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"Bal uhn" or "bayl uhn"?
On 7/29/2015 4:10 PM, rickman wrote:
On 7/29/2015 4:05 PM, Ian Jackson wrote: In message , rickman writes On 7/29/2015 3:14 PM, Ian Jackson wrote: In message , John S writes On 7/29/2015 1:16 PM, rickman wrote: Perhaps someone can explain the issue of current in the coax shield. Current gives rise to a magnetic field. But the current in the inner conductor is opposite and would create a magnetic field that would cancel the field of the outer conductor, no? What am I missing? Skin effect. The currents on the inside of the shield and on the outside of the shield see different things. They each have no idea what the other is doing. As for magnetic field, I must step aside. I can only report what the gurus say (nothing that I've found). Even though the coax shield is grounded at the shack end, both halves of the antenna get fed push-pull (in anti-phase) with the RF signal flowing on the outer skin of the inner conductor and the inner skin of the shield. However, at the antenna end, the returning RF on the shield side of the antenna doesn't know that it should stay on the inside of the shield. Because of the skin effect, it happily makes for the outside, whence it flows back to shack, and through the shack grounding connections. I am having trouble forming an image of this. What exactly is the source of the "returning RF"? Is this reflected RF at the impedance mismatch at the feedpoint? If so, the situation being discussed has no impedance mismatch, so no returning RF. Is the returning RF from the signal being radiated from the antenna inducing current in the shield? If so, doesn't the inner conductor also pick up the radiated signal? The RF (which is, of course, an AC signal) doesn't just flow out of the top end of the coax and into the two halves of the antenna. The fact that the antenna has a standing wave on it means that some RF is bouncing off the far ends of the antenna, and back to (and into) the top end of the coax. So you are saying that with a perfect match to an antenna with a real only impedance (the stated condition for this discussion) there will still be a reflected wave on the feed line? There is no reason why the returning RF current on the shield leg of the antenna should want to flow back on the inside of the shield - in fact, a combination of the Faraday shield effect and the skin effect encourages it to take the easy route on outside of the shield. I'm not at all clear on the location of current flow on the shield, but what about the current flow on the inner conductor? If the antenna reflects a balanced signal back into the cable isn't there also a current in the inner conductor which will create an opposing magnetic field? Maybe that is not the issue as some are talking about the problems created by the voltage drop to ground on the shield. Read this: http://eznec.com/Amateur/Articles/Baluns.pdf |
"Bal uhn" or "bayl uhn"?
On 7/29/2015 6:40 PM, Jeff Liebermann wrote:
On Wed, 29 Jul 2015 15:19:29 -0400, rickman wrote: I am having trouble forming an image of this. What exactly is the source of the "returning RF"? This might help: http://www.antennex.com/w4rnl/col0606/amod100.html The first few paragraphs are the applicable parts. Quoting a few tibits: "Fig. 1 presents one traditional way to portray the situation at the dipole feedpoint. Its general purpose is to show why the insertion of a balun is important as a precautionary measure in dipole construction." "However, the current from the braid has 2 paths: the right leg of the dipole in the figure and the outer side of the coaxial cable braid." Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. The rest of the article deals with modeling issues and problems. When modeling a balun, I use three conductors for the coax. The usual inner and outer conductors, which are assumed to handle only differential current and therefore do not radiate, and a mysterious 3rd conductor on the outside, which carries all the common mode current that does the radiating. And how is this third wire connected? Why do you see current in it? Is this just due to the voltage drop across the rest of the coax? If so, I would expect the current flow to be the same phase as the inner shield current. -- Rick |
"Bal uhn" or "bayl uhn"?
On 7/29/2015 7:13 PM, John S wrote:
On 7/29/2015 4:10 PM, rickman wrote: On 7/29/2015 4:05 PM, Ian Jackson wrote: In message , rickman writes On 7/29/2015 3:14 PM, Ian Jackson wrote: In message , John S writes On 7/29/2015 1:16 PM, rickman wrote: Perhaps someone can explain the issue of current in the coax shield. Current gives rise to a magnetic field. But the current in the inner conductor is opposite and would create a magnetic field that would cancel the field of the outer conductor, no? What am I missing? Skin effect. The currents on the inside of the shield and on the outside of the shield see different things. They each have no idea what the other is doing. As for magnetic field, I must step aside. I can only report what the gurus say (nothing that I've found). Even though the coax shield is grounded at the shack end, both halves of the antenna get fed push-pull (in anti-phase) with the RF signal flowing on the outer skin of the inner conductor and the inner skin of the shield. However, at the antenna end, the returning RF on the shield side of the antenna doesn't know that it should stay on the inside of the shield. Because of the skin effect, it happily makes for the outside, whence it flows back to shack, and through the shack grounding connections. I am having trouble forming an image of this. What exactly is the source of the "returning RF"? Is this reflected RF at the impedance mismatch at the feedpoint? If so, the situation being discussed has no impedance mismatch, so no returning RF. Is the returning RF from the signal being radiated from the antenna inducing current in the shield? If so, doesn't the inner conductor also pick up the radiated signal? The RF (which is, of course, an AC signal) doesn't just flow out of the top end of the coax and into the two halves of the antenna. The fact that the antenna has a standing wave on it means that some RF is bouncing off the far ends of the antenna, and back to (and into) the top end of the coax. So you are saying that with a perfect match to an antenna with a real only impedance (the stated condition for this discussion) there will still be a reflected wave on the feed line? There is no reason why the returning RF current on the shield leg of the antenna should want to flow back on the inside of the shield - in fact, a combination of the Faraday shield effect and the skin effect encourages it to take the easy route on outside of the shield. I'm not at all clear on the location of current flow on the shield, but what about the current flow on the inner conductor? If the antenna reflects a balanced signal back into the cable isn't there also a current in the inner conductor which will create an opposing magnetic field? Maybe that is not the issue as some are talking about the problems created by the voltage drop to ground on the shield. Read this: http://eznec.com/Amateur/Articles/Baluns.pdf Ok, I think I am getting it. When an unbalanced drive connects to a balanced line or antenna, the current does not fully flow into the antenna from the shield. The output of the shield becomes a sneak path routing it to ground. I suppose the balun works by giving the shield side of the connection a virtual zero ohm path so that the sneak path is short circuited. -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 7/29/2015 7:13 PM, John S wrote: On 7/29/2015 4:10 PM, rickman wrote: On 7/29/2015 4:05 PM, Ian Jackson wrote: In message , rickman writes On 7/29/2015 3:14 PM, Ian Jackson wrote: In message , John S writes On 7/29/2015 1:16 PM, rickman wrote: Perhaps someone can explain the issue of current in the coax shield. Current gives rise to a magnetic field. But the current in the inner conductor is opposite and would create a magnetic field that would cancel the field of the outer conductor, no? What am I missing? Skin effect. The currents on the inside of the shield and on the outside of the shield see different things. They each have no idea what the other is doing. As for magnetic field, I must step aside. I can only report what the gurus say (nothing that I've found). Even though the coax shield is grounded at the shack end, both halves of the antenna get fed push-pull (in anti-phase) with the RF signal flowing on the outer skin of the inner conductor and the inner skin of the shield. However, at the antenna end, the returning RF on the shield side of the antenna doesn't know that it should stay on the inside of the shield. Because of the skin effect, it happily makes for the outside, whence it flows back to shack, and through the shack grounding connections. I am having trouble forming an image of this. What exactly is the source of the "returning RF"? Is this reflected RF at the impedance mismatch at the feedpoint? If so, the situation being discussed has no impedance mismatch, so no returning RF. Is the returning RF from the signal being radiated from the antenna inducing current in the shield? If so, doesn't the inner conductor also pick up the radiated signal? The RF (which is, of course, an AC signal) doesn't just flow out of the top end of the coax and into the two halves of the antenna. The fact that the antenna has a standing wave on it means that some RF is bouncing off the far ends of the antenna, and back to (and into) the top end of the coax. So you are saying that with a perfect match to an antenna with a real only impedance (the stated condition for this discussion) there will still be a reflected wave on the feed line? There is no reason why the returning RF current on the shield leg of the antenna should want to flow back on the inside of the shield - in fact, a combination of the Faraday shield effect and the skin effect encourages it to take the easy route on outside of the shield. I'm not at all clear on the location of current flow on the shield, but what about the current flow on the inner conductor? If the antenna reflects a balanced signal back into the cable isn't there also a current in the inner conductor which will create an opposing magnetic field? Maybe that is not the issue as some are talking about the problems created by the voltage drop to ground on the shield. Read this: http://eznec.com/Amateur/Articles/Baluns.pdf Ok, I think I am getting it. When an unbalanced drive connects to a balanced line or antenna, the current does not fully flow into the antenna from the shield. The output of the shield becomes a sneak path routing it to ground. Or radiating it. I suppose the balun works by giving the shield side of the connection a virtual zero ohm path so that the sneak path is short circuited. Actually a balun works by giving the shield side of the connection a high impedance path so that the current is minimized. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
In message , rickman
writes On 7/29/2015 4:05 PM, Ian Jackson wrote: The RF (which is, of course, an AC signal) doesn't just flow out of the top end of the coax and into the two halves of the antenna. The fact that the antenna has a standing wave on it means that some RF is bouncing off the far ends of the antenna, and back to (and into) the top end of the coax. So you are saying that with a perfect match to an antenna with a real only impedance (the stated condition for this discussion) there will still be a reflected wave on the feed line? Maybe 'standing wave' is the wrong description. What I'm referring to is the approximately sinusoidal current and voltage distribution along the length of the antenna (high voltage at the ends, high current at the centre feedpoint). However, this does result from the outgoing AC wave meeting the wave bouncing back from the ends of the antenna. If you accept that both legs of the antenna have this 'waveform' (although where there's no balun, they are probably unequal), and the shape of the ;waveform; is the vectorial summation of the go-and-return RF signals, then it's pretty easy to see why a fair proportion of the returning signal should head down the outside of the shield. It's probably easier to visualise this than trying to work out why some of the forward-going RF signal (on the inside of the shield) should chose to do an about-turn at the antenna feedpoint, and immediately come back down on the outside of the shield - instead of flowing into the antenna wire. There is no reason why the returning RF current on the shield leg of the antenna should want to flow back on the inside of the shield - in fact, a combination of the Faraday shield effect and the skin effect encourages it to take the easy route on outside of the shield. I'm not at all clear on the location of current flow on the shield, but what about the current flow on the inner conductor? If the antenna reflects a balanced signal back into the cable isn't there also a current in the inner conductor which will create an opposing magnetic field? Maybe that is not the issue as some are talking about the problems created by the voltage drop to ground on the shield. -- Ian |
"Bal uhn" or "bayl uhn"?
