>>Even if
>>you do get say, 1 KOhm of impedance from all the inductances, that's NOT
>>a REAL resistance. That is a reactance.
>
>No it isn't. Think about your pi-network terminated in a 50 ohm load.
>The load impedance is transformed to say 2K across the tube - take away
>the dummy load and the 2K resisitive part disappears to infinity
>(ignoring network losses). Also, the 50 ohm resistor receives exactly
>the same amount of RF power as a 2K resistor wired across the tube
>would.
Ok, this is true. You are right. The reactive components do the job of
transforming impedances. I wasn't thinking straight. But my 50 Ohm load
is on the OUTPUT of that transforming network, not in the middle of it.
The supressor resistor is located a couple of inches or less from the
anode. The rest of the stray circuitry comes after it. So looking from
the tube out (which has a 2 KOhm impedance or so) how does that resistor
which is effectively IMMEDIATELY after the output of the tube get
transformed to some other value?
>
>The same applies to a 100 ohm suppressor resistor, when it is
>transformed to represent 1K in parallel with the tube. It will dissipate
>exactly as much power as a 1K resistor connected across the tube. In
>that sense the 1K is a real resistance.
Still, call me dense, but I don't understand how this works. I see no
reactive impedance transforming network between the output of the tube
and the supressor R. Just a staight piece of brass, copper, nichrome or
whatever your poison is.
73,
Jon
KE9NA
--------------------------------------------------------------------------
Jon Ogden
jono@webspun.com
www.qsl.net/ke9na
"A life lived in fear is a life half lived."
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