|
"Bal uhn" or "bayl uhn"?
On 8/1/2015 8:24 PM, wrote:
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. this is a pretty amazing revelation. So what are the assumptions to make this true? If there is no appreciable current flow in the coax, then the resistance of the wires is of no significance. Funny, when I make a loop antenna from coax, the Q still seems to be limited by the conductor resistance. Odd... 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 I don't see anything here that says there is no current flow in the coax. At the end of the coaxial structure, the electromagnetic field becomes a current flow in any conductors connected to the end of the coax. Where do the electrons come from that the current consists of? Does the wire end act as a capacitor? 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. That seems rather obvious. -- Rick |
"Bal uhn" or "bayl uhn"?
On 8/1/2015 8:31 PM, wrote:
rickman wrote: 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. I did; there is no current in the conductors of a coax where those conductors physically define a coaxial transmission line. https://en.wikipedia.org/wiki/Transm...#Coaxial_cable There is current in the conductors only after they no longer physically define a coaxial transmission line. Inside the coaxial structure everything is in the field and is balanced. Do you have any references that actually say there is no current flow in a coax? The reference given above doesn't even come close to saying this. -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
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. At RF, the coax braid is an impenetrable Faraday screen. That is what it is for, after all. The current simply can't go through the fractional mm of copper from inside to outside. 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. Which is why I said it is a high impedance path for the current on the outside of the screen. "Common mode" isn't really quite the right name for this current, as the inner is totally uninvolved, but it is certainly not differential, which satisfies your point. 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. -- Roger Hayter |
"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? Just to interject a further argument; I think I agree with you on this. You can't get coax outer current unless some current sink (eg an antenna element) is connected to the inner conductor side of the load resistor. But, consider a pefectly symmetrical dipole: if the potential on the centre conductor and the braid is exactly the same, how can the two antenna halves have different currents to allow some to flow down the outside of the braid? I am beginning to wonder if braid outside current with a symmetrical antenna *only* occurs when the coax outer interacts with the EM field of the antenna so as to actually alter the impedance of at least one of the antenna elements, or alter the two of them to a different extent, so that a common mode current is "left over" for the braid. It does seem likely that a long wire coming from the centre of the dipoe and not being absolutely symmetrical would have this effect. However, on this argument, you would not need a balun if your feeder was absolutely symmetrical! This theory seems eminently testable with antenna simulation programs. -- Roger Hayter |
"Bal uhn" or "bayl uhn"?
Jeff wrote:
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. So if what you are saying is correct then terminating a piece of coax with a 50 ohm resistor will not cause a perfect match and the load will not be 50 ohms due to the shunt impedance to earth of the coax braid. I think not. Jeff I agree I was probably wrong, and you are probably correct. -- Roger Hayter |
"Bal uhn" or "bayl uhn"?
On 8/2/2015 4:37 AM, Roger Hayter wrote:
rickman wrote: 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. At RF, the coax braid is an impenetrable Faraday screen. That is what it is for, after all. The current simply can't go through the fractional mm of copper from inside to outside. This is one of those cases where you have forgotten the details and underlying premises. A Faraday cage can't stop currents from flowing through the "cage". It stops fields from penetrating the cage by action of the resulting currents in the cage. Your car is largely a Faraday cage but you can still be electrocuted if a live wire is in contact with the chassis and you touch it while inside. There can be a high potential across different parts of the car from the current and that current can pass through you if you touch the cage. 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. Which is why I said it is a high impedance path for the current on the outside of the screen. "Common mode" isn't really quite the right name for this current, as the inner is totally uninvolved, but it is certainly not differential, which satisfies your point. Actually, I quoted you accurately above. You omitted the qualifier and that was my point. -- Rick |
"Bal uhn" or "bayl uhn"?