In article , rickman wrote:
Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. Don't ask "Why does current flow on the braid?". Ask "What would *stop* current from flowing on the braid?". Current flows on *all* paths that have less than an infinite impedance. That's its nature. Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the braid) will be nonzero, if the voltage at that point is nonzero (E != 0) and the impedance down the braid at that point is not infinite. The effect of a balun is to place a high "choking" impedance in series with the outside of the feedline braid, thus "choking off" the current flow. |
"Bal uhn" or "bayl uhn"?
On 7/29/2015 6:31 PM, rickman wrote:
On 7/29/2015 6:40 PM, Jeff Liebermann wrote: On Wed, 29 Jul 2015 15:19:29 -0400, rickman wrote: I am having trouble forming an image of this. What exactly is the source of the "returning RF"? This might help: http://www.antennex.com/w4rnl/col0606/amod100.html The first few paragraphs are the applicable parts. Quoting a few tibits: "Fig. 1 presents one traditional way to portray the situation at the dipole feedpoint. Its general purpose is to show why the insertion of a balun is important as a precautionary measure in dipole construction." "However, the current from the braid has 2 paths: the right leg of the dipole in the figure and the outer side of the coaxial cable braid." Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. The rest of the article deals with modeling issues and problems. When modeling a balun, I use three conductors for the coax. The usual inner and outer conductors, which are assumed to handle only differential current and therefore do not radiate, and a mysterious 3rd conductor on the outside, which carries all the common mode current that does the radiating. And how is this third wire connected? Why do you see current in it? Is this just due to the voltage drop across the rest of the coax? If so, I would expect the current flow to be the same phase as the inner shield current. Consider a half-wave dipole fed by coax. Pretend that the feed point of the antenna has the RF applied (which it does when you consider that the coax is just a medium to bring the RF up to that point). What does the the feed point see? It sees one antenna element on the coax center wire and two antenna elements on the shield (element + shield). The outside of the shield is now an antenna element going down toward the earth while the wire attached to the shield is the other actual element. So the 'balanced' antenna is now unbalanced due to the coax becoming one of the elements. This is easy to model in EZNEC. And, you can see the results of changing the length of the coax as well as the effects of installing impedances in the coax. |
"Bal uhn" or "bayl uhn"?
On 7/30/2015 2:01 PM, Dave Platt wrote:
In article , rickman wrote: Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. Don't ask "Why does current flow on the braid?". Ask "What would *stop* current from flowing on the braid?". Current flows on *all* paths that have less than an infinite impedance. That's its nature. Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the braid) will be nonzero, if the voltage at that point is nonzero (E != 0) and the impedance down the braid at that point is not infinite. The effect of a balun is to place a high "choking" impedance in series with the outside of the feedline braid, thus "choking off" the current flow. I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. The balun can have no effect on the impedance of the coax shield. Just as you ask, "What would *stop* current from flowing on the braid?" when connected to the antenna what will stop the current from flowing on the braid when connected to the balun? The only thing that will stop the current from flowing on the outside of the shield when connected to the balun is if the balun presents a much lower impedance path for the current than does the shield. The only way your suggestion makes sense is if the current actually comes *from* the antenna and the balun prevents that current from returning to the feed line. Maybe this is one of those pointless distinctions and both ways of looking at it are correct. -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 7/30/2015 2:01 PM, Dave Platt wrote: In article , rickman wrote: Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. Don't ask "Why does current flow on the braid?". Ask "What would *stop* current from flowing on the braid?". Current flows on *all* paths that have less than an infinite impedance. That's its nature. Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the braid) will be nonzero, if the voltage at that point is nonzero (E != 0) and the impedance down the braid at that point is not infinite. The effect of a balun is to place a high "choking" impedance in series with the outside of the feedline braid, thus "choking off" the current flow. I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. The balun can have no effect on the impedance of the coax shield. Just as you ask, "What would *stop* current from flowing on the braid?" when connected to the antenna what will stop the current from flowing on the braid when connected to the balun? You fail to understand the difference between a choke balun and a voltage balun. A voltage balun forces the voltages at the output to be equal. A choke balun provides a very high impedance to the shield path. The only thing that will stop the current from flowing on the outside of the shield when connected to the balun is if the balun presents a much lower impedance path for the current than does the shield. Nope, it is just the opposite. A high impedance path prevents the flow of current. I = E/Z From high school algebra, as Z appoaches zero, I approaches infinity, and as Z approaces infinity, I approaches zero. The only way your suggestion makes sense is if the current actually comes *from* the antenna and the balun prevents that current from returning to the feed line. Nope, the current comes from the end of the coax and the outside of the shield is a separate current path from the inside of the coax and the antenna. Maybe this is one of those pointless distinctions and both ways of looking at it are correct. Nope, there is a very big point to it. Read www.eznec.com/Amateur/Articles/Baluns.pdf for a detailed discussion with pictures showing the current flow. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
On 7/30/2015 4:28 PM, rickman wrote:
On 7/30/2015 2:01 PM, Dave Platt wrote: In article , rickman wrote: Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. Don't ask "Why does current flow on the braid?". Ask "What would *stop* current from flowing on the braid?". Current flows on *all* paths that have less than an infinite impedance. That's its nature. Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the braid) will be nonzero, if the voltage at that point is nonzero (E != 0) and the impedance down the braid at that point is not infinite. The effect of a balun is to place a high "choking" impedance in series with the outside of the feedline braid, thus "choking off" the current flow. I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. The balun can have no effect on the impedance of the coax shield. Just as you ask, "What would *stop* current from flowing on the braid?" when connected to the antenna what will stop the current from flowing on the braid when connected to the balun? The balun is an impedance that the RF sees as it starts to travel down the outside of the coax toward the transmitter. But you know about common mode currents, I think. The only thing that will stop the current from flowing on the outside of the shield when connected to the balun is if the balun presents a much lower impedance path for the current than does the shield. Think of a common mode choke. However it is mechanically implemented, its purpose is to provide a block to common mode currents and allow differential currents only. The only way your suggestion makes sense is if the current actually comes *from* the antenna and the balun prevents that current from returning to the feed line. Actually, your statement is the key. The feed point of the antenna IS the current source. One end of the current source is applied to the 'hot' side of the antenna and the other end is applied to the other element plus coax shield. As far as the feed point is concerned it has one element on one side and two elements on the other side. |
"Bal uhn" or "bayl uhn"?
