-----Original Message-----
From: Gilmer, Mike <mgilmer@gnlp.com>
To: 'AMPS' <amps@contesting.com>
To: <amps@contesting.com>
Date: 16 July 1999 16:48
Subject: [AMPS] Power Transfer
>
>I'll risk sticking my neck out again into the battle
>(I learned something last time)
>
>
>>>With half of the total P being dissipated in the
>>>generator, and half of the power being delivered
>>>to the load R, the efficiency is obviously 50%.
>
>
>Is it being suggested, in essence, that if an amplifier
delivers (as per
>Bird wattmeter) 1500W RF to the coax, that due to the power
transfer
>rule, the amp's tubes must be generating at least 3KW RF?
My HV and
>plate current suggest otherwise.
>
>Or am I missing something?
>
>Mike
No, Mike you are not. It's true to state that maximum
possible power transfer occurs when source and load
resistances are equal (there are also caveats about reactive
matching where rf is concerned, but basically the same rule
applies). The point is that it is not necessary to work with
equal source and load resistances.
As a simple example, consider your ac line - the source
resistance is probably a few hundreths of an ohm - how much
does the voltage sag when you turn on a kettle? The most
power you can possibly, theoretically, get out of the line
is when the load you attach has the same few hundreths of an
ohm - on a 200V line, you will get several 100kW in your
load and there will be the same dissipation in the source
resistance so the efficiency is 50%. In practice you can't
go that far - you put a 20 ohm load on the line for your 2kW
kettle, the power lost in the source resistance is tiny and
the efficiency is 99.99%.
Basically the same applies in high power audio and rf
amplifiers - the output impedance of the amplifier (the
equivalent to source resistance in the example above) is
often not equal to the load you attach to it.
A complicating factor is that when the PSU volts start to
droop, or the tube runs out of emission, the output
impedance changes.....
Steve
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