Hi Peter, Since you brought the conduction angle into the discussion, the
Pi/2 number equates to 1.57. The customary shortcut is use of 1.8 for the K
factor and I am assuming the difference between 1.57 and 1.8 is due to the
plate voltage swing limitation.
The ratio of 1.57 to 1.8 is .872 and applied to a typical 4000v scenario
that would equate to 3488v. That leaves 512v as a minimum Ep and that is
within the range of expectations.
Of course as you say, the tube calculator is the ultimate method to do his
and removes all of the ambiguities from the calculations.
Most of my experience is at VHF and UHF applications. At those frequencies,
we don't get a choice of operating Q at all. What you see is what you get
and to some extent simplifies design techniques. On the other hand, there
are no shortcuts to preserving the plate (and cathode) resonator Q if you
want any power output. Different skills come into play for that.
As always, I appreciate and enjoy your comments.
73,
Gerald K5GW
In a message dated 5/1/2010 8:56:23 P.M. Central Daylight Time,
df3kv@tonline.de writes:
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@tonline.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@tonline.de writes:
Peter, you are right
73
Peter
Original Message
From: ampsbounces@contesting.com [mailto:ampsbounces@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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