On 7/30/2015 5:46 PM, John S wrote:
On 7/30/2015 4:28 PM, rickman wrote: On 7/30/2015 2:01 PM, Dave Platt wrote: In article , rickman wrote: Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. Don't ask "Why does current flow on the braid?". Ask "What would *stop* current from flowing on the braid?". Current flows on *all* paths that have less than an infinite impedance. That's its nature. Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the braid) will be nonzero, if the voltage at that point is nonzero (E != 0) and the impedance down the braid at that point is not infinite. The effect of a balun is to place a high "choking" impedance in series with the outside of the feedline braid, thus "choking off" the current flow. I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. The balun can have no effect on the impedance of the coax shield. Just as you ask, "What would *stop* current from flowing on the braid?" when connected to the antenna what will stop the current from flowing on the braid when connected to the balun? The balun is an impedance that the RF sees as it starts to travel down the outside of the coax toward the transmitter. But you know about common mode currents, I think. There is something fundamentally wrong with our communications. Are you saying the balun is *part* of the coax? I have seen baluns made by wrapping the coax around a core. I have been assuming the balun was a transformer between the feed line and the antenna. The only thing that will stop the current from flowing on the outside of the shield when connected to the balun is if the balun presents a much lower impedance path for the current than does the shield. Think of a common mode choke. However it is mechanically implemented, its purpose is to provide a block to common mode currents and allow differential currents only. The only way your suggestion makes sense is if the current actually comes *from* the antenna and the balun prevents that current from returning to the feed line. Actually, your statement is the key. The feed point of the antenna IS the current source. One end of the current source is applied to the 'hot' side of the antenna and the other end is applied to the other element plus coax shield. As far as the feed point is concerned it has one element on one side and two elements on the other side. But that is now how it was presented. Another post indicated the current the flows on the inside of the coax shield splits at the antenna feed point and part flows down the outside of the shield. An easy way to distinguish the two cases is to remove the antenna element from the shield. Of course this is no longer a balanced antenna, but still, what happens to the current on the shield? I assume it still flows on the shield inside in an amount equal to the center conductor current and then flows to ground on the outside of the shield? -- Rick |
"Bal uhn" or "bayl uhn"?
On 7/30/2015 4:55 PM, rickman wrote:
On 7/30/2015 5:46 PM, John S wrote: On 7/30/2015 4:28 PM, rickman wrote: On 7/30/2015 2:01 PM, Dave Platt wrote: In article , rickman wrote: Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. Don't ask "Why does current flow on the braid?". Ask "What would *stop* current from flowing on the braid?". Current flows on *all* paths that have less than an infinite impedance. That's its nature. Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the braid) will be nonzero, if the voltage at that point is nonzero (E != 0) and the impedance down the braid at that point is not infinite. The effect of a balun is to place a high "choking" impedance in series with the outside of the feedline braid, thus "choking off" the current flow. I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. The balun can have no effect on the impedance of the coax shield. Just as you ask, "What would *stop* current from flowing on the braid?" when connected to the antenna what will stop the current from flowing on the braid when connected to the balun? The balun is an impedance that the RF sees as it starts to travel down the outside of the coax toward the transmitter. But you know about common mode currents, I think. There is something fundamentally wrong with our communications. Are you saying the balun is *part* of the coax? I have seen baluns made by wrapping the coax around a core. I have been assuming the balun was a transformer between the feed line and the antenna. The only thing that will stop the current from flowing on the outside of the shield when connected to the balun is if the balun presents a much lower impedance path for the current than does the shield. Think of a common mode choke. However it is mechanically implemented, its purpose is to provide a block to common mode currents and allow differential currents only. The only way your suggestion makes sense is if the current actually comes *from* the antenna and the balun prevents that current from returning to the feed line. Actually, your statement is the key. The feed point of the antenna IS the current source. One end of the current source is applied to the 'hot' side of the antenna and the other end is applied to the other element plus coax shield. As far as the feed point is concerned it has one element on one side and two elements on the other side. But that is now how it was presented. Another post indicated the current the flows on the inside of the coax shield splits at the antenna feed point and part flows down the outside of the shield. An easy way to distinguish the two cases is to remove the antenna element from the shield. Of course this is no longer a balanced antenna, but still, what happens to the current on the shield? I assume it still flows on the shield inside in an amount equal to the center conductor current and then flows to ground on the outside of the shield? Yes. Remember that the end of the coax connected to the antenna is now the generator (the source, as jimp said correctly). If the center of the coax carries X current, then the inside of the shield carries X current. It MUST go somewhere. So, with nothing there but the outside of the shield it runs down the outside of the shield with no 'knowledge' of what is happening inside the coax. The shield has become the other element of the antenna whether you like it or not and whether it goes to ground or not. Look up coaxial antenna. |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 7/30/2015 5:46 PM, John S wrote: On 7/30/2015 4:28 PM, rickman wrote: On 7/30/2015 2:01 PM, Dave Platt wrote: In article , rickman wrote: Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. Don't ask "Why does current flow on the braid?". Ask "What would *stop* current from flowing on the braid?". Current flows on *all* paths that have less than an infinite impedance. That's its nature. Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the braid) will be nonzero, if the voltage at that point is nonzero (E != 0) and the impedance down the braid at that point is not infinite. The effect of a balun is to place a high "choking" impedance in series with the outside of the feedline braid, thus "choking off" the current flow. I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. The balun can have no effect on the impedance of the coax shield. Just as you ask, "What would *stop* current from flowing on the braid?" when connected to the antenna what will stop the current from flowing on the braid when connected to the balun? The balun is an impedance that the RF sees as it starts to travel down the outside of the coax toward the transmitter. But you know about common mode currents, I think. There is something fundamentally wrong with our communications. Are you saying the balun is *part* of the coax? I have seen baluns made by wrapping the coax around a core. I have been assuming the balun was a transformer between the feed line and the antenna. There are two types of baluns; voltage baluns and current baluns. A voltage balun is usually a transformer and it forces the output voltage to be equal. A current balun is something that increases the impedance of the outside the shield path. The common forms of choke balun are simply wrapping the coax into a coil, wrapping the coax into a coil around a ferrite rod, wrapping the coax into a coil on a ferrite toroid, or large ferrite beads strung on the coax. See www.eznec.com/Amateur/Articles/Baluns.pdf -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
In article , rickman
wrote: I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. No, it's really rather different. The balun can have no effect on the impedance of the coax shield. Just as you ask, "What would *stop* current from flowing on the braid?" when connected to the antenna what will stop the current from flowing on the braid when connected to the balun? Because, properly done, there's nowhere for that current to come from. The only thing that will stop the current from flowing on the outside of the shield when connected to the balun is if the balun presents a much lower impedance path for the current than does the shield. No, Kirchhoff's current law comes to the rescue. In fact you want the (current) balun to have a very high impedance ot common mode current. The only way your suggestion makes sense is if the current actually comes *from* the antenna and the balun prevents that current from returning to the feed line. Now you're close. In one way of looking at it, if you connect a coaxial feedline directly to a dipole antenna, that's EXACTLY where the current on the outside of the coax braid comes from -- one side of the antenna! I'll try to explain the whole thing in a different way. First, the current flowing "up" the inner conductor of the coax equals the current flowing "down" the inside surface of the outer conductor of the coax, no matter what. This assumes the coax shielding is 100%. The field between the outside surface of the inner conductor and the inner surface of the outer conductor is completely contained insisde the coax, and this requires that those two currents be identical. This much happens no matter what (if anything) is connected to the top end of the coax. Now connect something to the top of the coax, and what happens to those two (equal) curents? It depends upon what is connected. Connect a physically small resistor (resistance of the resistor doesn't matter, and if you want, you can put a physically small inductor or a physically small capacitor in series with that resistor). Kirchhoff's current law requires the current into the one end of this load to equal the current out of the other end of the load. The current into the end of the load connected to the inner conductor of the coax is the same as the current flowing up that inner conductor (Kirchhoff's law again). The same amount of curent flows out of the other end of the load, which is connected to the shield. So the current flowing down the inside of the shield equals the current flowing out of the shield's end of the load, and there is none left to flow down the outside of the braid (Kirchhoff's law once more). If some current did flow down the outside of the braid, the curent into one end of the load couldn't equal the current out of the other end of the laod and Kirchhoff would be unhappy. Now connect an antenna to the coax, instead of a resistor/reactor load. There isn't a Kirchhoff's law for antennas. Since the two halves of the dipole aren't perfectly coupled together, it is quite possible for the current flowing into the left half of the dipole to be different from the current flowing out of the right half. Should that happen, the current flowing out of the right half of the antenna down into the coax shield (inner plus outer surfaces, some on each) will be different from the current flowing down just the inside of the shield. (Remember the current down the inside of the shield HAS to equal the current up the inner conductor.) Where does the difference come from? The difference is the current flowing down the OUTSIDE of the shield. Put a choke consisting of some bifilar turns of parallel conductor feedline, or better yet consisting of a piece of small diameter coax, wound around a toroid, between the top of the coax feedline and the antenna feedpoint, and the current flowing up one conductor of the choke (one wire, or the inner conductor if you use coax in the choke) has to equal the current flowing down the other conductor of the choke (the other wire, or the inner surface of the shield of the piece of coax in the choke). That means the the currents out of one conductor at the top of the choke and into the other conductor at the top of the choke have to be equal. If they were different, the difference (common mode current) would have to flow through the common mode impedance of the choke. If the choke is a good choke, that common mode impedance will be high. So such common mode current has to be low. Only to the extent that it isn't zero, can the current out of the one conductor differ from the current into the other conductor (at the top end of the choke). So the choke does its best to "force" the currents at its top end to be equal. That's why it's called a current balun (not a voltage balun). Even if the impedances of the two halves of the antenna are different, the current balun will do its best to make the current into one side of the antenna equal the current out of the other side. Since the currents at the BOTTOM of the choke in the two conductors are equal, the stupid coaxial feedline thinks it is feeding a resistor (or maybe a resistor and a reactor in series), not an antenna, and behaves accordingly; there is no current flowing down the outside of the coax shield because there is nowhere for such a current to come from. David, VE7EZM and AF7BZ -- David Ryeburn To send e-mail, change "netz" to "net" |
"Bal uhn" or "bayl uhn"?
