[Amps] Calculation of Tube Impedance

[TexasRF at aol.com]

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 at t-online.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 at aol.com [mailto:TexasRF at aol.com] 
Sent: Samstag, 1.  Mai 2010 19:07
To: df3kv at t-online.de; petervandaalen at kpnmail.nl;  amps at 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 at t-online.de  writes:

Peter, you are right

73
Peter

-----Original  Message-----
From: amps-bounces at contesting.com  [mailto:amps-bounces at 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 at texascellnet.com>
To:  "'Lee Buller'" <k0wa at swbell.net>; <amps at 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 at contesting.com
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