On 8/2/2015 4:37 AM, Roger Hayter wrote:
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? Just to interject a further argument; I think I agree with you on this. You can't get coax outer current unless some current sink (eg an antenna element) is connected to the inner conductor side of the load resistor. But, consider a pefectly symmetrical dipole: if the potential on the centre conductor and the braid is exactly the same, how can the two antenna halves have different currents to allow some to flow down the outside of the braid? This has been explained previously. A dipole is not balanced when it is connected to the coax. The shield outer surface presents a third element which makes the shield side of the dipole different from the center conductor side. In the case of the resistor the current flowing in one side must flow out the other, so it is balanced no matter what. The dipole has no such requirement. If you restrict the current running into one side and not the other it can do nothing about it. I am beginning to wonder if braid outside current with a symmetrical antenna *only* occurs when the coax outer interacts with the EM field of the antenna so as to actually alter the impedance of at least one of the antenna elements, or alter the two of them to a different extent, so that a common mode current is "left over" for the braid. It does seem likely that a long wire coming from the centre of the dipoe and not being absolutely symmetrical would have this effect. However, on this argument, you would not need a balun if your feeder was absolutely symmetrical! This theory seems eminently testable with antenna simulation programs. You seem to be thinking the antenna elements present some sort of current sink that *must* extract some amount of current no matter what. They are just loads no different from the shield outer surface. The current will flow according to the impedance seen by the current. There certainly will be some interaction, but if you use a triaxial cable and only connect the outer shield to ground at the transmitter, the antenna won't "see" the inner shield and the current will still flow on the outside of that shield. The symmetrical feeder argument does not make sense to me. Why wouldn't the impact on both antenna elements be identical? I haven't heard anyone say you *do* need a balun if the feed line is symmetrical. That would be a balbal transformer. -- Rick |
"Bal uhn" or "bayl uhn"?
On 8/2/2015 4:20 AM, Jeff wrote:
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. So if what you are saying is correct then terminating a piece of coax with a 50 ohm resistor will not cause a perfect match and the load will not be 50 ohms due to the shunt impedance to earth of the coax braid. I think not. Maybe you can draw a diagram and show yourself where the current flows. The earth connection on the shield outside is in parallel with the earth connection on the shield inside. You can't have any current flow on the outside because any current flowing on the shield inside balances with the current flowing in the center conductor. The resistor requires the two currents are equal with none left to travel down the shield outside. Although I am still waiting for jim to explain his idea that there is no current flow in the coax. If that is true it will make this a *very* simple picture. -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/1/2015 8:24 PM, wrote: 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. this is a pretty amazing revelation. So what are the assumptions to make this true? https://en.wikipedia.org/wiki/Transm...#Coaxial_cable Do you think there is significant current in the walls of a wave guide? If there is no appreciable current flow in the coax, then the resistance of the wires is of no significance. Funny, when I make a loop antenna from coax, the Q still seems to be limited by the conductor resistance. Odd... Not at all, it is just that you don't understand that in that case it is not a transmission line. 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 I don't see anything here that says there is no current flow in the coax. What it says is where the energy IS, not where it is NOT. If you want to understand the effect of ohmic resistance, read and understand the differential equations described he https://en.wikipedia.org/wiki/Transm....27s_equations At the end of the coaxial structure, the electromagnetic field becomes a current flow in any conductors connected to the end of the coax. Where do the electrons come from that the current consists of? Does the wire end act as a capacitor? Are you really asking this? Where do the electrons come from at the terminals of a receiving antenna? 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. That seems rather obvious. Then why do you keep asking about it? -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/1/2015 8:31 PM, wrote: rickman wrote: 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. I did; there is no current in the conductors of a coax where those conductors physically define a coaxial transmission line. https://en.wikipedia.org/wiki/Transm...#Coaxial_cable There is current in the conductors only after they no longer physically define a coaxial transmission line. Inside the coaxial structure everything is in the field and is balanced. Do you have any references that actually say there is no current flow in a coax? The reference given above doesn't even come close to saying this. What it says is where the energy IS, not where the energy is NOT. Read and understand the differential equations here and it becomes obvious: https://en.wikipedia.org/wiki/Transm....27s_equations -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/2/2015 4:20 AM, Jeff wrote: 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. So if what you are saying is correct then terminating a piece of coax with a 50 ohm resistor will not cause a perfect match and the load will not be 50 ohms due to the shunt impedance to earth of the coax braid. I think not. Maybe you can draw a diagram and show yourself where the current flows. The earth connection on the shield outside is in parallel with the earth connection on the shield inside. You can't have any current flow on the outside because any current flowing on the shield inside balances with the current flowing in the center conductor. The resistor requires the two currents are equal with none left to travel down the shield outside. Although I am still waiting for jim to explain his idea that there is no current flow in the coax. If that is true it will make this a *very* simple picture. https://en.wikipedia.org/wiki/Transm....27s_equations Read and understand the equations and it becomes obvious. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
On 8/2/2015 1:25 PM, wrote:
rickman wrote: On 8/1/2015 8:24 PM, wrote: 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. this is a pretty amazing revelation. So what are the assumptions to make this true? https://en.wikipedia.org/wiki/Transm...#Coaxial_cable Do you think there is significant current in the walls of a wave guide? You keep posting the same link which offers nothing to support your point. So I have to assume you don't have any reason to believe there is no current flow in the coax. You keep responding to questions with questions. If you actually understand this stuff, do you care to explain any of it? -- Rick |
"Bal uhn" or "bayl uhn"?
