Why aren't smoothing resistors needed in parallel with each of those 100uf/150v caps?
What sluckey said, plus the added bleeders would cause problems.
Look at the top secondary winding, connected to D1, D2, C26, and C27. There is an upper and lower wire coming out of the secondary of the drawing.
Imagine for a moment that the voltage on the secondary is such that the upper wire is momentarily positive and the lower wire is momentarily negative. The lower wire being negative reverse-biases D1, so no current flows through it at this moment. However, D2 is forward-biased, so current flows through the winding and D2 and charges C27 to the peak voltage of the winding.
Now imagine the secondary voltage polarity switches. The upper wire is now negative and the lower wire is positive. D2 is reverse-biased and no current flows through it. D1 is now forward-biased, so current flows through the winding and D1, and charges C26 to the peak voltage of the winding.
Because D2 was reverse-biased and there was no current path for the cap to discharge, C27 still has the full peak voltage of the winding across it, and now C26 also has the full peak voltage of the winding across it. So the voltage on the caps adds and you have Vrms * 1.414 * 2 across the pair of caps.
But that is true if no current gets pulled from the supply, just as our B+ in any other amp rises up to the peak voltage of you pull out the output tubes. So really, the amp will draw off some current from these caps and reduce the supply voltage somewhat. Now, if you were to throw bleeder resistors across the caps, they would need to be quite large values, because they would tend to drain the lower cap while the upper cap is charging and vice versa. You could use them, but it would tend to knock down the output voltage and possibly the current capability of the doubler supply. And doublers (and higher voltage-multiplier circuits) are already known to have lower current capability than full-wave supplies.
this amp has a really unusual approach to deriving the signal for the reverb driver and then returning it to mix with the clean signal. I can't quite wrap my brain around what is happening at the junction of R25, R26 & R27. I think that the clean signal is being attenuated, relative to the reverb signal, by R25 & R27. If that's really a 560K plate resistor (R26), it's the biggest one I've seen yet for a 12AX7. Is my guess that the big value is needed to isolate the reverb signal from the clean signal close to correct?
I'm not really familiar with this circuit, so I'd invite corrections. But I think it's not to isolate the clean and reverb, but to mix them.
In a Fender amp, the same point is used to split off and return the reverb signal. So the big 3.3M resistor diverts most of the signal down into the reverb circuit.
Here, the reverb is split off at a triode grid, and returned at its plate. So the tube does the isolating for you.
Forget R25 for a moment. Both Channel 1 and channel 2 have 68k plate resistors fed by a common 220k resistor, which then goes to the power supply. Yes this is a tapped plate load like the Fender brown circuit, and the total plate load for each channel looks like about 290k. But because they share the 220k resistor, each channel's output is mixed as they're feeding C12. R25 is taking this mixing to the next level by mixing in the reverb signal along with the dry signals that feed the coupling cap.
I'm not 100% on this, but it looks to me like R27 is being used as a common impedance to mix all 3 signals. That also means the reevrb recovery's plate load is about 800k! That would give the stage a lot of gain, but very little current variation. If I'm right about the common impedance mixing, then it's the small current variation that's important. That would keep the reverb signal from drowning out the dry signals. I note that there is not a Reverb pot in the recovery circuit (like Fender) but in the drive circuit. So it looks like the idea is to never allow the reverb to completely swamp the dry signals, even if the reverb drive (Reverb Depth) pot is full up.
That's my guess anyway...