CR · Operational amplifier

The Op-Amp Integrator

Change just one component in the inverting amplifier and a circuit that multiplied becomes one that integrates. Learn how the integrator — with a capacitor in place of the feedback resistor — outputs the running accumulation of its input.

A constant input makes a sloping output

Slide the input voltage Vin. The input is flat, but the output slides down at a steady slope — the larger the input, the steeper. Set the input so the output reaches the target of −4 V at the end of the window.

Input voltage VinVin = +1.5 V

Output endpoint and slope

Vout(end) = -1.5 V  ·  slope = −1.5/RC

Far from −4

Swap the feedback for a capacitor

The inverting amplifier made an input current Vin/R at the virtual ground and ran it through the feedback resistor Rf to set the output. The integrator puts a capacitor there instead. The input current is still Vin/R, but now that current charges the capacitor. Keep pouring current into a capacitor and its voltage builds with time, so the output voltage changes at a steady rate.

The output’s slope is proportional to the input

When the current into a capacitor is constant, its voltage rises in a straight line. With current Vin/R and capacitance C, the output changes at a slope of −Vin/(RC). Double the input and the slope doubles. The minus sign is inherited from the inverting amplifier. For a constant input the output is a sloped line — a ramp.

Integration is banking up area

In general the output is the time integral of the input, Vout = −(1/RC)∫Vin dt. Integration banks up the area under the input curve in time order. So a positive pulse followed by a negative one makes the output rise then fall — a triangle wave. In the s-domain the capacitor’s impedance 1/sC is exactly the 1/s of integration, and a differentiator is the mirror image with R and C swapped.

ObserveVout = −(1/RC) ∫Vin dt

The output is the integral of the input.

Choosefeedback = ?

A capacitor in the feedback slot.

Fill inslope = ?

The slope is proportional to the input.

On your ownZ_C = ?

Integration is 1/s in the s-domain.

Back to the first screen

The larger the input, the steeper the output ramp tilted, and it reached −4 V at the end of the window when the input was +4 V (RC = 1). A flat input became a sloped output because a constant current steadily charged the capacitor and pulled its voltage up in a straight line. The inverting amplifier that multiplied became, with one component, an integrator — letting the circuit bank up area on your behalf.

An integrator replaces the inverting amplifier’s feedback resistor with a capacitor. The input current Vin/R at the virtual ground charges the capacitor, so the output is the integral of the input, Vout = −(1/RC)∫Vin dt. For a constant input the output is a ramp of slope −Vin/(RC). In the s-domain the capacitor’s 1/sC corresponds to the 1/s of integration, and swapping R and C gives a differentiator.