I think it is Plate V divided by (Plate I times the "K" factor) or
Resistance Anode = V/(I x 1.8) some people use 1.87
Lee
K0WA
________________________________
From: "TexasRF@aol.com" <TexasRF@aol.com>
To: df3kv@t-online.de; petervandaalen@kpnmail.nl; amps@contesting.com
Sent: Sat, May 1, 2010 12:06:44 PM
Subject: Re: [Amps] Calculation of Tube Impedance
While it is the plate voltage swing that generates the rf power, it is
customarily accepted that the k factor takes care of that issue. Plate V times
plate I divided by k factor gives the complete answer.
In actual practice, either method gets you close enough with the
consequence of a slightly different loaded Q between the two methods.
73,
Gerald K5GW
In a message dated 5/1/2010 10:45:48 A.M. Central Daylight Time,
df3kv@t-online.de writes:
Peter, you are right
73
Peter
-----Original Message-----
From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com] On
Behalf Of Peter van Daalen
Isn't it the anode voltage swing instead of the anode voltage ?
Pse discard if I'm wrong.
73,
Peter, PE1E
----- Original Message -----
From: "Phil Clements" <philc@texascellnet.com>
To: "'Lee Buller'" <k0wa@swbell.net>; <amps@contesting.com>
Sent: Thursday, April 29, 2010 9:57 PM
Subject: Re: [Amps] Calculation of Tube Impedance
snip...(Peter )
In the case of a tube running
> AB2, (grounded grid) with an anode voltage of 4000 volts, and a current
of
> 1
> amp, the output impedance would be 2222 ohms. (E divided by the sum of 1
X
> 1.8) The 1.8 is the K factor for the class of operation.
>
> (((73)))
> Phil, K5PC
>
>
>
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