Op-Amp Calculator

In the inverting configuration, the input signal is applied to the inverting (−) input through the input resistor R_in, the feedback resistor R_f connects the output back to that same node, and the non-inverting (+) input is tied to ground.

To analyze the circuit, we apply the two golden rules of an ideal op-amp with negative feedback: no current flows into either input, and the op-amp adjusts its output until the two inputs are at the same voltage. Since the (+) input is grounded at 0 V, the (−) input must also rest at 0 V. This node is not physically connected to ground, yet it behaves as though it were, so we call it a virtual ground.

With the (−) node held at 0 V, the current flowing through R_in is simply V_in / R_in. None of that current can enter the op-amp, so all of it must flow on through R_f. Working out the voltage that R_f must develop gives us the closed-loop output:

V_out = −(R_f / R_in) · V_in

Notice that the op-amp's own raw gain (its open-loop gain, the amplification of the chip with no feedback) never appears in this result. That gain is huge but imprecise. Negative feedback trades it away so the closed-loop gain depends only on the resistor ratio, which you can set exactly.

Let us calculate the output for V_in = 1 V, R_in = 10 kΩ, and R_f = 22 kΩ:

• Gain: A_v = −R_f / R_in = −22 kΩ / 10 kΩ = −2.2 (6.85 dB).
• Output: V_out = −2.2 × 1 V = −2.2 V.
• Noise gain: 1 + R_f / R_in = 3.2.
• Bandwidth: with a 1 MHz GBW op-amp, 1 MHz / 3.2 ≈ 313 kHz.

Because −2.2 V falls comfortably within a ±15 V supply, the output is a clean, inverted copy of the input. If we increase the gain or the input until the ideal output would be, say, −20 V, the op-amp can no longer reach it and the waveform flattens at the −15 V rail. The plot above shows this clipping as it happens.

Switching the toggle to non-inverting rearranges the circuit. The signal now drives the (+) input directly, while R_in and R_f form a feedback network from the output back to the (−) input, with R_in returning to ground.

The same golden rules apply. Feedback holds the two inputs at equal voltage, so the (−) node must sit at V_in. R_in and R_f form a voltage divider from V_out, and that divider has to produce V_in at the (−) node. Solving for the output gives:

V_out = (1 + R_f / R_in) · V_in

Here the output stays in phase with the input, and the gain can never be less than 1. Using the same 10 kΩ and 22 kΩ resistors gives a gain of +3.2 (10.1 dB), so a 1 V input now produces +3.2 V.

Two practical differences set this configuration apart from the inverting one. First, the input impedance is enormous, because the signal connects straight to the op-amp's input rather than through a resistor, so it draws almost nothing from the source. Second, the minimum gain is unity, which is exactly the voltage follower. One quantity stays the same in both circuits: the noise gain, 1 + R_f/R_in. Two stages built from the same resistor pair therefore share the same bandwidth even though their signal gains differ.

Switching the toggle to voltage follower strips the circuit down to its simplest closed-loop form. There are no external resistors at all: the output is wired directly back to the inverting (−) input, and the signal drives the non-inverting (+) input. It is the non-inverting amplifier taken to its limit, with R_f = 0 and R_in = ∞.

Run the golden rules through it once more. Feedback forces the two inputs to the same voltage, and since the (−) input is connected straight to the output, the output must equal the (+) input. The result could not be plainer:

V_out = V_in

A gain of exactly 1 (0 dB) is useful when you think about impedance. The signal connects straight to the (+) input, which has very high impedance, so the source delivers almost no current and is barely loaded. The output, meanwhile, comes from the op-amp's low-impedance driver and can supply real current into the next stage. That is the whole point: the voltage follower takes a fragile, high-impedance signal (i.e: a sensor, the tap of a voltage divider, etc) and reproduces it with a stiff, low-impedance copy that can drive a load without sagging. And because its gain is just 1, the follower keeps the op-amp's full GBW.

Switching the toggle to summing turns the inverting amplifier into a mixer. Instead of one input resistor there are now several, and each one carries its own signal into the inverting (−) node. The non-inverting (+) input is still grounded, and a single feedback resistor R_f still connects the output back to the (−) node.

