Vout = -Rf × [ (V1/R1) + (V2/R2) + ... + (Vn/Rn) ]
The Core Formula and Symbol Definitions
Before applying the formula to a breadboard or PCB layout, you must understand the physical components each variable represents. Below is the definitive spec-sheet table for the inverting summing amplifier.| Symbol | Definition | Standard Unit | Typical Bench Range |
|---|---|---|---|
| Vout | Output voltage at the op-amp pin | Volts (V) | -13V to +13V (on ±15V rails) |
| Rf | Feedback resistor connecting output to inverting input | Ohms (Ω) | 1kΩ to 100kΩ |
| Vn | Voltage applied to the n-th input channel | Volts (V) | -5V to +5V |
| Rn | Input resistor for the n-th channel | Ohms (Ω) | 1kΩ to 100kΩ |
| n | Total number of input channels | Unitless | 2 to 8 (practical limit) |
Derivation and Underlying Assumptions
The formula is not arbitrary; it is a direct application of Kirchhoff’s Current Law (KCL) at the inverting input node (V-) of the op-amp.Step 1: Apply KCL. The sum of all currents entering the inverting node must equal zero. The currents from the input sources (I1, I2, In) and the feedback current (If) meet at this node.
I1 + I2 + ... + In + If = 0
Step 2: Apply the Virtual Ground assumption. Because the non-inverting input (V+) is tied to ground (0V) and the op-amp has massive open-loop gain, negative feedback forces the inverting input (V-) to also sit at 0V. This is the 'virtual ground'.
Step 3: Substitute Ohm's Law. Since V- = 0V, the current through any input resistor Rn is simply Vn / Rn. The feedback current is Vout / Rf.
(V1/R1) + (V2/R2) + ... + (Vout/Rf) = 0
Step 4: Isolate Vout. Move the feedback term to the other side and multiply by -Rf to arrive at the final formula.
Critical Assumptions: This derivation assumes an ideal op-amp with infinite input impedance (no current enters the op-amp pins) and infinite open-loop gain. In reality, a part like the Texas Instruments TL072 has a finite input impedance and a tiny input bias current. For high-precision DC summing, these non-ideal traits require a compensation resistor on the non-inverting pin, though for AC audio summing, the ideal formula holds sufficiently true.
Worked Examples with Unit Tracking
Abstract math fails on the bench if you drop a decimal. Here are two real-world scenarios with explicit unit tracking to ensure your calculations map to physical multimeter readings.Problem 1: Unity-Gain Audio Mixer
Scenario: You are summing three audio lines using a TL072. Rf = 10kΩ. R1 = R2 = R3 = 10kΩ. The instantaneous DC voltages at the inputs are V1 = 1.5V, V2 = -0.8V, and V3 = 0.3V.
- Write the formula: Vout = -Rf × [ (V1/R1) + (V2/R2) + (V3/R3) ]
- Substitute values: Vout = -10,000Ω × [ (1.5V / 10,000Ω) + (-0.8V / 10,000Ω) + (0.3V / 10,000Ω) ]
- Calculate branch currents (Volts / Ohms = Amps):
I1 = 0.00015A (0.15mA)
I2 = -0.00008A (-0.08mA)
I3 = 0.00003A (0.03mA) - Sum the currents: 0.15mA - 0.08mA + 0.03mA = 0.10mA (0.0001A)
- Multiply by -Rf: Vout = -10,000Ω × 0.0001A = -1.0V
Problem 2: Weighted 3-Bit DAC
Scenario: You are building a rudimentary DAC using an OPA2134. Rf = 40kΩ. The input resistors are weighted by powers of two: R1 = 10kΩ (MSB), R2 = 20kΩ, R3 = 40kΩ (LSB). The logic levels are V1 = 2.0V, V2 = 0V, V3 = 1.0V.
