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"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 |
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