On 7/30/2015 6:09 PM, wrote:
rickman wrote: On 7/30/2015 5:46 PM, John S wrote: On 7/30/2015 4:28 PM, rickman wrote: On 7/30/2015 2:01 PM, Dave Platt wrote: In article , rickman wrote: Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. Don't ask "Why does current flow on the braid?". Ask "What would *stop* current from flowing on the braid?". Current flows on *all* paths that have less than an infinite impedance. That's its nature. Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the braid) will be nonzero, if the voltage at that point is nonzero (E != 0) and the impedance down the braid at that point is not infinite. The effect of a balun is to place a high "choking" impedance in series with the outside of the feedline braid, thus "choking off" the current flow. I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. The balun can have no effect on the impedance of the coax shield. Just as you ask, "What would *stop* current from flowing on the braid?" when connected to the antenna what will stop the current from flowing on the braid when connected to the balun? The balun is an impedance that the RF sees as it starts to travel down the outside of the coax toward the transmitter. But you know about common mode currents, I think. There is something fundamentally wrong with our communications. Are you saying the balun is *part* of the coax? I have seen baluns made by wrapping the coax around a core. I have been assuming the balun was a transformer between the feed line and the antenna. There are two types of baluns; voltage baluns and current baluns. A voltage balun is usually a transformer and it forces the output voltage to be equal. A current balun is something that increases the impedance of the outside the shield path. The common forms of choke balun are simply wrapping the coax into a coil, wrapping the coax into a coil around a ferrite rod, wrapping the coax into a coil on a ferrite toroid, or large ferrite beads strung on the coax. See www.eznec.com/Amateur/Articles/Baluns.pdf Refer to figure 8. The balun is inserted between the feed line and the antenna. Clearly this example adds no impedance to the shield of the feed line. Rather it must present a very low impedance to the flow of current from the shield to the antenna element. You can also look at the illustrations in Appendix 1, both the voltage balun and the current balun. In both cases they show transformers which must present a low impedance path for the Io current as they call it. The text here does talk about a construction of the current balun from coax and the high impedance to current flowing on the outside of the shield *in the balun*. But when considering the feed line, the balun provides a low impedance to the current flowing from the inside of the feed line shield (Ii) which means it will not follow any other path. -- Rick |
"Bal uhn" or "bayl uhn"?
On 7/30/2015 6:05 PM, John S wrote:
On 7/30/2015 4:55 PM, rickman wrote: On 7/30/2015 5:46 PM, John S wrote: On 7/30/2015 4:28 PM, rickman wrote: On 7/30/2015 2:01 PM, Dave Platt wrote: In article , rickman wrote: Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. Don't ask "Why does current flow on the braid?". Ask "What would *stop* current from flowing on the braid?". Current flows on *all* paths that have less than an infinite impedance. That's its nature. Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the braid) will be nonzero, if the voltage at that point is nonzero (E != 0) and the impedance down the braid at that point is not infinite. The effect of a balun is to place a high "choking" impedance in series with the outside of the feedline braid, thus "choking off" the current flow. I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. The balun can have no effect on the impedance of the coax shield. Just as you ask, "What would *stop* current from flowing on the braid?" when connected to the antenna what will stop the current from flowing on the braid when connected to the balun? The balun is an impedance that the RF sees as it starts to travel down the outside of the coax toward the transmitter. But you know about common mode currents, I think. There is something fundamentally wrong with our communications. Are you saying the balun is *part* of the coax? I have seen baluns made by wrapping the coax around a core. I have been assuming the balun was a transformer between the feed line and the antenna. The only thing that will stop the current from flowing on the outside of the shield when connected to the balun is if the balun presents a much lower impedance path for the current than does the shield. Think of a common mode choke. However it is mechanically implemented, its purpose is to provide a block to common mode currents and allow differential currents only. The only way your suggestion makes sense is if the current actually comes *from* the antenna and the balun prevents that current from returning to the feed line. Actually, your statement is the key. The feed point of the antenna IS the current source. One end of the current source is applied to the 'hot' side of the antenna and the other end is applied to the other element plus coax shield. As far as the feed point is concerned it has one element on one side and two elements on the other side. But that is now how it was presented. Another post indicated the current the flows on the inside of the coax shield splits at the antenna feed point and part flows down the outside of the shield. An easy way to distinguish the two cases is to remove the antenna element from the shield. Of course this is no longer a balanced antenna, but still, what happens to the current on the shield? I assume it still flows on the shield inside in an amount equal to the center conductor current and then flows to ground on the outside of the shield? Yes. Remember that the end of the coax connected to the antenna is now the generator (the source, as jimp said correctly). If the center of the coax carries X current, then the inside of the shield carries X current. It MUST go somewhere. So, with nothing there but the outside of the shield it runs down the outside of the shield with no 'knowledge' of what is happening inside the coax. The shield has become the other element of the antenna whether you like it or not and whether it goes to ground or not. The point is that the current flows on the outside of the shield because the current follows the lowest impedance path it has to follow, same as in all other cases. If you connect a balun to the end of the coax with no antenna connection on the side from the shield, there will be no current flow to the antenna element. So all the inner shield current will *still* flow back down the outside of the shield showing that there is no need to "explain" anything about a reflected wave from the antenna end causing the current to flow down the coax. -- Rick |
"Bal uhn" or "bayl uhn"?
