If you need the direct answer right now: the inverting amplifier equation is Vout = -Vin × (Rf / Rin). This single formula dictates the closed-loop voltage gain for one of the most fundamental operational amplifier (op-amp) topologies on the workbench. But plugging numbers into a formula without understanding the boundary conditions is how you end up with a clipped, distorted signal and a fried feedback resistor. Below, we break down the derivation, track units through solved problems, and look at a real-world scenario where the textbook math meets silicon reality.
The Core Inverting Amplifier Equation and Symbol Definitions
The inverting amplifier uses negative feedback to force the op-amp into its linear operating region. The governing equation for the output voltage is:
Vout = -Vin × (Rf / Rin)
| Symbol | Parameter | Standard Unit | Definition & Bench Context |
|---|---|---|---|
| Vout | Output Voltage | Volts (V) | The voltage at the op-amp output pin relative to circuit ground. Constrained by the power supply rails. |
| Vin | Input Voltage | Volts (V) | The signal voltage applied to the input resistor. Can be DC or AC (RMS/peak). |
| Rf | Feedback Resistor | Ohms (Ω) | Connects the output pin back to the inverting input (-). Sets the upper bound of the gain. |
| Rin | Input Resistor | Ohms (Ω) | Connects the input signal to the inverting input (-). Sets the circuit's input impedance. |
| Av | Voltage Gain | V/V (Dimensionless) | The ratio Vout/Vin. Equal to -(Rf / Rin). The negative sign indicates a 180° phase shift. |
Derivation via Kirchhoff's Current Law (KCL)
To understand why the equation works, we apply KCL at the inverting input node (V-). We rely on two ideal op-amp assumptions: infinite input impedance (no current flows into the op-amp pins) and infinite open-loop gain (which forces V- to equal the non-inverting input V+). Since V+ is tied to ground (0V), V- becomes a virtual ground (0V).
- The current flowing through the input resistor is: Iin = (Vin - V-) / Rin. Since V- = 0V, this simplifies to Iin = Vin / Rin.
- The current flowing through the feedback resistor is: If = (V- - Vout) / Rf. Since V- = 0V, this simplifies to If = -Vout / Rf.
- Because infinite input impedance means no current enters the op-amp's inverting pin, KCL dictates that Iin = If.
- Equating the two: Vin / Rin = -Vout / Rf.
- Rearranging to solve for Vout yields the final equation: Vout = -Vin × (Rf / Rin).
For a deeper look at the ideal op-amp model and virtual ground concepts, the All About Circuits semiconductor textbook provides an excellent foundational breakdown.
Assumptions, Boundaries, and Unit Traps
The inverting amplifier equation is mathematically perfect, but silicon is not. Here is when the formula applies, where it breaks down, and the mistakes that will ruin your prototype.
When the Formula Applies (and Its Assumptions)
The equation holds true only when the op-amp is operating in its linear region. This requires:
- High Open-Loop Gain (AOL): Real op-amps like the TL072 or OPA2134 have an AOL of 100,000 V/V or more. If your closed-loop gain approaches the op-amp's open-loop gain, the equation loses accuracy.
- Sufficient Power Supply Headroom: Vout cannot exceed the supply rails. If you are running an LM741 on ±12V rails, the output will clip around ±10V due to internal transistor voltage drops.
- Frequency Limits: At high frequencies, the op-amp's gain-bandwidth product (GBWP) limits the actual gain. A 100kHz signal will not achieve the same gain as a 1kHz signal on a standard audio op-amp.
Realistic Answer Magnitudes
A realistic closed-loop voltage gain (Av) for a standard inverting amplifier sits between -0.1 and -100. If your calculation yields a gain of -5,000, you have almost certainly designed a comparator, not a linear amplifier, and the output will immediately slam into the supply rail.
Unit Mistakes That Break the Math
The most common bench error is mixing kilo-ohms (kΩ) and ohms (Ω) without converting. The ratio Rf / Rin is dimensionless, meaning both resistors must be in the same unit prefix.
The Trap: If Rf = 100 kΩ and Rin = 10 Ω, the gain is -10,000. If you carelessly plug "100" and "10" into your calculator without accounting for the 'k' prefix, you will calculate a gain of -10. Always convert everything to base Ohms (Ω) before calculating, or ensure both are explicitly in kΩ.
