A capacitor causes a phase change at specific frequencies.
This is the root of why a capacitor can cut frequencies.
So. Vb is the op amp's ground. Ground is the CIRCUIT zero volts.
A op amp wants to stay balanced with 4.5V DC (when we're on 9V supplies like batteries etc.) at the center. Everything the op amp is doing is AC across that point. 4.5V + 4.5V adds up to 9V. The circuit wants only AC at the input and output -- DC offset is bad for the other devices.
It's important to note at this point that 4.5V is a ground reference ONLY to the op amp when we're running the circuit on a positive-only power supply. If we had a split rail power supply, then ground would be Vb.
When we provide a DC bias to an op amp, we usually do so by referencing the non-inverting pin of the first op amp stage that sees audio to whatever 1/2 the supply voltage is (ground for split rail, 4.5V for typical guitar pedals). In a non-inverting amplifier, that's done with a big resistor. (Inverting stages can just directly couple the e.g. pin 3 to 4.5V). This is done so that, in the simplest possible terms, the op amp will actually work right.
I've been dancing around your actual question, but I promise I understand what you're asking. I just want you to get a good sense of what Vb actually IS, and why the op amp cares about it at all.
Let's pull back for a sec and look at the circuit as a whole. We have AC presented at the inputs and outputs, with capacitors to block DC voltage from being seen by any other circuit in line.
Between those capacitors, even though there's DC present almost everywhere you look, we're still playing with an AC signal. Remember, though that DC is only a voltage if it's relative to something else. When you measure DC on the pin of an op amp, you are usually doing so from the perspective of circuit ground -- you put your black lead on 0V and measure here and there. But what happens if you put your black lead on 4.5V? Well now suddenly there are very few places where you see any voltage at all while you're sticking close to the op amp and not going outside the boundaries set by those input and output capacitors. You will see 4.5V at pin 8 of a dual op amp and -4.5 at pin 4.
Some intriguing things are possible when you don't really have any DC messing with the signal you're playing with. You can DC couple op amp stages and avoid capacitors in between. (Some classic preamps go to great lengths to ensure that there are no capacitors anywhere, for instance.) Another thing you can do is reference things to Vb that you normally think of as going to ground ... but without the capacitor involved.
So we're finally going to tackle the specific circuits you mentioned.
Well, actually, hold up a sec. There's one more schematic you should go grab. Let's say ... the electra distortion. Look at the output. The diodes are back to back and they go to ground. After the output capacitor. The transistor should be biased to 4.5V (sound familiar?).
Compare this to the 8Ball, where the diodes go to Vb. What's the difference? Well, the op amp thinks Vb is its ground. So the op amp doesn't care that the diodes are going to Vb and it'll happily keep chugging away. Even without a capacitor.
Quiz time before you read further: What would happen if you referenced those diodes to circuit ground (0V) instead, without adding the blocking capacitor(s) like in the electra distortion? Hint: What will the OP AMP think 0V is?
Let's look at the Egghead. Without getting into the (surprisingly complicated) difference between sending diodes directly to Vb, what voltage are the diodes in the Egghead referenced to?
Well, you could just grab your multimeter and play around on the breadboard, but it's better to think about this a little more analytically. What are ALL pins of the op amp sitting at (except supply pins 8 and 4)? So what does the op amp think the voltage reference is for those feedback diodes?
By this point, hopefully it's clear that as far as the op amp is concerned, the diodes are all being referenced to the same voltage, so there isn't really a difference between D1 and D2 in the 8Ball compared with all the other diodes in those two pedals.
There's one other thing that's different between the op amp stages in the 8Ball and Egghead. This one's a little more complicated. It's the negative feedback capacitors, C3 in the Egghead (which goes to ground) and C4 in the 8Ball.
Well, let's do a quick review of how the gain works in a dual op amp. It's set by the resistance in the feedback loop of pins 1 and 2 divided by the resistance from pin 2 to ground.
You'll find if you move things around on your breadboard that if you change C4 in the 8Ball to go to ground instead of Vb, it sounds different. Probably gainier, but not massively so. Technically you don't even need C4 (meaning you could short it and the circuit would still work), but it's there for tone shaping.
We could stop at this point and leave it at "well, it sounds different between Vb and ground, so someone thought Vb sounded better." We could even go with the slightly more technical answer of "at some point someone doing the design thought it would make the PCB layout easier." Which is waaaaaaaay more common than most people think for circuit design choices.
But at this point, you might be able to figure out why it sounds different, which is just as important as knowing it does. What's the main thing that's different between actual ground and 4.5V "ground" in any of these circuits? The million dollar word starts with an i. Or is it a z?
