Did not take a complete set of voltage readings, to begin with. But, I noticed that the B+1 was lower that of the 300 The secondary AC of this PT was right at 418v, and the B+1 489vDC. The 300 had a secondary AC of 413v and a B+1 of 505vDC.
I don't know what "B+1" is. If the two amps have different power transformers, rectifiers, filter cap values, filter cap newness, or circuit current draw then the voltages won't necessarily match.
And when you mention "low B+" the first thing I saw in the schematic is the 1kΩ resistor that all the amp's B+ current passes through. But given your questions about bias, I don't think you want to add B+ voltage at this moment.
Anyway, my bias voltage leaving the diode is -35vDC and my bias current is right at 100mA.
In the meantime, if someone would, please do me a favor and do the math on this circuit. The PT secondary is still right at 418vAC, and the bias voltage measured -37vDC. ... my schematic of my Ampeg B-12-XY ... had a PT secondary of 417vAC, an identical bias supply circuit (34K instead of the 47K), and -52vDC out.
Here's the reply that Dave H. gave on the Music-Electronics forum:
I didn't open your math program, and you don't want to calculate. With due respect to Dave H., the math for a rectifier circuit very painful and is not as simple as what he presented. Ohm's Law stops working with the non-linear process of rectification, and every part in the circuit has an impact on the answer. The complexity is why you never see "gurus" present a simple formula for computing B+ voltage with a given a.c. input and rectifier tube, because there's so many factors involved and steps to do (you can find them in RDH4 if you really want).
So instead, let's look at your amp's bias circuit, and see what parts will change its output voltage. It will be faster for you to tack-solder parts and see what happens than to calculate anyway.
If the 0.1uF coupling caps going to the output tube grids are leaky, they will drag the bias voltage closer to 0v. So pull the output tube from the amp, disconnect the wire from the bias supply to the 120kΩ bias feed resistors, and measure the negative voltage on that wire. If it has gone up, you know the coupling caps need replacing.
If your 100uF cap is old and leaky (or has reduced capacitance-value), it will drag the bias voltage down. If you're doing frequent repairs, you have something similar on-hand (or should)... Swap the cap and see if the bias voltage goes up. If it does, the old cap needed replacing anyway.
Dave H. pointed out there is a voltage divider from the high voltage to the diode to reduce the raw a.c. input. There is also a 47kΩ after the diode to again reduce rectified voltage. If none of the above steps raised the bias voltage (because the caps were good), then you have a choice: you can raise bias voltage by fiddling any/all of the resistor values, but which one?
If you don't see signs of damage, and didn't notice low resistance values for any of these resistors while the power was off, then you can either raise the value of the 47kΩ, raise the value of the 56kΩ or lower the value of the 100kΩ resistor. Any/all of these changes will increase either the raw a.c. input or the rectified voltage. The fastest way for you is to lower the value of the 100kΩ. Get a 1MΩ pot, and use 2 of the terminals to create a rheostat, which you will tack in parallel with the 100kΩ. With the amp off, set the pot for maximum resistance between the terminals you used. With power applied, carefully turn the pot and monitor bias voltage, tube current or both, until you reach your desired idle current/bias voltage. Turn off, remove the pot from the circuit (without disturbing the wiper position), measure resistance between the terminals you used. Find the nearest standard-value to that resistance and solder that resistor across the 100kΩ. Bias done.
If you have to choose between higher & lower standard values, the lower value will result in a higher bias voltage and lower tube current (maybe the safer choice), and the higher value will result in less bias voltage and higher tube current.
You can calculate bias circuit values if you like, but I don't think you'll find a faster or cheaper approach than the one above.