Here's something I haven't figured out yet.

[pinkphiloyd]

I am, admittedly, still very new to all of this. But I've been obsessing over amps and effects for about 3 months now, and even I'm amazed at what I've learned, just devouring anything and everything I can find on the web. But I'm not quite sure what's going on here, yet. I think this is a general cathode follower question.

This is a lightning schematic, I've circled the area in question, basically the 2nd 12ax7. Okay, so, like I said. Cathode follower. I get that. Pulling your signal from the cathode of the second half of that tube. Because the output impedance is relatively low and so therefore it's good for driving that tone stack. I'm good so far. So...
What's up with the plate of the first half being tied to the grid of the second? Which is then tied to the plate of the first. Is this some sort of feedback pathway? Is signal from the plate of the second half feeding back into it's own grid through that 100K plate resistor? And if you wanted, could you route that signal somewhere else? Thanks in advance. A foolish newb in need of all the help he can get.

[rfgordon]

The signal from the first triode is coupled to the grid of the second, as you can see. Because the grid of the second triode is at the same DC potential as the plate on the first, the second plate needs to be at a higher DC potential. That's why the second plate is tied to the B+ rail.

[pinkphiloyd]

Okay, I think I follow you. But could you still acquire signal from the second plate (pin 6) if desired? Such as illustrated here?

[Firestorm]

In the CF, there isn't really any "signal" on the plate of the second triode because it's tied directly to the B+ rail, and hence is at AC ground.

[pinkphiloyd]

In the CF, there isn't really any "signal" on the plate of the second triode because it's tied directly to the B+ rail, and hence is at AC ground.

Okay...so, what is "output 1" in the above diagram? Now I AM confused.

[DonMoose]

That diagram could be considered a PI (concertina type?) if you bypassed the cathode resistor.

As the voltage on the grid goes up, the voltage on the plate (tied to output 1) will go down, because the plate current will rise. As the plate current rises, the cathode current will also rise, so the voltage across the cathode resistor will rise - and the voltage at output 2 will rise.

Hope this helps!

[Firestorm]

The difference between the CF and the split-load PI diagram you posted is the presence of a plate resistor. In a common cathode voltage amp, where the cathode is connected to the "common" (ground), voltage amplification occurs because the AC input signal on the grid causes a changing current to flow in the tube. When the signal swings positive, more current flows from the cathode to the plate and the plate supply. If there is a resistor in series with the plate, this rising current causes the voltage across the resistor to drop. When the signal swings negative, less current flows, so the voltage across the plate resistor rises. This changing voltage (inverted from the input signal) is the output signal and has gain. The opposite thing is happening on the cathode, where a signal is developed across the cathode resistor that is in phase with the input signal and has no gain (well technically it almost has unity gain; slightly less than 1.)

In the split-load PI, the resistors are chosen to produce roughly equal and opposite signals with roughly a gain of 1.