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"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 6:05 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 |
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