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Re: [Amps] Loaded output circuits

To: amps@contesting.com
Subject: Re: [Amps] Loaded output circuits
From: "Bill, W6WRT" <dezrat1242@yahoo.com>
Reply-to: dezrat1242@yahoo.com
Date: Sun, 30 Aug 2009 18:43:53 -0700
List-post: <amps@contesting.com">mailto:amps@contesting.com>
On Sun, 30 Aug 2009 11:27:39 -0600, John Lyles <jtml@losalamos.com> wrote:

>Once a matching network 
>is built in metal or discrete components like capacitors & inductors 
>then it has a fixed transformation ratio, PERIOD.

Correct, although W4TV disagrees.


> If the operating point 
>of the active device (called 'plate impedance' for tubes) varies, this 
>network is no longer optimal.

Correct


> Reduction of drive power into a linear 
>amplifier changes Rp. If the plate voltage is reduced the proper amount 
>(this is easily calculated) then the same network becomes optimal again. 

Correct, but not relevant to the original question. 


>If no change is made, however, the network may no longer be terminated 
>in the proper R. As Joe explained, the circulating current can rise 
>significantly. 

It can only rise if you RETUNE THE TANK CIRCUIT. The original question was about
reducing drive WITHOUT RETUNING THE TANK CIRCUIT.

Certain keys on my keyboard are showing excessive wear, especially the ones that
spell out WITHOUT RETUNING. What is it about the original question that makes
everybody want to reduce power, RETUNE and then answer the question?

73, Bill W6WRT
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