Basic Crack Theory (questions)

[Joe Garfield]

Edit: much of this is wrong because I didn’t understand the ‘direction’ of current and electron flow.

Hi All,
This is my first electronics project, but I’m a mechanical engineer so I’ve had an EE class and have a very general understanding of basic electrical components. I just bought some Crack and am now looking at the schematic as I’d like to learn as I build.

1) I feel confident I understand the power supply (increase voltage with coils, convert to DC with diodes, and keep the voltage stable with capacitors. I read that one of the three capacitors has the most influence on sound - I assume this is the one closest to the PT since it’s upstream of the resistors? And the effect on sound is how quickly and smoothly that cap loads and unloads....

2) I’m actually a bit mind blown here - I always thought we were amplifying the audio signal and listening to that, but it looks like what we actually listen to is the power supply! Power comes in (+V) and everything downstream of the 6080 is basically controlling what the 6080 sees as (-V). Woah!

3) My understanding was that tubes are voltage amplifiers. Is the 6080 actually amplifying any voltage, or is Vmax = Vplate?

4) I can’t really figure out how the 12AU7 is amplifying or controlling current, or at least how the current output gets combined with the voltage output of the 6080. I’ll keep looking and thinking, but any ‘high level’ or general comments is appreciated!

[Doc B.]

The 12AU7 is doing the amplification. The 6080 is a cathode follower, which in this case is more or less equivalent to an output transformer. It works as a buffer - lowers the output impedance so the amp can push enough current to drive high impedance headphones.

[Joe Garfield]

Ah, it's not 'gain and current stages' like a solid state amp I guess...

So is the audio signal path out from the volume pot, in to the 12AU7 grid, and out through the LED? I didn't think that was the case since LED is tied to ground, and the 'top' of the 12AU7 in the schematic view looks like the highest voltage so current would not flow out the top.

Sorry if I'm lacking even enough basics to ask the right questions. I'm just trying to get a general idea of how the circuit works without having to 'deep dive' every component.

[Paul Birkeland]

keep the voltage stable with capacitors. I read that one of the three capacitors has the most influence on sound - I assume this is the one closest to the PT since it’s upstream of the resistors? And the effect on sound is how quickly and smoothly that cap loads and unloads....

The capacitors are really there to rip down the power supply ripple as much as possible so you can't hear it at the output of the amplifier.  Since this is a class A circuit, the magnitude of DC voltage is quite stable on its own.  The last cap in the power supply is in the signal path.

4) I can’t really figure out how the 12AU7 is amplifying or controlling current, or at least how the current output gets combined with the voltage output of the 6080. I’ll keep looking and thinking, but any ‘high level’ or general comments is appreciated!

The 12AU7 has a quiescent current established by its bias voltage (the LED) and the plate loading resistor. (22.1K)  This equates to roughly 3mA of idle current in the Crack circuit driver stage.  When you apply AC signal voltage to the grid, you are dynamically changing the bias provided by the LED, since the grid will be driven closer and further away with AC signal voltage.  This change in bias voltage between the plate and cathode produces similar changes in current through the tube, and those small alterations in current appears across the 22.1K plate loading resistor, thusly amplification! 

[Paul Birkeland]

Ah, it's not 'gain and current stages' like a solid state amp I guess...

It's tough to draw comparisons to solid state amps unless you look at super simple FET stuff, but even then there will be a lot more going on than in something like the Crack.

So is the audio signal path out from the volume pot, in to the 12AU7 grid, and out through the LED? I didn't think that was the case since LED is tied to ground, and the 'top' of the 12AU7 in the schematic view looks like the highest voltage so current would not flow out the top.

We tend to talk about signal paths as loops.  The output of the first stage does come off the 12AU7 plate, not the cathode.  The cathode is at AC ground (approximately).

[Joe Garfield]

Thank you. It looks like I made incorrect assumptions about where the high voltages are. Maybe looking at the test page (voltage checks) of the manual will help me put this puzzle together.

[Joe Garfield]

Oh, energizing the plate ATTRACTS electron flow! I was missing the interaction of the heater circuit, and how tubes work in general, and (+) actually means greater electronegativity

[Joe Garfield]

Here's a nice introductory article on how vacuum tubes (thermionic valves) work:
https://new.engineering.com/eng/story/vacuum-tubes-the-world-before-transistors

This one (at the bottom) has some good info on how the second tube is used as a cathode follower:
https://vacuumtubes.net/How_Vacuum_Tubes_Work.htm

[Doc B.]

Oh, energizing the plate ATTRACTS electron flow! I was missing the interaction of the heater circuit, and how tubes work in general, and (+) actually means greater electronegativity

Buddha used to say "the solution for you - flowbs is to stand on your head when you look at a schematic"

[cddc]

Interesting talks 

I'm an electronics newbie too. Just curious about the LED biasing.

I read in the HF DV336 thread about the LED mod some guys did with their DV336's. I figured DV336 might use some resistors as biasing.

So just curious to know what's the advantage of using LED as biasing vs using a resistor? If one of the 12AU7 LED's burns out on my Crack someday, what the value of the biasing resistor should be if I were to replace the LED with a resistor? 

[Paul Birkeland]

There are components going on 30+ years with old red LEDs still working properly.  If one stops functioning, you'd just replace it.  If you replace it with only a resistor, gain goes down, output impedance goes up, and the distortion characteristics change a bit.  This effect may be somewhat mitigated by the Speedball being installed in the circuit. 

The LED itself has a very low dynamic impedance, which allows it to function much like a cathode resistor with a very large bypass cap.

[cddc]

Great, thank you very much for explaining the LED, PB!

It seems using LED provides lots of benefits, I guess that's why these DV336 guys did their LED mods.

It's also good to know that the LED's can last long enough so that I don't have to worry about the burnout issue.