power 66.9 watts, RL -1.74). Since mismatch loss is the "amount of power lost due to reflection", and is as if an "attenuator with a value of the mismatch loss where placed in series with the transmission line", then I would think that VSWR would *definitely* matter, and not just for highly lossy lines either. But here again, I'm probably not seeing the entire picture here. What am I missing?? Confused! -Bill |
18:1 all band dipole?
On Fri, 30 Mar 2007 02:43:38 GMT, Cecil Moore
wrote: Buck wrote: Cecil Moore wrote: Please explain your antenna and radio. I am assuming you have a solid state rig with an so-239 connector on it for the basic and then you do what? Hi Buck, Please check out my associated web page and then ask me anything that you don't understand. http://www.w5dxp.com/notuner.htm Contrary to what you may have been told, you can change the 50 ohm SWR seen by your transmitter by changing the length of the 450 ohm ladder-line. Ok, I have seen that. It isn't that the antenna isn't tuned, it is tuned with the twin-lead instead of a conventional tuner. I might have enough twin lead to build one of those. I'll have to consider it. It has always tweaked my interest. Thanks -- 73 for now Buck, N4PGW www.lumpuckeroo.com |
18:1 all band dipole?
Buck wrote:
Ok, I have seen that. It isn't that the antenna isn't tuned, it is tuned with the twin-lead instead of a conventional tuner. It has a tuned feeder instead of a conventional tuner. -- 73, Cecil http://www.w5dxp.com |
18:1 all band dipole?
On Fri, 30 Mar 2007 12:37:53 GMT, Cecil Moore
wrote: Buck wrote: Ok, I have seen that. It isn't that the antenna isn't tuned, it is tuned with the twin-lead instead of a conventional tuner. It has a tuned feeder instead of a conventional tuner. I am curious to know, have you measured the power both at the antenna and the radio to see what loss there might be? I wouldn't expect very much, personally. On a G5RV, it is taught that the twin-feedline is also part of the antenna itself, is that true in your model as well? thanks Buck -- 73 for now Buck, N4PGW www.lumpuckeroo.com |
All HF Band Feeline Tuned antenna
http://www.w5dxp.com/notuner.htm Cecil, Looking more at your antenna, I am making several observations. I posted the URL at the top for other readers to know what we are talking about. I kind of wish you would change the name of the antenna and not call it 'no-tuner' because I feel let down every time I look at it and see the feed-line tuner. I realize it isn't a conventional tuner. Don't get me wrong, I think it is a GREAT idea and I like it very much and would like to try it myself. However, I feel it is a little deceptive in the name. I have a few questions about the system. I may have asked some already in another thread, so please bear with me. This is a better place to discuss it rather than hijacking another thread. I see you use 450 ohm ladder line (or window-line as some call it.) I am wondering if the concept will also work with other impedance feedline such as 600, 300 or 75 ohm twin-line or even possibly with coax. **** **** I just looked over your program. I see that you have it setup to allow 300 or 450 ohm ladder line. I wonder if you have LLWL available for 600 and 75 ohm.... I will try to put this in XL. (I don't have a basic program.) OK, I just found the compiled program. I am interested in making one, but I only have 300 ohm. Does the feedline act as part of the antenna? I am sure it will, at least up to the 'no-tuner' if it acts like a G5RV, but do you know if the feedline radiates? Have you measured the power at both sides of the 'no-tuner' to see what loss there might be? I doubt there would be much considering that you are using window-line. I see a couple of the frequencies are above 1.5:1, which I am not comfortable going over with solid state rigs. Do you think that could be fine-tuned with the addition of a 1/2 foot section or maybe with that and a 1/4 foot section? Would it be safe to assume that I can create a mono-band dipole, maybe even multi-band - if I am lucky, by fixing the length of the dipole and the feedline such that the increment gives me the imax at the balun for the desired frequency(s)? Finally, you have a 1:1 choke at the feeder. I see that the better quality coax, the more toroids are needed. Would there be a problem with using a foot of RG-58 with the fewer toroids and then connect that to the high-quality line? I suspect that by then the common-mode currents will be gone and there would be minimum loss in such a short strand of lower-grade coax for HF. I doubt there would be a noticeable signal loss. Does the 'no-tuner' feedline need to be spread out. I see from the photo that your 16 foot section is one large loop, I figure it must be close to a four-foot diameter loop. Is there a similar system that would work with a vertical? Thank you, I appreciate your taking time to answer this. I hope many of us can learn from it. Buck N4PGW -- 73 for now Buck, N4PGW www.lumpuckeroo.com |
18:1 all band dipole?
