We seem to have lost the thread somewhere along the line.
The original question I posed was to do with the possibility of making an
output network which ALWAYS provided the correct match to the tube,
irrespective of power output. As you will see from the various
calculations, at low power output the RL required is considerably higher
than at full output. I realise that the question is a bit academic and
probably too difficult to achieve in practice. But the concept of
adjusting the tank network so that it always provides the optimum RL for
the power level in use fascinates me and I wondered if anyone had attempted
it.
Why does it fascinate me? Well it has practical applications for an
amplifier which has to be run from a limited power budget where the
efficiency is critical.
The average amplifier suffers severe inefficiencies when operated at a
power output well below its maximum. Also I intuitively believe that the
linearity must be improved if the anode RL is correctly matched to the
output network. Anyone come across references to this topic?
John ZS5JF
----------
From: Rich Measures <measures@vc.net>
To: John Fielding <johnf@futurenet.co.za>; Amps Contesting
<amps@contesting.com>
Subject: Re: Fw: [AMPS] Anode load line calculations.
To: <amps@contesting.com>
Date: 21 December 1998 08:52
>Both - but what's the difference? An amplifier set up to give 100W using
a
>2000V supply or a 1kW amp using the same ht have the same requirement.
? The requirement is seemingly set by the conditions under which the
amplifier is tuned up. Assuming 60% effiency, if a 1666w input, w. 2000v
anode supply, 1000w-out amplifier is, , tuned up at 1666w input, approx
0.833a of anode current is required. Therefore RL is roughly. 2000v/(2 x
0.833a) = 1200 ohms and the Pi-network tank circuit must be tuned to
match 50 ohms to 1200 ohms. Assuming that the drive level is susequently
reduced until the output is 100w, the anode-V swing is reduced and the
anode-I swing is also reduced. In my opinion, when the output of this
amplifier is 100w, RL does not become 22k-ohms. . However, if it were
decided to tune up the amplifier for operation at 100w out, RL would
increase to roughly 22k-ohms.
>The point I was trying to make, and which you are conveniently ignoring,
is
>that the RL in either case is radically different if you use the formula I
>posted previously.
>
Semi-agreed. . However, the RL formula is an approximation that applies
for the power level at which an amplifier is tuned. When drive is
reduced to a tenth, RL does not increase ten-fold. .
- later, John
......................................................................
>> From: Rich Measures <measures@vc.net>
>> To: John Fielding <johnf@futurenet.co.za>; Amps Contesting
><amps@contesting.com>
>> Subject: Re: Fw: [AMPS] Anode load line calculations.
To: <amps@contesting.com>
>> Date: 21 December 1998 03:39
>>
>>
>> >
>> >But Rich you misunderstand what I am trying to tell you and the
>others!!!!
>> >
>> >The anode voltage swing WILL be the 80% or so of the dc supply voltage
-
>> >but it is now developed across a much higher impedance of 22800 ohms.
>>
>> ? John: Are we talking about a 100w amplifier using a 2000v anode
>> supply, delivering 100w - - or are we talking about a 1000w amplifier
>> using a 2000v anode supply, delivering 100w because of reduced drive?
>> - thanks
>> >.......
>>
>>
>> Rich...
>>
>> R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
>>
>>
>> --
>> FAQ on WWW: http://www.contesting.com/ampfaq.html
>> Submissions: amps@contesting.com
>> Administrative requests: amps-REQUEST@contesting.com
>> Problems: owner-amps@contesting.com
>> Search: http://www.contesting.com/km9p/search.htm
Rich...
R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
----------
--
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