On 8/2/2015 1:28 PM, wrote:
rickman wrote: On 8/1/2015 8:31 PM, wrote: rickman wrote: 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. I did; there is no current in the conductors of a coax where those conductors physically define a coaxial transmission line. https://en.wikipedia.org/wiki/Transm...#Coaxial_cable There is current in the conductors only after they no longer physically define a coaxial transmission line. Inside the coaxial structure everything is in the field and is balanced. Do you have any references that actually say there is no current flow in a coax? The reference given above doesn't even come close to saying this. What it says is where the energy IS, not where the energy is NOT. What is the sound of one hand clapping... -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/2/2015 4:37 AM, Roger Hayter wrote: rickman wrote: 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. At RF, the coax braid is an impenetrable Faraday screen. That is what it is for, after all. The current simply can't go through the fractional mm of copper from inside to outside. This is one of those cases where you have forgotten the details and underlying premises. A Faraday cage can't stop currents from flowing through the "cage". It stops fields from penetrating the cage by action of the resulting currents in the cage. Your car is largely a Faraday cage but you can still be electrocuted if a live wire is in contact with the chassis and you touch it while inside. There can be a high potential across different parts of the car from the current and that current can pass through you if you touch the cage. DC certainly. VLF probably, certainly 60Hz! But any RF where the skin effect depth is much less than the thickness, no. This is probably iteself a field effect, but said field is inseparable from an RF voltage. 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. Which is why I said it is a high impedance path for the current on the outside of the screen. "Common mode" isn't really quite the right name for this current, as the inner is totally uninvolved, but it is certainly not differential, which satisfies your point. Actually, I quoted you accurately above. You omitted the qualifier and that was my point. Fair enough. -- Roger Hayter |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/2/2015 1:25 PM, wrote: rickman wrote: On 8/1/2015 8:24 PM, wrote: 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. this is a pretty amazing revelation. So what are the assumptions to make this true? https://en.wikipedia.org/wiki/Transm...#Coaxial_cable Do you think there is significant current in the walls of a wave guide? You keep posting the same link which offers nothing to support your point. So I have to assume you don't have any reason to believe there is no current flow in the coax. You keep asking the same question over and over when the answer is in the link I keep posting. Obviously you have not read and understood the link. You keep responding to questions with questions. If you actually understand this stuff, do you care to explain any of it? Since you don't seem to be able to deal with the self discovery methhod of instruction... There is no significant current in the walls of a wave guide. There is no significant current in the conductors inside of a coax transmission line. That such is true is shown by the equations in the link I have posted many times now. Do you see any error in any of those equations? -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/2/2015 1:28 PM, wrote: rickman wrote: On 8/1/2015 8:31 PM, wrote: rickman wrote: 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. I did; there is no current in the conductors of a coax where those conductors physically define a coaxial transmission line. https://en.wikipedia.org/wiki/Transm...#Coaxial_cable There is current in the conductors only after they no longer physically define a coaxial transmission line. Inside the coaxial structure everything is in the field and is balanced. Do you have any references that actually say there is no current flow in a coax? The reference given above doesn't even come close to saying this. What it says is where the energy IS, not where the energy is NOT. What is the sound of one hand clapping... Do you see any error in the equations in the link I provided that show where the energy IS? -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
On 8/2/2015 2:20 PM, Roger Hayter wrote:
rickman wrote: On 8/2/2015 4:37 AM, Roger Hayter wrote: rickman wrote: 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. At RF, the coax braid is an impenetrable Faraday screen. That is what it is for, after all. The current simply can't go through the fractional mm of copper from inside to outside. This is one of those cases where you have forgotten the details and underlying premises. A Faraday cage can't stop currents from flowing through the "cage". It stops fields from penetrating the cage by action of the resulting currents in the cage. Your car is largely a Faraday cage but you can still be electrocuted if a live wire is in contact with the chassis and you touch it while inside. There can be a high potential across different parts of the car from the current and that current can pass through you if you touch the cage. DC certainly. VLF probably, certainly 60Hz! But any RF where the skin effect depth is much less than the thickness, no. This is probably iteself a field effect, but said field is inseparable from an RF voltage. But the case we are talking about is not a cage keeping the RF out, it is a case of keeping the RF *in*. Will the skin effect somehow prevent the current of the inner shield surface from reaching the outer surface at the discontinuity? -- Rick |
"Bal uhn" or "bayl uhn"?
