Is a wire dipole a series circuit or not?

Discussion in 'General Electronics Chat' started by KL7AJ, Jul 21, 2009.

  1. KL7AJ

    Thread Starter AAC Fanatic!

    Nov 4, 2008
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    In a recent thread on simple series circuit the question was asked about why the current in each component of a series circuit HAD to be equal. Lots of good answers given. But it also raised a good question about how you explain a simple dipole antenna (assuming at perfect resonance) where the current is NOT equal in each segment of the wire, but follows a classic sinusoidal current distribution.

    Now, most of us old R.F. geezers grasp this pretty intuitively....we immediately recognize that any time you DO have a circuit like this that doesn't seem to follow ohm's law, you have an antenna....almost by definition. We know this, and it's pretty much second nature, but it still doesn't explain where the alternate paths actually exist...by ohm's law (or even KCL), the currents leaving each section of wire have to equal the currents entering each section of wire...but we know that they don't....we can measure this "aberration" with an r.f. current meter or light bulb. There HAS to be a parallel path somewhere...or actually MANY of them. What gives?

    We know that the mysterious entity Radiation Resistance is actually modeled by a single resistor in series with the antenna's reactive components. But this equivlalent circuit is actually misleading.

    Radiation resistance actually manifests itself as an infinite series of resistors IN PARALLEL with the "real conductor', i.e., the wire. These resistors are progressively smaller in value as you approach the center of the antenna from either end. We actually DO have multiple parallel paths, but these paths are space itself.

    Interestingly enough, if the antenna is self resonant, we will find that the current and voltage at ANY point along the antenna are precisely in phase! In other words, the power factor at any point on the antenna is 1, even though the phasing between voltage and current, relative to DISTANCE is 90 degrees.

    It's stuff like this that makes me wonder if we should be teaching antenna theory BEFORE teaching lumped constant theory.....the latter will be very easy to grasp afterwards. :)

    eric
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    We had two hams in the Electronic Instrument Lab, but still referred to radio as "fm". After sweating a J pole antenna out of 1" copper pipe where only the top 3' was active in a continuous length of 8', well it was strange to see it actually function as something besides a dead short.

    By the way, a good heat gun will sweat pipe as well as a torch.
     
  3. Tesla23

    Active Member

    May 10, 2009
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    Thought provoking post, but I'd suggest that the reason for the changing current along the dipole is the changing charge distribution along the dipole. If the current density is sinusoidal along the antenna (and exp(jωt) in time) then the charge density is cosinusoidal along the antenna (and also exp(jωt) in time). This can be derived from the equation of continuity. This is the charge distribution that terminates the E lines emanating from the antenna.

    Here is a nice animation of the E lines from a half-wave dipole (from here), note that many terminate on the antenna, for this to occur you need an oscillating charge distribution on the antenna.


    [​IMG]

    Circuit theory concepts really fall apart when the circuit elements have lengths that are appreciable fractions of a wavelength.
     
  4. KL7AJ

    Thread Starter AAC Fanatic!

    Nov 4, 2008
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    Cool animation! I think it belongs in the Ebook!

    eric
     
  5. steveb

    Senior Member

    Jul 3, 2008
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    I agree on both comments.

    Kudos to KL7AJ. This is an excellent thought provoking question. It belongs in every text book on EM field theory. I never considered this before.

    Also, your answer makes sense to me. In that other thread, I made a comment that if the current is not equal in a wire, then charge would have to build up somewhere. I then said that charge can't build up in a good conductor, hence the current is equal. However, this is of course only valid in the context of circuit theory.

    For an antenna, the device length is comparable to the wavelength. This means that propagation delays are significant and one end of the antenna can not immediately respond to what the other end is doing. Also, charges are significantly accelerating when there is radiation. Hence, if there is delay and acceleration, charge buildup on a conductor is the result, no matter how good the conductivity is.
     
  6. Tesla23

    Active Member

    May 10, 2009
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    One other example that comes to mind is a standing wave on a transmission line. Here you don't have any radiation (well not if the line is shielded - e.g. coax), but you have a conductor where neither the voltage is constant (minimum at the nodes) nor the current (maximum at the nodes).

    As Steve says - kudos for making us think.
     
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