The non inverting amplifier equation defines the closed-loop voltage gain of an operational amplifier where the input signal is applied to the high-impedance non-inverting (+) terminal. The direct formula for output voltage is:
Vout = Vin × (1 + (Rf / Ri))
Alternatively, expressed as voltage gain (Av):
Av = 1 + (Rf / Ri)
The Core Non Inverting Amplifier Equation and Symbol Definitions
Every variable in the equation maps to a specific physical component or measurable node in the circuit. Below is the definitive spec-sheet table for the formula.
| Symbol | Name | Standard Unit | Definition & Practical Constraints |
|---|---|---|---|
| Vout | Output Voltage | Volts (V) | The amplified signal at the op-amp output pin. Bounded by the supply rails minus the saturation voltage drop. |
| Vin | Input Voltage | Volts (V) | The source signal applied to the non-inverting (+) terminal. Must remain within the op-amp's common-mode input range. |
| Av | Voltage Gain | Dimensionless (V/V) | The ratio of Vout to Vin. In a non-inverting topology, Av is always ≥ 1. |
| Rf | Feedback Resistor | Ohms (Ω) | Connects the output pin to the inverting (-) terminal. Typically 10kΩ to 100kΩ to balance noise and bias current errors. |
| Ri | Input/Ground Resistor | Ohms (Ω) | Connects the inverting (-) terminal to ground. Sets the gain ratio alongside Rf. |
Derivation, Assumptions, and Realistic Magnitudes
The non inverting amplifier equation relies on the virtual short principle. In an ideal op-amp with negative feedback, the open-loop gain is infinite, forcing the voltage difference between the inverting (-) and non-inverting (+) inputs to zero. Therefore, V- = V+ = Vin.
The inverting node acts as a voltage divider between Vout and ground. Applying Kirchhoff's Current Law (KCL) at the inverting node (assuming zero input bias current):
- (V- - 0) / Ri + (V- - Vout) / Rf = 0
- Substitute V- with Vin: (Vin / Ri) + (Vin / Rf) - (Vout / Rf) = 0
- Multiply through by Rf: Vin(Rf/Ri) + Vin - Vout = 0
- Rearrange for Vout: Vout = Vin × (1 + Rf/Ri)
- Ideal vs. Real: This assumes infinite open-loop gain and zero input bias current. For precision DC work with BJT-input op-amps like the NE5532, bias currents flowing through Rf and Ri create offset voltages. FET-input op-amps like the OPA2134 minimize this.
- Realistic Gain Magnitude: Av can never be less than 1. If you need attenuation (Av < 1), you must use a passive voltage divider before a unity-gain buffer, not this topology.
- Resistor Bounds: Keep Rf + Ri between 10kΩ and 200kΩ. Values below 1kΩ force the op-amp output stage to source excessive current, triggering thermal shutdown or clipping. Values above 1MΩ make the circuit highly susceptible to parasitic capacitance, destroying high-frequency bandwidth.
Worked Examples with Strict Unit Tracking
Abstract math hides real-world failure modes. These two examples track units explicitly to show both ideal math and physical limitations.
Problem 1: Standard Linear Amplification
Given: An audio preamplifier using a TL072 op-amp on ±15V rails. Vin = 120 mV (RMS). Rf = 47 kΩ, Ri = 10 kΩ.
Find: Vout in Volts.
- Convert to base units: Vin = 0.120 V. Rf = 47,000 Ω, Ri = 10,000 Ω.
- Calculate Gain (Av): Av = 1 + (47,000 / 10,000) = 1 + 4.7 = 5.7 V/V.
- Calculate Vout: Vout = 0.120 V × 5.7 = 0.684 V (or 684 mV RMS).
- Reality Check: The TL072 output swing is typically ±13V on ±15V rails. 0.684 V is well within the linear region. The equation holds true.
Problem 2: Rail Saturation and Clipping
Given: A sensor interface using an LM358 on a single +5V supply (GND and +5V). Vin = 1.2 V DC. Rf = 100 kΩ, Ri = 10 kΩ.
Find: The actual measured Vout.
- Calculate Ideal Gain: Av = 1 + (100 kΩ / 10 kΩ) = 1 + 10 = 11 V/V.
- Calculate Ideal Vout: Vout = 1.2 V × 11 = 13.2 V.