On 8/2/2015 1:31 PM, wrote:
rickman wrote: On 8/2/2015 4:20 AM, Jeff wrote: 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. So if what you are saying is correct then terminating a piece of coax with a 50 ohm resistor will not cause a perfect match and the load will not be 50 ohms due to the shunt impedance to earth of the coax braid. I think not. Maybe you can draw a diagram and show yourself where the current flows. The earth connection on the shield outside is in parallel with the earth connection on the shield inside. You can't have any current flow on the outside because any current flowing on the shield inside balances with the current flowing in the center conductor. The resistor requires the two currents are equal with none left to travel down the shield outside. Although I am still waiting for jim to explain his idea that there is no current flow in the coax. If that is true it will make this a *very* simple picture. https://en.wikipedia.org/wiki/Transm....27s_equations Read and understand the equations and it becomes obvious. This discussion is getting very tiresome. You keep posting links that do not support your statements. Where does this web page say there is no current flowing in a transmission line? "The telegrapher's equations (or just telegraph equations) are a pair of linear differential equations which describe the voltage and current on an electrical transmission line with distance and time. " That sure looks to me like it says current *does* flow. This section includes the equation describing the current. So what are you going on about? -- Rick |
"Bal uhn" or "bayl uhn"?
On 8/2/2015 1:31 PM, wrote:
rickman wrote: On 8/2/2015 4:20 AM, Jeff wrote: 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. So if what you are saying is correct then terminating a piece of coax with a 50 ohm resistor will not cause a perfect match and the load will not be 50 ohms due to the shunt impedance to earth of the coax braid. I think not. Maybe you can draw a diagram and show yourself where the current flows. The earth connection on the shield outside is in parallel with the earth connection on the shield inside. You can't have any current flow on the outside because any current flowing on the shield inside balances with the current flowing in the center conductor. The resistor requires the two currents are equal with none left to travel down the shield outside. Although I am still waiting for jim to explain his idea that there is no current flow in the coax. If that is true it will make this a *very* simple picture. https://en.wikipedia.org/wiki/Transm....27s_equations Read and understand the equations and it becomes obvious. Better yet, read down a few paragraphs you will find an animation of the current flow.... https://en.wikipedia.org/wiki/Transmission_line#Short -- Rick |
"Bal uhn" or "bayl uhn"?
On 8/2/2015 2:24 PM, wrote:
rickman wrote: On 8/2/2015 1:28 PM, wrote: rickman wrote: On 8/1/2015 8:31 PM, wrote: rickman wrote: 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. I did; there is no current in the conductors of a coax where those conductors physically define a coaxial transmission line. https://en.wikipedia.org/wiki/Transm...#Coaxial_cable There is current in the conductors only after they no longer physically define a coaxial transmission line. Inside the coaxial structure everything is in the field and is balanced. Do you have any references that actually say there is no current flow in a coax? The reference given above doesn't even come close to saying this. What it says is where the energy IS, not where the energy is NOT. What is the sound of one hand clapping... Do you see any error in the equations in the link I provided that show where the energy IS? We aren't discussion "energy" we are discussing current. Nothing you have provided says there is no current flow. In fact your own references clearly explain how to calculate the current and even provide an illustration. See my other post. -- Rick |
"Bal uhn" or "bayl uhn"?
On 8/2/2015 2:22 PM, wrote:
rickman wrote: On 8/2/2015 1:25 PM, wrote: rickman wrote: On 8/1/2015 8:24 PM, wrote: 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. this is a pretty amazing revelation. So what are the assumptions to make this true? https://en.wikipedia.org/wiki/Transm...#Coaxial_cable Do you think there is significant current in the walls of a wave guide? You keep posting the same link which offers nothing to support your point. So I have to assume you don't have any reason to believe there is no current flow in the coax. You keep asking the same question over and over when the answer is in the link I keep posting. Obviously you have not read and understood the link. You keep responding to questions with questions. If you actually understand this stuff, do you care to explain any of it? Since you don't seem to be able to deal with the self discovery methhod of instruction... There is no significant current in the walls of a wave guide. There is no significant current in the conductors inside of a coax transmission line. That such is true is shown by the equations in the link I have posted many times now. Do you see any error in any of those equations? Yes, I see an error in how you interpret the equations. Now, instead of discussing this in many places, why don't you respond to my post where I refer you to your own reference? -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/2/2015 2:22 PM, wrote: rickman wrote: On 8/2/2015 1:25 PM, wrote: rickman wrote: On 8/1/2015 8:24 PM, wrote: 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. this is a pretty amazing revelation. So what are the assumptions to make this true? https://en.wikipedia.org/wiki/Transm...