Hi, Paul.
After some further study, I have some follow-up questions for whenever you have time.
Input stage question:
I found the Youtube channel of "Uncle Doug," a retired teacher who repairs guitar amps, does great explainers (and has entertaining pets). He's got a series on grid leak, stopper, cathode bias, and plate resistors that's really informative, but I don't see leak or stopper components in the Crack's input. It occurred to me that the volume pot might actually serve those purposes, with the part of the resistor that's on the input side of the pot's contact as the stopper, and part that's on the ground-referenced side as the leak.
1. Am I right?
2. If so, don't the values change when the volume does?
3. If not, why no grid leak or stopper in the circuit?
Output stage questions:
Online examples of cathode followers always explain how they function as preamp outputs or drivers for tone/effects loops in guitar amps. Applied as the output of an OTL headphone amp, there are still some things I just can't wrap my head around...
First, that output resistor. You said:
Without those resistors the output side of the coupling caps will float up and present a high DC voltage at the headphone jack, which could damage your headphones or shock you when you plug them in.
I racked my brain trying to understand how DC gets onto that jack if I remove that resistor (don't worry, I won't leave it out of my build). It might be that I'm misunderstanding how the coupling cap functions. I *think* DC volts see a capacitor as an open switch and AC volts see it as a closed switch. And if that's the case, there's no DC continuity between ground, the jack, and anything with +VDC on it. What am I missing?
Second, (since I've now learned enough to become dangerous), I have more detailed questions about how a cathode follower works.
You said I guessed right about the 12AU7 stage being a signal voltage amplifier, with the 6080 amplifying current in order to lower the overall output impedance so those signal volts can drive the cans. I also noted the current gain stage looks like an inverted version of the voltage gain stage. But more I learn, the more I notice dissimilarities... First, there's nothing that looks like a grid leak or stopper resistor for the 6080. Is that because the 12AU7 is serving those functions? If so, does the 6080's leaking grid current result in a slight decrease in DC volts on the 12AU7's plate? Or does the whole 12AU7 circuit act as a signal ground reference for the 6080?
The reason I ask is because my understanding of triode operation - at idle - is that bias current flows through the tube because the plate has a very high + DC voltage with reference to the cathode, while the cathode has a slight + DC voltage with reference to the grid (which has no volts at idle because it's referenced to ground via the leak resistor). Then...
the 12AU7 just has to drive the grid of the 6080, which is almost like driving nothing at all.
WHEN THAT HAPPENS.... why is amplified current available on the cathode side of the 6080?
AND...is the 6080's cathode resistor ALSO essentially acting as plate resistor as far as the VAC signal is concerned?
AND... wouldn't that also essentially make the coupling capacitor for the output double as a bypass capacitor for the 6080 cathode resistor?!
Recognizing that I don't actually need to know any of this to build the kit, I TRULY want to thank you for your efforts to satisfy my curiosity.
Cheers, Mike