The gain for an inverting amplifier is the ratio of the feedback resistor to the input resistor, expressed as a negative value because the output signal is phase-shifted 180 degrees relative to the input. In a real circuit, this topology scales and flips the polarity of an input voltage, allowing you to step down high-voltage sensor signals or invert audio phases without loading the source. Beginners commonly confuse the closed-loop gain (which is set by your external resistors and dictates the actual circuit behavior) with the op-amp's internal open-loop gain (which is typically >100,000 and largely irrelevant to the final output).
The Core Formula and Virtual Ground
To understand why the formula works, you have to look at the concept of 'virtual ground.' In a standard inverting configuration, the non-inverting input (+) is tied directly to physical ground (0V). Because an ideal op-amp has infinite open-loop gain, the feedback loop forces the inverting input (-) to match the non-inverting input. Therefore, the inverting pin sits at 0V, even though it isn't physically connected to ground.
Think of the circuit like a rigid mechanical seesaw balanced on a fulcrum placed exactly at the 0V mark. The input resistor is the distance from the fulcrum to where you push down, and the feedback resistor is the distance to the output side. If you push down on the input side, the output side must push up proportionally based on the ratio of those distances. This is why the output is inverted: a positive input voltage forces current through $R_{in}$ toward the virtual ground, and the op-amp must pull its output negative to sink that same current through $R_f$ to maintain the 0V balance.
According to foundational design principles outlined by Texas Instruments, this virtual ground node is what gives the inverting amplifier its distinct, predictable input impedance, which is simply equal to the value of $R_{in}$.
Worked Example: Designing a -10x Sensor Preamp
Let us build a real circuit. Suppose you have a pressure sensor that outputs a positive 0.25V DC signal at maximum load, but your microcontroller's ADC requires a -2.5V signal to register that same maximum load. You need a gain of -10.
- Target Gain: $A_v = -10$
- Select $R_{in}$: Choose a standard 10kΩ resistor. This provides a reasonable 10kΩ input impedance, which won't load down most sensor outputs.
- Calculate $R_f$: Since $A_v = -R_f / R_{in}$, we plug in the numbers: $-10 = -R_f / 10,000$. Solving for $R_f$ gives us 100,000Ω, or 100kΩ.
- Verify Current Flow: With 0.25V applied to $R_{in}$ (10kΩ), Ohm's law dictates a current of $I = V / R = 0.25V / 10,000Ω = 25μA$.
- Verify Output Voltage: That same 25μA must flow through $R_f$ (100kΩ). The voltage drop across $R_f$ is $V = I imes R = 25μA imes 100,000Ω = 2.5V$. Since current flows from the virtual ground (0V) toward the output to balance the input current, the output pin sits at $0V - 2.5V = -2.5V$.
The math holds up perfectly. The op-amp outputs exactly -2.5V.
Where You Meet This in Practice
The inverting amplifier is not just a textbook exercise; it is a workhorse topology on the bench and in commercial products.
- Audio Phase Inversion: In mixing consoles and guitar effects pedals, an inverting amplifier with a gain of exactly -1 (where $R_f = R_{in}$) is used to flip the phase of an audio signal to cancel noise or align multi-miked drum tracks.
- Transimpedance Amplifiers: If you replace $R_{in}$ with a current source like a photodiode, the circuit becomes a transimpedance amplifier. (Transimpedance is a circuit configuration that converts an input current to an output voltage, commonly measured in volts per ampere or ohms). This is exactly how optical receivers and laser rangefinders convert tiny light-induced currents into readable voltages.
- Active Filters: Multiple-feedback (MFB) low-pass and band-pass filters rely heavily on inverting op-amp stages to achieve high Q-factors without using bulky inductors.
- DAC Output Buffering: Many legacy R-2R digital-to-analog converters output a negative voltage swing relative to a reference. An inverting stage flips this back to a positive 0-5V range for microcontrollers.
Practical Resistor Selection and Bandwidth Trade-offs
While the math says you can use 1Ω and 10Ω to get a gain of -10, or 10MΩ and 100MΩ to do the same, real-world physics will punish you for both extremes. Choosing the right resistor decade is critical for managing thermal noise, bias currents, and power consumption. Analog Devices emphasizes that ignoring op-amp non-ideal traits leads to massive DC offset errors in high-gain circuits.
| Resistor Range ($R_{in}$) | Pros | Cons & Failure Modes | Best Op-Amp Pairings |
|---|---|---|---|
| 10Ω - 100Ω | Extremely low Johnson thermal noise. | Draws massive current; exceeds typical op-amp output drive limits (usually max 20-30mA); wastes power. | High-current audio drivers (e.g., LM49990). |
| 1kΩ - 100kΩ | The 'sweet spot'. Balances low noise, low power, and manageable bias current offsets. | Stray PCB capacitance can cause high-frequency ringing if layout is poor. | General purpose (LM358, TL072, NE5532). |
| 1MΩ - 10MΩ | Draws almost zero current; ideal for battery-powered high-impedance sensors. | High thermal noise; input bias currents create massive DC voltage offsets across $R_f$. | JFET/CMOS inputs only (TL072, LMC6482). |
Finally, remember the Gain-Bandwidth Product (GBWP). If your op-amp has a GBWP of 1MHz, and you configure it for a closed-loop gain of -100, your maximum usable bandwidth drops to just 10kHz. Always check the datasheet's GBWP specification before finalizing your resistor ratio.
Frequently Asked Questions
Why is the gain for an inverting amplifier negative?
The negative sign in the gain equation does not mean the circuit destroys or loses signal amplitude. It denotes a 180-degree phase shift. If you feed a sine wave that swings positive into the input, the output sine wave will swing negative at that exact same moment. In DC terms, a positive input voltage yields a negative output voltage.
Does the op-amp open-loop gain affect the inverting amplifier closed-loop gain?
In an ideal theoretical model, no. The closed-loop gain is dictated entirely by the external resistors. In reality, yes, but only at the margins. If your op-amp's open-loop gain drops at high frequencies (which all of them do), your closed-loop gain will start to deviate from the ideal resistor ratio. This is why you cannot use a standard audio op-amp to achieve a gain of -100 at 5MHz; the internal open-loop gain has already rolled off below 100 at that frequency.
How do I calculate the input impedance of an inverting amplifier?
Unlike the non-inverting amplifier (which boasts near-infinite input impedance), the input impedance of an inverting amplifier is simply the value of the input resistor, $R_{in}$. Because the inverting pin is held at a virtual ground (0V), the source 'sees' only $R_{in}$ connecting to ground. If your sensor requires a high-impedance load, you must use a large $R_{in}$ value or buffer the signal first.
What happens if the calculated output exceeds the op-amp supply rails?
The op-amp will clip. If you are powering an LM358 with a single +5V supply and ground, and your math dictates an output of -2.5V, the output will simply slam against the 0V rail and stay there. To get a true negative output voltage, your op-amp must be powered by a dual power supply (e.g., +9V and -9V) or you must bias your virtual ground to a mid-supply reference voltage (like 2.5V) instead of true 0V.
