Op-Amp Gain Calculator

Calculate voltage gain, bandwidth, and output voltage for inverting and non-inverting op-amp configurations.

Unity gain-bandwidth product
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Feedback Sets Closed-Loop Op-Amp Gain

An operational amplifier has enormous open-loop gain, but practical linear circuits use negative feedback to set a predictable closed-loop gain. For an ideal inverting amplifier, Vout/Vin=−Rf/Rin. For an ideal non-inverting amplifier, gain is 1+Rf/Rg. Negative feedback drives the input difference close to zero while input currents are approximately zero in the ideal model.

Real op-amps have finite gain-bandwidth product, slew rate, input bias currents, offset voltage, output-current limits, and supply-rail constraints. A requested closed-loop gain can therefore be mathematically correct yet impossible at a given frequency or output amplitude.

Inverting: Av=−Rf/Rin,   non-inverting: Av=1+Rf/Rg
SymbolMeaningWhy it appears / units
AvClosed-loop voltage gainDimensionless ratio Vout/Vin.
RfFeedback resistanceΩ; connects output to the feedback input network.
GBWGain-bandwidth productHz; approximate gain-times-bandwidth limit for many compensated op-amps.

The minus sign for an inverting amplifier means a 180° polarity inversion, not negative amplification magnitude. As closed-loop gain increases, available bandwidth usually decreases for a fixed gain-bandwidth product.

Check the requested output voltage against the supply rails. Compute the ideal closed-loop gain first, then multiply by the input amplitude. If that output exceeds the available swing, the circuit will saturate and the linear gain formula no longer predicts the waveform amplitude.

Worked Examples

Example 1: Inverting: Rf=100kΩ, Rin=10kΩ, Vin=0.5V
Av=−100/10=−10
Result: Vout=−5V (inverted!)
Classic inverting amplifier, 20dB gain
Example 2: Non-inverting: Rf=100kΩ, Rin=10kΩ, GBP=1MHz
Av=1+10=11, BW=1MHz/11
Result: Av=11×, BW=90.9kHz
Gain-bandwidth tradeoff: higher gain=lower BW
Example 3: Inverting gain of −10
Rin=10kΩ, Rf=100kΩ
Result: Av=−10
A 0.2V input ideally produces −2V if the output rails allow it.
Example 4: Bandwidth estimate
GBW=1MHz, closed-loop gain=20
Result: bandwidth≈50kHz
This first-order estimate shows the gain-bandwidth tradeoff.

Common Mistakes

⚠️
Ignoring supply rails

An op-amp cannot produce arbitrarily large output voltage. Clipping occurs when the required output exceeds its usable output swing.

⚠️
Treating virtual ground as a physical ground connection

The inverting node may be near 0V because of feedback, but it is not directly grounded and can move when the ideal assumptions fail.

⚠️
Ignoring gain-bandwidth and slew rate

High gain at high frequency or large fast signals may exceed the op-amp’s dynamic limits even when resistor ratios are correct.

Frequently Asked Questions

Gain-bandwidth product?
GBP = A_v × BW = constant for a given op-amp. LM741: 1 MHz. TL071: 3 MHz. TL082: 4 MHz. OPA657: 1.6 GHz. To get 100× gain (40dB): bandwidth = GBP/100.
Virtual ground in inverting amplifier?
The inverting input is held at 0V by negative feedback ('virtual ground'). No current flows into the op-amp input. All current through Rin flows through Rf. This simplifies analysis enormously.
Why are op-amp input currents often treated as zero?
The ideal model assumes infinite input impedance. Real devices draw small bias currents, whose importance depends on op-amp type and resistor values.
What does virtual short mean?
With strong negative feedback in linear operation, the op-amp drives its output so V+≈V−. The inputs are not physically shorted; the approximation fails during saturation or without appropriate feedback.
Why does bandwidth fall as gain rises?
For many internally compensated op-amps, the closed-loop gain multiplied by bandwidth is approximately constant over a useful range, so higher gain leaves less bandwidth.
Can an op-amp output reach its supply voltage exactly?
Not always. Rail-to-rail devices can approach the rails under specified loads, but every op-amp has output-swing limits that depend on supply voltage and output current.
Why can an op-amp output differ from the ideal closed-loop gain?
The ideal gain relation assumes the amplifier remains in its linear operating region. A real output cannot exceed its supply-rail capability and is limited by bandwidth, slew rate, input common-mode range, output current, and finite open-loop gain. Large requested gain at high frequency can therefore clip or roll off.

Formula Explorer connections

Interpretation: This formula links charge, voltage, current, resistance, capacitance, power or circuit time response. Assumption: Confirm DC versus AC conditions, RMS versus peak values, component topology and steady-state versus transient behavior. Ideal components may be assumed.

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