On Fri, 30 Mar 2007 09:59:04 -0400, Buck
wrote: On Fri, 30 Mar 2007 12:37:53 GMT, Cecil Moore wrote: Buck wrote: Ok, I have seen that. It isn't that the antenna isn't tuned, it is tuned with the twin-lead instead of a conventional tuner. It has a tuned feeder instead of a conventional tuner. I am curious to know, have you measured the power both at the antenna and the radio to see what loss there might be? I wouldn't expect very much, personally. On a G5RV, it is taught that the twin-feedline is also part of the antenna itself, is that true in your model as well? thanks Buck Please refer to my new thread. Thanks -- 73 for now Buck, N4PGW www.lumpuckeroo.com |
All HF Band Feeline Tuned antenna
After reading your webpage more closely......
On Fri, 30 Mar 2007 10:42:26 -0400, Buck wrote: http://www.w5dxp.com/notuner.htm I see you use 450 ohm ladder line (or window-line as some call it.) I am wondering if the concept will also work with other impedance feedline such as 600, 300 or 75 ohm twin-line or even possibly with coax. **** Does the feedline act as part of the antenna? I am sure it will, at least up to the 'no-tuner' if it acts like a G5RV, but do you know if the feedline radiates? That answer would be 'no'. I see that the purpose of the feedline length is to get the maximum current at the center point of the dipole. I see a couple of the frequencies are above 1.5:1, which I am not comfortable going over with solid state rigs. Do you think that could be fine-tuned with the addition of a 1/2 foot section or maybe with that and a 1/4 foot section? Based on the fact that the antenna is being fed maximum current at the center, I would guess that this should work. Buck N4PGW -- 73 for now Buck, N4PGW www.lumpuckeroo.com |
All HF Band Feeline Tuned antenna
Buck wrote:
I see you use 450 ohm ladder line (or window-line as some call it.) I am wondering if the concept will also work with other impedance feedline such as 600, 300 or 75 ohm twin-line or even possibly with coax. **** Might as well go into some detail here. The feedpoint impedances encountered in a 130 foot dipole may range from ~50 ohms to ~5000 ohms. In order to limit the maximum SWR the feedline Z0 should be ~SQRT(50)(5000) = ~500 ohms. Thus the choice of 450 ohm line. Given that the feedpoint impedances may range from ~50 to ~5000 ohms, here are the maximum SWRs that may be expected for the different Z0s. Since the antenna system is fed at a current maximum point, I have included the impedance at the current maximum point which needs to be between 25 ohms and 100 ohms in order to avoid foldback. Z0 SWRmax Imax Impedance 600 12:1 50 ohms 450 11:1 41 ohms 300 17:1 18 ohms 75 67:1 1 ohm 50 100:1 0.5 ohms It should be readily apparent why coax is a no-no for this antenna system. Even 300 ohm twinlead will result in a 50 ohm SWR of 50/18 = 2.8:1, high enough to cause foldback. I am interested in making one, but I only have 300 ohm. Let's take 40m as an example. The dipole is one wavelength and will have a 300 ohm SWR of 16:1 on 7.2 MHz according to EZNEC. 300/16 = 18.75 ohms at the current maximum point resulting in a 50 ohm SWR of 2.7:1 enough to cause foldback without an antenna tuner. Such is life. I do use 300 ohm line on my 20m rotatable dipole and it does work 20m-10m but there are probably 10m frequencies where the 50 SWR is too high. Does the feedline act as part of the antenna? I am sure it will, at least up to the 'no-tuner' if it acts like a G5RV, but do you know if the feedline radiates? I monitor the current balance in my feedlines. They are so well balanced that there is no hint of common-mode current on the coax side of my current/choke/balun. Balanced currents radiate a negligible amount. Have you measured the power at both sides of the 'no-tuner' to see what loss there might be? I doubt there would be much considering that you are using window-line. No I haven't measured the losses. I have trusted Owen's transmission line calculator for that data. I see a couple of