I don't really understand analog electronics very well and I wish this was a bit better explained for a general audience.

I think +18V is enough for some quantum physics to happen and suck an electron through the first transistor even though no current is "supposed" to flow, and because it's coming from the base, that turns the second transistor "on" a little bit and "tries" to "brown out" the second transistor's collector causing the voltage at the first capacitor to drop.

This is where I get very lost and confused, because to me if you're trying to enhance the "brownout" to 5V, why in the world would you put a capacitor there?

I know the capacitor not connected to ground is called "AC coupling" and it somehow passes high frequencies but not low ones or something like that, but I don't really understand it at all. It's really hard for me to understand how putting a large capacitor near such a tiny signal doesn't simply cause that signal to get lost in the capacitor's ocean of capacity for absorbing electrons.

Then I tried to look at what the 74AL304 is doing, it's apparently a standard inverter, but I think it's trying to be used in an "analog" way here. The inverter is "trying" to calculate "A = ~A" but since the connection from input to output is a resistor, you can't think in digital terms and you have to look at its analog characteristics.

I found a datasheet on TI's website but it doesn't have much information about its analog behavior, I'm thinking that to begin to understand what it does, I'd need a voltage sweep curve or a diagram of the internal transistors of the 74ALS04.

My intuition is that the electric force is really strong, so when an electron comes onto one side of the capacitor, it forces an electron off the other side, so if you black-box the capacitor, it "acts like a wire": Current that goes in one side comes out the other side. (I'll run with this for this post, but now I have another unanswered question, "Okay then, if electron in side A = electron out side B, how is a capacitor different from a wire?")

So the 74ALS04 is "trying" to raise its output when the input is low, and lower its output when the input is high. Therefore if you sweep the input voltage, at some point it must switch from "trying" to raise, and begin "trying" to lower. By connecting output and input, a too-low or too-high voltage drives the inverter toward the crossover point, and forces it to stay there.

Now if you suddenly force an electron into the input of the inverter, it has to react by sending the output in the opposite direction. The circuit designer seems to have assumed this occurs in an amplifying way, at least in this circuit.

It's still very confusing for me to imagine the input and output of the inverter at different voltages, because then current is flowing through the resistor, but where does that current go? Maybe it goes back into the capacitor?

Okay, so a tiny voltage drop at the inverter's input causes it to surge current to its output. A tiny amount of the surge goes back through the 27K resistor which eventually raises the input voltage so the surge shuts itself down, but most of the surge goes through the low-resistance path and flows into the next stage's capacitor (how many stages you need depends on the amplification factor vs. how small the input signal is), then after the two stages you have a final "normal" inverter gate cleans up the signal so you have a digital output.

I still don't understand AC coupling. Anyone care to take a crack at explaining why the capacitors are necessary, and what happens if you replace them with wires?