Op-Amp Differentiator

An op-amp differentiator produces an output proportional to the rate of change of its input voltage. It is used in edge-detection circuits, FM demodulators, and waveshaping stages where slope information matters.

Core Concept

The differentiator swaps the positions of R and C compared to the integrator. The input capacitor C1 passes only changing signals. A constant DC input produces no current through C1, so the output is zero. A rapidly changing input drives a large current through C1 and Rf, producing a large output spike.

The inverting summing point at the op-amp input sits at virtual ground. The current through C1 equals C1 * dVin/dt. This current all flows through Rf, producing Vout = -Rf * C1 * dVin/dt. The TL071 is commonly used because its high slew rate handles fast edges without distortion.

High-frequency noise is amplified strongly because gain rises with frequency. A small series resistor R1 (100 Ω to 1 kΩ) at the input limits the high-frequency gain to -Rf/R1 and prevents oscillation. This is called the practical differentiator and is the version used in actual designs.

Key Equations

Ideal differentiator output:

Vout = -Rf * C1 * (dVin/dt)

Where Rf is feedback resistance in ohms, C1 is input capacitance in farads, and dVin/dt is the rate of change of input in V/s.

Gain magnitude at frequency f:

|Av| = 2pifRfC1

Gain increases at 20 dB/decade, giving a high-pass characteristic. The phase shift is +90 degrees. For the practical differentiator with series resistor R1:

Maximum gain (at high frequencies) = -Rf/R1

The critical frequency above which gain is flat: fc = 1/(2piR1C1)

Given:
  Vin = triangular wave, slope = 4000 V/s (rising edge)
  Rf = 10 kΩ = 10,000 Ω
  C1 = 0.01 µF = 0.01 × 10^-6 F

Why this formula:
  Vout = -Rf * C1 * (dVin/dt) for ideal differentiator

Formula:
  Vout = -Rf * C1 * (dVin/dt)

Substitution:
  Vout = -10,000 × 0.01×10^-6 × 4000

Calculation:
  = -10,000 × 0.00000001 × 4000
  = -10,000 × 0.00004
  = -0.4 V

Final Answer:
  Output voltage during rising slope = -0.4 V (inverted)

Exam Tip: GATE tests the sign and frequency response. The output is negative (inverting) for a positive-slope input. Gain = 2pifRfC1 increases with frequency, making the differentiator noise-sensitive. Students often write the integrator gain formula here. Also note: a triangular wave input produces a square wave output, not spikes. Spikes appear only for square wave inputs at the transition edges.

Key Properties

• Input element is capacitor C1; feedback element is resistor Rf.
• Vout = -RfC1*(dVin/dt). Output is proportional to the slope of the input.
• Gain rises at 20 dB/decade: the circuit has a high-pass frequency response.
• Phase shift is +90 degrees, opposite to the integrator.
• Practical differentiators add R1 (100 Ω to 1 kΩ) in series with C1 to limit gain at high frequencies and prevent oscillation.
• TL071 (FET-input) is preferred because its high slew rate (13 V/µs) handles fast input transitions.
• Triangular wave input produces square wave output. Square wave input produces spike output.

Quick Revision

• C1 is at input, Rf is in feedback.
• Vout = -Rf * C1 * dVin/dt.
• Gain = 2pifRfC1, rises with frequency (high-pass).
• Phase = +90 degrees.
• Add R1 in series with C1 for stability (practical differentiator).
• Triangular in, square out. Square in, spike out.
• fc = 1/(2piR1*C1) is the gain-limiting frequency.
• Exam trap: Students confuse the differentiator with the integrator. Remember: capacitor at INPUT means differentiator. Capacitor in FEEDBACK means integrator.