- Substitute values: Vout = -40,000Ω × [ (2.0V / 10,000Ω) + (0V / 20,000Ω) + (1.0V / 40,000Ω) ]
- Calculate branch currents:
I1 = 0.0002A (0.2mA)
I2 = 0A
I3 = 0.000025A (0.025mA) - Sum the currents: 0.2mA + 0 + 0.025mA = 0.225mA (0.000225A)
- Multiply by -Rf: Vout = -40,000Ω × 0.000225A = -9.0V
Rearranged Forms for Component Selection
On the bench, you rarely know all the variables upfront. Usually, you know your target Vout and your available input voltages, and you need to pick resistors from your E24 series bins. Here are the rearranged forms solving for the critical design variables.- Solving for Feedback Resistor (Rf):
Rf = -Vout / [ (V1/R1) + (V2/R2) + ... + (Vn/Rn) ]
Use case: You have fixed input resistors and need to set a specific overall gain. - Solving for a Specific Input Resistor (Rx):
Rx = Vx / [ (-Vout/Rf) - Σ(Vi/Ri) ] (where the sum excludes channel x)
Use case: Dialing in the exact weight for a specific sensor channel in a multi-sensor array. - Solving for an Unknown Input Voltage (Vx):
Vx = Rx × [ (-Vout/Rf) - Σ(Vi/Ri) ]
Use case: Reverse-engineering an unknown input signal based on the measured output and known circuit values.
Bench Realities: Unit Mistakes and Magnitude Checks
The math is flawless; human execution is not. When your physical circuit refuses to match your spreadsheet, check these two failure modes.The Unit Mismatch Trap: The most common error is mixing base units and kilo-units. If Rf is 47kΩ (47,000Ω) and R1 is 4,700Ω, the ratio is 10. If you accidentally plug '47' and '4700' into the formula without converting both to Ohms or both to kΩ, your calculated gain will be off by a factor of 1,000. Rule: Always convert all resistors to base Ohms and all voltages to base Volts before calculating, or consistently use kΩ and mA throughout the entire equation.
The Magnitude and Rail-Clipping Reality Check: A realistic answer magnitude is strictly bounded by your op-amp's power supply. If you power an LM741 with ±15V rails, the maximum realistic output magnitude is roughly ±13V (due to internal transistor voltage drops). If your formula calculates Vout = -18.5V, your physical multimeter will read approximately -13.5V. The op-amp has hit the negative supply rail and is clipping. To fix this, you must either increase the supply voltage, reduce the input signal amplitudes, or lower the Rf value to reduce the circuit's gain. For rail-to-rail op-amps like the TLC2272, the output can swing within millivolts of the supply rails, but the fundamental limit remains.
Frequently Asked Questions
Can the summing amplifier formula be used for non-inverting configurations?
Technically yes, but practically no. A non-inverting summing amplifier requires a complex resistor network at the non-inverting pin, and the resulting formula involves superposition and voltage dividers that are highly sensitive to resistor tolerances. The inputs also interact with one another (changing V1 alters the gain of V2). In professional op-amp circuit design, engineers almost exclusively use the inverting summing configuration for its clean virtual-ground isolation, and simply add a second inverting stage with unity gain if a positive output polarity is required.
How does input bias current affect the summing amplifier formula in practice?
Real op-amps require a tiny DC current to bias their internal transistors. In bipolar op-amps like the LM324, this input bias current flows through the feedback resistor Rf, creating an unwanted DC offset voltage at the output that the ideal formula does not predict. To cancel this, you add a compensation resistor (Rcomp) between the non-inverting pin and ground. The value for Rcomp is the parallel equivalent of all input resistors and the feedback resistor: Rcomp = Rf || R1 || R2 || ... || Rn. If you use a JFET or CMOS op-amp (like the TL072 or OPA1612), the bias current is in the picoamp range, and this compensation step is usually unnecessary.
What happens to the formula if one of the input resistors is removed or open?
If an input resistor is removed (open circuit), that specific channel simply drops out of the summation. The current for that branch becomes 0A. The remaining channels continue to sum normally according to the formula, provided the combined output does not exceed the op-amp's supply rails. This is why summing amplifiers are highly fault-tolerant in multi-sensor industrial applications; a dead sensor simply reads as zero contribution rather than crashing the entire summing node.
Why is my calculated summing amplifier output voltage slightly off from the measured value?
Minor deviations (1% to 5%) are caused by resistor tolerance and the op-amp's input offset voltage (Vos). Standard 5% carbon film resistors will introduce significant gain errors in a weighted DAC. For precision summing, use 1% or 0.1% metal film resistors. Additionally, the op-amp's finite open-loop gain means the virtual ground is not exactly 0.000V, but perhaps 0.001V. For a deeper dive into component-level error sources, consult All About Circuits' operational amplifier theory guide. If your error is massive (>20%), check for parasitic oscillation by probing the output with an oscilloscope; a missing bypass capacitor on the op-amp supply pins is the usual culprit.
