An inverting amplifier uses a current balance

Current through two resistors

Connect input to V− through Rin, output to V− through Rf, and V+ to a reference voltage Vref. Assume negligible input current and a stable unsaturated loop. Then V− ≈ Vref, so KCL at the inverting node gives $(V_{in}-V_{ref})/R_{in}+(V_o-V_{ref})/R_f=0$. Rearranging yields $V_o=V_{ref}-(R_f/R_{in})(V_{in}-V_{ref})$.

For a dual-supply example with Vref = 0, Rin = 10 kΩ, Rf = 40 kΩ, and Vin = 0.2 V, output is −0.8 V. Input current is 20 μA. This input resistance is approximately Rin, so a high-source-resistance sensor would be loaded. A virtual ground is not a physical ground connection: feedback makes the node close to ground while resistor currents flow into and out of it.

A single-supply example

On an assumed 3.3 V supply, use Vref = 1.65 V and the same ratio 4. If Vin = 1.75 V, output is 1.25 V. If Vin = 1.55 V, output is 2.05 V. The amplifier inverts deviations around the reference, rather than producing negative absolute output voltages that the supply cannot support.

If an input excursion of +0.5 V around Vref requests output −0.35 V, the circuit cannot satisfy the ideal equation on that supply. Check output swing, input common-mode range, and required load current before calling the design complete.

Reference quality enters the signal

Rewriting the equation gives $V_o=(1+R_f/R_{in})V_{ref}-(R_f/R_{in})V_{in}$. With ratio 4, a 1 mV reference shift produces a 5 mV output shift if Vin stays fixed in absolute voltage. Reference noise and drift therefore matter. If the sensor and amplifier share a tracking reference, analyze their correlation rather than automatically treating both reference contributions as independent.

Record the signal gain −4 and the noise gain 5 separately. The latter affects bandwidth estimates for a compensated voltage-feedback op amp. One resistor ratio serves two related but different calculations.