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[AMPS] MFJ-259 and matching circuits

To: <amps@contesting.com>
Subject: [AMPS] MFJ-259 and matching circuits
From: W4EF@pacbell.net (Michael Tope)
Date: Wed, 3 Feb 1999 11:31:48 -0000
Rich and Jon,

Would it not be possible to tune the circuit for min VSWR
using the exciters VSWR meter, as Rich suggested (real 
world operating conditions), then turn off the amp, terminate
its input port in 50 ohms and measure the impedance of the filter
looking back from the cathode with the MFJ-259B (cathode disconnected
of course)? This impedance should be approximately equal to the 
complex conjugate of the large signal cathode impedance. It 
would be interesting to see if this number comes close to Jon's
estimate of 110 ohms -j (2*pi*f*27pf)^-1.  

73 de Mike W4EF 

P.S. Somebody smarter than I could probably figure the error
bars for the impedance based upon the VSWR at the amplifier
input.
----------
From:   Rich Measures[SMTP:measures@vc.net]
Sent:   Wednesday, February 03, 1999 6:28 AM
To:     'amps'
Subject:        RE: [AMPS] MFJ-259 and matching circuits




>>?  Assuming that a g-g input is like a resistor is bound to bring some, 
>
>OK for all of you who doubt it then.  What other assumption can one make?  
>The tube is spec'd for an "average" driving impedance of 110 Ohms with an 
>input capacitance of 27 pF.  Last time I looked, Ohms was a resistance 
>value (hence using a resistor) and pF was a capacitance value (hence using 
>a capacitor).  Sure there could be some inductance in there too.  However, 
>the resistors and cap had some lead length which add a little bit of 
>inductance as well.


>So......if you have all 3 passive sorts of components covered.....what 
>else could it be???
>
?  The cathode is like a diode that is driven with pos. and neg. 
potentials.  Heavy current flows on neg.  No current flows on pos.  A 
resistor this is not.  

>Now, I KNOW that in a GG circuit, the input impedance varies over the 
>drive cycle.  But you have to have some sort of impedance that you use to 
>create your input match.  A pi-net or an L-net or a T-net match one 
>impedance to another in addition to being like the venerable flywheel and 
>storing energy.  So one end is 50 Ohms, pray tell to all you out there 
>what the other end is?  Well, I used what the data sheets say is the 
>driving impedance which equals 110 Ohms.  So on paper, you use 110 Ohms in 
>parallel with 27 pF (which on any data sheet you find is spec'd as the 
>input capacitance).  You absorb the 27 pF into the value of C2 in your 
>Pi-net and then solve the equation for a pi matching network.  

?  110 ohms could be close.  In this case, the swr meter in the radio 
will work as well as the MFJ-259.   
>
>Now if what they taught me about matching networks in college is all 
>wrong, I would like to know.  

?  In college we learned how to match resistive loads.  A cathode is not. 
 

>I KNOW that there are some additional real 
>world effects inside the tube that I can't completely account for by 
>simulating it, but it should get one relatively close.  Otherwise what 
>good is working out networks by any form of calculation.  Should all of 
>our engineering be hit or miss?
>
?  Not at all.  Set the input pi network Q for 2.  (XC1= 25-ohms).  With 
full drive, adj. L and C2 for a match to the actual cathode.  
>
-  later, Jon


Rich...

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


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