On 7/30/2015 10:52 PM, David Ryeburn wrote:
In article , rickman wrote: I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. No, it's really rather different. You aren't explaining anything. The balun can have no effect on the impedance of the coax shield. Just as you ask, "What would *stop* current from flowing on the braid?" when connected to the antenna what will stop the current from flowing on the braid when connected to the balun? Because, properly done, there's nowhere for that current to come from. Uh, the current comes from the shield inside. The connection from the shield inside and the shield outside is always there. The current continues to flow on the inside because of the interaction with the inner conductor current. On reaching the balun this interaction is disrupted and current can flow back down the outside of the shield if a better path is not provided. The balun provides that path by the equal current requirement. In essence the impedance to the shield current is virtually zero. The only thing that will stop the current from flowing on the outside of the shield when connected to the balun is if the balun presents a much lower impedance path for the current than does the shield. No, Kirchhoff's current law comes to the rescue. In fact you want the (current) balun to have a very high impedance ot common mode current. I think that is the same as saying the impedance to a differential current is low which is what I am saying, no? The only way your suggestion makes sense is if the current actually comes *from* the antenna and the balun prevents that current from returning to the feed line. Now you're close. In one way of looking at it, if you connect a coaxial feedline directly to a dipole antenna, that's EXACTLY where the current on the outside of the coax braid comes from -- one side of the antenna! Why does it come from the antenna? If you disconnect the side of the antenna connected to the shield, the current will flow back down the outside of the shield, no? The current comes from the shield inside path, not the antenna. I'll try to explain the whole thing in a different way. You aren't saying anything new. This has all been said before and does not contradict what I am saying... First, the current flowing "up" the inner conductor of the coax equals the current flowing "down" the inside surface of the outer conductor of the coax, no matter what. This assumes the coax shielding is 100%. The field between the outside surface of the inner conductor and the inner surface of the outer conductor is completely contained insisde the coax, and this requires that those two currents be identical. This much happens no matter what (if anything) is connected to the top end of the coax. Now connect something to the top of the coax, and what happens to those two (equal) curents? It depends upon what is connected. Connect a physically small resistor (resistance of the resistor doesn't matter, and if you want, you can put a physically small inductor or a physically small capacitor in series with that resistor). Kirchhoff's current law requires the current into the one end of this load to equal the current out of the other end of the load. The current into the end of the load connected to the inner conductor of the coax is the same as the current flowing up that inner conductor (Kirchhoff's law again). The same amount of curent flows out of the other end of the load, which is connected to the shield. So the current flowing down the inside of the shield equals the current flowing out of the shield's end of the load, and there is none left to flow down the outside of the braid (Kirchhoff's law once more). If some current did flow down the outside of the braid, the curent into one end of the load couldn't equal the current out of the other end of the laod and Kirchhoff would be unhappy. Now connect an antenna to the coax, instead of a resistor/reactor load. There isn't a Kirchhoff's law for antennas. Since the two halves of the dipole aren't perfectly coupled together, it is quite possible for the current flowing into the left half of the dipole to be different from the current flowing out of the right half. Should that happen, the current flowing out of the right half of the antenna down into the coax shield (inner plus outer surfaces, some on each) will be different from the current flowing down just the inside of the shield. (Remember the current down the inside of the shield HAS to equal the current up the inner conductor.) Where does the difference come from? The difference is the current flowing down the OUTSIDE of the shield. Nothing here indicates any current is coming *from* the antenna which you say further above. "that's EXACTLY where the current on the outside of the coax braid comes from -- one side of the antenna!" In fact, you seem to be saying exactly what I am saying, the current comes *from* the inside of the shield and some returns on the outside of the shield. You seem to be talking about current flow *from* the inner conductor and *into* the outer conductor, so maybe this is not significant terminology since it is just to indicate the polarity of the current rather than the source. Put a choke consisting of some bifilar turns of parallel conductor feedline, or better yet consisting of a piece of small diameter coax, wound around a toroid, between the top of the coax feedline and the antenna feedpoint, and the current flowing up one conductor of the choke (one wire, or the inner conductor if you use coax in the choke) has to equal the current flowing down the other conductor of the choke (the other wire, or the inner surface of the shield of the piece of coax in the choke). That means the the currents out of one conductor at the top of the choke and into the other conductor at the top of the choke have to be equal. If they were different, the difference (common mode current) would have to flow through the common mode impedance of the choke. If the choke is a good choke, that common mode impedance will be high. So such common mode current has to be low. Only to the extent that it isn't zero, can the current out of the one conductor differ from the current into the other conductor (at the top end of the choke). So the choke does its best to "force" the currents at its top end to be equal. That's why it's called a current balun (not a voltage balun). Even if the impedances of the two halves of the antenna are different, the current balun will do its best to make the current into one side of the antenna equal the current out of the other side. Since the currents at the BOTTOM of the choke in the two conductors are equal, the stupid coaxial feedline thinks it is feeding a resistor (or maybe a resistor and a reactor in series), not an antenna, and behaves accordingly; there is no current flowing down the outside of the coax shield because there is nowhere for such a current to come from. If you give this just a little bit of thought, you will realize that the impedance to the current flowing from the shield inner side into the choke has to be much lower than the impedance the shield outer side presents. That path is still there and if the balun did not present a significantly lower impedance path, current would still flow on the shield outside. You can explain it with balanced currents which is not invalid. But nothing you have said contradicts what I have said. -- Rick |
"Bal uhn" or "bayl uhn"?
On 7/31/2015 1:15 AM, rickman wrote:
On 7/30/2015 10:52 PM, David Ryeburn wrote: In article , rickman wrote: I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. No, it's really rather different. You aren't explaining anything. Now you are being argumentative, Rick. Do you want to continue the dialog or not? |
"Bal uhn" or "bayl uhn"?
On 7/31/2015 12:19 PM, John S wrote:
On 7/31/2015 1:15 AM, rickman wrote: On 7/30/2015 10:52 PM, David Ryeburn wrote: In article , rickman wrote: I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. No, it's really rather different. You aren't explaining anything. Now you are being argumentative, Rick. Do you want to continue the dialog or not? That is BS. Saying something is "different" with no explanation is not "discussion" and is not useful to a "dialog". I'm just pointing that out. In fact, much of what is being said here is talking past the points I have made. Rather than try to understand what is going on, most here seem to just repeat the standard explanation without thinking it through. One of the references discusses the case of a balun made by wrapping the coax around a core, but when discussing a separate balun made of wires they say, "When constructed of twisted-pair line, the effect on imbalance current is the same and for the same reasons, but operation is more difficuit to visualize". That is a cop-out. A pair of wires is very clearly different from a coax. Anytime you don't wish to reply to my posts you are free to refrain. -- Rick |
"Bal uhn" or "bayl uhn"?
In article , rickman
wrote: On 7/31/2015 12:19 PM, John S wrote: On 7/31/2015 1:15 AM, rickman wrote: On 7/30/2015 10:52 PM, David Ryeburn wrote: In article , rickman wrote: I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. No, it's really rather different. You aren't explaining anything. Now you are being argumentative, Rick. Do you want to continue the dialog or not? Worse than argumentative. I DID explain, but I can't read for him. I give up. Between us, his word is last. Anyone else who wishes to show me where my argument involving resistors, Kirchhoff's law, fields inside coaxial cables, etc. is incorrect, I'll be glad to read. I'm always ready to learn. In the meantime I'll connect my current baluns the way I described, and they'll work the way I described. He can do what he wants. David, VE7EZM (now), AF7BZ (now), and W8EZE (1949-1967) -- David Ryeburn To send e-mail, change "netz" to "net" |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 7/31/2015 12:19 PM, John S wrote: On 7/31/2015 1:15 AM, rickman wrote: On 7/30/2015 10:52 PM, David Ryeburn wrote: In article , rickman wrote: I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. No, it's really rather different. You aren't explaining anything. Now you are being argumentative, Rick. Do you want to continue the dialog or not? That is BS. Saying something is "different" with no explanation is not "discussion" and is not useful to a "dialog". I'm just pointing that out. It was nothing more than a knee jerk response without regard to everything that followed. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 7/30/2015 6:09 PM, wrote: rickman wrote: On 7/30/2015 5:46 PM, John S wrote: On 7/30/2015 4:28 PM, rickman wrote: On 7/30/2015 2:01 PM, Dave Platt wrote: In article , rickman wrote: Yes, I read that, but it doesn't really explain this current. Later they make the statement, "the current on the braid outside side is the sum of currents other than transmission line currents on the entire coaxial cable structure". This is pretty clear, but still does not explain the source, or maybe I should say "why" the current flows on the braid and not the antenna. Don't ask "Why does current flow on the braid?". Ask "What would *stop* current from flowing on the braid?". Current flows on *all* paths that have less than an infinite impedance. That's its nature. Remember, I = E / R (or, for AC/RF, I = E / Z). "I" (current on the braid) will be nonzero, if the voltage at that point is nonzero (E != 0) and the impedance down the braid at that point is not infinite. The effect of a balun is to place a high "choking" impedance in series with the outside of the feedline braid, thus "choking off" the current flow. I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. The balun can have no effect on the impedance of the coax shield. Just as you ask, "What would *stop* current from flowing on the braid?" when connected to the antenna what will stop the current from flowing on the braid when connected to the balun? The balun is an impedance that the RF sees as it starts to travel down the outside of the coax toward the transmitter. But you know about common mode currents, I think. There is something fundamentally wrong with our communications. Are you saying the balun is *part* of the coax? I have seen baluns made by wrapping the coax around a core. I have been assuming the balun was a transformer between the feed line and the antenna. There are two types of baluns; voltage baluns and current baluns. A voltage balun is usually a transformer and it forces the output voltage to be equal. A current balun is something that increases the impedance of the outside the shield path. The common forms of choke balun are simply wrapping the coax into a coil, wrapping the coax into a coil around a ferrite rod, wrapping the coax into a coil on a ferrite toroid, or large ferrite beads strung on the coax. See www.eznec.com/Amateur/Articles/Baluns.pdf Refer to figure 8. The balun is inserted between the feed line and the antenna. Clearly this example adds no impedance to the shield of the feed line. Rather it must present a very low impedance to the flow of current from the shield to the antenna element. You can also look at the illustrations in Appendix 1, both the voltage balun and the current balun. In both cases they show transformers which must present a low impedance path for the Io current as they call it. The text here does talk about a construction of the current balun from coax and the high impedance to current flowing on the outside of the shield *in the balun*. But when considering the feed line, the balun provides a low impedance to the current flowing from the inside of the feed line shield (Ii) which means it will not follow any other path. If you want to argue about it, argue with the author of the article. There is little that you said that has anything to do with what I said. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
On 7/31/2015 1:55 PM, David Ryeburn wrote:
In article , rickman wrote: On 7/31/2015 12:19 PM, John S wrote: On 7/31/2015 1:15 AM, rickman wrote: On 7/30/2015 10:52 PM, David Ryeburn wrote: In article , rickman wrote: I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. No, it's really rather different. You aren't explaining anything. Now you are being argumentative, Rick. Do you want to continue the dialog or not? Worse than argumentative. I DID explain, but I can't read for him. I give up. Between us, his word is last. Anyone else who wishes to show me where my argument involving resistors, Kirchhoff's law, fields inside coaxial cables, etc. is incorrect, I'll be glad to read. I'm always ready to learn. In the meantime I'll connect my current baluns the way I described, and they'll work the way I described. He can do what he wants. I replied to your post in detail. The comment above was about the single line response which was not useful. If you want to take an attitude fine. But please don't act like I was not communicating. I explained to you my points and now you choose to complain. I never said baluns don't work. So your comment about your equipment is not relevant to the discussion. -- Rick |
"Bal uhn" or "bayl uhn"?
