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[TowerTalk] trap loss

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Subject: [TowerTalk] trap loss
From: ae110@rgfn.epcc.Edu (Ed E Jensen)
Date: Mon, 3 Mar 97 11:03:26 MST

Hi all,

Using XL = 2pi*f*L and R=XL/Q, I get a little less than 1 ohm of
loss resistance in the traps using the measurements posted here.  This
neglects radiation from the trap.  

Power loss = I**R and we know R but what is I?  

IF we knew the current in the element we could multiply it
by Q (240 in this case) to get the circulating current in the trap.  Now
there is a about 4.5 amps at the center of a 50 ohm antenna getting 1000
watts but what is it half way out where the 10m trap is?  If the trap were 
not there it would be down to .707*I or about 3 amps.  But using P=(QI)**2(R) 
gives (240*3)**2*1 = 518kW, not too likely.  

So what is the current in the element where it connects to the trap?
Anybody got a RF ammeter they can insert in an element?

73, Ed

Ed Jensen, K5ED, El Paso, TX (

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