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Re: Topband: Shunt fed towers and common mode chokes

To: <topband@contesting.com>
Subject: Re: Topband: Shunt fed towers and common mode chokes
From: "Tom W8JI" <w8ji@w8ji.com>
Reply-to: Tom W8JI <w8ji@w8ji.com>
Date: Fri, 7 Dec 2012 10:29:47 -0500
List-post: <topband@contesting.com">mailto:topband@contesting.com>
Right. 5K Ohms is a good rule of thumb for choking Z to prevent noise coupling from feedline to antenna, and if the antenna is reasonably close to balance, is also enough from the point of view of dissipation.
This is where we can all get into trouble. We need to carefully read, 
consider, and understand the particular situation or problem, and not just 
apply a universal rule to systems that have extremes of requirements. The 
result is always far from optimal results, almost always wasteful, and 
sometimes even deleterious.
One size or method always gains popularity because it is an easy sell, no 
one has to think because a complex extremely wide range of problems are 
portrayed as simple with a simple universal answer anyone can duplicate. 
Everyone loves a simple rule........even if 99% of the time it is 
impossible, wasteful, or wrong.
This problem shows how not considering common mode voltages or impedances 
can get us deep into trouble.
In this case the original poster has a two-dipole 80-meter antenna attached 
high on a tower. He needs to isolate common mode 160 meter tower antenna 
currents and voltages from that array.
1.) He clearly stated electrical length (and by definition that also means 
impedance) of the transmission lines to the switch was critical, because 
open circuit reactance of the line tunes the unused element as a director.
2.) Worse yet, the antenna connection or isolation point is almost certainly 
at a very high voltage point.
Assuming this array is at 100 feet on a typical 130-foot tower, and we 
somehow can use universal 5000-ohm chokes, EZNEC predicts the following for 
each choke:
Frequency = 1.8 MHz

Load 1        Voltage = 1582 V at 90.73 deg.
             Current = 0.3164 A at 90.73 deg.
             Impedance = 5000 + J 0 ohms
             Power = 500.6 watts

Load 2        Voltage = 1582 V at 90.73 deg.
             Current = 0.3164 A at 90.73 deg.
             Impedance = 5000 + J 0 ohms
             Power = 500.6 watts

             Total applied power = 1500 watts

             Total load power = 1001 watts
             Total load loss = 4.781 dB


Get ready for some serious core heating!!!!!

It isn't just this system either. Many systems will either fail to work (from heat) or seriously waste materials if we simply decide 5000 ohms fixes everything.
73 Tom 

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