The summing amplifier (often called an inverting adder) is a fundamental operational amplifier circuit that outputs a voltage proportional to the weighted, inverted sum of its input voltages. Whether you are building a 3-channel audio mixer, a digital-to-analog converter (DAC), or a sensor signal combiner, the core summing amp equation dictates your resistor choices and output limits.
The direct answer for the output voltage of an inverting summing amplifier with n inputs is:
Vout = -Rf × [ (V1 / R1) + (V2 / R2) + ... + (Vn / Rn) ]
This formula assumes an ideal op-amp operating in its linear region. Below, we break down the derivation, define every variable, and walk through real-world bench examples with strict unit tracking.
The Core Formula and Symbol Definitions
The equation is derived directly from Kirchhoff's Current Law (KCL) applied to the inverting input node. Because the non-inverting input is tied to ground (0V) and the op-amp has massive open-loop gain, the inverting input becomes a virtual ground (0V). The current flowing through each input resistor must equal the current flowing through the feedback resistor, as no current enters the op-amp's high-impedance input pins.
| Symbol | Parameter | Standard Unit | Definition & Bench Notes |
|---|---|---|---|
Vout |
Output Voltage | Volts (V) | The final voltage at the op-amp output pin. Inverted relative to positive DC inputs. |
Rf |
Feedback Resistor | Ohms (Ω) | Connects the output pin to the inverting input. Sets the global gain multiplier. |
Vn |
Input Voltage (n) | Volts (V) | The signal voltage applied to the n-th input branch. Can be AC or DC. |
Rn |
Input Resistor (n) | Ohms (Ω) | Sets the specific weight (gain) for the n-th input. Gain for input n = Rf / Rn. |
n |
Number of Inputs | Integer | Total count of input branches tied to the virtual ground node. |
Real-World Component Constraints and Assumptions
The mathematical formula assumes an ideal op-amp, but physical silicon imposes hard limits. The equation only applies when the op-amp is operating in its linear region. If the calculated Vout exceeds the power supply rails (minus the headroom voltage), the op-amp saturates and the equation fails.
Furthermore, input bias currents can create offset voltages at the virtual ground if your resistor values are too high. According to Texas Instruments' op-amp specification guidelines, JFET or CMOS input op-amps are preferred for high-impedance summing networks to minimize DC offset errors.
| Op-Amp Model | Supply Range | Max Linear Output Swing | Input Bias Current | Best Summing Application |
|---|---|---|---|---|
| LM741 | ±10V to ±18V | ±14V (on ±15V rails) | 80 nA (typ) | Legacy educational builds; avoid for precision DC. |
| TL072 | ±7V to ±18V | ±13.5V (on ±15V rails) | 2 pA (typ) | Audio mixers, synth modules (low noise, JFET input). |
| OPA2134 | ±2.5V to ±18V | ±12V (on ±15V rails) | 5 pA (typ) | High-fidelity audio, precision instrumentation. |
| LM358 | 3V to 32V (Single) | VCC - 1.5V | 20 nA (typ) | Single-supply sensor summing; suffers from crossover distortion. |
| MCP6002 | 1.8V to 6.0V | Rail-to-Rail (~5mV loss) | 1 pA (typ) | Battery-powered 3.3V microcontroller DAC interfacing. |
Rearranged Forms for Component Selection
On the bench, you rarely solve for Vout first. Usually, you know your desired output and input signals, and you need to calculate resistor values. Here are the algebraically rearranged forms of the summing amp equation:
- Solving for Feedback Resistor (Rf):
Rf = -Vout / [ (V1/R1) + (V2/R2) + ... + (Vn/Rn) ]
Use when: You have fixed input resistors and need to set the master gain to hit a specific target voltage. - Solving for a Specific Input Resistor (Rk):
Rk = Vk / [ (-Vout / Rf) - Σ(Vi / Ri) ](where i ≠ k)
Use when: Designing a mixer where one channel needs a specific gain relative to the others. - Solving for Required Input Voltage (Vk):
Vk = Rk × [ (-Vout / Rf) - Σ(Vi / Ri) ](where i ≠ k)
Use when: Debugging a circuit to find what input signal is required to force a specific output, given known resistors.
Worked Example 1: 3-Channel Audio Mixer
Scenario: You are building a sub-mixer for a synthesizer rig using a TL072 op-amp on ±15V rails. You need to sum three signals: a low-level microphone (needs 10x gain), a line-level synth (needs unity gain), and a hot drum machine (needs 0.5x attenuation).
