Op-Amp Integrator

An op-amp integrator produces an output voltage that is the mathematical integral of its input. Inside signal processing chains, it converts square waves into triangular waves and sits at the heart of analog computers and waveform generators.

Core Concept

The integrator uses a capacitor in the feedback path instead of a resistor. As current flows through R1, it charges Cf. The output voltage equals the accumulated charge on the capacitor, which is the integral of the input over time.

The virtual ground at the inverting input keeps one end of Cf at 0V. All input current flows into Cf, so Vout tracks the running integral. The LM741 or TL071 op-amp must have a bandwidth well above the signal frequency for accurate integration.

At DC, the capacitor is an open circuit. This gives infinite gain at zero frequency, making the output saturate if there is any DC offset at the input. A resistor in parallel with Cf (typically 100 kΩ to 1 MΩ) limits low-frequency gain and prevents saturation in practical circuits.

Key Equations

Output voltage of the ideal integrator:

Vout(t) = -(1 / R1Cf) * integral[Vin(t) dt] + Vout(0)

Where R1 is the input resistor in ohms, Cf is the feedback capacitor in farads, and Vout(0) is the initial capacitor voltage. The negative sign comes from the inverting topology.

The integration time constant is:

tau = R1 * Cf

For a square wave input of amplitude Vm and frequency f, the output triangular wave has peak amplitude:

Vpeak = Vm / (4 * f * R1 * Cf)

The gain magnitude at frequency f is:

|Av| = 1 / (2pifR1Cf)

Given:
  Vin = square wave, amplitude = 2V (peak)
  R1 = 10 kΩ = 10,000 Ω
  Cf = 0.1 µF = 0.1 × 10^-6 F
  Frequency f = 1 kHz

Why this formula:
  For a square wave input, the output triangular peak is Vm / (4fR1Cf)

Formula:
  Vpeak = Vm / (4 * f * R1 * Cf)

Substitution:
  Vpeak = 2 / (4 × 1000 × 10000 × 0.1×10^-6)

Calculation:
  Denominator = 4 × 1000 × 10000 × 0.0000001
             = 4 × 1000 × 0.001
             = 4 × 1
             = 4
  Vpeak = 2 / 4 = 0.5V

Final Answer:
  Output triangular wave peak amplitude = 0.5V

Exam Tip: GATE often asks what happens to the output if Vin has a DC component. With no feedback resistor, the integrator output ramps to saturation because the capacitor charges continuously at DC. Also remember the gain formula |Av| = 1/(2pifR1Cf): gain decreases with increasing frequency, opposite to a differentiator. Do not confuse tau = R1Cf with the time constant of an RC low-pass filter; the integrator gives an inverted output.

Key Properties

• The circuit performs mathematical integration: a constant input produces a linearly ramping output.
• Gain magnitude is 1/(2pifR1Cf), so it falls at 20 dB/decade as frequency rises, giving a low-pass characteristic.
• A parallel resistor Rf (100 kΩ to 1 MΩ) across Cf limits DC gain to -Rf/R1 and prevents output saturation.
• The phase shift of the ideal integrator is -90 degrees at all frequencies.
• The TL071 is preferred over LM741 for integrators because its lower input offset voltage reduces output drift.
• For a triangular input, the output is a parabolic waveform. For a sinusoidal input, the output is also sinusoidal but shifted by -90 degrees and scaled by 1/(2pifR1Cf).

Quick Revision

• Feedback element is a capacitor Cf; input element is a resistor R1.
• Vout = -(1/R1Cf) * integral of Vin dt.
• Square wave in gives triangular wave out.
• Gain = 1/(2pifR1Cf), decreases with frequency.
• Phase shift = -90 degrees (ideal integrator).
• Add Rf parallel to Cf to prevent saturation at DC.
• tau = R1*Cf determines integration speed.
• Exam trap: Students swap the gain formula with the differentiator. For the integrator, gain goes DOWN with frequency (low-pass). For the differentiator, gain goes UP (high-pass). Getting this backwards is the most common GATE error on this topic.