Op-Amp Fundamentals: Inverting and Non-Inverting Amplifiers

The two ideal-op-amp rules

An op-amp has two inputs, $V_+$ and $V_-$ , and (with negative feedback) two assumptions make circuit analysis almost algebra-free:

No current flows into either input terminal (infinite input impedance).
The op-amp drives its output to force $V_+ = V_-$ , the virtual short, even though the two inputs aren't actually connected.

Inverting amplifier

The input signal drives $R_{in}$ into the $V_-$ terminal, with a feedback resistor $R_f$ from output back to that same terminal, while $V_+$ is grounded. Since $V_-$ is forced to $0$ V (a "virtual ground") and no current enters the input, the current through $R_{in}$ must equal the current through $R_f$ , giving:

\text{Gain} = \dfrac{V_{out}}{V_{in}} = -\dfrac{R_f}{R_{in}}

Worked example: inverting amplifier

$R_{in}=1\text{k}\Omega$ , $R_f=10\text{k}\Omega$ , $V_{in}=0.5\text{V}$ :

V_{out} = -\dfrac{10\text{k}}{1\text{k}} \times 0.5 = -5\text{V}

Non-inverting amplifier

The input signal now drives $V_+$ directly, and $R_1$ plus $R_f$ form a voltage divider from output back to $V_-$ , forcing $V_- = V_{in}$ by the virtual-short rule:

\text{Gain} = \dfrac{V_{out}}{V_{in}} = 1+\dfrac{R_f}{R_1}

always positive (no inversion) and always at least 1, even if $R_f=0$ . That special case, with $R_f = 0$ and $R_1$ left out, is the voltage follower: gain exactly 1, used as a buffer because it draws almost no current from the source.

Worked example: non-inverting amplifier

$R_1=1\text{k}\Omega$ , $R_f=9\text{k}\Omega$ , $V_{in}=0.5\text{V}$ :

V_{out} = \left(1+\dfrac{9\text{k}}{1\text{k}}\right) \times 0.5 = 10 \times 0.5 = 5\text{V}

Same gain magnitude (10) and output size as the inverting example above, but positive, and it took a resistor ratio of 9 rather than 10 to get there, a direct consequence of the $+1$ in the formula.

Inverting vs non-inverting: the differences

Inverting and non-inverting amplifiers compared

	Inverting	Non-inverting
Signal goes to	$V_-$ through $R_{in}$	$V_+$ directly
Gain	$-\dfrac{R_f}{R_{in}}$	$1 + \dfrac{R_f}{R_1}$
Output phase	Flipped (180°)	Same as input
Smallest gain magnitude	Any, including below 1	1 (voltage follower)
Input impedance	About $R_{in}$	Very high

Worked example: a summing amplifier

Add a second input resistor to the inverting amplifier and the virtual ground adds the currents from both inputs:

V_{out} = -R_f\left(\dfrac{V_1}{R_1} + \dfrac{V_2}{R_2}\right)

With $R_1 = R_2 = R_f = 10\text{ k}\Omega$ , $V_1 = 1$ V and $V_2 = 2$ V:

V_{out} = -10\text{k}\left(\dfrac{1}{10\text{k}} + \dfrac{2}{10\text{k}}\right) = -3\text{ V}

Common mistakes

Dropping the minus sign. An inverting amplifier's output is always opposite in sign to its input.
Forgetting the $+1$ . A non-inverting amplifier with $R_f = 9\text{ k}\Omega$ and $R_1 = 1\text{ k}\Omega$ has a gain of 10, not 9.
Ignoring the supply rails. The output can't go beyond the supply voltages. A gain of $-10$ on a 2 V input asks for $-20$ V, but on a $\pm 12$ V supply the output saturates near $-12$ V instead.
Using the virtual short without negative feedback. $V_+ = V_-$ only holds when the output is fed back to the inverting input. Without feedback (a comparator), the output just swings to a rail.

Where this comes up in exams

Op-amp circuits are listed in the GATE 2027 EC (amplifiers and summers, under Analog Circuits), EE (operational amplifiers) and IN (op-amp applications including the adder) syllabi. They are not in the JEE Main 2026 or NEET (UG) 2026 syllabi.

Checked against GATE 2027 EC syllabus, GATE 2027 EE syllabus, GATE 2027 IN syllabus, JEE (Main) 2026 syllabus (NTA) and NEET (UG) 2026 syllabus (NMC, published by NTA). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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