Why do op-amps have labeled terminals


For some type of op-amp circuits, the inverting terminal is placed on top, while on others it's at the bottom (i.e. inverting vs non-inverting amplifier). I don't understand the convention – how do you decide what to label each terminal? Aren't either labels valid since both terminals are equivalent?

Best Answer

An individual op-amp will have one non-inverting input (usually denoted with a + symbol) and one inverting input (usually denoted with a -).

They are very much not equivalent. As their description makes apparent, one inverts it's input value, and the other does not.

Now, with regard to the drawn symbol, which is on top is generally a function of what will make the schematic clearer and/or easier to draw. They're both valid ways to draw an op-amp, though the circuit has to accommodate which connection is where.

You decide what to label each terminal based on the part datasheet, which will tell you which physical pin maps to what function of the device.

Fundamentally, an op-amp's output is the difference between the input pins, multiplied by the op-amps gain (ignoring the non-idealities of op-amps for the moment).

\$ V_{OUT} = GAIN * (V_{+} - V_{-})\$

Now, let's take the two simplest possible configurations:

  • - input is grounded (e.g. 0V), input signal is on + input:
    Circuit behaviour is: \$ V_{OUT} = GAIN * (V_{+} - 0)\$, which simplifies to \$ V_{OUT} = GAIN * V_{+}\$

  • + input is grounded (e.g. 0V), input signal is on - input:
    Circuit behaviour is: \$ V_{OUT} = GAIN * (0 - V_{-})\$, which simplifies to \$ V_{OUT} = GAIN * -V_{-}\$

In the latter case, the input voltage is \$-V_{-}\$, which has the effect of inverting the input voltage on the - across the voltage at the + pin (in this case ground, so the term simplifies out). The fact that you invert the sign of the input is why the inverting input is called the inverting input \$-\$.

I don't know how you're analyzing an op-amp circuit that you are not seeing a difference when you swap the inputs, but you're apparently doing it wrong.