[AMPS] Impedance of amplifiers

[Michael Tope]

Rich, et al:

Not knowing anything about this, I would guess that
a first approximation would be to take the output power
of the amp in watts and then work backwards thru the pi 
network using simple linear circuit analysis to calculate 
the input current phasor. Since all this current has to 
flow thru the blocking cap, you get your "first order" 
answer from this. Of course, this ignores any circulating 
currents due to harmonics which may be significant as the 
PI network should look like a decreasingly low impedance 
as the harmonic number increases (viz. the shunt tuning cap). 

73 de Mike, W4EF......

"I am not an amp builder, but I play one on the internet."




----------
From: 	Rich Measures[SMTP:measures@vc.net]
Sent: 	Sunday, July 18, 1999 12:44 AM
To: 	W8JI@contesting.com; amps@contesting.com
Subject: 	Re: [AMPS] Impedance of amplifiers




>
>> ?  For c. five years, the ARRL Handbook had the wrong Pi-L tank values - 
>> -  despite being repeatedly told about the problem.  .
>
>Is it easy to explain what is wrong about the values? 
>
?  The error was in C2,  Eventually it was fixed. 

>  > since 1995, the Handbook has had an inaccurate method of calculating RF
>> current in a DC blocking cap.  My guess is that it will remain there until
>> sometime around 2040.  .  
>
>What is the correct method?

? Calculate  the difference between instantaneous anode potential at max. 
conduction and anode supply potential.  Divide by 2^0.5 to convert peak v 
to RMS v. .  Disconnect the C1 side of the blocking cap.  Measure 
anode-C. with tube in socket  Calculate capacitive-reactance in ohms.  
Current=volts/ohms.    

?  later, Tom


Rich...

R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures  


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