#Coaxial_cable Do you think there is significant current in the walls of a wave guide? You keep posting the same link which offers nothing to support your point. So I have to assume you don't have any reason to believe there is no current flow in the coax. You keep asking the same question over and over when the answer is in the link I keep posting. Obviously you have not read and understood the link. You keep responding to questions with questions. If you actually understand this stuff, do you care to explain any of it? Since you don't seem to be able to deal with the self discovery methhod of instruction... There is no significant current in the walls of a wave guide. There is no significant current in the conductors inside of a coax transmission line. That such is true is shown by the equations in the link I have posted many times now. Do you see any error in any of those equations? Yes, I see an error in how you interpret the equations. Now, instead of discussing this in many places, why don't you respond to my post where I refer you to your own reference? What error do you see in the equations? -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/2/2015 2:24 PM, wrote: rickman wrote: On 8/2/2015 1:28 PM, wrote: rickman wrote: On 8/1/2015 8:31 PM, wrote: rickman wrote: 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. I did; there is no current in the conductors of a coax where those conductors physically define a coaxial transmission line. https://en.wikipedia.org/wiki/Transm...#Coaxial_cable There is current in the conductors only after they no longer physically define a coaxial transmission line. Inside the coaxial structure everything is in the field and is balanced. Do you have any references that actually say there is no current flow in a coax? The reference given above doesn't even come close to saying this. What it says is where the energy IS, not where the energy is NOT. What is the sound of one hand clapping... Do you see any error in the equations in the link I provided that show where the energy IS? We aren't discussion "energy" we are discussing current. This shows that either you did not read the equations or you do not understand them. What happens in a coaxial transmission line is described by mathematics, not arm waving bafflegab. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/2/2015 1:31 PM, wrote: rickman wrote: On 8/2/2015 4:20 AM, Jeff wrote: 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. So if what you are saying is correct then terminating a piece of coax with a 50 ohm resistor will not cause a perfect match and the load will not be 50 ohms due to the shunt impedance to earth of the coax braid. I think not. Maybe you can draw a diagram and show yourself where the current flows. The earth connection on the shield outside is in parallel with the earth connection on the shield inside. You can't have any current flow on the outside because any current flowing on the shield inside balances with the current flowing in the center conductor. The resistor requires the two currents are equal with none left to travel down the shield outside. Although I am still waiting for jim to explain his idea that there is no current flow in the coax. If that is true it will make this a *very* simple picture. https://en.wikipedia.org/wiki/Transm....27s_equations Read and understand the equations and it becomes obvious. This discussion is getting very tiresome. You keep posting links that do not support your statements. Where does this web page say there is no current flowing in a transmission line? What part of the equations tell where the energy is and now where the energy is not are you having trouble understanding? "The telegrapher's equations (or just telegraph equations) are a pair of linear differential equations which describe the voltage and current on an electrical transmission line with distance and time. " That sure looks to me like it says current *does* flow. This section includes the equation describing the current. So what are you going on about? I am not "going on about" anything, it is you that fails to understand what the referenced equations mean. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/2/2015 1:31 PM, wrote: rickman wrote: On 8/2/2015 4:20 AM, Jeff wrote: 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. So if what you are saying is correct then terminating a piece of coax with a 50 ohm resistor will not cause a perfect match and the load will not be 50 ohms due to the shunt impedance to earth of the coax braid. I think not. Maybe you can draw a diagram and show yourself where the current flows. The earth connection on the shield outside is in parallel with the earth connection on the shield inside. You can't have any current flow on the outside because any current flowing on the shield inside balances with the current flowing in the center conductor. The resistor requires the two currents are equal with none left to travel down the shield outside. Although I am still waiting for jim to explain his idea that there is no current flow in the coax. If that is true it will make this a *very* simple picture. https://en.wikipedia.org/wiki/Transm....27s_equations Read and understand the equations and it becomes obvious. Better yet, read down a few paragraphs you will find an animation of the current flow.... In the special cases of a shorted and open parallel transmission line, not a coaxial transmission line. The physics of a parallel line are different than the physics of a coaxial line. If you can't understand pictures, I guess there is no hope of you understanding mathematics. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
On 8/2/2015 2:58 PM, wrote:
