Hi Gerald,
Sure, it is not critical at all within limits.
But the results will be quite different when tubes with high screen voltage
are selected for an amplifier.
Just to answer the question by Peter, PE1E ; indeed the more correct rough
formula is: voltage swing / Ia*k.
The k factor is NOT taking care of the voltage swing but is the factor
considering the conduction angle (Pi/2 for AB1)
To have it perfect the tube curves plus the exact formulas are necessary to
figure it all out.
The Eimac performance computer will get it accurate enough also.
73
Peter
_____
From: TexasRF@aol.com [mailto:TexasRF@aol.com]
Sent: Samstag, 1. Mai 2010 19:07
To: df3kv@t-online.de; petervandaalen@kpnmail.nl; amps@contesting.com
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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> Amps mailing list
> Amps@contesting.com
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