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"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"?
|
"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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