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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: (Jon Ogden)
Date: Mon, 15 Feb 1999 20:23:14 -0600

In order to match the high impedance of the 4-1000A to a lower impedance 
so that I have better loaded Q in my tank circuit, I decided to take the 
advice many people had given me and that is to make an L-network to step 
the impedance down to a more workable level.  This L-network would 
consist of the output capacitance of the tube as the shunt (parallel) 
element and a small inductor as the series element.

OK, so at 5KV, the 4-1000A's plate impedance is 4600 Ohms = R1 (from 
Eimac Specs)
The Cplate of the 4-1000A is 7.6 pF (from the ARRL handbook)

Let's choose L (the series inductor) to be 1 uH (a suggested value by 

What secondary impedance do we obtain?


In an L-net Q=R1/Xp

At 28 MHz: Xp=1/(2*pi*28E6*7.6E-12)= 747.91 Ohms

Therefore Q=4600/747.91 = 6.15

R2 = Xs/Q  (R2 = output impedance)

At 28 MHz: Xs=2*pi*28E6*1E-6 = 175.9 Ohms

R2 = 175.9/6.15 = 28.6 Ohms

Now 28.6 Ohms is unrealistically small in order to properly get a good 
match with realistic components.

OK, so what would a good R2 be?  Let's try for 2500 Ohms as this will 
give decent Qs with the component values I have for the tank circuit.


Q is still 6.15 since neither R1 nor Xp have changed.

Xs = R2*Q = 2500*6.15 = 15375 Ohms

At 28 MHz: Ls = Xs/(2*pi*28E6) = 15375/(2*pi*28E6) = 87.4 uH

This won't work either since it is a HUGE coil.

OK, what gives?  Am I missing something here or doing a calculation 

Everyone says they've seen this "trick" done in the ARRL handbook and 
other texts.  Can someone be specific as to what handbook (what year)?  I 
sure can't find it yet in the 1999 issue or in my 1989 issue.

This idea won't work unless I am totally missing the concept of how to do 

So what am I missing or what have I misunderstood that folks have 



Jon Ogden
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