OK,
In order to match the high impedance of the 41000A 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 Lnetwork to step
the impedance down to a more workable level. This Lnetwork 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 41000A's plate impedance is 4600 Ohms = R1 (from
Eimac Specs)
The Cplate of the 41000A is 7.6 pF (from the ARRL handbook)
Let's choose L (the series inductor) to be 1 uH (a suggested value by
others).
What secondary impedance do we obtain?
OK:
In an Lnet Q=R1/Xp
At 28 MHz: Xp=1/(2*pi*28E6*7.6E12)= 747.91 Ohms
Therefore Q=4600/747.91 = 6.15
R2 = Xs/Q (R2 = output impedance)
At 28 MHz: Xs=2*pi*28E6*1E6 = 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.
Therefore:
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
wrong?
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
this.
So what am I missing or what have I misunderstood that folks have
suggested?
73,
Jon
KE9NA

Jon Ogden
KE9NA
http://www.qsl.net/ke9na < CHECK IT OUT! It's been updated!!!!!
"A life lived in fear is a life half lived."

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