Non-inverting amplifier (op-amp)

The non-inverting amplifier is the second of the two foundational op-amp Negative feedback topologies. The input signal $v_I$ drives the non-inverting input directly; the feedback resistor $R_2$ runs from the output back to the inverting input, and $R_1$ runs from the inverting input to ground. Its closed-loop gain is $1+R_2/R_1$ — always positive, always at least one, and set entirely by external resistors.

Deriving the gain

Same toolkit as the Inverting amplifier (op-amp): the two golden rules of the Ideal op-amp model plus KCL.

Step 1 — golden rule 2. Negative feedback forces $v_{+}=v_{-}$. Here the signal drives the non-inverting input directly, so $v_{+}=v_I$, and therefore $v_{-}=v_I$. Note the inverting node is not grounded this time — there is no virtual ground, just a virtual short tracking the input.

Step 2 — golden rule 1 + KCL at $v_{-}$. No current enters the inverting input. The inverting node connects to two things: $R_1$ going down to ground, and $R_2$ coming back from $v_O$. The current down through $R_1$ is $(v_{-}-0)/R_1$; the current arriving from the output through $R_2$ is $(v_O-v_{-})/R_2$. With no current into the op-amp, these must balance:

$\frac{v_{-}}{R_1}=\frac{v_O-v_{-}}{R_2}$

Substitute $v_{-}=v_I$:

$\frac{v_I}{R_1}=\frac{v_O-v_I}{R_2}$

Step 3 — solve. Multiply out: $R_2{v}_I=R_1(v_O-v_I)$, so $R_2{v}_I+R_1{v}_I=R_1{v}_O$, giving

$\boxed{\frac{v_O}{v_I}=1+\frac{R_2}{R_1}}$

The gain is positive — no phase inversion — and cannot be made less than unity (set $R_2=0$ or $R_1=\infty$ and you get exactly $1$, the Voltage follower (op-amp)). With $R_2=99\text{k}\Omega$ and $R_1=1\text{k}\Omega$ the gain is $+100$. Intuition for the "$1+$": the inverting node is held at $v_I$ by the virtual short, so the resistor string $R_1$–$R_2$ from ground to $v_O$ is a Voltage divider whose tap must equal $v_I$; solving the divider for $v_O$ gives the extra $1$.

Input resistance — the advantage

The source drives the non-inverting input directly, and golden rule 1 says no current flows into it. So the source sees the op-amp’s own input resistance, which is ideally infinite:

$R_{in}=R_{in,\text{op-amp}}\to \infty$

Nothing is drawn from the source — no Voltage divider loading, no signal lost. This is the configuration’s principal advantage and the reason it is preferred for buffering high-impedance sources where the Inverting amplifier (op-amp)‘s finite $R_{in}=R_1$ would attenuate the signal. The trade-off: you cannot get gain below 1, and the inverting node sees the common-mode input swing (relevant for Common-mode rejection ratio at high gain). For a pure buffer with no gain, the degenerate case is the Voltage follower (op-amp).