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|17th July 2004, 09:24 AM||#11|
Join Date: Apr 2004
i was encouraged to use both units of the double triode in the 6EM7.
also, i had suggestions that the B+ would not be sufficient enough for a DRD design.
and i misinterpreted Jeremy Epstein's Free Lunch calculations. the Free Lunch resistor should be from the plate of the driver to the cathode of the output tube, not in series with the cathode resistor as in previous schematics that i posted.
so here goes, based on the new requirements and my limited understanding of DRD and the 6EM7 tube.
B+ of 800v.
300B operated at 60mA, 450v and -100v bias
voltage at cathode = 350v
voltage at grid = 250v
giving cathode resistor of 7085R and free lunch resistor of 9184R at the cathode of the 300b.
6EM7 low mu unit 2 operated at 10mA, 150v and -30v bias
voltage at cathode = 100v
voltage at grid = 70v
giving cathode resistor of 1000R and free lunch resistor of 50000R at the cathode of 6EM7 unit 2.
6EM7 high mu unit 1 operated at 0.6mA, 69v and -1v bias
voltage at cathode = 1v
giving cathode resistor of 1666R at cathode of 6EM7 unit 1.
the 1666R resistor here creates a 1v at the cathode.
question is, how do i apply a 1v bias at the grid?
does the schematic look correct so far?
|18th July 2004, 01:43 AM||#12|
Join Date: Jun 2003
"As I recall, this is the "choked monkey" design. Utterly horrible as you waste half the power in that 3k 25W resistor below the 300B."
The AC return path for the 300B is through the 70uF capacitor, not the 3K resistor.
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