Audited 31 Jul 2026·Last updated 28 Oct 2026·5 citations·Tier 1·0 uses

Op Amp Gain Calculator — Inverting, Non-Inverting, Buffer and Difference

Work out op-amp closed-loop gain, output voltage, noise gain and bandwidth for inverting, non-inverting, follower and difference stages from Rf, Ri and GBW.

Op Amp Gain Calculator

Configuration
Only the RATIO of Rf to Ri sets the gain, but the absolute values matter in practice: too small and the op-amp has to drive the feedback network, too large and bias current, noise and stray capacitance start to dominate. Values from about 1 kΩ to 100 kΩ are the usual compromise.
Ri unit
The resistor from the output back to the inverting input. Together with Ri it sets both the signal gain and the noise gain.
Rf unit
Zero and negative values are allowed. In the difference configuration this field is the inverting input, V₋.
V
The non-inverting input of the difference amplifier. Ignored in the other three configurations.
V
From the first page of the op-amp's datasheet, often listed as unity-gain bandwidth. A 741 is about 1 MHz; general-purpose and precision parts run from about 1 to 10 MHz, and high-speed parts reach hundreds of MHz.
GBW unit
A symmetric supply is assumed. Remember that a real op-amp does not swing all the way to its rails — a classic part on ±15 V clips at about ±13 V.
V
Closed-loop voltage gain
10
Signed closed-loop gain: 1 + Rf/Ri non-inverting, −Rf/Ri inverting, 1 for the follower and Rf/Ri for the difference amplifier; a minus sign means the output is inverted.
Output voltage
5 V
Gain in decibels
20 dB
Gain magnitude
10
Voltage actually amplified
0.5 V
Noise gain
10
Closed-loop bandwidth
100,000 Hz
Feedback factor β
0.1
Reading of the result
Non-inverting: the signal enters the + input, so the output is in phase and the input impedance is very high. Signal gain and noise gain are equal here, which is why this configuration gets the most bandwidth for a given gain. 5 V sits comfortably inside the ±12 V supply, using 41.67 % of the available swing. Note that a real op-amp does not swing all the way to its rails: Fiore works his examples with a ±15 V supply giving roughly ±13 V of usable output, so this comparison is optimistic by about that margin unless the part is specified rail-to-rail. This page also models only the small-signal bandwidth — 100000 Hz here. A large signal is additionally limited by slew rate, and the smaller of the two limits is the one that governs.

Background.

Put a 9 kΩ resistor from an op-amp's output back to its inverting input, a 1 kΩ resistor from that input to ground, feed the signal into the other input, and the stage amplifies by ten. Replace the chip with a different part from another manufacturer and it still amplifies by ten. That indifference to the device is why operational amplifiers run so much of analogue electronics. An op-amp's own open-loop gain is enormous — often more than 100,000 — but it drifts with temperature and varies from one unit to the next. Negative feedback throws nearly all of it away and replaces it with a ratio of two resistors you can buy to one percent or better.

A resistor ratio is only the start of a working stage, though. Before anything is soldered, four things need answering: what the gain is for the way the parts are wired, how far the output will travel for the signal you feed in, how much of the op-amp's bandwidth survives at that gain, and whether the result fits between the supply rails. All four come out of one calculation here, for the four resistive configurations that cover most practical circuits: non-inverting, inverting, the voltage follower and the single-op-amp difference amplifier.

Those four are less interchangeable than their formulas suggest. Wire the same 1 kΩ and 9 kΩ pair as an inverting stage and the gain becomes −9, one unit smaller and opposite in sign, yet the bandwidth stays exactly where it was. Bandwidth follows the noise gain, which is 1 + Rf/Ri in both layouts, rather than the gain the signal sees. Push that to unity gain and an inverter made from two equal resistors has only half the bandwidth of a plain follower built on the same chip. Showing noise gain beside signal gain is meant to make that trade visible before it turns up on an oscilloscope.

What is op amp gain calculator?