the frequencies are above 1.5:1, which I am not comfortable going over with solid state rigs. Do you think that could be fine-tuned with the addition of a 1/2 foot section or maybe with that and a 1/4 foot section? The disadvantage of this method is that it lacks one dimension of tuning necessary to achieve 50 ohms. Unless you do one more impedance transformation, no amount of fine-tuning the ladder- line length will get any closer. Would it be safe to assume that I can create a mono-band dipole, maybe even multi-band - if I am lucky, by fixing the length of the dipole and the feedline such that the increment gives me the imax at the balun for the desired frequency(s)? One such example is at: http://www.w5dxp.com/HEDZ.htm This antenna works on 75m and 40m with a fixed length of ladder- line. Finally, you have a 1:1 choke at the feeder. I see that the better quality coax, the more toroids are needed. Would there be a problem with using a foot of RG-58 with the fewer toroids and then connect that to the high-quality line? Just use RG-58 entirely unless you are running high power. Does the 'no-tuner' feedline need to be spread out. I see from the photo that your 16 foot section is one large loop, I figure it must be close to a four-foot diameter loop. I have my 16 foot loop coiled in a 4 turn spiral around a piece of fiberglas fishing pole. Is there a similar system that would work with a vertical? Verticals are not usually fed with ladder-line so probably more trouble than it is worth. -- 73, Cecil, w5dxp.com |
VSWR doesn't matter? But how about "mismatch loss"?
Yes, and it should hurt , because we are using English and text to show something that s/b shown in pictures . Im E.E. , KC7CC, and more.. i can simply show you with pictures. It will be instantly clear . ---BTW------ Im doin ARM computers , I will give a free Op System . It will NOT use English nor ASCII . No arbitrary definitions ... No C++ , No Linux , No M$ , no more "Free Lunch" ..... We use Coax for its isolation from nearby absorbers .. parallel line is much lower loss but absorbs into other objects close . We do not use caps , but stubs . But they are tuned ( freq dependent ) . The fast way to follow this , is to draw a picture , then edit it as you go . English will only get you a college degree and a free lunch ( Liberals ) . __________________________________ Now my head *really* hurts! This is a VERY confusing subject, to say the least. (And I also thought antenna tuners actually had a *loss* due to their limited Q...I think I'm going to change careers now and just become a pet groomer; or perhaps simply give up completely and work at Radio Shack). -Bill |
VSWR doesn't matter? But how about "mismatch loss"?
Bob Myers wrote:
"Cecil Moore" wrote in message ... while at the same time this overall trend results in a very localized DECREASE in entropy (increase in order; in this case, the evolution of complex systems on Earth). How about considering the localized W5DXP system? :-) -- 73, Cecil http://www.w5dxp.com |
VSWR doesn't matter? But how about "mismatch loss"?
"Dave Oldridge" wrote in message 9... If the lossless transmission line (obviously no such animal exists) were tuned with a lossless tuner, then VSWR would not matter at all. Dave, I'm pretty sure I know what you mean here, but it should be noted that this isn't entirely true. The line would have to have a couple of other characteristics besides being simply "lossless" for VSWR not to matter at all. The problem, of course, is that a VSWR of other than 1:1 implies (by definition!) that the voltages and currents along the length of the line are not constant; there are cyclic variations in each, with maxima and minima located at half-wavelength intervals (that's the whole "standing wave" thing in the first place, right?). Particularly in high-power situations, it is possible for the maxima to exceed the ratings of the source or of the line itself. Bob M. (KC0EW) |
VSWR doesn't matter? But how about "mismatch loss"?
