Ok, there has been a LOT of discussion lately about tube input impedance
in a grounded grid amp. So some of you are tired of it, big deal. I
belong to this mail reflector becuase I am trying to improve my skills
and increase my knowledge.
My question is on what is the value of the cathode current in a GG amp.
According to Ohms Law, Pin = (Ic^2)*Z. Where Pin=Input Power, Ic=cathode
current and Z=input impedance.
>From Eimac's data, the driving impedance of a 41000A is 110 Ohms at a
plate voltage of 5 KV. Therefore, with Pin=90 Watts I would calculate Ic
to be the SQRT(90/110)=.90 A. This seems like a big value but I am
assuming that what goes in to an amp must come out so if the plate is
running then at 650 mA (per spec) then the total grid current would be
250 mA. But maybe this is a wrong assumption (to add the currents)
becuase the Eimac data shows the screen grid current to be 55 mA and the
grid current to be 115 mA or a total of 170 mA. So therefore, I am short
80 mA somewhere. Is that 80 mA effectively dissipated as heat or what.
Or am I completely flawed in my assumption of what goes in must come out.
I have a feeling I am.
No flames please. I am genuinely trying to learn some stuff here and I
don't feel like going and digging through old text books.
Thanks for the help.
73,
Jon
KE9NA

Jon Ogden
KE9NA
http://www.qsl.net/ke9na
"A life lived in fear is a life half lived."

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