The analysis is the inverting amplifier's, repeated once per input. Feedback holds the (−) node at a virtual ground of 0 V, so the current arriving from the i-th input is simply V_i / R_i. None of these currents can enter the op-amp, so all of them merge at the node and flow on together through R_f. Summing the currents and working out the voltage R_f develops gives:

V_out = −R_f · (V₁/R₁ + V₂/R₂ + ⋯ + V_n/R_n)

The virtual ground is what keeps the inputs from interfering with one another. Each input resistor terminates at a node that is always 0 V, so the current one input contributes does not depend on what any other input is doing. You can add channels, remove them, or change one level without disturbing the rest: exactly the behaviour you want from a mixer.

If you make all the input resistors equal to a single value R, the expression collapses to a clean inverted average-by-count:

V_out = −(R_f / R) · (V₁ + V₂ + ⋯ + V_n)

Take three inputs of V₁ = 1 V, V₂ = 2 V, and V₃ = 3 V, each through R = 10 kΩ, with R_f = 10 kΩ:

• Output: V_out = −(10 kΩ / 10 kΩ) · (1 + 2 + 3) V = −6 V.
• Weighting: with equal resistors every input is scaled by the same −R_f/R = −1, so the stage adds the inputs and inverts the result.
• Unequal resistors: halving R₁ to 5 kΩ would double the first input's weight to −2, giving V_out = −(2·1 + 2 + 3) = −7 V, while leaving the other two channels untouched.

The noise gain that sets the bandwidth uses the parallel combination of all the input resistors: 1 + R_f / (R₁ ∥ R₂ ∥ ⋯). More inputs lower that parallel resistance and raise the noise gain, so a mixer with many channels gives up more bandwidth than a single-input inverting stage with the same R_f.

Switching the toggle to difference wires both op-amp inputs to a signal. One input, V₁, drives the inverting (−) input through R₁ with the feedback resistor R_f returning from the output; the other, V₂, drives the non-inverting (+) input through R₂, which forms a divider with a resistor R_g to ground. The circuit is a bridge of four resistors around the op-amp, and its job is to amplify the difference between the two inputs while ignoring whatever they have in common.

The cleanest way to see this is superposition. With V₂ grounded the stage is a plain inverting amplifier, so V₁ contributes −(R_f/R₁)·V₁. With V₁ grounded, the R₂–R_g divider sets the (+) input voltage and the stage acts as a non-inverting amplifier on it. When the resistors are balanced (R₁ = R₂ = R_in and R_f = R_g), the two contributions combine into a single clean expression:

V_out = (R_f / R_in) · (V₂ − V₁)

The output is the amplified difference of the two inputs, with a gain of R_f/R_in and no minus sign out front. The calculator assumes this balanced case, so a single input-resistor value and a single feedback-resistor value set both sides of the bridge.

The payoff of balancing the resistors is common-mode rejection. Any voltage that appears on both inputs at once (a shared ground offset, picked-up noise, the bias on a sensor) sits in the V₂ − V₁ term as zero and never reaches the output. Set V₁ equal to V₂ in the calculator and the output waveform collapses to a flat line, however large that common level is. This is why the difference amplifier is the front end of choice for current-sense measurements and bridge sensors, where a small signal of interest rides on a much larger shared voltage. In a real circuit the rejection is limited by how well the resistor ratios match: a fraction of a percent mismatch is enough to let some of the common-mode signal leak through, which is why precision difference amplifiers use trimmed, ratio-matched resistor networks.

• Only the ratio sets the gain. R_in = 10 kΩ with R_f = 4.7 kΩ gives the same gain as 100 kΩ with 47 kΩ. Choose the absolute values to balance source loading against noise and bias-current error.
• The inverting input impedance equals R_in. Unlike the non-inverting configuration, the inverting amplifier loads the source with R_in. Keep this in mind when driving the stage from a high-impedance sensor.
• Add a bias-compensation resistor equal to R_in ∥ R_f in series with the (+) input. It cancels the offset voltage produced by input bias current, which matters most when the resistors are large.
• Watch the supply rails. The output can only swing within the supply, and for AC signals it is the peak, not the average, that clips. Real op-amps fall short of the rails unless they are rail-to-rail types.