 Solution

In an operational-amplifier negative feedback loop, an operational amplifier has large open-loop gain $A$, with $V_o=A(V_{+}-V_{-})$. Feeding a fraction $\beta V_o$ back to the inverting input opposes an output increase. For a non-inverting signal input,Its sensitivity to open-loop gain changes is reduced by $d\ln A_{\mathrm{cl}}/d\ln A=1/(1+A\beta )$. Thus resistor ratios can set a reproducible gain despite device variation; feedback also suppresses internal distortion and reduces output impedance. Gain is traded for usable bandwidth, subject to the amplifier's frequency response.

For the non-inverting connection, place $R_f$ between output and inverting input and $R_g$ between that input and ground. Then $\beta =R_g/(R_f+R_g)$, giving gain $1+R_f/R_g$. For an inverting connection, ground the non-inverting input, inject the signal through $R_{\mathrm{in}}$ and return $R_f$ from output to the inverting node. Small differential input and negligible input current then give gain $-R_f/R_{\mathrm{in}}$. The near-zero node is a virtual ground, not a physical ground connection.

The feedback must remain stable at the frequencies where loop gain matters. Extra phase lag can turn the intended opposition into oscillation; adequate phase margin, supply headroom, input common-mode range, output-current capability and slew rate are required. Feedback neither prevents clipping nor removes all input noise.^[ref]