On 7/31/2015 12:13 PM, rickman wrote:
On 7/31/2015 12:19 PM, John S wrote: On 7/31/2015 1:15 AM, rickman wrote: On 7/30/2015 10:52 PM, David Ryeburn wrote: In article , rickman wrote: I can't say I agree with your "choking" impedance idea. The coax connects to the balun in the same way it connects to the antenna. No, it's really rather different. You aren't explaining anything. Now you are being argumentative, Rick. Do you want to continue the dialog or not? That is BS. Saying something is "different" with no explanation is not "discussion" and is not useful to a "dialog". I'm just pointing that out. If you had continued to read the entire post, it may have made more sense to you. His explanation followed your contentious comment. In fact, much of what is being said here is talking past the points I have made. Rather than try to understand what is going on, most here seem to just repeat the standard explanation without thinking it through. One of the references discusses the case of a balun made by wrapping the coax around a core, but when discussing a separate balun made of wires they say, "When constructed of twisted-pair line, the effect on imbalance current is the same and for the same reasons, but operation is more difficuit to visualize". That is a cop-out. A pair of wires is very clearly different from a coax. Then we are all in error concerning the points of your query. You posted "Perhaps someone can explain the issue of current in the coax shield. Current gives rise to a magnetic field. But the current in the inner conductor is opposite and would create a magnetic field that would cancel the field of the outer conductor, no? What am I missing?" Was this resolved? What is it that you want? Please summarize your point(s) again for those of us who have forgotten them or wish not to wade back through the posts and ferret them out. Better yet, start a new thread. Anytime you don't wish to reply to my posts you are free to refrain. Of course. You don't need to remind me of my options. |
"Bal uhn" or "bayl uhn"?
Jeff wrote:
I give up. Between us, his word is last. Anyone else who wishes to show me where my argument involving resistors, Kirchhoff's law, fields inside coaxial cables, etc. is incorrect, I'll be glad to read. I'm always ready to learn. In the meantime I'll connect my current baluns the way I described, and they'll work the way I described. He can do what he wants. Conservation of energy also shows that with a perfectly matched dipole all of the energy applied to it will be radiated, or lost as heat in the elements. So by extension there can be no power flowing on the coax outer. Jeff So, clearly, the energy supplied from the feeder and conducted down the coax outer nerver reaches the antenna. Just as two resistors in parallel share the energy supplied by a voltage applies across them. -- Roger Hayter |
"Bal uhn" or "bayl uhn"?
Jeff wrote:
On 01/08/2015 10:54, Roger Hayter wrote: Jeff wrote: I give up. Between us, his word is last. Anyone else who wishes to show me where my argument involving resistors, Kirchhoff's law, fields inside coaxial cables, etc. is incorrect, I'll be glad to read. I'm always ready to learn. In the meantime I'll connect my current baluns the way I described, and they'll work the way I described. He can do what he wants. Conservation of energy also shows that with a perfectly matched dipole all of the energy applied to it will be radiated, or lost as heat in the elements. So by extension there can be no power flowing on the coax outer. Jeff So, clearly, the energy supplied from the feeder and conducted down the coax outer nerver reaches the antenna. Just as two resistors in parallel share the energy supplied by a voltage applies across them. No, there will be no energy "supplied from the feeder and conducted down the coax outer" as it will all be conducted into the antenna and radiated if the antenna is perfectly balanced. The power flowing up the coax, when perfectly matched, will only 'see' the antenna load, not the coax outer as and element in parallel. Jeff If it was perfectly balanced before you connected the feeder, it will no longer be perfectly balanced once the shunt impedance to earth of the coax braid is connected to *one* side of it. Dare I say that that is why you need a balun? This asymmetrical shunt impedance is the source of the whole question. -- Roger Hayter |
"Bal uhn" or "bayl uhn"?