- Apply Physical Limits: The LM358 is powered by a +5V rail. Furthermore, the LM358 is not rail-to-rail on the high side; its output transistor requires about 1.5V of headroom. Maximum Vout = 5.0 V - 1.5 V = 3.5 V.
- Actual Result: The op-amp saturates. The multimeter will read ~3.5 V DC, and the signal will be severely clipped. The equation predicts the attempted output, but physics limits the actual output.
Rearranged Forms for Component Selection
On the bench, you rarely solve for Vout. You usually have a required gain and need to pick E24 series resistors. Here are the algebraically rearranged forms of the non inverting amplifier equation:
- Solve for Feedback Resistor (Rf):
Rf = Ri × (Av - 1)
Use case: You have a 10kΩ ground resistor and need a gain of 12. Rf = 10k × 11 = 110kΩ. - Solve for Ground Resistor (Ri):
Ri = Rf / (Av - 1)
Use case: You have a 47kΩ feedback resistor and want a gain of 5. Ri = 47k / 4 = 11.75kΩ (use standard 12kΩ). - Solve for Required Input Voltage (Vin):
Vin = Vout / (1 + (Rf / Ri))
Use case: Reverse-engineering the maximum input signal before clipping occurs. - Solve for Gain (Av):
Av = Vout / Vin
Common Unit and Magnitude Mistakes
When the math on your breadboard doesn't match the math on your screen, one of these three errors is usually the culprit:
The ratio Rf/Ri is dimensionless, meaning you can use kΩ for both without converting to base Ohms. However, if Rf = 47 kΩ and Ri = 470 Ω, plugging in "47" and "470" yields a ratio of 0.1. You must align prefixes: 47,000 / 470 = 100, or 47k / 0.47k = 100.
Beginners frequently calculate Rf/Ri and stop, forgetting the "+ 1" in the formula. A 10kΩ/10kΩ divider does not yield a gain of 1; it yields a gain of 2. (If you need a gain of exactly 1, use a unity-gain buffer with no resistors).
The equation assumes the op-amp can "see" the input voltage. If you use an LM324 on a single 5V supply and apply 4.5V to the non-inverting input, the input stage saturates before amplification even begins. The equation will predict a massive Vout, but the physical chip will output garbage. Always check the datasheet's common-mode input voltage range.
For deeper reading on op-amp precision design and input stage limitations, consult the Analog Devices MT-044 Tutorial and the All About Circuits semiconductor guide. Specific output swing limits can be verified in manufacturer datasheets, such as the Texas Instruments LM358 datasheet.
Frequently Asked Questions
Why is the non inverting amplifier equation gain never less than 1?
Because the input signal is applied directly to the non-inverting terminal, the op-amp must drive its output high enough to force the inverting terminal (via the voltage divider) to match the input. Even if Rf is 0Ω (a short) and Ri is infinite (open), the output is directly wired to the inverting input, creating a unity-gain buffer where Vout = Vin. The physical topology prevents signal attenuation; to attenuate, you must place a passive resistor divider before the non-inverting input.
How does input offset voltage affect the non inverting amplifier equation?
Real op-amps have an internal mismatch called input offset voltage (Vos), which acts as a tiny DC voltage source in series with the non-inverting input. The non inverting amplifier equation amplifies this error by the exact same gain as your signal: Verror = Vos × (1 + Rf/Ri). If you are designing a high-gain DC amplifier (e.g., Av = 1000) using an LM741 with a typical Vos of 2mV, your output will have a 2V DC error before you even apply a signal. For high-gain DC applications, you must select precision op-amps with Vos in the microvolt range, like the OPA2277.
Can I use the non inverting amplifier equation for AC signals and high frequencies?
Yes, but only up to the op-amp's bandwidth limit. The equation calculates the ideal closed-loop gain, but real op-amps suffer from Gain-Bandwidth Product (GBWP) limitations. If an op-amp has a GBWP of 1 MHz and you set your resistor ratio for a gain of 100 (40 dB), your maximum usable frequency drops to 10 kHz (1 MHz / 100). Beyond this frequency, the actual gain will roll off at -20 dB/decade, and the equation will over-predict your Vout. Always verify that your required gain multiplied by your maximum signal frequency is less than the op-amp's GBWP.
