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Re: [Amps] Calculation of Tube Impedance

Subject: Re: [Amps] Calculation of Tube Impedance
Date: Sun, 2 May 2010 09:52:48 EDT
List-post: <">>
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.
Gerald K5GW
In a message dated 5/1/2010 8:56:23 P.M. Central Daylight Time, 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.




From: [] 
Sent: Samstag, 1.  Mai 2010 19:07
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 
of a slightly different loaded Q between the two  methods.


Gerald  K5GW

In a message dated 5/1/2010  10:45:48 A.M. Central Daylight Time,  writes:

Peter, you are right


-----Original  Message-----
From:  [] On
Behalf Of Peter van  Daalen

Isn't it the anode voltage swing instead of the anode voltage  ?
Pse discard if I'm wrong.

Peter, PE1E

----- Original  Message ----- 
From: "Phil Clements" <>
To:  "'Lee Buller'" <>; <>
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  

> 1
> amp, the output impedance would be 2222 ohms. (E  divided by the sum of 1 
> 1.8) The 1.8 is the K factor for the class  of operation.
> (((73)))
> Phil,  K5PC
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