Operational amplifier circuits
The ideal op-amp and its golden rules, inverting, non-inverting, summing, difference, integrator and differentiator circuits, and practical limits (GBW, slew rate, saturation, offset), with bridge, bandwidth/slew-rate and integrator examples.
Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.
Why it matters
Strain gauges, thermocouples, load cells and current-sense resistors produce millivolt signals that must be amplified, offset and scaled before an ADC can read them. Op-amp circuits do that job, and they also build the error amplifiers, integrators and buffers inside every analog controller and sensor interface in a mechatronic system. A handful of standard circuits, analysed with two simple rules, covers most of what you will meet.
Key ideas
What an op-amp is. A high-gain differential voltage amplifier: v_out = A_OL·(v₊ − v₋), with open-loop gain A_OL typically 10⁵–10⁶. It runs from a supply (for example ±15 V or a single 5 V rail) and its output cannot go beyond that supply; it saturates a little short of the rails unless it is a rail-to-rail type.
Ideal op-amp model. Infinite open-loop gain, infinite input impedance, zero output impedance, infinite bandwidth, zero offset. Two "golden rules" follow when there is negative feedback and the output is not saturated:
- No current flows into either input.
- The output adjusts so that v₊ = v₋ (virtual short). If v₊ is grounded, v₋ is a virtual ground.
These rules are invalid without negative feedback (open loop or positive feedback): the op-amp then acts as a comparator and the output sits at a rail.
Why negative feedback. It trades excess gain for a closed-loop gain fixed by resistor ratios, independent of the uncertain A_OL. It also widens bandwidth, reduces distortion, and moves input and output impedances in the useful direction.
Standard linear circuits.
- Inverting amplifier: input through R1 to v₋, R_f from output to v₋, v₊ grounded. Gain −R_f/R1; input impedance equals R1.
- Non-inverting amplifier: input to v₊, divider R_f and R1 from output to ground feeds v₋. Gain 1 + R_f/R1 (never below 1); input impedance very high.
- Voltage follower (buffer): output tied to v₋, input to v₊. Gain +1, used to stop a high-impedance source being loaded.
- Summing amplifier: several inputs, each through its own resistor, into the virtual ground of an inverting stage; output is a weighted, inverted sum. Used for mixing, offsetting and simple DACs.
- Difference amplifier: amplifies v₂ − v₁ and rejects the common-mode voltage, provided the resistor ratios are matched. The three-op-amp instrumentation amplifier adds two input buffers for very high input impedance and sets gain with one resistor; it is the standard bridge/strain-gauge front end.
- Integrator: capacitor in the feedback path of an inverting stage; output is the negative time integral of the input. A large resistor across C prevents drift into saturation from offset. Used in PI controllers and ramp generators.
- Differentiator: capacitor at the input, resistor in feedback; output proportional to −dv_in/dt. It amplifies high-frequency noise, so a series input resistor is added in practice.
Practical limits.
- Gain-bandwidth product (GBW): for a voltage-feedback op-amp with a single dominant pole, closed-loop bandwidth ≈ GBW / (noise gain), where noise gain = 1 + R_f/R1 for both inverting and non-inverting stages.
- Slew rate (SR): maximum rate of change of output (V/µs). A sine of peak V_p needs SR ≥ 2π·f·V_p, otherwise it distorts into a triangle.
- Output saturation: the output cannot exceed the supply rails, whatever the formula says.
- Input offset voltage and bias currents: produce DC errors, multiplied by the noise gain; integrators accumulate them.
- CMRR: a real op-amp responds slightly to common-mode input; resistor mismatch in a difference amplifier usually limits CMRR more than the op-amp does.
Formulas
v_out = A_OL · (v₊ − v₋) (open loop)
A_v = −R_f / R1 (inverting)
A_v = 1 + R_f / R1 (non-inverting); A_v = 1 (follower)
v_out = −R_f · (v₁/R₁ + v₂/R₂ + … + v_n/R_n) (inverting summer)
v_out = (R₂/R₁) · (v₂ − v₁) (difference amplifier with R₃ = R₁, R₄ = R₂)
v_out = −(1/(R·C)) · ∫ v_in dt + v_out(0) (integrator)
v_out = −R·C · dv_in/dt (differentiator)
f_CL = GBW / (1 + R_f/R1) (closed-loop −3 dB bandwidth)
f_max = SR / (2π · V_p) (full-power bandwidth)
- v: voltages (V); R: resistances (Ω); C: capacitance (F); t: time (s); A_v: closed-loop voltage gain (V/V); A_OL: open-loop gain (V/V); GBW: gain-bandwidth product (Hz); SR: slew rate (V/s); V_p: output peak (V); f: frequency (Hz).
