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Broadband tube driver circuit needed
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. |
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