A noninverting stage preserves input polarity

Feedback divider sets the gain

Connect Vin to V+, Rf from output to V−, and Rg from V− to Vref. With an ideal stable linear loop and negligible input currents, KCL gives $V_o=V_{ref}+(1+R_f/R_g)(V_{in}-V_{ref})$. The signal and noise gain are both $1+R_f/R_g$ for this simple topology. A follower is the special case of direct output feedback and unity gain.

Let Vref = 1.65 V, Rg = 10 kΩ, and Rf = 100 kΩ. Gain is 11. A sensor input of 1.67 V produces an ideal output of 1.87 V, a 220 mV deviation above reference. At 1.63 V input, output is 1.43 V. A ±20 mV input range therefore maps to 1.43–1.87 V, comfortably inside an assumed 0.2–3.1 V output range.

Loading still exists

The ideal noninverting input has infinite resistance, but a real op amp has input resistance, capacitance, and bias current. A source with 10 kΩ resistance and 10 nA bias current can acquire 100 μV of input error, which gain 11 turns into approximately 1.1 mV output error. The sign depends on current direction. Include this term separately from input offset voltage.

An output buffer provides voltage isolation but draws energy from its supply. It cannot deliver unlimited load current. At 1.87 V into 10 kΩ to ground, load current is 0.187 mA. Into 100 Ω it is 18.7 mA, a very different device requirement. Feedback does not make output-current limits disappear.

Choose ratios and absolute values

Changing Rg/Rf from 10 kΩ/100 kΩ to 1 kΩ/10 kΩ preserves ideal gain but increases feedback-network current and changes noise and bias-current effects. For this input, the current through Rg is (1.67 − 1.65)/10 kΩ = 2 μA. Keep a connection table and distinguish current to the reference from current to the load.

The reference must be able to handle those feedback currents. ADI's single-supply biasing note explains practical reference and decoupling considerations.