- All closed-loop formulas assume an ideal op-amp with negative feedback and an output inside the supply rails.
Worked examples
Example 1 (standard: bridge signal with a difference amplifier). A strain-gauge bridge gives v₁ = 2.00 V and v₂ = 2.05 V. A difference amplifier uses R₁ = R₃ = 10 kΩ and R₂ = R₄ = 100 kΩ, on ±12 V supplies. Find v_out.
v_out = (R₂/R₁)·(v₂ − v₁).R₂/R₁ = 100 kΩ/10 kΩ = 10.v_out = 10 × (2.05 − 2.00) V = 10 × 0.05 V = 0.5 V.- The 2 V common-mode voltage is rejected (ideal, matched resistors); 0.5 V is well inside the rails.
v_out = 0.50 V.
Example 2 (GATE level: bandwidth and slew rate). A non-inverting amplifier uses R1 = 1 kΩ and R_f = 99 kΩ with an op-amp of GBW = 1 MHz and SR = 0.5 V/µs. (a) Find the gain and the closed-loop bandwidth. (b) What is the highest frequency at which it can deliver an undistorted 10 V peak sine?
A_v = 1 + R_f/R1 = 1 + 99/1 = 100.f_CL = GBW/A_v = 1 × 10⁶ Hz / 100 = 10 kHz.- Slew limit:
f_max = SR/(2π·V_p) = (0.5 × 10⁶ V/s)/(2π × 10 V) = 7.96 kHz. - The slew limit (7.96 kHz) is lower than the small-signal bandwidth (10 kHz), so it governs at full output.
A_v = 100, f_CL = 10 kHz, f_max ≈ 7.96 kHz for 10 V peak.
Example 3 (integrator). An integrator has R = 10 kΩ and C = 0.1 µF, starts at v_out(0) = 0, and receives a +1 V step for 2 ms (supplies ±12 V).
RC = 10⁴ Ω × 10⁻⁷ F = 1 ms.v_out = −(1/RC)·v_in·t = −(1/10⁻³ s) × 1 V × 2 × 10⁻³ s = −2 V, falling linearly.
v_out = −2.0 V after 2 ms (a ramp of −1 V/ms).
Common mistakes
- Using the virtual-short rule in a circuit with no negative feedback (a comparator).
- Forgetting the minus sign of the inverting, summing and integrator stages.
- Writing the non-inverting gain as R_f/R1 instead of 1 + R_f/R1.
- Reporting an output beyond the supply rails; the real output saturates.
- Dividing GBW by the signal gain of an inverting stage (R_f/R1) instead of the noise gain (1 + R_f/R1).
- Confusing bandwidth limits with slew-rate limits; large outputs are usually slew limited.
- Assuming a difference amplifier rejects common mode when its resistor ratios are not matched.
For GATE ME
Questions are mostly direct applications of the ideal model: output of an inverting, non-inverting, summing or difference circuit for given resistors and inputs; integrator output for a step or square input; occasionally closed-loop bandwidth from GBW or a saturation check against the supply. Practise writing KCL at the inverting node with the virtual-short rule; it solves almost every circuit.
Quick check
- An inverting amplifier has R1 = 2 kΩ and R_f = 10 kΩ. What is its gain?
- A non-inverting amplifier needs a gain of 11 with R1 = 1 kΩ. What R_f is required?
- Why is the inverting input called a virtual ground in an inverting amplifier?
- An op-amp with GBW = 2 MHz is used at a noise gain of 20. What is the closed-loop bandwidth?
- An inverting summer has R_f = 10 kΩ, inputs 0.5 V through 10 kΩ and −0.2 V through 5 kΩ. What is v_out?
Answers: 1. −5. 2. 10 kΩ. 3. Negative feedback forces v₋ = v₊ = 0 V, yet no current flows into the input. 4. 100 kHz. 5. −0.1 V.
Interview questions
All Electrical Circuits and Electronics interview questionsTry answering each one aloud before you open it.
1.What is an operational amplifier and what are its basic characteristics?Concept
An operational amplifier, or op-amp, is a high-gain electronic voltage amplifier with a differential input and, usually, a single-ended output. Its basic characteristics include high input impedance, low output impedance, and high gain. Op-amps are used in various configurations to perform mathematical operations such as addition, subtraction, integration, and differentiation.
