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CR · Operational amplifier

The Op-Amp Integrator

Change just one part of an inverting amplifier and a circuit that multiplied its input by a fixed number becomes one that integrates the input. In this unit you learn how the integrator, with a capacitor in place of the feedback resistor, turns the accumulated amount of its input into an output voltage.

A flat input makes a slanted output

The light-blue horizontal line is the input voltage, and the gold line is the output voltage over the same time window. The input is flat, yet the output slants down at a constant slope, and the larger the input, the steeper that slope. Set the input so the output at the end of the window reaches the target of −4V on the green dashed line.

Input voltage VinVin = +1.5 V
Output at window end and slope
Vout(end) = -1.5 V · slope = −1.5/RC
Far from −4

Put a capacitor in the feedback spot

In the inverting amplifier the virtual ground fixed the input current at Vin/R, and that current, passing through the feedback resistor, set the output. The integrator puts a capacitor in that spot. The input current is still Vin/R, but now instead of passing through a resistor it piles up charge on the capacitor. As the charge builds, the voltage across the capacitor grows with time, so the output voltage follows not the input's present value but the total that has flowed in so far.

The output slope is proportional to the input

When a constant current flows into a capacitor, its voltage climbs in a straight line. With current Vin/R and capacitance C, the output voltage changes with slope −Vin/(RC). Double the input and the slope doubles too. The minus sign is inherited from the inverting amplifier, so a positive input makes the output tilt downward, like the gold line in the plot. A constant input therefore gives a slanted straight line at the output, a ramp.

Integration piles up area

In the general case, where the input changes with time, the output is the input integrated over time, Vout = −(1/RC)∫Vin dt. Integration piles up, in time order, the area between the input curve and the horizontal axis, and the faintly shaded band under the input line in the plot is exactly that area. Feed in a square wave that alternates between positive and negative and the output becomes a triangle wave that keeps going down and back up. In the s-domain, the capacitor's impedance 1/sC matches the 1/s that stands for integration, and swapping the places of the resistor and capacitor gives a differentiator.

ObserveVout = −(1/RC) ∫Vin dt
The output is the integral of the input.
Choosefeedback = ?
A capacitor sits in the feedback spot.
Fill inslope = ?
The slope is proportional to the input.
On your ownZC = ?
Integration is 1/s in the s-domain.

Back to the first screen

The larger the input, the steeper the gold ramp tilted, and the output reached −4V at the end of the window when the input was +4V (RC = 1). The flat input turned into a slanted output because a constant current kept filling the capacitor and pushed its voltage up in a straight line. The area of the band under the input line matched the size of the output end point for the same reason: integration piles up area. One changed part turned a multiplying inverting amplifier into an integrator, and the circuit took over the job of accumulating area.

An integrator is an inverting amplifier with its feedback resistor replaced by a capacitor. Thanks to the virtual ground, the input current Vin/R charges the capacitor, and the output becomes the integral of the input, Vout = −(1/RC)∫Vin dt. For a constant input the output is a ramp with slope −Vin/(RC). In the s-domain the capacitor's 1/sC corresponds to the 1/s of integration, and swapping the resistor and capacitor gives a differentiator.