I have a question; I noticed someone say that "a tube amp" (in this case the Crack OTL) is usually better thought of as a current amplifier than a voltage amplifier. Is this so?
Which half of the Crack are you referring to? (12AU7 front half or 6080 output half?) Speedball or loading resistors? (This changes the answer to the question a bit, lol)
Are they generalizing tubes incorrectly
When you lump all tube amps into one generalization, I would say so.
An old school veteran Air Force tube radio technician told me that in audio it's typically a voltage gain situation But I think he said both voltage and current gain are happening.
This will largely depend on the circuit and tubes used. Some tubes (the 12AX7 for instance) are good at voltage amplification but not so great with current amplification. Other tubes (the 6080 for instance) are useless as voltage amplifiers but marvelous current amplifiers. A tube like the 6C45PI does both really well. On the circuit side, I've seen and listened to 6080's and 12AX7's setup as both current and voltage amplifiers.
I did some of my own research that lead to some basic circuit designs with tube voltage gain and tube current gain stages explained separately then combined for an OTC power amp, further confusing the issue for me.
Yeah, I would leave the transformer out for now, especially since the discussion is mostly around the Crack.
Is the crack, as an OTL headphone amp mainly utilizing voltage gain? or both current and voltage stages?
One of each. The 12AU7 stage is far easier to understand, as it's just a common grounded cathode amplifier. From a DC perspective, the 12AU7 grid is at ground, the cathode voltage is set (or limited if you will) by the 1.57V LED, and the plate is loaded with a 22.1K resistor. The combination of the 1.5V LED, the 22.1K resistor, the B+ available, and the characteristics of the 12AU7 establish what DC voltage appears at the plate.
On the AC side, signal is applied to the grid of the 12AU7 and modulates the quiescent current flowing through the stage. If you examine the plate curves for a 12AU7 and put a dot at 165V (our B+) on the voltage axis and a dot on 7.5mA on the current axis (7.5ma is the 165V available divided by 22.1K plate load from Ohm's Law), then you will see that when you move along that line between the 1.5V of bias to the left 1.5V (to the 0V line, though in reality we won't quite get there) and to the right 1.5V (to the imaginary line of 3V of bias), then the plate voltage changes. The first image posted shows this. For 3V of peak-to-peak voltage at the input, we get about 38V peak-to-peak at the output (so a voltage amplification factor of about 12.5). You'll notice that the current swings by +/- a ma or two as well. When you install the Speedball, this load line becomes nearly horizontal, swinging very little current but more voltage (available voltage swing will still be limited by the 1.5V of bias from our LED). So if we put 2V AC into a stock Crack, we expect to see about 25V coming out of the 12AU7 gain stage, but without the current available to really be that useful.
The 6080 is setup as a cathode follower, which has a fixed plate voltage (the plate is tied to the filtered high voltage supply directly) and a load under the cathode. Back when we looked at the 12AU7, when we applied signal to the grid, we were moving around between the 0V and 3V curve on the charts since the cathode voltage wasn't changing (again, thinking of AC signals, not DC bias). With the 3K cathode resistor and no bypass capacitor, the cathode's AC voltage isn't tied down. When the incoming signal swings positive, more current flows through the cathode resistor and when the incoming signal swings negative, less current flows through the cathode resistor. The reason that this type of stage is called a "Cathode Follower" is because the cathode voltage follows the grid voltage.
In terms of the charts, we can draw a load line for the 6080 and see everything line up nicely in terms of where the DC voltages will settle when we run the amplifier, but what about the AC voltages? Since the cathode is following the grid, the AC voltage available at the cathode can only be as much as the AC voltage at the grid (though in reality it is a bit less), so ultimately our AC load line ends up depending a whole lot on the loading impedance, and the cathode follower itself can be treated as a current amplifier.
I've intentionally completely glossed over how feedback/output impedance play into this since you didn't ask and it's a long conversation.
