The Inverting Op Amp Equation: Direct Answer & Core Formula

The inverting op amp equation defines the closed-loop voltage gain of an operational amplifier configured with negative feedback through a resistor network. The direct answer for the output voltage is:

Vout = -Vin × (Rf / Rin)

This formula relies on the concept of a virtual ground at the inverting input terminal. Because the non-inverting input is tied to physical ground (0V) and the op amp's massive open-loop gain forces the differential input voltage to near zero, the inverting input sits at a virtual 0V. All input current flows through Rin and directly into Rf, generating the inverted output voltage.

Symbol Definition and Typical Bench Ranges
Symbol Parameter Unit Typical Bench Range
Vout Output Voltage Volts (V) Bounded by supply rails (e.g., ±13.5V on a ±15V supply)
Vin Input Voltage Volts (V) -10V to +10V (must not exceed common-mode limits)
Rf Feedback Resistor Ohms (Ω) 1 kΩ to 1 MΩ
Rin Input Resistor Ohms (Ω) 1 kΩ to 1 MΩ
Av Closed-Loop Voltage Gain V/V (Dimensionless) -0.1 to -1000

When the Formula Applies (and When It Breaks)

The inverting op amp equation assumes an ideal operational amplifier. This means infinite open-loop gain, infinite input impedance, zero output impedance, and infinite bandwidth. In reality, the formula holds true within a few percent of error under specific conditions, but it breaks down if you ignore physical limitations.

Core Assumptions

  • Linear Operation: The op amp must not be saturated. If the math predicts a 20V output but your supply rails are ±12V, the formula fails because the output will hard-clip at approximately ±10.5V.
  • Negative Feedback Dominance: The loop gain must be high enough to maintain the virtual ground. At very high frequencies, the op amp's open-loop gain drops, and the virtual ground assumption degrades.

Unit Mistakes That Break the Math

Warning: The most common bench mistake is mixing prefix multipliers without scaling. If Vin is in millivolts (mV) and you expect Vout in Volts, you must convert. Furthermore, the ratio Rf / Rin is dimensionless. You can use kΩ for both, or MΩ for both, and the units cancel. Do not mix kΩ and Ω without adjusting the decimal.

Realistic Answer Magnitudes

A realistic closed-loop gain (Av) for a standard inverting stage sits between -1 and -100. Gains beyond -1000 are rarely practical in a single stage due to the Gain-Bandwidth Product (GBWP) limitation and increased noise. If your calculation yields a gain of -5,000, you need to cascade two stages rather than forcing a 5 MΩ feedback resistor, which will act as an antenna for 60Hz mains hum.

Rearranged Forms for Quick Bench Calculations

When debugging a PCB or sizing components on the fly, you rarely need to solve for Vout. Keep these rearranged forms handy for your bench notebook:

  • Solving for Feedback Resistor: Rf = -Rin × (Vout / Vin)
  • Solving for Input Resistor: Rin = -Rf × (Vin / Vout)
  • Solving for Required Input: Vin = -Vout × (Rin / Rf)
  • Solving for Gain Magnitude: |Av| = Rf / Rin

Worked Examples with Unit Tracking

Let's run through two common scenarios, tracking units explicitly to prevent scaling errors.

Problem 1: Predicting Output Voltage

Given: An audio preamp stage uses an Rin of 2.2 kΩ and an Rf of 22 kΩ. The input signal from a DAC is Vin = 150 mV peak.

Find: Vout in Volts.

  1. Convert Vin to base units: 150 mV = 0.150 V.
  2. Write the equation: Vout = -Vin × (Rf / Rin)
  3. Substitute values: Vout = -0.150 V × (22,000 Ω / 2,200 Ω)
  4. Cancel units and solve ratio: 22k / 2.2k = 10 (Dimensionless gain of -10).
  5. Final calculation: Vout = -0.150 V × 10 = -1.50 V.

Result: The output swings to -1.50V peak, perfectly within standard ±12V audio rails.

Problem 2: Sizing Resistors for a Target Gain and Swing

Given: You need to invert and amplify a sensor signal (Vin(max) = 50 mV) to drive an ADC that expects a maximum of -2.5V. Your supply rails are ±5V.