Rearranged Forms for the Workbench
When designing a circuit, you rarely know all variables upfront. Here are the rearranged forms of the inverting amplifier equation to solve for any missing parameter:
| Target Variable | Rearranged Equation | Use Case |
|---|---|---|
| Feedback Resistor (Rf) | Rf = -(Vout × Rin) / Vin | You have a fixed input resistor and need a specific output voltage swing. |
| Input Resistor (Rin) | Rin = -(Vout × Rf) / Vin | You need to set a specific input impedance to match a source (e.g., 10kΩ for audio). |
| Input Voltage (Vin) | Vin = -(Vout × Rin) / Rf | Reverse-engineering the required sensor output to achieve a target ADC voltage. |
| Voltage Gain (Av) | Av = -Rf / Rin | Quickly determining the amplification factor without calculating absolute voltages. |
Solved Problems: From Textbook to Breadboard
Let's run through two practical calculations, strictly tracking units to ensure no prefix errors slip through.
Problem 1: Calculating Output Voltage
Given: A sensor outputs a DC voltage (Vin) of 250 mV. The circuit uses an input resistor (Rin) of 4.7 kΩ and a feedback resistor (Rf) of 22 kΩ. Find Vout.
- Convert to base units: Vin = 0.25 V; Rin = 4,700 Ω; Rf = 22,000 Ω.
- Apply the equation: Vout = -Vin × (Rf / Rin)
- Substitute values: Vout = -0.25 V × (22,000 Ω / 4,700 Ω)
- Calculate the ratio: 22,000 / 4,700 = 4.6808 (Dimensionless gain)
- Final multiplication: Vout = -0.25 V × 4.6808 = -1.17 V
Bench check: The output is -1.17V. This is well within the linear range of a standard op-amp running on ±5V or ±12V rails.
Problem 2: Sizing the Feedback Resistor for a Target Gain
Given: You are conditioning a 0.2V peak AC signal from a microphone capsule. You need an inverted output of -5.0V peak to drive an ADC. Your input resistor (Rin) is fixed at 10 kΩ to provide a suitable load for the mic. Find the required Rf.
- Identify knowns: Vin = 0.2 V; Vout = -5.0 V; Rin = 10,000 Ω.
- Select rearranged form: Rf = -(Vout × Rin) / Vin
- Substitute values: Rf = -(-5.0 V × 10,000 Ω) / 0.2 V
- Handle the double negative: Rf = (5.0 V × 10,000 Ω) / 0.2 V
- Calculate: Rf = 50,000 V·Ω / 0.2 V = 250,000 Ω (or 250 kΩ)
Bench check: A 250 kΩ resistor is a standard E24 value. The gain is -25, which is perfectly reasonable for an audio preamp stage.
Real-World Bench Scenario: When Math Meets Silicon
Equations assume ideal components. Here is a scenario demonstrating what happens when you trust the inverting amplifier equation without checking the datasheet.
The Setup
You are building a guitar preamp using an NE5532 dual op-amp. The guitar pickup outputs a 1V peak AC signal (Vin). You want a gain of -10 to drive the next pedal in the chain. You power the NE5532 with a standard 9V battery split into a virtual ground, giving you ±4.5V supply rails. You choose Rin = 10 kΩ and Rf = 100 kΩ.
The Numbers
According to the inverting amplifier equation:
- Vout = -1Vpeak × (100 kΩ / 10 kΩ)
- Vout = -10Vpeak
The Outcome
You wire it up on the breadboard, plug in the guitar, and hook an oscilloscope to the output. Instead of a clean, amplified sine wave, the tops and bottoms of the waveform are brutally flattened off (clipped), and the audio sounds like heavy fuzz distortion.
What Went Wrong
The math didn't fail; the assumptions did. The equation calculated a required output swing of ±10V. However, your power supply rails are only ±4.5V. Furthermore, the NE5532 is not a rail-to-rail output op-amp. According to the Texas Instruments NE5532 datasheet, the typical output voltage swing under a 10kΩ load is about 2V below the supply rails.
This means your maximum possible output swing is roughly ±2.5V. The op-amp hit its physical ceiling at 2.5V and hard-clipped the rest of the 10V signal.
The Fix: You have two choices. Either reduce the gain to -2 (change Rf to 20 kΩ) so the output stays within the ±2.5V linear window, or upgrade your power supply to ±12V rails to give the NE5532 the headroom it needs to swing ±10V. For more on op-amp limitations and output swing headroom, the Electronics Tutorials op-amp guide covers practical non-ideal characteristics extensively.
Mastering the inverting amplifier equation means knowing how to calculate the theoretical ideal, but having the bench experience to recognize when the silicon taps out. Always check your supply rails, verify your resistor prefixes, and keep an eye on the gain-bandwidth product when pushing frequencies.
