> "parallel" input stage, ...shared anode resistor, ...separate cathode resistors
Does this count toward my grade?
Will it earn me a bonus in my paycheck?
If I know the answer, will the babes fall all over me?
I think you are just thinking-up hypothetical problems to amuse yourself.
First: if the idea is "more", why combine a "more" stage with a "less" stage? If they are very different, the more-stage will overwhelm the less stage. Probably not worth the 2 watts of heat. Two tubes in *cascade* will do much-much "more" than two tubes in one stage. And give more places to hang stuff. (Volume and tone-pots, tone-shapers, overload loss networks.)
Obviously you "can" compute the bias. Probably you can compute it on the curves. It is not "two separate valves" because the added constraint is that both plates must sit at the same voltage.
That amused me, for about 20 seconds.
The answer didn't come, and I didn't think it worth 21 seconds. Because even if you could predict "1.23mA, 0.678mA, 97V", what does that mean for THE MUSIC? Is it good for Polka? Ska? Funk? Bakersfield?
And why compute when you can BUILD? And build faster, since there isn't a cook-book analysis handy. 10 minutes of solder in an idle amp will not only tell you V and mA, it may give a fair start on discovering what The Sound is and how it might fit into THE MUSIC.
There is a cheap trick, especially for high-Mu triodes. You can do it on a matchbook cover. You can figure the approximate equivalent resistance from cathode resistor and Mu. You can then figure the approximate bias point from plate resistor and supply voltage. Knowing that both sections must get the =same= plate voltage, you can parallel the equivalent resistances, work out the plate voltage. Also total current. Now go back to the curves, plot the two cathode resistor lines, the estimated plate voltage, and read each plate current (they should sum to near the estimated total current).
But I do not see the point.
Topic: Preamp paralleled operation - extra gain (Read 4585 times)