I would not bother with dc heaters or elevating the heaters for this amp. But, I don't understand why the dc voltage would only be 5.8VDC with the bridge circuit you posted. 6.3VAC into a bridge should give 8.9V unloaded. Surely the filaments are not pulling that 8.9V down to 5.8V. I suspect something may not be wired correctly?
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About dropping the B+... BUILD IT FIRST!
Leave the choke right where it is. Put the dropping resistor between the rectifier V8 pin 3 and the STBY switch. You may even have room to mount the resistor directly on the socket for V8.
Now, you need to build the circuit IAW the schematic first so you'll know exactly what your voltages are. Then you will know how much you want to drop the voltage. Then do the math and add the appropriate size resistor.
Here's how to do it using my voltage readings as an example. My B+ into the filter is currently 350v and let's say I want to drop it to 325V (like the original).
Step 1. Determine how much total B+ current you have. This is very easy to do. Just use two gator clip leads to connect a 1Ω resistor across the STBY switch. Measure the millivolt drop directly across that resistor. IN my amp, I have exactly 100mV. That means there is 100mA (0.1A) flowing thru that 1Ω resistor and that is the total B+ current drawn by my amp at idle.
Step 2. Do the math. I want a resistor that will drop 25V and there will be 100mA flowing through it. So, R = E/I, R = 25V/0.1A, R = 250Ω. And the power dissipated by that resistor will be P = EI, P = 25V x 0.1A, P = 2.5W. Double that for safety so you would use a 5 Watt resistor. I'd be more comfortable using a 10 watter.
So, in my example, I need a 250Ω 10W resistor to drop my B+ from 350V to 325V. Now just unclip the 1Ω resistor and clip in the 250Ω resistor to check voltages before you actually solder anything.
Don't throw this info away! You might need it some day. Did I say, BUILD IT FIRST? If not, I meant to.
