In an inverting amplifier, why is the virtual ground not a real ground?

In an inverting amplifier, why is the virtual ground not a real ground?

• A. It is a real ground connected via a resistor.
• B. The inverting input is held at the same potential as the noninverting input by negative feedback.
• C. There is no current flowing into the inverting input.
• D. The op-amp fixes a constant potential regardless of inputs.

Answer: (B)

The key idea is that the op-amp uses negative feedback to force the inverting input to match the noninverting input voltage. When the noninverting input is connected to ground, that means the inverting input is held at 0 V as well. But it’s not a real ground, because there’s no low-impedance connection to earth at that node. The op-amp adjusts its output so that the voltages at both inputs are equal, and because the inputs themselves draw essentially no current, all the current from the input resistor flows through the feedback resistor to the output. So the node sits at ground potential without being a true ground.

The key idea is that the op-amp uses negative feedback to force the inverting input to match the noninverting input voltage. When the noninverting input is connected to ground, that means the inverting input is held at 0 V as well. But it’s not a real ground, because there’s no low-impedance connection to earth at that node. The op-amp adjusts its output so that the voltages at both inputs are equal, and because the inputs themselves draw essentially no current, all the current from the input resistor flows through the feedback resistor to the output. So the node sits at ground potential without being a true ground.