>? Assuming that a gg input is like a resistor is bound to bring some,
OK for all of you who doubt it then. What other assumption can one make? The
tube is spec'd for an "average" driving impedance of 110 Ohms with an input
capacitance of 27 pF. Last time I looked, Ohms was a resistance value (hence
using a resistor) and pF was a capacitance value (hence using a capacitor).
Sure there could be some inductance in there too. However, the resistors and
cap had some lead length which add a little bit of inductance as well.
So......if you have all 3 passive sorts of components covered.....what else
could it be???
Now, I KNOW that in a GG circuit, the input impedance varies over the drive
cycle. But you have to have some sort of impedance that you use to create your
input match. A pinet or an Lnet or a Tnet match one impedance to another in
addition to being like the venerable flywheel and storing energy. So one end
is 50 Ohms, pray tell to all you out there what the other end is? Well, I used
what the data sheets say is the driving impedance which equals 110 Ohms. So on
paper, you use 110 Ohms in parallel with 27 pF (which on any data sheet you
find is spec'd as the input capacitance). You absorb the 27 pF into the value
of C2 in your Pinet and then solve the equation for a pi matching network.
Now if what they taught me about matching networks in college is all wrong, I
would like to know. I KNOW that there are some additional real world effects
inside the tube that I can't completely account for by simulating it, but it
should get one relatively close. Otherwise what good is working out networks
by any form of calculation. Should all of our engineering be hit or miss?
73,
Jon
KE9NA

Jon Ogden
jono@enteract.com
www.qsl.net/ke9na
"A life lived in fear is a life half lived."

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