On 8/1/2015 9:09 AM, Roger Hayter wrote:
Jeff wrote: On 01/08/2015 10:54, Roger Hayter wrote: Jeff wrote: Conservation of energy also shows that with a perfectly matched dipole all of the energy applied to it will be radiated, or lost as heat in the elements. So by extension there can be no power flowing on the coax outer. Jeff So, clearly, the energy supplied from the feeder and conducted down the coax outer nerver reaches the antenna. Just as two resistors in parallel share the energy supplied by a voltage applies across them. No, there will be no energy "supplied from the feeder and conducted down the coax outer" as it will all be conducted into the antenna and radiated if the antenna is perfectly balanced. The power flowing up the coax, when perfectly matched, will only 'see' the antenna load, not the coax outer as and element in parallel. Jeff If it was perfectly balanced before you connected the feeder, it will no longer be perfectly balanced once the shunt impedance to earth of the coax braid is connected to *one* side of it. Dare I say that that is why you need a balun? This asymmetrical shunt impedance is the source of the whole question. Not trying to criticize anyone here. I just want to make the observation that much of the theory that is applied to the systems discussed here are generalization which only apply under certain conditions. Often these conditions are "ideal" and even in the best of circumstances won't match reality perfectly... but are good enough. Then a discussion starts where something outside the assumptions is explored and the "ideal" assumptions are forgotten. The discussion goes in many directions because the general rules are applied when instead, the topic needs to be considered at a more fundamental level. So look at the assumptions (or conditions) that exist in this case and see which ones apply and which don't. I'm pretty sure that the condition of an impedance match at the feed point does not exist among others. That was my point when I suggested that at the junction of the coaxial feed line and the balun the same condition exists of there being a path for current on the outside of the shield. This can divert current flow from the inside of the shield unless... the current path through the balun from the shield is a very low impedance. In order to just eliminate any idea of the balun "preventing" an unbalanced current flow back from the antenna, replace the antenna with a matched resistor. Then there will be no reflection of any kind at the load and the only point that needs to be considered is the junction of the feed line and the balun. -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/1/2015 9:09 AM, Roger Hayter wrote: Jeff wrote: On 01/08/2015 10:54, Roger Hayter wrote: Jeff wrote: Conservation of energy also shows that with a perfectly matched dipole all of the energy applied to it will be radiated, or lost as heat in the elements. So by extension there can be no power flowing on the coax outer. Jeff So, clearly, the energy supplied from the feeder and conducted down the coax outer nerver reaches the antenna. Just as two resistors in parallel share the energy supplied by a voltage applies across them. No, there will be no energy "supplied from the feeder and conducted down the coax outer" as it will all be conducted into the antenna and radiated if the antenna is perfectly balanced. The power flowing up the coax, when perfectly matched, will only 'see' the antenna load, not the coax outer as and element in parallel. Jeff If it was perfectly balanced before you connected the feeder, it will no longer be perfectly balanced once the shunt impedance to earth of the coax braid is connected to *one* side of it. Dare I say that that is why you need a balun? This asymmetrical shunt impedance is the source of the whole question. Not trying to criticize anyone here. I just want to make the observation that much of the theory that is applied to the systems discussed here are generalization which only apply under certain conditions. Often these conditions are "ideal" and even in the best of circumstances won't match reality perfectly... but are good enough. Then a discussion starts where something outside the assumptions is explored and the "ideal" assumptions are forgotten. The discussion goes in many directions because the general rules are applied when instead, the topic needs to be considered at a more fundamental level. So look at the assumptions (or conditions) that exist in this case and see which ones apply and which don't. I'm pretty sure that the condition of an impedance match at the feed point does not exist among others. That was my point when I suggested that at the junction of the coaxial feed line and the balun the same condition exists of there being a path for current on the outside of the shield. This can divert current flow from the inside of the shield unless... the current path through the balun from the shield is a very low impedance. In order to just eliminate any idea of the balun "preventing" an unbalanced current flow back from the antenna, replace the antenna with a matched resistor. Then there will be no reflection of any kind at the load and the only point that needs to be considered is the junction of the feed line and the balun. Well I agree with your first point. Not least, the conduction path along the coax outer is more like a radiating element than a lumped component, because of its length. But your second point is unhelpful in some circumstances. For instance, if the type of balun is the inductive coil of the feeder with or without ferrites, then there simply *is no* current path down the outside of the feeder from the junction of the balun and the feeder, Except from the outer of the cable in the balun coil, and it is this that is decoupled by the inductance. Secondly, even if you connect a resistor across the end ot the feeder, consider that the inner conductor just goes to the resistor, but the outer conductor sees the resistor and the outer side of the braid in parallel. So you will get RF (and therefore some radiation) on the outer of the coax even if you just connect a resistor across the end. This would actually be quite a simple lab experiment, at UHF or higher. Compare the amount of RF on the outer with a bare surface mount 50ohm or with one of the screened 50 ohm terminations (which does not allow any signal to get to the outer). Or compare the SWR which will be near 1.0 with the screened load and might be very different with the unscreened, sim,ply because the coax outer is shunting one side of the load. -- Roger Hayter |
"Bal uhn" or "bayl uhn"?
On 8/1/2015 1:38 PM, Roger Hayter wrote:
But your second point is unhelpful in some circumstances. For instance, if the type of balun is the inductive coil of the feeder with or without ferrites, then there simply *is no* current path down the outside of the feeder from the junction of the balun and the feeder, Except from the outer of the cable in the balun coil, and it is this that is decoupled by the inductance. Your description is not clear to me. Secondly, even if you connect a resistor across the end ot the feeder, consider that the inner conductor just goes to the resistor, but the outer conductor sees the resistor and the outer side of the braid in parallel. So you will get RF (and therefore some radiation) on the outer of the coax even if you just connect a resistor across the end. Ok, let's discuss this. You are describing a circuit that is just the coax and a terminating resistor. You seem to be saying that current will flow on the outer surface of the shield. If that were true, where does it come from? In this simple circuit the current on the shield inner surface matches the current on the inner conductor. So there is no source for current to flow on the shield outer surface. This would actually be quite a simple lab experiment, at UHF or higher. Compare the amount of RF on the outer with a bare surface mount 50ohm or with one of the screened 50 ohm terminations (which does not allow any signal to get to the outer). Or compare the SWR which will be near 1.0 with the screened load and might be very different with the unscreened, sim,ply because the coax outer is shunting one side of the load. Ok, can anyone do this? -- Rick |
"Bal uhn" or "bayl uhn"?
Here is a more basic question. What are the assumptions to be able to
say the current on the shield inner layer equals the current in the inner conductor of a coax? I'd be willing to bet I can construct a circuit where this is not true. -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/1/2015 1:38 PM, Roger Hayter wrote: But your second point is unhelpful in some circumstances. For instance, if the type of balun is the inductive coil of the feeder with or without ferrites, then there simply *is no* current path down the outside of the feeder from the junction of the balun and the feeder, Except from the outer of the cable in the balun coil, and it is this that is decoupled by the inductance. Your description is not clear to me. I am looking at the junction of the feeder with the balun, and the only source of current on the outside of the feeder is connected by a very high inductance to the source of signal at the antenna end. Secondly, even if you connect a resistor across the end ot the feeder, consider that the inner conductor just goes to the resistor, but the outer conductor sees the resistor and the outer side of the braid in parallel. So you will get RF (and therefore some radiation) on the outer of the coax even if you just connect a resistor across the end. Ok, let's discuss this. You are describing a circuit that is just the coax and a terminating resistor. You seem to be saying that current will flow on the outer surface of the shield. If that were true, where does it come from? In this simple circuit the current on the shield inner surface matches the current on the inner conductor. So there is no source for current to flow on the shield outer surface. I am beginning to think you may be right! Sorry. But it doesn't affect the argument if there is an antenna, however symmetrical, connected to both conductors. This would actually be quite a simple lab experiment, at UHF or higher. Compare the amount of RF on the outer with a bare surface mount 50ohm or with one of the screened 50 ohm terminations (which does not allow any signal to get to the outer). Or compare the SWR which will be near 1.0 with the screened load and might be very different with the unscreened, sim,ply because the coax outer is shunting one side of the load. Ok, can anyone do this? It would be very interesting! -- Roger Hayter |
"Bal uhn" or "bayl uhn"?
