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RE: OK . . . So . . . Working Through This . . .

Actually, one of the things that I like about the LTP (differential amplifier) is that it is fairly easy to understand. Referring to your original schematic - the 18K resistor that is common to both cathodes has 125 volts across it, so the current through it is 7 mA. This is split equally (in theory) between the two 6SN7 triodes, so the current through each plate resistor is 3.5 mA. So the drop across the 33K is about 115 volts, which explains why the plate voltage is 420 - 115 = 305 volts. But the grids are at potential of +120 volts, so the plate is about 180 volts higher than the grid. If we move the grids to a zero volts and the supply to 300 volts and leave the plate resistors and current the same, the plates will be at 185 volts and the grid to plate voltage will be the same - 185 volts.

In the proposed design, we need to get about 125 volts across the 18K resistor common to the cathodes. The grids will be about 5 volts negative with respect to the cathode, or in other words, the cathodes will be 5 volts positive with respect to the grids (0 volts). So we need to connect the bottom (previously grounded) end of this resistor to -120.

“So just another R-C section or does this supply need to be negative voltage?”

It needs to be a negative supply, but only needs to supply about 7 mA of current.

“Also, would this require the use of fixed bias or could I still use cathode bias as planned?”

You would not have to go to fixed bias, but it would be a possibility.



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