input matching to grounded grid amplifier

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Opcom:
The job is to use a pi network to match a typical 50 Ohm exciter to the '50 Ohm' cathode of a tube.

I read this:
"(www.somis.org/D-amplifiers4.html)  Most of the tuned input and tuned output circuits in HF amplifiers are pi-networks. There are a number of ways to define the Q of a pi-network. In what follows, Q is defined as the input impedance of the pi-network divided by the reactance of the input, shunt element--typically a capacitor. This definition of Q is the one used by Eimac® in Care and Feeding of Power Grid Tubes.  Eimac® recommends using a pi input network Q of c. 5 for Class AB2 grounded-grid operation. To arrive at a Q of 2, the reactance [X] of the input capacitor, C1, is minus j50 ohm÷2=minus j25 ohm. Using C=1÷[25(2Pi x f)], approximately 220pF of input capacitance is needed for a Q of 2 on the 10m band. In actual practice, however, 220pF may be far from the value that produces a satisfactory SWR with a particular model transceiver and a particular length of coax. It may be possible to find a length of coax that would ameliorate this problem on 10m--but there are eight other bands to contend with below 30MHz. Since band switching different lengths of coax is hardly practicable, it would be useful if the input capacitors were adjustable in a grounded-grid amplifier's tuned input circuit"

Excel came up with these values but the above 'instructions' are incomplete as are most articles.
F KHzL uHC pFQ @50 Ohms289900.1372202215780.1703202.2141280.2704702.171140.55910236801.11700218682.233001.9He mentions the pi network, but does his paragraph about this mean that if the input and output Z are both 50 Ohms that the output C is the same as C in the above table? No complete method I could put in excel was given.

The blocking question is: "Since most pi networks are calculated by input C, partial L, other section pf partial L, and output C, then is the calculated value of L, above, the whole inductor in the pi network?

Would it be as well to try one of the 'pi network calculators' for this?
Got to make the cathode tuning circuit before the amp building can continue.
This article http://ham.homelinux.org/schemi-pdf/68hb199.pdf gives values for the 3-1000 which has an input Z of 65-67 ohms, close to 50 Ohms, but does not give the inductances, only turns on a specified slug tuned forms form which may or may not be available, and it was to be run at 65W drive, not 100-400 so those slug tuned forms may not be appropriate.

DMOD:
The L should be the total series L in the C-Shunt L-Series C-Shunt Pi circuit.

For a loaded Q of 4.6, here is what my Pi-Net equations in MatLab gave me for 3.5 MHz:

Zin = 50 ohms, Zout = 67 - j1700 (This is taking into account the 27 pF input capacitance)

Q1 = 4.6201


C1 =  420 pF


L = 1.1 uH


C2 (tubegrid) = 370 pF


For a loaded Q of 4.6, here is what my Pi-Net equations in MatLab gave me for 7.2 MHz:

Zin = 50 ohms, Zout = 67 - j1700 (This is taking into account the 27 pF input capacitance)

Q1 = 4.6201


C1 =  200 pF


L = 0.5 uH


C2 (tubegrid) = 180 pF


Phil - AC0OB

Opcom:
Hmm I got 420pF and 1.1uH resonate at 7.404KHz
and
Xc=51.2 Ohms.

For some reason I get Q=0.98, calculated per the somis.org article of Q=R/X. R being the '50 Ohm' input and X being the input C reactance. There are many reasons I'm messed up on this apparently believing articles on the internet may be one of them.

An old DOS program that's worked well before on tube pi outputs gives me this:
Q=5
Zin=50
Zout=67
Input C reactance=10
output c reactance=-11
L reactance=21.5
F=3.5MHz
Input C=4542pF
Inductance=0.979uH
Output C= -3944pF

It does not take into account 'j' either.. or does not mention it.

Where does the -j1700 Ohms come from?

Otherwise it looks like it is oddly close to a decimal point off from your capacitance calculations.

The summary is I don't know how to do this correctly. I think I've been here before looking for the correct math.

DMOD:
I am using the Wingfield Pi-Net equations.

I have to input an initial Qin = 10 in order to get a loaded Q near 5.

The -j1700 comes from the tube's input capacitance
Xc = 1/(6.28X27.2e-12 FaradsX3.5e6 Hz).

So the tube's input impedance at 3.5 MHz is 67 - j1700.

At 7.2 MHz it is = 67 -j813 ohms.


Phil - AC0OB

W1ITT:
What makes this all troublesome is that the input impedance of a grounded grid amplifier varies somewhat with drive power.  I have used an employer's high power network analyzer setup, feeding as much as 150 watts into a directional coupler to overcome adjacent transmitting antennas at shortwave broadcast sites.  It occurred to me that it would be fun to use that setup to quantify the impedance change with some of the popular grounded grid tubes, across a range of drive power levels.  We usually employ the high power network analyzer in a swept configuration, but it could just as easily be used in a CW mode.
The point of this is that, despite good calculations, it's a moving target.  And with a mode like SSB where the drive varies from nothing to 100%  it is even more troublesome.   Be prepared to tweak and optimize at your preferred drive power level.  If the Q of the input network is kept low enough, things will probably work out OK.  Sometimes "what works" wins out over engineering.

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