Assuming that each triode is biased to run in Class A and that they were running the 12AU7 at full plate dissipation (2.75W) and the plate idle voltage is 300, then you can work out that the plate current (which in a triode is equal to the tube current) is 2.75/300 = 0.00916A or 9.16mA.
I would do this differently. We can't assume the 12AU7 would idle at maximum dissipation at 11.5v of bias.
Details below.
I would like to know how to calculate correct cathode bias resistor for 12au7 push-pull amp?
From Brimar datasheet operating conditions are: anode volts 300, grid volts -11,5 and output load 10kohm
Brimar did the hard work for you... they gave you 2 good operating conditions. Now you just have to inject some background knowledge.
While you
can make a push-pull circuit that is plate-loaded with resistors, let's assume Brimar wanted you to use an output transformer to load your 12AU7. How can we know this? The data sheet paragraph talks about using this push-pull stage as a driver for a Class AB2 or Class B output stage (made of other tubes). Because class AB2 (and often class B) stages draw grid current, they need a driver with a low output impedance (to source the current) which tends to rule out resistance loading.
So you're using an output transformer, and B+ is very nearly equal the 300v plate voltage. We know bias is 11.5v, so how do we figure current?
Brimar gave you 2 graphs of plate current when plate voltage and bias are known; they're on pages 7 & 8.
When you look at the usual plate curves on page 7, you see a grid voltage line for -10v and -15v. You
could use a ruler, place it perpendicular to the -10v and -15v lines, and draw new -11v through -14v grid lines. At this point, you'd look for where the vertical 300v plate voltage line runs through a point halfway between the -11v and -12v lines, then read off current. But you know what? The graph on page 8 has the intermediate lines...
So go to the graph on page 8 (generally called the "mutual characteristics"), and look at the bottom axis of the graph. First vertical line left of -10v is the -11v line; halfway between this and one more vertical line to the left is -11.5v. Now move vertically from this point up to the 300v plate voltage curve. At this intersection, follow right to read your plate current of 11.3mA
per triode.
Well, 11.3mA * 300v = 3.39w, which is well over the plate dissipation rating Brimar published. Either this is safe (because Brimar
did publish this as a operating condition) and the plate dissipation rating is conservative, or you should back-track and adhere to Tubeswell's suggestion.
But assuming this is safe, you will have the current of 2 triodes in a shared cathode resistor, or 11.3mA * 2 = 22.6mA. You need 11.5v across the cathode resistor, so 11.5v/0.0226A = 509Ω. Power dissipation would be over 1/4w, so I'd use a 1w resistor, just because. 509Ω isn't a standard value, but 510Ω is in a 5% resistor.
You might try a set of values in the 5% series, starting at Tubeswell's 680Ω and working downward to 620Ω, 560Ω and 510Ω, if you're not sure the 680Ω is giving everything you want and if you watch for redplating. 1w 5% resistors in these values shouldn't be hard to find in a metal oxide or metal film variety.
Topic: Push-Pull 12au7 (Read 8690 times)