rickman wrote: On 8/2/2015 1:31 PM, wrote: rickman wrote: On 8/2/2015 4:20 AM, Jeff wrote: 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. So if what you are saying is correct then terminating a piece of coax with a 50 ohm resistor will not cause a perfect match and the load will not be 50 ohms due to the shunt impedance to earth of the coax braid. I think not. Maybe you can draw a diagram and show yourself where the current flows. The earth connection on the shield outside is in parallel with the earth connection on the shield inside. You can't have any current flow on the outside because any current flowing on the shield inside balances with the current flowing in the center conductor. The resistor requires the two currents are equal with none left to travel down the shield outside. Although I am still waiting for jim to explain his idea that there is no current flow in the coax. If that is true it will make this a *very* simple picture. https://en.wikipedia.org/wiki/Transm....27s_equations Read and understand the equations and it becomes obvious. Better yet, read down a few paragraphs you will find an animation of the current flow.... In the special cases of a shorted and open parallel transmission line, not a coaxial transmission line. The physics of a parallel line are different than the physics of a coaxial line. If you can't understand pictures, I guess there is no hope of you understanding mathematics. Ok, I am throwing in the towel. You have out "jimmed" me. Enjoy... -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/2/2015 2:58 PM, wrote: rickman wrote: On 8/2/2015 1:31 PM, wrote: rickman wrote: On 8/2/2015 4:20 AM, Jeff wrote: 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. So if what you are saying is correct then terminating a piece of coax with a 50 ohm resistor will not cause a perfect match and the load will not be 50 ohms due to the shunt impedance to earth of the coax braid. I think not. Maybe you can draw a diagram and show yourself where the current flows. The earth connection on the shield outside is in parallel with the earth connection on the shield inside. You can't have any current flow on the outside because any current flowing on the shield inside balances with the current flowing in the center conductor. The resistor requires the two currents are equal with none left to travel down the shield outside. Although I am still waiting for jim to explain his idea that there is no current flow in the coax. If that is true it will make this a *very* simple picture. https://en.wikipedia.org/wiki/Transm....27s_equations Read and understand the equations and it becomes obvious. Better yet, read down a few paragraphs you will find an animation of the current flow.... In the special cases of a shorted and open parallel transmission line, not a coaxial transmission line. The physics of a parallel line are different than the physics of a coaxial line. If you can't understand pictures, I guess there is no hope of you understanding mathematics. Ok, I am throwing in the towel. You have out "jimmed" me. Enjoy... You mean I answered your bafflegab with fact and mathematics for which you have no reply. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/2/2015 2:20 PM, Roger Hayter wrote: rickman wrote: On 8/2/2015 4:37 AM, Roger Hayter wrote: rickman wrote: 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. At RF, the coax braid is an impenetrable Faraday screen. That is what it is for, after all. The current simply can't go through the fractional mm of copper from inside to outside. This is one of those cases where you have forgotten the details and underlying premises. A Faraday cage can't stop currents from flowing through the "cage". It stops fields from penetrating the cage by action of the resulting currents in the cage. Your car is largely a Faraday cage but you can still be electrocuted if a live wire is in contact with the chassis and you touch it while inside. There can be a high potential across different parts of the car from the current and that current can pass through you if you touch the cage. DC certainly. VLF probably, certainly 60Hz! But any RF where the skin effect depth is much less than the thickness, no. This is probably iteself a field effect, but said field is inseparable from an RF voltage. But the case we are talking about is not a cage keeping the RF out, it is a case of keeping the RF *in*. Will the skin effect somehow prevent the current of the inner shield surface from reaching the outer surface at the discontinuity? There should not be a discontinuity in the shield where the balun joins the feeder. (I suspect little would leak even if there was a small opening because of the symmetry with the centre conductor field, but that is more like the 50 ohm resistor load case.) -- Roger Hayter |
"Bal uhn" or "bayl uhn"?
On 8/3/2015 4:40 AM, Jeff wrote:
So if what you are saying is correct then terminating a piece of coax with a 50 ohm resistor will not cause a perfect match and the load will not be 50 ohms due to the shunt impedance to earth of the coax braid. I think not. Maybe you can draw a diagram and show yourself where the current flows. The earth connection on the shield outside is in parallel with the earth connection on the shield inside. You can't have any current flow on the outside because any current flowing on the shield inside balances with the current flowing in the center conductor. The resistor requires the two currents are equal with none left to travel down the shield outside. Indeed, as you state there is no current flow on the outer of the coax; the same situation exists for a perfectly balanced dipole attached directly to the coax, as far as the coax is concerned the situation is no different, all it sees is a perfectly matched load, just like the resistor, all of the power is dissipated in the antenna and none flows anywhere else including the coax outer. But there is a difference. With the resistor load the current flowing from the inner lead of the coax *must* (by Kirchoff's 1st law) balance the current flowing on the shield inner surface. So there is no current remaining to flow on the shield outer surface. No diagram needed. For the case of the dipole, the load is no longer balanced with the shield outer surface connected to one side and there is no requirement for the current in the two arms to be equal. So in this case current flows on the shield outer surface in an amount inversely proportional to the impedance seen by the current in the antenna element and the shield. -- Rick |
"Bal uhn" or "bayl uhn"?
