This was a mistake I made twice already, so I thought it would be good to write this for when anyone else makes it too. Op-amps have two golden rules:

  1. Zero current flows into the minus and plus terminals, else called the inverting and non-inverting inputs
  2. If there is negative feedback, the op-amp will ensure that the voltages at the minus and plus terminals are equal

And from these rules a misunderstanding can begin. Nothing is off when looking at an op-amp in the non-inverting configuration.

Non-inverting amplifier

As one expects, zero current flows into this circuit by rule one. That means the input impedance (the load this amplifier circuit presents) is infinite. In reality, this would be in the megaohms, but they more or less do not load their inputs.

However, that is not true for the inverting configuration! The key here is rule two.

Inverting amplifier

There is negative feedback here, so the voltages at the plus and minus terminals are equal. Since the plus terminal is tied to ground, the minus terminal must be at zero volts. This is called “virtual ground”. The reason why it isn’t “real” ground is because the current cannot end there (remember rule one, zero current flows into the minus terminal!). But the point is this: from the perspective of the input, the other side of R1 is still at zero volts, so by Ohm’s Law just as much current flows as would flow if the other end was real ground. In other words, all the input sees is a load resistor to ground, real or not. Therefore, the input impedance of the inverting amplifier circuit is exactly R1, a far-cry from infinite!

Apparent load seen by V_in

What the input sees: R1 appears as a loading resistor

In reality, this more or less does not change: inverting amplifiers present a load equal to R1 to their inputs.

Does this contradict rule one? Absolutely not. I did not keep calling the minus and plus terminals “inputs” in this article because my mistake started by confusing the circuit’s input with the op-amp’s inputs. Zero current flows into the terminals, and zero current flows into a non-inverting amplifier, but current does flow into an inverting one. What an amplfier does has everything to do with the configuration, less so with the op-amp itself.

So why does this matter? Input impedance is critical for designing audio amplifiers, for example. If I use an inverting circuit and make R1 too low, I can draw more current than DACs or other audio equipment can handle, and this can cause distortion or even damage. In these cases, it is best to use a non-inverting configuration or to choose a safe, high R1 like 10k. This also applies to summing amplifiers, which are also an inverting configuration.


There’s another way to look at it if you’re familiar with linear circuit design. If I model the input as an independent voltage source…

Independent voltage source applied

I can convert R1 and that source into a Norton equivalent, where R1 suddenly becomes trivial.

Transimpedance amplifier

Transformed inverting amplifier where $V_{out} = -i_{in}*R_f$ and $i_{in} = \frac{V_{in}}{R_1}$

And this reveals a new way to look at inverting amplifiers: a current-to-voltage converter, also called a transimpedance amplifier. Transimpedance (and inverting) amplifiers are fundamentally based on current, and all R1 does is set the current. Because of this, inverting amplifiers can never draw zero current without also having a zero output, whereas a non-inverting one can. In other words, inverting amplifiers can never have infinite input impedance, even ideally.