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

"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.

"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.

QRP HomeBuilder web site is at
http://www.qrp.pops.net/
and they have coil winding software

QRP HomeBuilder web site is at
http://www.qrp.pops.net/
and they have coil winding software