And we explained early on that those bias resistors will have zero effect on the B+ by themselves. The only way they affect it is by controlling power tube current. Tube current affects B+ voltage. This would be the same if the bias were a totally separate power supply.
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tubeswell, thank you for your clear explanation. However, my question was about how those resistors affect B+, not the bias voltage.
They won't affect B+. In the bias supply circuit, you have a total of 441.7 kOhm in series between HT and ground. That is a high impedance (almost effectively an 'infinite' resistance) from the point of view of the load on the PT. A working bias supply draws only about 1mA or so of current (at the best of times). Assuming its actually drawing 1mA, then 1mA across 441.7k = 441.7V (However, your schematic hasn't given us any voltage information. so my number crunching is hypothetical at this point)
They won't affect B+. In the bias supply circuit, you have a total of 441.7 kOhm in series between HT and ground. That is a high impedance (almost effectively an 'infinite' resistance) from the point of view of the load on the PT.
Thank you, this helps put things into perspective.
If you take the power tubes out while tweaking the bias voltage, you will see that the plate node rises to B+ or nearly so (and that the high voltage rises across all nodes without the idle current through the tubes). Adjusting the bias components without the power tubes should have zero effect on B+.
When you have a controllable, reliable range of voltages on the bias supply, then put the power tubes back in. Adjusting bias at that point will affect B+ because the idle current through the power tubes will draw down B+. Don't red-plate the tubes!
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