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Re: [Amps] every db lost re Tubes vs. Solid State

To: "amps@contesting.com" <amps@contesting.com>
Subject: Re: [Amps] every db lost re Tubes vs. Solid State
From: Jim Brown <jim@audiosystemsgroup.com>
Reply-to: jim@audiosystemsgroup.com
Date: Wed, 02 May 2012 09:45:17 -0700
List-post: <amps@contesting.com">mailto:amps@contesting.com>
On 5/1/2012 2:05 PM, Larry Benko wrote:
> Jim,
>
> I have made the exact measurement I mentioned below as an exercise.  
> Are you questioning the 2:1 SWR or the fact that a 2"1 SWR is a 
> mismatch loss of .55dB?

Both.  See the family of curves in the ARRL Handbook for excess 
attenuation due to mismatch, which has been in every edition of the 
handbook since I've been buying it (the 50s), and which I've confirmed 
by cranking the equations. The curve for a 2:1 SWR shows an excess 
attenuation of 0.1 dB if the matched loss is 0.4dB, 0.2dB if the matched 
loss is 1dB, 0.4dB if the matched loss is 10dB. And remember, the 
connector is quite small as a fraction of a wavelengh at HF, which is 
what this discussion is about, and not much more in the lower half of 
the VHF spectrum.  If you've measured more than that, I suspect you are 
simply seeing the standing waves at the point of the measurement.

Very small values of attenuation can be VERY difficult to measure.  
Several years ago, I tried to measure the attenuation of Commscope 3227 
from 1 MHz to 1 GHz   Getting good data below a few MHz requires a VERY 
long sample -- even with 1,000 ft, there is some mismatch between the 50 
ohm resistive terminations of the HP gear and the complex Zo of the 
cable at those frequencies, which is NOT 50+j0.

Another point. While we like to ASSUME that input stages have an ideal 
resistive impedance that matches the cable we're using, many do not.  
Ditto for output stages -- indeed, the Zo of most output stages is 
nowhere near an ideal match for the cable.

73, Jim K9YC


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