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Re: [TowerTalk] Fwd: current balance in ladder line?

To: towertalk@contesting.com
Subject: Re: [TowerTalk] Fwd: current balance in ladder line?
From: jimlux <jimlux@earthlink.net>
Date: Wed, 19 Jan 2011 12:07:10 -0800
List-post: <towertalk@contesting.com">mailto:towertalk@contesting.com>
On 1/19/11 9:53 AM, Hans Hammarquist wrote:
>
> You can, if you want, view the dipol as two pieces, each with one pole at the 
> feeder and the other pole at ground. If the the two part are equal, as we 
> intend to have them, the two "hot" poles, will have the same current. If, on 
> the other hand, we have an off-center feed the impedances will be unequal and 
> the only way to keep the current in he two poles equal is to "force" it with 
> an isolating balun such as a current choke.
>
> The current between the antenna wire and ground is, in scientific term, 
> called "displacement current". You actually have a current without a 
> conductor. That's the same type of current you have through a capacitor.
>
> Hans - N2JFS
>

Only if the feedline is part of the circuit.  If you had a asymmetric 
dipole, and a tiny battery powered transmitter at the feedpoint, the 
currents in the (only) two terminals of the transmitter would be equal 
and opposite.

What you describe with displacement currents is where the "antenna" 
includes the feedline, and you have the circuit loop is ground: rig: 
feedline: antenna: displacement current to ground.

It's when you start putting the feedline  (or really, any conductor in 
the vicinity) into the system that the issues with asymmetry arise.




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