Electronic – Basic question regarding op-amp output voltage

operational-amplifier

When we model an operational amplifier, we say that the output voltage \$V_{out}\$ is proportional to the difference between the inputs \$V_\pm\$, \$V_{out}=A(V_+-V_-)\$. I've read that this voltage is always with respect to ground.

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However, when we add a double supply to the op-amp, how is ground defined? Does the circuit take it to be \$V_-\$? Or maybe it is \$(V_+-V_-)/2\$?

Best Answer

"Ground" in this context is really a misnomer. It's only a reference. For example, take this circuit that you probably know:

schematic

simulate this circuit – Schematic created using CircuitLab

and connect the "ground" to another control instead:

schematic

simulate this circuit

You can think of it as being the exact same circuit with the exact same behavior, provided that the reference control stays constant. It simply centers on this new reference, and there's now no need for an explicit ground.

The actual rules for opamps are:

  1. The output goes as far as it needs to, to keep the two inputs equal.
  2. If the "+" input is higher than the "-" input, it goes up; if lower, it goes down.
  3. If the output hits one or the other supply rail, it stops there.
    (actually a volt or two short unless it's explicitly a rail-to-rail design)

You'll notice that there's no mention of "ground" in any of that. Only comparing the two inputs to each other and the output to the supply rails. That's it!

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