What is the difference between positive, negative and common?
To put it as simply as possible .... it depends on where you put your multimeter probes.
The following diagram shows the same battery setup, repeated 3 times. Each of these has been labelled with voltages. All of them are correct
in themselves (you cannot mix them). The trick is - you pick
one point as your reference point and call that zero volts. Once you've done that, connect the black lead from your digital multimeter to that spot and leave it there. Use the red lead for your voltage measurements. (Note: This is safe with digital meters. Analogue meters usually require special care.)
Example:
I always thought that if is do this I will simply double the voltage.
So say the power supply's are 6V each, if I connect a bulb across the V+ and common or V- and common I would have 6V going through the bulb right?
Absolutely correct.
And if I connect the bulb across V+ and V- I would get 12V?
Again, absolutely correct.
So what is all this positive and negative voltage?
If I wanted to get pedantic, I would ask you - if you connect that bulb across the V+ and V-, have you put +12v across it or -12v across it?
The answer is, as I originally stated, it depends on where you put your multimeter probes.
So, if they are all the same thing, then you might ask - "Which is the best?" The answer to that is "Whichever is the most convenient" - which leads us to an answer for this question:
Its especially confusing in Op-Amps. Why would I amplify a voltage to a negative?
There are several ways to try and explain this - but I will choose this one:
Because the mathematics is easier.If you take the case of a sine wave, mathematically, it cycles between + 2 and -2, so setting a reference point on a circuit where the signal follows the same movement makes it a lot easier.
For example, say we have an amplifier circuit that takes a sine wave input signal and produces an output that is twice the magnitude....
1. Using the middle example of mine, lets say the input signal varies from +7v and +11v i.e. 4v peak to peak.
OK - let's multiply everything by 2!
This gives an output signal varying from +14v to +22v i.e. 8v peak to peak. But that can't be right. The high voltage of 22v is impossible. The maximum is 18v. If we were to put the signals on a scope, we would see that our maths is wrong.
Our maths needs some "tweaking".
The problem is that the amplifier does not amplify signals relative to our 0v reference - it amplifies them relative to somewhere between the two power rails. With no input signal, we can see this is around the +9v mark and our scope observations confirm this.
We can now re-do the maths using this "offset" of 9v ... and it looks like this:
- Input signal varies between +7v and +11V (4v p-p)
- Subtract the offset and the "input signal" varies from -2v and +2v (still 4v p-p)
- We multiply the signal by the amplification factor of 2
- The "output signal" varies between -4v and +4v (8v p-p)
- We add back the offset and the output signal varies from 5v to 13v (8v p-p) (confirmed by scope observation)
2.Now let's try it with the first example.
- Input signal varies between -2v and +2v (4v p-p)
- We multiply the signal by the amplification factor of 2
- The output signal varies between -4v and +4v (8v p-p)
Much easier.