Closed-loop gain is the ratio of an amplifier's output voltage to its input voltage once negative feedback is connected. With enough open-loop gain inside the loop, the op-amp drives its output to whatever level keeps its two inputs at almost the same voltage. In the non-inverting circuit that means the feedback divider Ri and Rf must return exactly the input voltage to the − input, so the output must be (1 + Rf/Ri) times larger. In the inverting circuit the + input is grounded, the − input sits at a virtual ground, and the current through Ri has nowhere to go but through Rf, so the output is −Rf/Ri times the input. The voltage follower is the non-inverting circuit with the divider removed, giving a gain of exactly 1. The difference amplifier combines both paths and multiplies V₊ − V₋ by Rf/Ri. Noise gain, 1 + Rf/Ri, is the gain the feedback loop itself works against, and it is that figure, not the signal gain, that sets the bandwidth.

How to use this calculator.

  1. Pick the configuration. Non-inverting suits high-impedance sources and gets the most bandwidth for a given gain; inverting gives a virtual-ground summing node and a defined input resistance; the follower converts impedance at unity gain; the difference amplifier subtracts one voltage from another.
  2. Enter Ri and Rf, each with its own unit. The gain depends only on their ratio, but keep both somewhere between about 1 kΩ and 100 kΩ. The follower ignores these two fields entirely.
  3. Enter the input voltage. For an AC signal, use the peak value rather than RMS when you want the rail check to mean anything: 1 V RMS is about 1.414 V peak. In difference mode this field is V₋ and the second field is V₊.
  4. Copy the gain-bandwidth product from the datasheet's first page, where it may be called unity-gain bandwidth, and set its unit. Use the typical figure at your supply voltage if the datasheet gives several.
  5. Enter one rail of a symmetric supply, so ±12 V is entered as 12. For a single-supply stage biased at mid-supply, half the supply voltage is the closest equivalent.
  6. Read the signed gain, the noise gain beside it, the bandwidth, and the verdict, which states how much of the rail-to-rail swing the output uses.

The formula.

A_v = 1 + R_f/R_i · A_v = −R_f/R_i · A_noise = 1 + R_f/R_i · f₂ = GBW ⁄ A_noise

Signal gain depends on the configuration. Non-inverting: Av = 1 + Rf/Ri. Inverting: Av = −Rf/Ri. Follower: Av = 1. Difference: Vout = (Rf/Ri) × (V₊ − V₋), with the gain shown as Rf/Ri and the voltage actually amplified shown separately as V₊ − V₋. These are Equations 4.1, 4.2 and 6.1 in James Fiore's Operational Amplifiers and Linear Integrated Circuits. The output is Av multiplied by the amplified voltage, and the decibel figure is 20·log₁₀|Av|, taken on the magnitude because a negative number has no logarithm; the sign stays in the signed gain.

Noise gain is 1 + Rf/Ri for every configuration that uses resistors, and 1 for the follower. Its reciprocal is the feedback factor β, the fraction of the output returned to the inverting input. Bandwidth follows from the way most op-amps are compensated: open-loop gain rolls off at 20 dB per decade, so gain multiplied by frequency stays roughly constant at the gain-bandwidth product. The closed-loop response stays flat until the falling open-loop gain meets the noise gain, which happens at f₂ = GBW ÷ A_noise (Fiore Eq. 5.1), the point where the response is 3 dB down. Use the linear noise gain there, never the decibel value: 1 MHz divided by 20 gives 50 kHz, half the true answer for a gain of 10.

The op-amp model is ideal except for that one finite parameter: infinite open-loop gain at low frequency, infinite input impedance, zero output impedance, no offset, bias current or noise. The rail verdict compares |Vout| with the rail you entered — clipping at or above it, a warning from 90% upward, comfortable below — and assumes the output could reach the rail, which most parts cannot. Slew rate, offset, common-mode and supply rejection, output current limits and stability margin are all outside the calculation.

A worked example.

