On Sep 5, 2005, at 5:22 AM, m.ford wrote:
>
> ----- Original Message -----
> From: "Gary Schafer" <garyschafer@comcast.net>
> To: "'Steven Cook'" <sccook1@cox.net>; "'Phil Clements'"
> <philc@texascellnet.com>; <Amps@contesting.com>
> Sent: Monday, September 05, 2005 1:20 AM
> Subject: Re: [Amps] CW and High Voltage
>
>
>> Best efficiency is obtained when the plate tune and load caps are
>> tuned
>> for
>> maximum output with a given amount of drive. Tuning up at maximum
>> power
>> with
>> full drive as is normally done gives best efficiency at that power
>> level.
>> Now if you reduce drive, power out drops and efficiency also drops.
>
> True for any machine be it electrical, mechanical or chemical.
> However, I am
> having difficulty with the term "full drive". Is it defined as the
> full
> output of the exciter or the maximum input rating of the tube or is it
> just an
> arbitrary term
> term used by hams?
I define "full drive" as what is required to bring the cathode-I up to
the point where linearity is on the verge of decreasing. For a young
3-500Z, this is c. 420mA. If the anode-V is c. 3000, this roughly 70w
per cathode.
> Most of my experience with testing amplifier efficiency was
> done at input levels up to the 1db compression point.
>
> Mike k1ern
>
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>
>
Richard L. Measures, AG6K, 805.386.3734. www.somis.org
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