Op Amp Gain Calculator converts input voltage and feedback resistor values into output voltage, closed-loop gain, feedback factor, noise gain, and resistor power dissipation for both amplifier types.
Calculate Op-Amp Gain and Output Voltage for Inverting and Non-Inverting Circuits
This calculator computes closed-loop gain, output voltage, feedback network behavior, and resistor power dissipation for inverting and non-inverting op-amp circuits from the input voltage and the two feedback resistors. Electronics engineers, audio circuit designers, and students use it to size feedback resistors for sensor conditioning, audio preamps, and instrumentation front ends before committing values to a schematic.
Entering Configuration, Input Voltage, and Feedback Resistors
Select Non-Inverting or Inverting configuration, then enter the input voltage (Vin) in V, the feedback resistor (Rf) in kΩ, and the input resistor (Rin) in kΩ. The calculator returns output voltage (Vout) in V, gain in V/V and dB, the feedback factor (β), noise gain, the inverting-node voltage, feedback current, and resistor power dissipation.
How Op-Amp Output Voltage Is Calculated
Both configurations rely on the ideal op-amp assumptions of infinite input impedance and a virtual short between the two input terminals — the foundation used throughout Texas Instruments’ Op Amps for Everyone application handbook and Sedra & Smith’s Microelectronic Circuits textbook, the standard references for feedback amplifier analysis.
For the non-inverting configuration:
$$V_{out} = V_{in}\left(1+\frac{R_f}{R_{in}}\right)$$
For the inverting configuration:
$$V_{out} = -V_{in}\frac{R_f}{R_{in}}$$
Gain in decibels is $A_v(dB) = 20\log_{10}(A_v)$. A common input mistake is applying the non-inverting circuit’s “+1” term to an inverting circuit, or dropping the negative sign on an inverting circuit’s gain — the two configurations use different formulas, not just a sign flip on the same one.
The feedback factor is $\beta = R_{in}/(R_f+R_{in})$, and the inverting input sits at $V_- = V_+$ by the virtual-short assumption — equal to Vin for the non-inverting case, or 0 V (virtual ground) for the inverting case. Because no current flows into the ideal op-amp’s input pin, the same current flows through Rf and Rin: $I_f = (V_{out}-V_-)/R_f$. Resistor power dissipation follows from $P = I^2R$.
Op-Amp Feedback Network
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Noise Gain vs. Signal Gain
Noise gain — $1/\beta$, or $1+R_f/R_{in}$ — is not the same as the closed-loop signal gain in an inverting circuit. For the non-inverting configuration the two are identical, but for the inverting configuration the signal gain is only $-R_f/R_{in}$ while the noise gain the feedback loop actually sees is still $1+R_f/R_{in}$.
This matters because usable bandwidth is set by the op-amp’s gain-bandwidth product divided by noise gain, not by the smaller-looking inverting signal gain — an inverting stage with a signal gain of only 1 can still carry a noise gain of 2, cutting the usable bandwidth in half compared to what a naive calculation would suggest.
Common Mistakes When Calculating Op-Amp Gain
- Applying the non-inverting formula’s “+1” to an inverting circuit, or omitting the negative sign on an inverting circuit’s output.
- Mixing resistor units — entering one resistor in Ω and the other in kΩ — which throws the gain off by a factor of 1000.
- Treating the calculated gain as valid at all frequencies, ignoring that the op-amp’s finite gain-bandwidth product limits usable bandwidth as gain increases.
Common Questions About Op-Amp Gain Calculations
What’s the difference between inverting and non-inverting op-amp gain?
Non-inverting gain is $1+R_f/R_{in}$ and always ≥1 with the output in phase with the input; inverting gain is $-R_f/R_{in}$, can be less than 1, and inverts the signal’s phase by 180°.
Why is the inverting node voltage not always 0 V?
It’s 0 V only in the inverting configuration, where the non-inverting input is grounded. In the non-inverting configuration, the virtual short sets V- equal to Vin, not 0 V.
What does the feedback factor β represent?
β is the fraction of Vout fed back to the inverting input through the Rin/Rf divider. A smaller β means more feedback is used up setting a higher gain, which is why noise gain (1/β) rises as β falls.
Why do Rf and Rin dissipate different power despite carrying the same current?
Power is $I^2R$, so with equal current the resistor with the higher value dissipates proportionally more — Rf (10 kΩ) dissipates five times what Rin (2 kΩ) does here, matching their 5:1 resistance ratio.
Does a higher gain always mean a better circuit?
No — higher gain reduces usable bandwidth through the noise gain and the op-amp’s gain-bandwidth product, and it amplifies input offset and noise proportionally, so gain should be set no higher than the application needs.