Given Values:
- Feedback Resistor:
Rf = 100 kΩ(100,000 Ω) - Input 1 (Mic):
V1 = 0.1 V,R1 = 10 kΩ(10,000 Ω) - Input 2 (Synth):
V2 = 1.0 V,R2 = 100 kΩ(100,000 Ω) - Input 3 (Drums):
V3 = 2.0 V,R3 = 200 kΩ(200,000 Ω)
Step-by-Step Solution:
- Calculate individual input currents (Ohm's Law at virtual ground):
I1 = V1 / R1 = 0.1 V / 10,000 Ω = 0.00001 A(10 µA)
I2 = V2 / R2 = 1.0 V / 100,000 Ω = 0.00001 A(10 µA)
I3 = V3 / R3 = 2.0 V / 200,000 Ω = 0.00001 A(10 µA) - Sum the currents (KCL):
Itotal = 10 µA + 10 µA + 10 µA = 30 µA(0.00003 A) - Apply the summing amp equation for Vout:
Vout = -Rf × Itotal
Vout = -100,000 Ω × 0.00003 A
Vout = -3.0 V
Result: The output is -3.0 V. Because the TL072 can swing to ±13.5V on ±15V rails, this circuit operates perfectly in the linear region without clipping. As detailed in the All About Circuits op-amp textbook, the virtual ground ensures these input currents sum linearly without interacting with one another.
Worked Example 2: 4-Bit Binary-Weighted DAC
Scenario: You are designing a rudimentary 4-bit Digital-to-Analog Converter using an LM358 on a single 12V supply. The digital logic outputs either 0V (LOW) or 5V (HIGH). You want the Most Significant Bit (MSB) to have a weight of 8, and the Least Significant Bit (LSB) to have a weight of 1.
Given Values:
Rf = 10 kΩ(10,000 Ω)- Bit 3 (MSB):
R1 = 10 kΩ, Input state = HIGH (V1 = 5 V) - Bit 2:
R2 = 20 kΩ, Input state = LOW (V2 = 0 V) - Bit 1:
R3 = 40 kΩ, Input state = HIGH (V3 = 5 V) - Bit 0 (LSB):
R4 = 80 kΩ, Input state = LOW (V4 = 0 V)
Step-by-Step Solution:
- Eliminate zero-voltage inputs:
SinceV2andV4are 0V, their current contribution is 0A. We drop them from the equation. - Calculate active branch currents:
I1 = 5 V / 10,000 Ω = 0.0005 A(500 µA)
I3 = 5 V / 40,000 Ω = 0.000125 A(125 µA) - Sum and multiply by Rf:
Itotal = 500 µA + 125 µA = 625 µA(0.000625 A)
Vout = -10,000 Ω × 0.000625 A
Vout = -6.25 V
Result: The mathematical output is -6.25 V. However, the LM358 is on a single 12V supply (0V to 12V). An LM358 cannot output a negative voltage on a single supply. In reality, the output will slam into the negative rail and read approximately 0.02 V (the LM358's low-side saturation limit). To make this DAC work, you must either provide a negative supply rail (e.g., -5V) or add a DC offset bias to the non-inverting pin.
Common Unit Mistakes and Magnitude Sanity Checks
When calculating summing amplifier networks on the bench, two specific errors account for 90% of debugging headaches:
The 1000x Unit Trap
The most frequent mistake is mixing kilo-ohms (kΩ) and ohms (Ω) without converting. If Rf is 100 kΩ and R1 is 10 kΩ, the gain is 10. But if you accidentally type 100,000 and 10 into your calculator, you get a gain of 10,000. Rule of thumb: Convert all resistors to base Ohms, or keep all resistors in kΩ. As long as the ratio is consistent, the kΩ prefixes cancel out in the division step. The same applies to mixing millivolts (mV) and Volts (V) for input signals.
Magnitude Sanity Checks (Rail Saturation)
What does a realistic answer magnitude look like? An op-amp cannot output a voltage greater than its power supply.
Bench Reality Check: If your summing amp equation spits outVout = -42 V, but you are powering a TL072 with standard ±15V bench supplies, your math is theoretically correct but physically impossible. The real-world answer is that the op-amp is in hard saturation, clipping at approximately -13.5V. Always compare your calculatedVoutagainst the 'Max Linear Output Swing' column in your op-amp's datasheet.
By respecting the virtual ground assumptions, tracking your units rigorously, and checking your final magnitude against the silicon's physical supply rails, the summing amp equation becomes a highly reliable tool for analog signal processing.
