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[AMPS] L-Net calculations: Am I doing something wrong?

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Subject: [AMPS] L-Net calculations: Am I doing something wrong?
From: (Rich Measures)
Date: Wed, 17 Feb 1999 14:43:06 -0800

>>>>?  RL depends on the ZSAC.  

?  (Zero Signal Anode Current)

>>>>With higher idle current, RL decreases.  
>>>>Using RL=DCV/2 x I-anode, at 0.8A, RL=3.125k-ohms.   Designing for a Q of 
>>>>15 at 28MHz, a Pi-L with an intermediate Z of 300-ohms, XC1=208 ohms,  
>>>>C1=26pF.  L1=1.5uH.  L2=0.64uH.  C-load=129pF.  Subtracting 10pF for the 
>>>>C-anode + stray C, C-tune is 18pF.  The min. C of a UCS-300 is 5pF.  Have 
>>>>I missed something?
>OK, so I owe you a pizza!  
?   ok.  If you are ever in the area, we could do lunch.  

>Some guys use a 1.8 number as the multiplication factor for I-anode.  

?  With an 8877, or a similar triode, so would I.

>No matter, it still is much less than the 4600 Ohms I was using.
>Using your formula, the 4600 Ohms assumes an I-anode of c.500 mA.
>OK, so answer this then: Why am I having problems matching on 10m?  

?  I would have to measure some things to figure it out.  

>Why does my power output top out at 800 Watts?  Once I get over 60 watts of 
>drive, I get no more power out no matter what I do.  

?  Sounds like a tuned input problem.  What is C1?

>In looking at everything I was assuming it was becasue Q was too high.
>What light can you shed on this?
>Also, how do you determine your peak anode current.  

?  look at the constant-current characteristic curves.  For good 
linearity, pick the highest peak-current curve that has a defined knee at 
the anode/plate potential equal to the screen-supply voltage.  For the 
8166/4-1000A, the highest  such curve is 2.6 peak-amps.  Divide 
peak-current by 3 to obtain avg. current.  This is how I arrived at the 
c. 0.8A figure.  

>The current will 
>change depending on what amount of drive you have applied to the amp.  So 
>I what I think you are saying is that you would need to drive the amp so 
>that the plate current is at 0.8 A.  But then if you decrease the drive, 
>you would change the impedance and hence the match.  Correct?
?  yes x 3.  The guy who wants to take it easy on his linear amp. and 
tunes up with reduced drive, gets a mistune every time.  .  . 

-  cheers, Jon


R. L. Measures, 805-386-3734, AG6K,  

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