Example

A sensor delivers 0.5 V and the stage after it needs about 5 V, so the example uses a non-inverting amplifier with Ri = 1 kΩ and Rf = 9 kΩ, a 1 MHz op-amp and ±12 V rails. The ratio is 9,000 ÷ 1,000 = 9, so the gain is 1 + 9 = 10, which is 20·log₁₀(10) = 20 dB. The output is 10 × 0.5 = 5 V, in phase with the input. Noise gain is also 1 + 9 = 10, the feedback factor β is 1 ÷ 10 = 0.1, and the closed-loop bandwidth is 1,000,000 ÷ 10 = 100,000 Hz, or 100 kHz. On the supply side, 5 ÷ 12 = 41.67% of the available swing, well under the 90% warning line of 10.8 V. The same two resistors wired as an inverter give −9, an output of −4.5 V and 19.0849 dB, but the noise gain is still 10 and the bandwidth is still 100 kHz: less gain for the same bandwidth. That 100 kHz is also a small-signal figure. For a 10 V peak at 100 kHz the output must move at 2π × 100,000 × 10 ≈ 6.28 V/µs, so a part rated at 0.5 V/µs would start distorting a 10 V peak at about 7.96 kHz, long before the bandwidth limit.

gain Bandwidth Product1
supply Rail Voltage12
gbw UnitMHz
input Resistance Unitkilohm
second Input Voltage0.2
configurationnon-inverting
feedback Resistance9
feedback Resistance Unitkilohm
input Voltage0.5
input Resistance1

Frequently asked questions.

Why does the inverting amplifier show a negative gain?
Because the output genuinely moves in the opposite direction to the input: when the input rises by 0.1 V, the output of a −9 stage falls by 0.9 V. For a sine wave that is a 180° phase shift. Fiore explains the sign as a consequence of referencing the output to ground. The decibel figure drops the sign, since only the magnitude has a logarithm.
Can a non-inverting stage have a gain below one?
No. Its gain is 1 + Rf/Ri, and the smallest value Rf/Ri can approach is zero, so the gain never falls below unity. For attenuation, use an inverting stage with Rf smaller than Ri — 1 kΩ against 10 kΩ gives −0.1, or −20 dB — or put a resistive divider in front of a follower if the polarity must be kept.
Why is noise gain shown separately from signal gain?
Because noise gain, 1 + Rf/Ri, is what the feedback loop actually works against, and it sets both bandwidth and stability. In a non-inverting stage the two numbers are identical. In an inverting stage the signal gain is Rf/Ri but the noise gain is one unit larger, so it gets less bandwidth than a non-inverting stage of the same gain. Op-amp input noise and offset voltage are also multiplied by the noise gain.
Is a unity-gain inverter as fast as a voltage follower?
No, it has half the bandwidth. A follower has a noise gain of 1 and gets the op-amp's whole gain-bandwidth product. An inverter with two equal resistors has a signal gain of −1 but a noise gain of 2, so on a 1 MHz part it reaches 500 kHz against the follower's 1 MHz. If you need speed at unity gain and the inversion is optional, use the follower.
How do I choose actual resistor values once I know the ratio?
Scale the pair so that both land roughly between 1 kΩ and 100 kΩ. Very low values force the op-amp to drive real current into its own feedback network, which eats output swing and can reach its current limit. Very high values let input bias current turn into offset voltage, add resistor thermal noise, and form an unwanted pole with stray capacitance across Rf. Precision DC stages tend toward the low end; battery-powered ones toward the high end.
My result says 'comfortably inside the rails' — will the output really reach that voltage?
Not necessarily. The check compares your output with the full rail voltage, and most op-amps cannot drive that far. Fiore's examples treat a ±15 V supply as giving about ±13 V of usable output on a classic part. Rail-to-rail output devices get much closer, often within tens of millivolts at light load, but the gap widens as load current rises. Check the output-swing specification in the datasheet at your load.
What limits a large signal at high frequency if the bandwidth looks fine?
Slew rate, the fastest the output can change, quoted in volts per microsecond. A sine wave of peak Vp at frequency f needs a slope of 2πfVp, so the highest undistorted frequency is SR ÷ (2πVp). That limit falls as amplitude rises and is independent of the small-signal bandwidth, so a stage can be flat to 100 kHz on millivolt signals and still distort a 10 V peak at a few kilohertz.
Is a single op-amp difference amplifier accurate enough for a bridge sensor?
Usually not for small signals riding on a large common-mode voltage. The formula assumes both resistor pairs match exactly, and any mismatch lets common-mode voltage leak into the output; Fiore notes that high common-mode rejection may need matching better than 0.01%. The two inputs also load the source unequally. A three-op-amp instrumentation amplifier with trimmed on-chip resistors is the usual fix.
Why use a voltage follower when it does not amplify anything?
Because it changes impedance. Its input draws almost no current, so it can read a high-impedance source such as a voltage divider, a pH electrode or a filter node without disturbing it, while its output can drive a cable, an ADC input or a low-value load. The voltage stays the same; the ability to supply current is what it adds.
Is it better to get a gain of 100 from one stage or from two stages of 10?
For bandwidth, two stages usually win. On a 1 MHz part a single stage with a noise gain of 100 reaches only 10 kHz. Two non-inverting stages of 10 each reach 100 kHz individually, and in cascade the combined −3 dB point is about 64.4 kHz, since each stage contributes some roll-off. The price is a second op-amp, a little more noise and offset, and more board space.

