+/- 9V

[bachevelle]

I just realized that I may have made a mistake on my first pedal. (the begginer project) Foolishly I didn't notice that the -9V lead went up to the input jack. I had assumed that was all part of the on/off switching, and since I breadboarded this thing, I left that part out. The problem is, I connected the -9V to the ground rail on my breadboard. I can take a wild stab now that this ISN'T correct. The question is, it worked, there was alot of noise but it did work. Is there any chance that I damaged anything? I know that this part of the question would be better suited for the begginer project forum, rest assured there is more to come. From the battery leads, is this "true" -9V? Where does it go from the terminal on the input jack? Also, I have a couple of 741s and 82s lying around, any suggestions on pedals that use these? One last thing on the +/- V,  I've seen some schematics with say +50VDC on the upper rail, and -50VDC on the lower rail, this is why I was wondering about the -9V from the battery, and where it went. I noticed on the pin-out of the 741s, there is V+ and V-. To sum it up....is the power from a 9V battery truely "bipolar?" (if that's the right word) To achieve + and - Voltages for an op-amp?

[calpolyengineer]

There is nothing wrong with the negative terminal of the battery going to the ground rail, that is what it does anyways because the input jack is grounded. What it seems to me is that you don't quite get the concept of voltage (no offense). Voltage is a relative measurement, not a definate quantity. For instance a So if you take your reference point (called ground) to be the negative terminal of the battery, then the positive terminal is 9V above that. If however you take ground to be the positive terminal then the negative terminal is 9V below it, or -9V. To answer your question, batteries are not bipolar. Because grounding is all relative, most of the time there is a voltage divider which will put out 0V (ground), 4.5V (V/2), and 9V (V). What you do is put ground into the -Vcc of the opamp, 9V to the +Vcc, and then feed 4.5V into the input signal to raise the signal 4.5VDC above the ground so that relative to the input, the +Vcc is +4.5V and the -Vcc is -4.5V. Another way to do this is with a "floating" ground in which you use another voltage divider, but you call your ground at the midpoint of the divider. In order to get real +/- 9V, you have to use two batteries (or use a charge dump, but that is getting pretty complex) and take your ground in between them, then the unconnected + terminal will be 9V above ground and the unconnected - terminal will be -9V.

-Joe

[bachevelle]

So why would you want -V anyway? You are absolutely right though, I'm still just trying to wrap my head around this. How can you be negative from ground? If Voltage can be considered "potential," and ground would seem to be no "potential", how can you have less "potential" than none? I'm retarded. I just don't seem to understand.

[calpolyengineer]

Ground isn't no potential though, it is 0V potential relative to itself. Ground is only where you reference it. So there is true earth ground, but the ground for a circuit might be many volts above (or below) this earth ground. However, inside the circuit, all of the voltages are relative to the circuit's ground.

Also you have to think of potential not as a quantity but as a magnitude of a relationship. The reason a DMM will tell you whatever voltage is because it reads the difference from the red lead with respect to the black lead. If you put the red lead on the + end of a 9V battery and the black on the -, the DMM will read 9V because the difference in potential from the black lead to the red lead is +9V. Now if you put the black lead on the + and the red lead on the -, it will read -9V because the - terminal is 9V below the + terminal.

Hopefully you know a little bit about thermodynamics because voltage can be thought of similarly to entropy. Entropy is commonly called a measure of the disorder of a system. But you cannot say that a system has X units of entropy, you can only measure the CHANGE in entropy. The same is basically true for voltage. You can't say that a particular point has X volts, you have to say that it is X volts different than my reference point.

In order to understand voltage as a potential, they usually teach a water pipe analogy. What they say is that there is a resevoir at the top of a hill and a pipe that runs down the hill. The equivalent to voltage is the pressure in the pipe at the bottom of the hill. With this analogy you can see that they are referencing the pressure in the pipe relative to the pressure at the resevoir (which they call zero). You can also see that a resevoir placed at sea level would have a potential of zero, but so would a resevoir placed on top of Mt Everest, but there is also a potential difference to one relative to the other (0V is not always 0V).

Anyways, back to the analogy. Since the pressure is higher at the bottom, they would say that there is a positive "voltage" difference. Now to understand negative voltage, do the opposite. A resevoir at the bottom of a ditch with a pre-filled pipe running up the side of the ditch. If you take a measurement of the pressure at the end of the pipe, you would expect it to be lower than the pressure at the resevoir because gravity would pull the water down the pipe. In this case since the pressure at the end of the pipe is less than the pressure at the resevoir (which is still "ground") then the potential would be negative.

I hope that gives you a pretty comprehensive understanding of voltage. I think I just about exhausted every possible way I know to describe voltage.

-Joe