On 8/3/2015 5:10 AM, Jeff wrote:
This has been explained previously. A dipole is not balanced when it is connected to the coax. The shield outer surface presents a third element which makes the shield side of the dipole different from the center conductor side. In the case of the resistor the current flowing in one side must flow out the other, so it is balanced no matter what. The dipole has no such requirement. If you restrict the current running into one side and not the other it can do nothing about it. No, a dipole is still balanced, the coax outer does not create a 3rd element other than by coupling. If the coax is taken off at right angles the coupling will be low as similar to both elements of the dipole. The dipole will never be perfectly balanced but can be a very close approximation. Ok, don't call the shield an element. A balanced antenna does not imply balanced current. The load of the antenna element on the coax will be equal, but the coax also has a parallel load from the shield outer surface. The two loads in parallel result in a different voltage on the end of the coax shield than on the end of the coax inner conductor. This different voltage causes the different current flow in the antenna element. A good illustration showing this can be seen near the top of the web page linked below. It your contention about the coax acting as a 3rd element were true then there would be severe distortion of both the impedance of the dipole and to it radiation pattern. This is not seen in practice and it is also demonstrable that there is little current flow on the coax outer when the dipole is well matched to the coax. I can't say for myself what happens in practice, but others here and on the web say there *is* severe distortion in the antenna pattern. See this link about halfway down the page. This is a EZNEC+ simulation. ymmv http://www.tomthompson.com/radio/EHa...ommonMode.html -- Rick |
"Bal uhn" or "bayl uhn"?
On 8/3/2015 4:49 AM, Jeff wrote:
This is one of those cases where you have forgotten the details and underlying premises. A Faraday cage can't stop currents from flowing through the "cage". It stops fields from penetrating the cage by action of the resulting currents in the cage. Your car is largely a Faraday cage but you can still be electrocuted if a live wire is in contact with the chassis and you touch it while inside. There can be a high potential across different parts of the car from the current and that current can pass through you if you touch the cage. If there are high different potentials between different parts of the car it is not a Faraday Cage. You are applying some rule you learned without understanding it. By ohms law there will be potential differences around the Faraday cage. Also assuming that it is a Faraday cage then touching it whist the cage is connected to a live conductor will not cause a shock, as everything is at the same potential. Only true if the conductor is perfect with no resistance. -- Rick |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/3/2015 4:49 AM, Jeff wrote: This is one of those cases where you have forgotten the details and underlying premises. A Faraday cage can't stop currents from flowing through the "cage". It stops fields from penetrating the cage by action of the resulting currents in the cage. Your car is largely a Faraday cage but you can still be electrocuted if a live wire is in contact with the chassis and you touch it while inside. There can be a high potential across different parts of the car from the current and that current can pass through you if you touch the cage. If there are high different potentials between different parts of the car it is not a Faraday Cage. You are applying some rule you learned without understanding it. By ohms law there will be potential differences around the Faraday cage. Also assuming that it is a Faraday cage then touching it whist the cage is connected to a live conductor will not cause a shock, as everything is at the same potential. Only true if the conductor is perfect with no resistance. Pedantically true but practically irrelevant; for a real world cage the voltage differences will be small fractions of a volt. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
Jeff wrote:
This has been explained previously. A dipole is not balanced when it is connected to the coax. The shield outer surface presents a third element which makes the shield side of the dipole different from the center conductor side. In the case of the resistor the current flowing in one side must flow out the other, so it is balanced no matter what. The dipole has no such requirement. If you restrict the current running into one side and not the other it can do nothing about it. No, a dipole is still balanced, the coax outer does not create a 3rd element other than by coupling. If the coax is taken off at right angles the coupling will be low as similar to both elements of the dipole. The dipole will never be perfectly balanced but can be a very close approximation. It your contention about the coax acting as a 3rd element were true then there would be severe distortion of both the impedance of the dipole and to it radiation pattern. This is not seen in practice and it is also demonstrable that there is little current flow on the coax outer when the dipole is well matched to the coax. Jeff The coax shield does create a 3rd element but the effect of it highly depends on on the length of the coax and whether or not the shield is grounded somewhere along the way. The effect can be anywhere from negligable and barely measurable to extremely significant and can be seen with an antenna modeling program. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
Jeff wrote:
So if what you are saying is correct then terminating a piece of coax with a 50 ohm resistor will not cause a perfect match and the load will not be 50 ohms due to the shunt impedance to earth of the coax braid. I think not. Maybe you can draw a diagram and show yourself where the current flows. The earth connection on the shield outside is in parallel with the earth connection on the shield inside. You can't have any current flow on the outside because any current flowing on the shield inside balances with the current flowing in the center conductor. The resistor requires the two currents are equal with none left to travel down the shield outside. Indeed, as you state there is no current flow on the outer of the coax; the same situation exists for a perfectly balanced dipole attached directly to the coax, as far as the coax is concerned the situation is no different, all it sees is a perfectly matched load, just like the resistor, all of the power is dissipated in the antenna and none flows anywhere else including the coax outer. Jeff That current flows on the outside of the shield is observable and measurable. How much flows is totally dependant on how long the shield is and whether or not it is grounded somewhere along the way. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
|
"Bal uhn" or "bayl uhn"?
