Date: Tue, 13 Dec 2022 17:36:44 -0500
From: Steve Bookout <steve@nr4m.com>
To: amps@contesting.com
Subject: [Amps] L-PI amp network
I cycled all of the unknown tubes thru
this amp.? They all were similar, taking 22 to 27 watts of drive for the
same 1500 watts out;? 550 ma to 600ma @.030 ma grid.
That was the background for my real question for the group.
Based upon GM3SEK's PI/PI-L calculator, I'm working with a plate load
impedance of ~3000 ohms.? I know from my own experience, and that of
others, that it's a pain to get the plate inductor 'right' so that it
all works and tunes, as it should and make power with reasonable
efficiency.
I have never used an L-Pi, but I basically understand the idea.
IS THERE A RECOMMENDED TOOL OUT THERE, WHICH WILL HELP ME FIGURE OUT MY
TWO "L's"?
Using GM3SEK's calculator, I see there is a line, 'Lead inductance
(total from tube to tank, but excluding suppressor)', which I have
played with.? Adding my 'L' inductance there, say '.6uh', does reduce
the inductance associated with PI, and it does raise the value of the C1
capacitor. ? ( I am using a 100 pf /15 KV vacuum variable and right now
as I write this, I don't remember it's minimum C, but it's in the range
of what I think you should see; maybe 5 pf?)? Using this, I was able to
come up with about 15 pf, vs 4 pf in a standard PI configuration.
Suggestions on if this would be an accurate way to figure this out? ?
Or, another/better way? ? And, what should I be looking for as maybe a
ratio of the 'L' vs the 'L' in the PI network. ? As I recall, I think
I'm looking at about 1.7 uh.
I would appreciate any constructive comments, or even precautionary
ones.
73 de Steve, NR4M
## Steve, I have been using the GM3SEK spreadsheet for a long time now,
since it 1st came out. It's what I use to design the L-PI networks. It
only works on tubes that DON'T require a suppressor.
## assume 1.5 kw out..and a tank eff of 60% on 10m. 1500/.60 = 2.5
kw DC
input.
2500w / 4000vdc = 625 ma required plate current. (28.6 mhz)
## Use 1.8 for the 'K'. 4000 / (1.8 x .625) = 3555 ohms plate load Z.
With an overall network Q of 19 , line 39 (input q = 17.2 output q =
1.8)
Enter 0 for all the parasitic lps/rps on lines 46+47.
Enter 10 pf on line 42
Enter 5 pf on line 43
Enter 3 pf on line 53
Enter 50 uh on line 51
Enter 0 on line 48
Enter 0 on line 52
Here's what u get.
C1 = 9.6 pf
L = 1.26 uh
C2 = 198.5 pf
An overall Q of 19 sucks. 11.825 amps of current through the tank
coil..which is a bunch.
I would use 3/8" OD tubing...or 1/2" to 5/8" wide strap coil. With
tubing
flattened at each end, then enough width to punch a hole big enough
for a
1/4-20 brass screw. DON'T use SS hardware. It will turn black on
you..and ditto with plane jane ferrous steel machine screws.
## Ok, let's use the L-PI config, which is really a C-L-C-L-C
We will use the tube's anode to grid C (+ strays) plus the 1st coil, to
form a step down L network..... stepped down low enough, that a
normal PI
net can be designed around, with a lot lower loaded q.
Enter .8 uh on line 48. Lower the overall Q on line 39 to 11.
Ideally, you want this 1st coil to be a little less than the main coil.
Yes, u have to keep juggling with the overall q on line 39, each time u
increment the 1st coil on line 39. Start with .1 uh..and keep
increasing,
while at the same time decreasing the overall q on line 39.
Ok, here is what we get. Note, the overall network Q on line 39 is
= to
the (input Q + output Q).
Overall network Q of 11 (28.6 mhz)
C1 = 12.7 pf ( input Q of 9.5)
L = .91 uh
C2 = 171.6 pf. (output Q of 1.5)
Note the 1st coil's value (.8uh) is slightly less than the main coil.
Again, use 3/8" tubing or 1/2" to 5/8" wide strap coils for BOTH
coils.
Yes, use ur 100 pf vac cap. You won't have any issues, and harmonic
suppression is typ higher when a vac cap is used. Air caps have the
issue
with the rotor wiper contact. I'm assuming ur vac cap goes down to 5
pf.
Harmonic suppression with the L-PI is superb. With the lower Q, you
won't
have to constantly be re-tuning when u qsy.
Jim VE7RF
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