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

Terry Gaiser w6ru@lightspeed.net
Mon, 15 Feb 1999 21:29:12 -0800


Jon,
I tried this idea with a YC-156 tube I have been experimenting with to
try to get it to play on 10 meters....its Cout is about 36PF. This is
the idea of a series inductor between the tube and C1 ??? To step down
the plate load X... so that you would need more C, less Xc at C1 ???
Talk about laying an egg! I would sure like to talk to someone that has
done this successfully if they exist.

Good Luck,
Terry W6RU

Jon Ogden wrote:
> 
> OK,
> 
> 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
> others).
> 
> What secondary impedance do we obtain?
> 
> OK:
> 
> 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.
> 
> 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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