2.Explain the concept of virtual short in an ideal operational amplifier.Concept
With negative feedback and an unsaturated output, an op-amp with very high open-loop gain needs only a microvolt-level difference between its inputs to produce any normal output voltage, so v₊ ≈ v₋. The inputs are at the same potential, yet no current flows between them because the input impedance is very high, hence 'virtual' short. When v₊ is grounded, the inverting input sits at 0 V and is called a virtual ground. The rule fails without negative feedback, for example in a comparator, where the output saturates.
3.Why is negative feedback used in operational amplifier circuits?Application
Negative feedback is used in operational amplifier circuits to stabilize the gain, increase bandwidth, reduce distortion, and improve linearity. By feeding a portion of the output back to the inverting input, the overall gain of the circuit becomes less dependent on the op-amp's open-loop gain, making the circuit's performance more predictable and reliable.
4.What happens if an operational amplifier is used without feedback?Application
In open loop the output is A_OL·(v₊ − v₋), and with A_OL around 10⁵ even a fraction of a millivolt of input difference drives the output to one of the supply rails. The op-amp therefore cannot amplify linearly, and offsets and noise alone decide where the output sits. This mode is used deliberately as a comparator, though a dedicated comparator IC is faster and recovers from saturation better.
5.Explain the difference between inverting and non-inverting amplifier configurations.Concept
In an inverting amplifier configuration, the input signal is applied to the inverting input, and the output is 180 degrees out of phase with the input. The gain is determined by the ratio of the feedback resistor to the input resistor. In a non-inverting amplifier configuration, the input signal is applied to the non-inverting input, and the output is in phase with the input. The gain is determined by the ratio of the sum of the feedback resistor and the input resistor to the input resistor.
6.Why is an op-amp used in a voltage follower configuration?Application
An op-amp is used in a voltage follower configuration to provide a buffer between circuits. This configuration has a gain of 1, meaning the output voltage follows the input voltage. It is used to prevent loading effects, as it has high input impedance and low output impedance, allowing it to drive heavy loads without affecting the input signal.
7.What is the purpose of a summing amplifier, and how does it work?Concept
A summing amplifier produces an output proportional to a weighted sum of several input voltages. Each input is connected through its own resistor to the inverting input, which is held at virtual ground by negative feedback, so each input contributes an independent current v_k/R_k. These currents add and flow through R_f, giving v_out = −R_f·(v₁/R₁ + v₂/R₂ + …), an inverted weighted sum. It is used for signal mixing, adding an offset to a sensor signal and in binary-weighted DACs.
8.Calculate the output voltage of an inverting amplifier with a feedback resistor of 10 kΩ and an input resistor of 2 kΩ, given an input voltage of 1 V.Numerical
The gain of an inverting amplifier is given by the formula: Gain = -Rf/Rin. Here, Rf = 10 kΩ and Rin = 2 kΩ. Therefore, Gain = -10 kΩ / 2 kΩ = -5. The output voltage (Vout) is the product of the gain and the input voltage (Vin): Vout = Gain × Vin = -5 × 1 V = -5 V.
9.What is the effect of bandwidth on the performance of an operational amplifier?Application
A typical voltage-feedback op-amp has a dominant pole, so its open-loop gain falls at 20 dB/decade and the gain-bandwidth product (GBW) is roughly constant. With feedback, the closed-loop bandwidth is about GBW divided by the noise gain (1 + R_f/R1), so a higher-gain stage has proportionally less bandwidth; a 1 MHz op-amp at gain 100 gives about 10 kHz. For large output swings the slew rate imposes a separate limit, f_max = SR/(2π·V_p), which is often the tighter one.
10.Determine the output voltage of a non-inverting amplifier with a feedback resistor of 8 kΩ and an input resistor of 2 kΩ, given an input voltage of 2 V.Numerical
The gain of a non-inverting amplifier is given by the formula: Gain = 1 + (Rf/Rin). Here, Rf = 8 kΩ and Rin = 2 kΩ. Therefore, Gain = 1 + (8 kΩ / 2 kΩ) = 1 + 4 = 5. The output voltage (Vout) is the product of the gain and the input voltage (Vin): Vout = Gain × Vin = 5 × 2 V = 10 V.
Finished this topic? Mark it so your progress, study plan and readiness keep up.
Stuck on something here?