References& sources.

  1. [1]James M. Fiore, Operational Amplifiers and Linear Integrated Circuits: Theory and Application, §4.2 'Inverting and Noninverting Amplifiers' (Engineering LibreTexts). Source of the two core gain equations: Eq. 4.1 Av = 1 + Rf/Ri for the non-inverting configuration and Eq. 4.2 Av = −Rf/Ri for the inverting one, with the printed explanation that 'There is a change in polarity because we reference the output signal to ground'. Also the source of the differential-amplifier statement at §4.2.8 and Eq. 4.11 — 'the output voltage is found by multiplying the differential input signal by Rf/Ri' — and of the input-impedance contrast between the two configurations. Open access; retrieved 2026-07-29.
  2. [2]James M. Fiore, Operational Amplifiers and Linear Integrated Circuits, §5.3 'Gain-Bandwidth Product' (Engineering LibreTexts). Source of Eq. 5.1, f₂ = GBW/A_noise, and of the noise-gain definition A_noise = 1 + Rf/Ri. Contains the statement this page's noise-gain output exists to make visible: 'Noise gain is the same as ordinary voltage gain for the noninverting voltage amplifier, but is one unit larger than the inverting amplifier's ordinary gain (Rf/Ri)', and therefore 'for the same gain, noninverting amplifiers will exhibit a higher break frequency than inverting types'. Also the warning 'Never use the gain in dB form for this calculation!'. Open access; retrieved 2026-07-29.
  3. [3]James M. Fiore, Operational Amplifiers and Linear Integrated Circuits, §5.4 'Slew Rate and Power Bandwidth' (Engineering LibreTexts). Source of the slew-rate definition ('the rate of change of voltage versus time', dv/dt), of the power-bandwidth relation f_max = Slew Rate/(2πVp), of the instruction 'To be on the conservative side, set Vp to the op amp's clipping level', and of the worked figure of 'a ±15 V supply yielding approximately ±13 V clipping levels' — which is the basis for this page's statement that a real op-amp does not swing to its rails. Open access; retrieved 2026-07-29.
  4. [4]James M. Fiore, Operational Amplifiers and Linear Integrated Circuits, §6.2 'Instrumentation Amplifiers' (Engineering LibreTexts). Source of Eq. 6.1, Vout = (Rf/Ri)(V_b − V_a), the exact difference-amplifier expression implemented here, together with the limitations that make this page's difference mode an idealisation: 'the input impedances are not isolated; indeed, the impedance of one input may very well be a function of the signal present on the other input', and 'Any mismatch in gain between the two inputs will result in a degradation of CMRR... For high CMRR systems, resistor matching may need to be better than 0.01%'. Open access; retrieved 2026-07-29.
  5. [5]James M. Fiore, AC Electrical Circuit Analysis: A Practical Approach, §10.3 'Bode Plots' (Engineering LibreTexts). Cited for the decibel and break-frequency conventions this page's gain-in-dB and closed-loop-bandwidth outputs rest on: Eq. 10.15 fc = 1/(2πRC), Eq. 10.14 the lag-network gain −10·log₁₀(1 + f²/fc²), and the statement that break frequencies are 'also known as the half-power points, or corner frequencies'. Open access; retrieved 2026-07-29.

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Method
A_v = 1 + R_f/R_i · A_v = −R_f/R_i · A_noise = 1 + R_f/R_i · f₂ = GBW ⁄ A_noise
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