Find: Standard 1% resistor values for Rin and Rf.

  1. Determine required gain: Av = Vout / Vin = -2.5 V / 0.050 V = -50.
  2. Choose a baseline Rin. To avoid loading the sensor, we select Rin = 1.00 kΩ (Standard E96 value).
  3. Rearrange for Rf: Rf = -Rin × Av.
  4. Substitute: Rf = -1.00 kΩ × (-50) = 50.0 kΩ.
  5. Select nearest 1% E96 standard value: 49.9 kΩ.
  6. Verify actual gain: -49.9k / 1.00k = -49.9.
  7. Verify max output: 0.050 V × -49.9 = -2.495 V.

Result: Use Rin = 1.00 kΩ and Rf = 49.9 kΩ. The output hits -2.495V, safely below the ±5V rail clipping threshold.

Decision Path: Picking Your Resistor Values and Op Amp

The equation gives you the ratio, but it doesn't tell you the absolute resistance values. Choosing between 10Ω/100Ω and 1MΩ/10MΩ for a gain of -10 drastically alters circuit performance. Use this decision tree to lock in your bill of materials.

Resistor and IC Selection Decision Tree
Condition / Constraint Action / Rule Concrete Pick
IF signal source impedance is high (>10 kΩ) THEN Rin must be ≥100 kΩ to prevent signal attenuation via voltage divider loading. Rin = 100 kΩ, scale Rf up.
IF signal frequency is high (>100 kHz) THEN keep resistors low (<5 kΩ) to minimize parasitic capacitance RC low-pass filtering. Rin = 1 kΩ, use high GBWP op amp.
IF minimizing DC offset from input bias current THEN keep parallel equivalent resistance (Rin || Rf) low, or add a compensation resistor to the non-inverting pin. Target Rin between 1k and 10k.
IF general-purpose DC/audio bench use (Default) THEN target the 10 kΩ to 100 kΩ range for optimal thermal noise and bias current balance. See Default Recommendation Below
The Default Bench Recommendation:
For 90% of general-purpose DC and audio inverting amplifier builds, stop overthinking and use Rin = 10.0 kΩ (1% metal film, e.g., Yageo MFR-25 series). Scale Rf to set your gain. Pair this network with a TL072 dual op amp if you have dual supplies (±12V), or an LM358 if you are constrained to a single 5V-12V supply. If you need ultra-low noise for high-end audio, swap the TL072 for an OPA2134.

Real-World Non-Idealities: What the Equation Ignores

While the inverting op amp equation is the foundation of your design, real silicon introduces errors that the basic formula ignores. According to precision design guidelines from Texas Instruments, you must account for three major non-idealities in precision circuits:

1. Gain-Bandwidth Product (GBWP)

The equation assumes infinite bandwidth. In reality, an op amp's open-loop gain rolls off at 20 dB/decade. If your chosen IC (like the TL072) has a GBWP of 3 MHz, and you configure it for a closed-loop gain of -100 (40 dB), your usable bandwidth shrinks to just 30 kHz. If you need a gain of -100 at 100 kHz, the basic equation will fail you; you must select an op amp with a GBWP > 10 MHz, such as the OPA1641.

2. Input Bias Current and Offset Voltage

The equation assumes zero current flows into the op amp inputs. In bipolar op amps (like the LM358), input bias currents flow through Rf, generating an unwanted DC offset voltage at the output (Voffset = Ibias × Rf). If Rf is 1 MΩ and Ibias is 50 nA, you inherit a 50 mV DC error. For high-impedance designs, use JFET or CMOS input op amps where bias currents are in the picoamp range.

3. Slew Rate Limiting

The formula predicts instantaneous voltage changes. However, if you input a 1V step into a gain-of-10 circuit, the output must swing 10V. If the op amp's slew rate is 0.5 V/µs (typical for an LM358), it will take 20 µs to reach the final value, turning your crisp step into a slow ramp. Always verify that the required dV/dt of your Vout is well below the datasheet slew rate specification.

By anchoring your design in the core inverting op amp equation and systematically applying these real-world constraints, you transition from theoretical textbook math to reliable, working hardware. For deeper analysis on virtual ground limitations and feedback networks, All About Circuits provides excellent schematic breakdowns of these edge cases.