-----Original Message-----
From: R. Measures [mailto:r@somis.org]
Sent: Monday, September 05, 2005 11:37 AM
To: m.ford
Cc: Gary Schafer; amps@contesting.com
Subject: Re: [Amps] CW and High Voltage
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
>
Richard L. Measures, AG6K, 805.386.3734. www.somis.org
I would agree with Rich. Also "full drive" may be all the exciter can put
out as is the case with many amp / exciter setups. In the case of a pair of
3-500s 100 watts of drive would indeed be full drive for the setup. As Rich
explains, those tubes will require around 140 watts for full performance. If
the exciter is capable of 400 watts output, then 140 watts would be full
drive with the setup noted.
73
Gary K4FMX
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