On 8/1/2015 3:29 PM, Roger Hayter wrote:
rickman wrote: On 8/1/2015 1:38 PM, Roger Hayter wrote: But your second point is unhelpful in some circumstances. For instance, if the type of balun is the inductive coil of the feeder with or without ferrites, then there simply *is no* current path down the outside of the feeder from the junction of the balun and the feeder, Except from the outer of the cable in the balun coil, and it is this that is decoupled by the inductance. Your description is not clear to me. I am looking at the junction of the feeder with the balun, and the only source of current on the outside of the feeder is connected by a very high inductance to the source of signal at the antenna end. How is the inner surface of the shield not connected to the outer surface of the shield? Secondly, even if you connect a resistor across the end ot the feeder, consider that the inner conductor just goes to the resistor, but the outer conductor sees the resistor and the outer side of the braid in parallel. So you will get RF (and therefore some radiation) on the outer of the coax even if you just connect a resistor across the end. Ok, let's discuss this. You are describing a circuit that is just the coax and a terminating resistor. You seem to be saying that current will flow on the outer surface of the shield. If that were true, where does it come from? In this simple circuit the current on the shield inner surface matches the current on the inner conductor. So there is no source for current to flow on the shield outer surface. I am beginning to think you may be right! Sorry. But it doesn't affect the argument if there is an antenna, however symmetrical, connected to both conductors. Actually it does. With an antenna connected currents can flow from the inner and shield conductors unequally. It will only be unequal if there is an alternate path for the current. That alternate path will be the outer surface of the shield. This would actually be quite a simple lab experiment, at UHF or higher. Compare the amount of RF on the outer with a bare surface mount 50ohm or with one of the screened 50 ohm terminations (which does not allow any signal to get to the outer). Or compare the SWR which will be near 1.0 with the screened load and might be very different with the unscreened, sim,ply because the coax outer is shunting one side of the load. Ok, can anyone do this? It would be very interesting! -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/1/2015 3:29 PM, Roger Hayter wrote: rickman wrote: On 8/1/2015 1:38 PM, Roger Hayter wrote: But your second point is unhelpful in some circumstances. For instance, if the type of balun is the inductive coil of the feeder with or without ferrites, then there simply *is no* current path down the outside of the feeder from the junction of the balun and the feeder, Except from the outer of the cable in the balun coil, and it is this that is decoupled by the inductance. Your description is not clear to me. I am looking at the junction of the feeder with the balun, and the only source of current on the outside of the feeder is connected by a very high inductance to the source of signal at the antenna end. How is the inner surface of the shield not connected to the outer surface of the shield? At the point the balun joins the feeder, the only way RF can get from inside the coax braid (skin effect deep) to the outside of the coax braid is to go all the way up to the antenna and down the outer surface of the balun. This is a high inductance path. This is easy to see if the balun is continuous with the feeder, but even if it is not the join is like the case with the matched load which I have now agreed there is no 'spare' currrent to cross from the inside to the outside, as the inside current has to match the inner conductor current. Secondly, even if you connect a resistor across the end ot the feeder, consider that the inner conductor just goes to the resistor, but the outer conductor sees the resistor and the outer side of the braid in parallel. So you will get RF (and therefore some radiation) on the outer of the coax even if you just connect a resistor across the end. Ok, let's discuss this. You are describing a circuit that is just the coax and a terminating resistor. You seem to be saying that current will flow on the outer surface of the shield. If that were true, where does it come from? In this simple circuit the current on the shield inner surface matches the current on the inner conductor. So there is no source for current to flow on the shield outer surface. I am beginning to think you may be right! Sorry. But it doesn't affect the argument if there is an antenna, however symmetrical, connected to both conductors. Actually it does. With an antenna connected currents can flow from the inner and shield conductors unequally. It will only be unequal if there is an alternate path for the current. That alternate path will be the outer surface of the shield. This would actually be quite a simple lab experiment, at UHF or higher. Compare the amount of RF on the outer with a bare surface mount 50ohm or with one of the screened 50 ohm terminations (which does not allow any signal to get to the outer). Or compare the SWR which will be near 1.0 with the screened load and might be very different with the unscreened, sim,ply because the coax outer is shunting one side of the load. Ok, can anyone do this? It would be very interesting! -- Roger Hayter |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/1/2015 1:38 PM, Roger Hayter wrote: But your second point is unhelpful in some circumstances. For instance, if the type of balun is the inductive coil of the feeder with or without ferrites, then there simply *is no* current path down the outside of the feeder from the junction of the balun and the feeder, Except from the outer of the cable in the balun coil, and it is this that is decoupled by the inductance. Your description is not clear to me. Secondly, even if you connect a resistor across the end ot the feeder, consider that the inner conductor just goes to the resistor, but the outer conductor sees the resistor and the outer side of the braid in parallel. So you will get RF (and therefore some radiation) on the outer of the coax even if you just connect a resistor across the end. Ok, let's discuss this. You are describing a circuit that is just the coax and a terminating resistor. You seem to be saying that current will flow on the outer surface of the shield. If that were true, where does it come from? In this simple circuit the current on the shield inner surface matches the current on the inner conductor. So there is no source for current to flow on the shield outer surface. The inside and the outside of the shield are connected together at the point where the resistor connects to them. The source of the current is the electromagnetic field that propagates inside the coax. As the shield is another current path, some current will flow down it. How much depends on the length of the shield in wavelengths which determines the impedance of that path. This would actually be quite a simple lab experiment, at UHF or higher. Compare the amount of RF on the outer with a bare surface mount 50ohm or with one of the screened 50 ohm terminations (which does not allow any signal to get to the outer). Or compare the SWR which will be near 1.0 with the screened load and might be very different with the unscreened, sim,ply because the coax outer is shunting one side of the load. Ok, can anyone do this? Anyone with appropriate test equipement. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
rickman wrote:
Here is a more basic question. What are the assumptions to be able to say the current on the shield inner layer equals the current in the inner conductor of a coax? I'd be willing to bet I can construct a circuit where this is not true. Inside the transmission line the energy is carried in the electromagnetic field between the conductors, not in the conductors. https://en.wikipedia.org/wiki/Transm...#Coaxial_cable Nothing external to the transmission line can chage this. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
On 8/1/2015 4:31 PM, Roger Hayter wrote:
rickman wrote: On 8/1/2015 3:29 PM, Roger Hayter wrote: rickman wrote: On 8/1/2015 1:38 PM, Roger Hayter wrote: But your second point is unhelpful in some circumstances. For instance, if the type of balun is the inductive coil of the feeder with or without ferrites, then there simply *is no* current path down the outside of the feeder from the junction of the balun and the feeder, Except from the outer of the cable in the balun coil, and it is this that is decoupled by the inductance. Your description is not clear to me. I am looking at the junction of the feeder with the balun, and the only source of current on the outside of the feeder is connected by a very high inductance to the source of signal at the antenna end. How is the inner surface of the shield not connected to the outer surface of the shield? At the point the balun joins the feeder, the only way RF can get from inside the coax braid (skin effect deep) to the outside of the coax braid is to go all the way up to the antenna and down the outer surface of the balun. I think this is our first point of disagreement. There is nothing to to stop the current flowing on the shield inside surface from moving to the shield outside surface other than a tiny amount of resistance in the shield wire. Unless the current flow sees a lower impedance path to follow through the balun, it will travel back on the shield outside surface. This is a high inductance path. I think you are applying this term without appreciating the full meaning. It is a high impedance path for common mode currents, but a low impedance path for differential currents. Since the current in the shield inner surface balances the current on the center conductor, it is a very low impedance path for the full current on the shield. If it were accurate to say the balun was "a high impedance path" without the qualifications, the balun would prevent the desired signal from reaching the load. This is easy to see if the balun is continuous with the feeder, but even if it is not the join is like the case with the matched load which I have now agreed there is no 'spare' currrent to cross from the inside to the outside, as the inside current has to match the inner conductor current. I won't argue that any of this is correct. It does not conflict in any way with what I have said. -- Rick |
"Bal uhn" or "bayl uhn"?
|
"Bal uhn" or "bayl uhn"?
On 8/1/2015 4:52 PM, wrote:
rickman wrote: Here is a more basic question. What are the assumptions to be able to say the current on the shield inner layer equals the current in the inner conductor of a coax? I'd be willing to bet I can construct a circuit where this is not true. Inside the transmission line the energy is carried in the electromagnetic field between the conductors, not in the conductors. https://en.wikipedia.org/wiki/Transm...#Coaxial_cable Nothing external to the transmission line can chage this. Um, do you want to answer the question about the assumptions required to assume equal currents in the two conductors of a coax? If not that's fine. -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/1/2015 4:47 PM, wrote: rickman wrote: On 8/1/2015 1:38 PM, Roger Hayter wrote: But your second point is unhelpful in some circumstances. For instance, if the type of balun is the inductive coil of the feeder with or without ferrites, then there simply *is no* current path down the outside of the feeder from the junction of the balun and the feeder, Except from the outer of the cable in the balun coil, and it is this that is decoupled by the inductance. Your description is not clear to me. Secondly, even if you connect a resistor across the end ot the feeder, consider that the inner conductor just goes to the resistor, but the outer conductor sees the resistor and the outer side of the braid in parallel. So you will get RF (and therefore some radiation) on the outer of the coax even if you just connect a resistor across the end. Ok, let's discuss this. You are describing a circuit that is just the coax and a terminating resistor. You seem to be saying that current will flow on the outer surface of the shield. If that were true, where does it come from? In this simple circuit the current on the shield inner surface matches the current on the inner conductor. So there is no source for current to flow on the shield outer surface. The inside and the outside of the shield are connected together at the point where the resistor connects to them. The source of the current is the electromagnetic field that propagates inside the coax. As the shield is another current path, some current will flow down it. How much depends on the length of the shield in wavelengths which determines the impedance of that path. I would like to clarify this point. You are saying that some of the current that flow to the load on the shield inside surface will flow back on the shield outside surface. That means the current in the inner conductor will no longer equal the current in the shield inner surface, right? There is no current in the shield inner surface, the energy is in the ELECTROMAGNETIC FIELD between the inner and outer conductors. To be nit pickingly precise, there is some small current in the inner surface of the shield and the center wire, but for real coax that surface current is insignificant. You can look at coax as a wave guide if that makes it easier to understand, though the mode is different than the mode in what is normally called wave guide. https://en.wikipedia.org/wiki/Transm...#Coaxial_cable At the end of the coaxial structure, the electromagnetic field becomes a current flow in any conductors connected to the end of the coax. One of those conductors is always the outside of the shield because of the physical structure of coax. The sum of the currents in the outside of the shield plus all other conductors connected to the outside shield is equal to the sum of the currents of all the conductors connected to the center wire. -- Jim Pennino |
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