Operational Amplifier Ground – What Reference-Potential Does an Op-Amp Use as Ground?


Wikipedia states on operational amplifiers, that, simply speaking, they provide an output voltage at their output which is the difference of the two input voltages, multiplied by some (very very large) number.
Since voltages are just electric potential differences, I'd like to know what point in the circuit is the reference point for the output of the operational amplifier?

For the inputs, it essentially doesn't matter. Since the behaviour of the operational amplifier only depends on the difference of the two inputs, it doesn't matter what ground they correspond to.

My simplest guess would be that the output voltage of the operational amplifier has the same common ground all the other devices in the electric circuit have. In that case: How does the operational amplifier know of this common ground, if it doesn't have a connection to the common ground?

I need to know what the reference point is to even make any sense of the statement "the output voltage is A times the input difference". If I don't know what the reference point is, then the output voltage is just a number without meaning.

Best Answer

It doesn't know nor care. Opamp's internal circuitry works like this:

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Uout is near 0V when U1 < U2 and Uout is near the full supply voltage Us when U1 > U2. Just around the case U1=U2 there's a transition zone A. Its width in practical opamps is well below one millivolt. Nobody guarantees that the zone is linear or symmetrically around the zero, but normally opamps are used with a feedback circuitry which forces the opamp to output such Uout that U1-U2 is inside the transition zone.

People often divide Us to two parts in series. The upper part is said to be the positive supply and the lower part is said to be the negative supply. The midpoint is said to be the ground and all voltages in application designs are referenced to it. But internally the opamp references all to one of the poles of the supply voltage Us.

In low cost IC designs the internal circuitry cannot accept U1 and U2 to be whatever, they must be between 0 and +Us and there's some margin needed. The margin is needed at least in one end of the range 0...+Us to leave some operating room for the internal circuitry.

Many buyers want no margin at 0V, they want that U1 and U2 can be 0V. If the internal circuitry is designed properly (=PNP input transistors), the usable range for U1 and U2 is from zero to 1...1,5V less than Us. One old and well known IC design LM124 is like that. Analyzing its internal schematic reveals that it actually uses the plus pole of the Us as its internal reference point, but that makes no difference to my drawings and equations.