Help with understanding Cmoy assembly

Compare the schematic to the layout and to get an idea of what is going on (how things are connected) - with 2 ways of representing it (schematic vs physical layout), it's usually easier to see things more clearly.

Working with the image linked below:

http://tangentsoft.net/audio/cmoy-tutorial/bitmaps/mini-cmoy--150-ann-board-1.jpg

Which one is the ground jumper?

The one in the middle that connects the two strips that run down the center of the board?

the best way to understand or learn what is going on is to disregard the physical circuit and try to learn from only the schematic.

which jumper is the ground jumper you mention? i'm assuming based on what you call it it's connecting something to ground. if you are finding certain parts of the circuit confusing or dont understand why something is the way it is i can try explaining it more.

i meant that more as a don't rely on the physical pictures to understand what the circuit's doing if you want to learn the design, but by all means ask questions. as mister x said, if you're looking at someone else's picture of a circuit and dont know what a particular piece's purpose is, then going back and forth between the picture and the schematic can help you see what part of the schematic that piece is, and then from there try to understand its purpose.

you probably mean voltages, not current. to use an old analogy, current is like water; it goes in one direction or the other, but the flow is never negative.

with this circuit, there isn't quite a negative voltage either. the "negative" voltage comes from the virtual ground. everything needs to be connected to the same ground in order to have the same reference point; a voltage is a difference in potential between two points. if your circuit isn't properly grounded some parts may be off doing their own thing and not behaving how you'd expect it to with the rest of the circuit. since the opamps require a positive and negative voltage for power, the solution is the virtual ground. let's say you're using a 9v battery. that means the difference in potential between the positive and negative ends of the battery is 9v; for ease of calculations and such we say that the ground of the battery is 0. by introducing the virtual ground, which by design is halfway in between the positive and negative (4.5v in this case), and then using this plane as the reference point for everything else, we in effect now see the + side of the battery at 4.5v and the - side at -4.5v, with the virtual ground as 0v and everything else in the circuit grounded to that.

a physical representation is a helpful tool for when you are first starting to learn, but eventually you will move on to mainly just using schematics :)

also this isn't a good example for a helpful learning physical representation of a circuit since it emphasizes compactness so it's not very clear with what's going on.

that's a way to look at it.

disclaimer: i dont remember too much about how opamps work so i may be kind of incorrect on the following, but im pretty sure it's correct.

audio signals are ac signals (think sine wave) that are centered on the signal ground (for stereo audio, the channels are left/right/ground). the opamp's maximum output voltages are the +/- voltages powering it. If it didn't have the negative voltage and was instead connected to ground there, it wouldn't be able to output any negative voltages. when the audio signal goes above or below the maximum the opamp can put out, it just puts out the max. this is what clipping is; the signal is being clipped off at the voltages that try to go above the max.

let's say the negative side of the battery is 0v. by tieing everything to the virtual ground (including the ground channel from the audio), we are in effect level shifting the audio signal up by 4.5v (with our 9v battery example). this makes it so that the ac signal is now centered on 4.5v and wont clip (assuming the signal's magnitude never exceeds 4.5). if we had instead tied the ground to the negative side of the battery, since the opamp can only output from 0 to 9 volts and the sine wave is centered on 0v, you would only be able to output the positive half of the waveform. it's all a game of reference points; with it centered at 4.5, we can output both the positive and negative parts of the signal, and then by passing 4.5 as the ground for the output, your headphones will use that as the reference point that the audio signal is centered around. if you had passed on the 0v ground from the battery instead, your headphones would not be happy.

Something like that.

http://tangentsoft.net/elec/vgrounds.html