"Bob P" wrote in message
y.com...
Want to have broadband (7-28 Mhz) 1 Watt (50 Ohms) class A solid state
amplifier driving grid of 6L6 tube to about 2 or 3 mA grid current, tube
is
biased to about -65 Volts for class C operation with 550 Volts on the
plate.
I don't want to have any tuned circuits in the 6L6 grid if I can avoid it.
As of now, the 6L6 has a 6CL6 driver feeding tuned grid circuit, easily
gets
2-3 mA.
My goal is to eliminate the driver tube and the tuned grid circuit.

Anyone have knowledge of toroid(s) that might be required, turns ratio,
material, etc.?
Do I need a more driving power ? Probably could get 2 Watts out of my
driver
circuit if need be.

thanks
Bob
kb8tl at yahoo dot com





Hmm. 6L6 grid capacitance = 12pF (that's for a 6L6G - plain 6L6 is 10).
Broadband that up to 28MHz, you need the parallel R to be no more than 450
ohms (and that's for 3dB down at 28MHz).

Overall voltage swing 2*65V; call it 140V, for an RMS of 50V. (50V)^2 /
(450*ohms) = 5.6W. Ignoring the grid drive . Assume that the coupling
circuit is 80% efficient, then you need about 7W.

The problem is that danged input capacitance. You can almost call out the
bandwidth from the constraints. You _should_ be able to make a circuit that
can be fixed-tuned for each band, you should be able to make it about 2MHz
wide if you size the coils right.