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W5DXP > ANTENN 20.05.05 21:38l 37 Lines 1297 Bytes #999 (0) @ WW
BID : 4287817D$1_2
Read: GUEST DF5NK
Subj: Re: High Impedance Feedlines = Lower Loss - Why ?
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Sent: 050515/1805Z @:G8LMC.AMPR.ORG HamServ V2.66
Cecil Moore wrote:
> Reg Edwards wrote:
>
>> The complete equation is -
>>
>> Attenuation = R/2/Ro + G*Ro/2 Nepers
>>
>> where G is the conductance of the dielectric, which is small for
>> materials such as polyethylene and Teflon. And 1 Neper = 20/Ln(10) =
>> 8.686 dB.
>
>
> Reg, I didn't disagree with your equation. I disagreed with this
> statement of yours:
>
>> The number one reason for attenuation being higher is because the
>> conductor diameter is smaller and, as a consequence, its resistance is
>> higher.
As an illustrated example: Assume a parallel feedline made from #24
wire and having a characteristic impedance of 600 ohms.
What size would the wire in 50 ohm coax have to be to equal the HF
matched line loss of the #24 600 ohm line? (The wire in the coax has
to be 12 times as conductive as the wire in the parallel feedline in
order to offset the effect of Z0.)
A rough estimate indicates that the #24 600 ohm line has approximately
the same matched line loss as RG-213 with its #13 wire.
--
73, Cecil http://www.qsl.net/w5dxp
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