Integrator, differentiator and precision rectifiers

Why it matters

Integrators turn a flow-rate signal into total volume, a current into charge, and a square wave into a ramp for a function generator; they are also the heart of dual-slope ADCs and PI controllers. Differentiators turn position into velocity and detect edges. Precision rectifiers make it possible to rectify a 10 mV AC signal for an AC voltmeter or envelope detector, which an ordinary diode with a 0.6 V drop simply cannot do.

Key ideas

Ideal inverting integrator. Input resistor R to the virtual-ground node, capacitor C from output to that node. The input current v_i/R has nowhere to go except into C, so the capacitor voltage, and hence the output, is the running integral: v_o = −(1/RC)∫v_i dt + v_o(0). A constant input produces a linear ramp; a square wave produces a triangle wave. In the frequency domain the gain is −1/(jωRC): falling at 20 dB/decade, with unity gain at f = 1/(2πRC).

Why the ideal integrator fails in practice. At DC its gain is the op-amp's open-loop gain, so the tiny offset voltage and bias current are integrated too and the output drifts into saturation. The practical (lossy) integrator puts a large resistor R_f across C. Below f_a = 1/(2πR_fC) it is an inverting amplifier with gain −R_f/R; well above f_a it integrates. Choose f_a at least ten times below the lowest signal frequency. Integrators also need a reset switch across C in many instruments.

Ideal differentiator. Capacitor C at the input, resistor R in feedback: v_o = −RC·dv_i/dt. Its gain rises at 20 dB/decade, so it amplifies high-frequency noise and, combined with the op-amp's own roll-off, tends to ring or oscillate.

Practical differentiator. Add a small resistor R_s in series with C (and often a small capacitor across R). Then the circuit differentiates only up to f_b = 1/(2πR_sC) and behaves as an inverting amplifier of gain −R/R_s above it. Design rule: signal frequencies below f_b; f_b well above the highest signal frequency but low enough to stop noise.

Precision rectifiers. Putting the diode inside the op-amp's feedback loop divides its forward drop by the open-loop gain, so the circuit rectifies millivolt signals.

• Precision half-wave (super diode): a non-inverting buffer with the diode in the feedback path; output follows positive inputs and is zero for negative ones.
• Inverting precision half-wave: two diodes, one closing the loop for each polarity, so the op-amp output never saturates; this improves speed.
• Precision full-wave (absolute-value) circuit: a half-wave stage plus a summing amplifier gives v_o = |v_i| (with a weighting of 2 on the half-wave output).
• Speed is limited by slew rate: when the input crosses zero the op-amp output must jump by about two diode drops, so high-frequency signals are rectified inaccurately.

Formulas

v_o(t) = −(1/RC)·∫ v_i dt + v_o(0) — ideal integrator; R in Ω, C in F, RC in s.

Δv_o = −V_in·Δt / (RC) — change in output for a constant input V_in over Δt.

|A(f)| = 1 / (2π·f·R·C) — integrator gain magnitude.

f_a = 1 / (2π·R_f·C) — lower limit of integration in the practical integrator; DC gain −R_f/R.

v_o(t) = −R·C·dv_i/dt — ideal differentiator.

|A(f)| = 2π·f·R·C — differentiator gain magnitude; unity at f = 1/(2π·R·C).

f_b = 1 / (2π·R_s·C) — upper limit of differentiation in the practical differentiator; high-frequency gain −R/R_s.

v_o = |v_i| — ideal full-wave precision rectifier; effective diode drop V_γ/A.

Worked examples

Example 1 (standard: integrator). An integrator has R = 10 kΩ and C = 0.1 µF. Its input is a ±1 V square wave at 1 kHz (no DC). Find RC, the peak-to-peak output, and the lower limit f_a if R_f = 1 MΩ is placed across C.

1. RC = 10⁴ × 10⁻⁷ = 1 ms.
2. Each half-period lasts Δt = 0.5 ms with a constant input of 1 V.
3. Δv_o = V_in·Δt/RC = 1 × 0.5 ms/1 ms = 0.5 V; the output ramps down 0.5 V, then up 0.5 V: a triangle of 0.5 V peak-to-peak.
4. f_a = 1/(2πR_fC) = 1/(2π × 10⁶ × 10⁻⁷) = 1.59 Hz; DC gain −R_f/R = −100. The 1 kHz signal is far above f_a, so it is integrated accurately.

Answer: RC = 1 ms; triangle of 0.5 V peak-to-peak; f_a ≈ 1.6 Hz.

Example 2 (GATE level: practical differentiator). A differentiator has C = 0.1 µF, feedback R = 10 kΩ and series R_s = 1 kΩ. The input is a 2 V peak-to-peak, 100 Hz triangular wave. Find the unity-gain frequency, f_b, and the output waveform.

1. Unity-gain frequency: f = 1/(2πRC) = 1/(2π × 10⁻³) = 159 Hz.
2. f_b = 1/(2πR_sC) = 1/(2π × 10³ × 10⁻⁷) = 1.59 kHz. The 100 Hz input (and its first few harmonics) is below f_b, so the circuit differentiates.
3. Slope of the triangle: it rises 2 V in half a period (5 ms), so dv_i/dt = 2/0.005 = 400 V/s.
4. v_o = −RC·dv_i/dt = −10⁻³ × 400 = −0.4 V during the rising half, +0.4 V during the falling half.

Answer: unity gain at 159 Hz, f_b ≈ 1.59 kHz, output a ±0.4 V square wave (0.8 V peak-to-peak), inverted relative to the slope.

Common mistakes

• Forgetting the minus sign of the inverting integrator and differentiator.
• Leaving out the initial condition v_o(0) when integrating.
• Getting RC wrong by a factor of 1000 (kΩ × µF = ms, not s), which leads to impossible outputs of thousands of volts.
• Ignoring saturation: an integrator with a constant input ramps only until it hits the supply rail.
• Using the ideal integrator without R_f and being surprised by drift to a rail.
• Thinking a precision rectifier has exactly zero drop at all frequencies; slew rate limits it.

For GATE IN

Expect output waveforms of integrators and differentiators for step, square, triangle and sine inputs, time to saturation, corner frequencies of the practical forms, and transfer characteristics of precision half-wave and full-wave rectifiers. Practise sketching input and output waveforms on the same time axis with correct amplitude and sign.

Quick check

1. R = 100 kΩ, C = 1 µF, constant input +1 V, v_o(0) = 0. v_o after 50 ms?
2. Same integrator: time to reach −13 V?
3. A differentiator has RC = 1 ms and input slope 2 V/ms. Output?
4. Why does a precision rectifier work for 10 mV signals?

Answers: 1. −0.5 V. 2. 1.3 s. 3. −2 V. 4. The diode is inside the feedback loop, so its drop is divided by the op-amp's open-loop gain.