Inverting amplifier output and saturation

For an ideal inverting amplifier, V_out = −(R_f/R_in) V_in.

Worked answer.

Step 1: Gain. For an ideal op-amp in the inverting configuration, the inverting input is a virtual ground. The current through R_in equals the current through R_f, which gives
A_v = −R_f / R_in = −100 kΩ / 10 kΩ = −10.

Step 2: Output for 0.2 V. V_out = −10 × 0.2 = −2 V. This is well within the ±12 V supply, so the op-amp works linearly.

Step 3: Check with currents. Current through R_in = 0.2 V / 10 kΩ = 20 µA. The same 20 µA flows through R_f (an ideal op-amp draws no input current), so the drop across R_f is 20 µA × 100 kΩ = 2 V. Since the inverting node sits at 0 V, the output is at −2 V. This matches.

Step 4: Input of 1.5 V. The ideal formula predicts −15 V, but the output cannot go beyond the supply rails. It saturates near the negative rail. For a general-purpose op-amp such as the 741 on ±12 V, the output swing is typically limited to roughly ±10 to ±11 V; a rail-to-rail output op-amp gets closer to ±12 V. Either way, the output clips and is no longer −10 × V_in. The virtual-ground assumption also fails, because the feedback loop is no longer in control.

Input resistance. The input resistance seen by the source is R_in = 10 kΩ, because the inverting input is at virtual ground. This is one drawback of the inverting configuration compared to the non-inverting one, whose input resistance is very high.

Common mistake. Giving +10 as the gain, forgetting the 180° phase inversion.

Likely follow-up. "What resistor should you put on the non-inverting input?" A resistor equal to R_in ∥ R_f (about 9.1 kΩ) to ground reduces the output offset caused by input bias currents.