"Cecil Moore" wrote in message . net... bw wrote: Entropy increases Seems to me that evolution violates that principle. Not at all; "entropy increases" is with respect to the total entropy of a closed system. But in the case of "evolution," the only closed system which makes any sense to consider is the entire solar system, or at the very least the Sun-Earth system. It is entirely pemissible for an overall increase in entropy to occur (i.e., the Sun slowly loses energy to the rest of the universe) while at the same time this overall trend results in a very localized DECREASE in entropy (increase in order; in this case, the evolution of complex systems on Earth). Bob M. -- 73, Cecil http://www.w5dxp.com |
VSWR doesn't matter? But how about "mismatch loss"?
Bill ...
I would bet that most of the "confustion" comes from the conditions people put on their answers to the question. Some postulate a "tuner" between generator and load. Some postulate a specific internal impedance of the generator. Some postulate a specific length of feeder line (either lossless or resistive, which is another parameter in and of itself). Some postulate lots of other stuff, almost all of which is valid in the context of their answer. What exactly do we mean when we say that we have a "100 watt transmitter"? What we are actually saying is that the transmitter will cause a resistor of a specific value to dissipate 100 watts of energy when tied to the transmitter output port and the transmitter keyed. Let's not muddy the waters up by asking if we are talking about peak power, PE power, average power, or whatnot. Let's just assume an unmodulated carrier putting out a constant power into the resistor that gets just as hot as when 100 dc watts (E*I) is pumped into it from a battery. What value resistor? Whatever the designer/engineer/manufacturer specifies. 32 ohms? Sure. 50 ohms? No problem. 300 ohms? Certainly. Any competent engineer can give you a specified power into a specified resistive load. The crux of the question becomes, "What happens if my transmitter is specified into a 50 ohm resistor and I put a 100 ohm resistor as the load? How much "loss" do I get (or another way of asking that same question is how much power is dissipated in the 100 ohm resistor)?" The answer is that it is impossible to tell without making the measurement. That may seem like a "wiggle" answer, but the truth of it is that the output stage design of the transmitter will dictate how it handles a "VSWR load". In some output stages, the output voltage will increase to the point of nearly driving 100 watts into the 2:1 VSWR resistor. Some will shut themselves down with a protection circuit. Some will go into parasitic oscillation. Some will fry the output devices. Now increase the magnitude (and probably the sign also) of the problem to toss in a complex impedance instead of a resistive load and the confusion factor goes up rapidly. What DOES that output stage do when the load has an inductive component? Or a capacitive component? And both a resistive component and an imaginary component that varies with frequency? The simple answer to your question outside the "laboratory environment" where everything is nicely matched and the internal impedances are set "just so" is that there IS NO SINGLE RIGHT ANSWER to this simple question. And that is most probably the cause of your confusion. Jim "billcalley" wrote in message oups.com... But here again, I'm probably not seeing the entire picture here. What am I missing?? Confused! -Bill |
VSWR doesn't matter? But how about "mismatch loss"?
"Bob Myers" wrote in news:MmdPh.1809
: "Dave Oldridge" wrote in message 9... If the lossless transmission line (obviously no such animal exists) were tuned with a lossless tuner, then VSWR would not matter at all. Dave, I'm pretty sure I know what you mean here, but it should be noted that this isn't entirely true. The line would have to have a couple of other characteristics besides being simply "lossless" for VSWR not to matter at all. Not if there's a lossless tuner. The problem, of course, is that a VSWR of other than 1:1 implies (by definition!) that the voltages and currents along the length of the line are not constant; there are cyclic variations in each, with maxima and minima located at half-wavelength intervals (that's the whole "standing wave" thing in the first place, right?). Particularly in high-power situations, it is possible for the maxima to exceed the ratings of the source or of the line itself. In which case it's not truly "lossless" (and after breakdown becomes VERY lossy). -- Dave Oldridge+ ICQ 1800667 |
VSWR doesn't matter? But how about "mismatch loss"?
Cecil Moore wrote:
bw wrote: Entropy increases Seems to me that evolution violates that principle. Duh, life always violates it locally, but makes the sum total of entropy higer than it would have been. You aren't the idiot you appear, I hope. tom K0TAR |
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