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"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/3/2015 1:39 PM, wrote: rickman wrote: On 8/3/2015 4:49 AM, Jeff wrote: This is one of those cases where you have forgotten the details and underlying premises. A Faraday cage can't stop currents from flowing through the "cage". It stops fields from penetrating the cage by action of the resulting currents in the cage. Your car is largely a Faraday cage but you can still be electrocuted if a live wire is in contact with the chassis and you touch it while inside. There can be a high potential across different parts of the car from the current and that current can pass through you if you touch the cage. If there are high different potentials between different parts of the car it is not a Faraday Cage. You are applying some rule you learned without understanding it. By ohms law there will be potential differences around the Faraday cage. Also assuming that it is a Faraday cage then touching it whist the cage is connected to a live conductor will not cause a shock, as everything is at the same potential. Only true if the conductor is perfect with no resistance. Pedantically true but practically irrelevant; for a real world cage the voltage differences will be small fractions of a volt. "Your car is largely a Faraday cage but you can still be electrocuted if a live wire is in contact with the chassis and you touch it while inside." Why don't we conduct an experiment. You are saying it will only be a small fraction of a volt, we'll put you in the car. lol I notice you did not address what I actually wrote. Yet another red herring from the red herring master; I would never call a car a Faraday cage for a number of reasons. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
rickman wrote:
On 8/3/2015 1:44 PM, wrote: Jeff wrote: This has been explained previously. A dipole is not balanced when it is connected to the coax. The shield outer surface presents a third element which makes the shield side of the dipole different from the center conductor side. In the case of the resistor the current flowing in one side must flow out the other, so it is balanced no matter what. The dipole has no such requirement. If you restrict the current running into one side and not the other it can do nothing about it. No, a dipole is still balanced, the coax outer does not create a 3rd element other than by coupling. If the coax is taken off at right angles the coupling will be low as similar to both elements of the dipole. The dipole will never be perfectly balanced but can be a very close approximation. It your contention about the coax acting as a 3rd element were true then there would be severe distortion of both the impedance of the dipole and to it radiation pattern. This is not seen in practice and it is also demonstrable that there is little current flow on the coax outer when the dipole is well matched to the coax. Jeff The coax shield does create a 3rd element but the effect of it highly depends on on the length of the coax and whether or not the shield is grounded somewhere along the way. The effect can be anywhere from negligable and barely measurable to extremely significant and can be seen with an antenna modeling program. Let me summarize... a) No current on shield outer surface b) Current on shield outer surface C) Any conceivable combination of the above Your answer is c). Thanks for playing! Nope, just yet another knee jerk response. If you had actually read and understood what I wrote, the only possible answer is b. The only thing in question is the magnitude of the current; that there is a current is a given. -- Jim Pennino |
"Bal uhn" or "bayl uhn"?
On 8/3/2015 2:25 PM, wrote:
rickman wrote: On 8/3/2015 1:39 PM, wrote: rickman wrote: On 8/3/2015 4:49 AM, Jeff wrote: This is one of those cases where you have forgotten the details and underlying premises. A Faraday cage can't stop currents from flowing through the "cage". It stops fields from penetrating the cage by action of the resulting currents in the cage. Your car is largely a Faraday cage but you can still be electrocuted if a live wire is in contact with the chassis and you touch it while inside. There can be a high potential across different parts of the car from the current and that current can pass through you if you touch the cage. If there are high different potentials between different parts of the car it is not a Faraday Cage. You are applying some rule you learned without understanding it. By ohms law there will be potential differences around the Faraday cage. Also assuming that it is a Faraday cage then touching it whist the cage is connected to a live conductor will not cause a shock, as everything is at the same potential. Only true if the conductor is perfect with no resistance. Pedantically true but practically irrelevant; for a real world cage the voltage differences will be small fractions of a volt. "Your car is largely a Faraday cage but you can still be electrocuted if a live wire is in contact with the chassis and you touch it while inside." Why don't we conduct an experiment. You are saying it will only be a small fraction of a volt, we'll put you in the car. lol I notice you did not address what I actually wrote. Yet another red herring from the red herring master; I would never call a car a Faraday cage for a number of reasons. Neither is a coax shield. -- Rick |
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