How it works
An inverting summing amplifier with inputs V1, V2, V3 through R1, R2, R3 and feedback Rf gives Vout = −Rf(V1/R1 + V2/R2 + V3/R3). When all resistors equal R, Vout = −(V1 + V2 + V3). The integrator replaces Rf with capacitor C: Vout = −(1/RC)∫Vin dt; with R = 10 kΩ and C = 0.1 μF, the time constant is 1 ms, making it useful for generating ramp waveforms from step inputs. The differentiator swaps R and C: input capacitor C and feedback R give Vout = −RC × dVin/dt — it amplifies high-frequency noise and is rarely used without input limiting. The difference amplifier with equal resistors R gives Vout = V2 − V1, rejecting common-mode signals.
Key points to remember
Key op-amp application formulas: summing amplifier Vout = −Rf(V1/R1 + V2/R2 + ...), integrator Vout = −(1/RC)∫Vin dt, differentiator Vout = −RC(dVin/dt). The precision rectifier uses an op-amp to overcome the 0.7 V diode drop, rectifying signals as small as millivolts. An instrumentation amplifier built from three op-amps has differential gain set by a single resistor RG: gain = 1 + 2R/RG, offering very high CMRR (>100 dB) ideal for strain gauge and thermocouple signal conditioning. Voltage-to-current converter (Howland pump) produces Iout = Vin/R regardless of load — used in 4–20 mA industrial transmitters. Always check output saturation limits (typically ±(VCC − 1.5 V) for most op-amps) before finalising gain calculations.
Exam tip
The examiner always asks you to derive the output of an inverting summing amplifier — write KCL at the virtual ground node, sum all input currents equalling the feedback current, and simplify to get Vout = −Rf(V1/R1 + V2/R2 + V3/R3).