Active Low Pass Filter

A phone equalizer boosts bass while cutting treble. The active low-pass filter behind that bass control uses an op-amp to amplify and shape the frequency response with a sharp cut-off no passive RC stage can match.

Core Concept

A passive RC filter gives only -20 dB/decade roll-off and its output impedance affects the load. Adding an op-amp solves both problems. The active low-pass filter uses the op-amp to buffer the output, maintaining low output impedance regardless of load current.

The most common topology is the Sallen-Key configuration. Two RC sections feed the non-inverting input of the op-amp, and feedback from the output back to the junction between the two stages boosts the Q factor near fc. This achieves a second-order Butterworth response with -40 dB/decade roll-off using one TL071 or LM741.

For a first-order active LPF, a single RC network feeds the non-inverting input and the op-amp provides gain. Gain A = 1 + Rf/R1. The cut-off frequency is still determined entirely by the RC network and is independent of the gain-setting resistors.

Key Equations

Cut-off frequency (1st order or Sallen-Key equal-R equal-C): fc = 1 / (2πRC)

Transfer function (1st order, non-inverting): H(s) = A / (1 + s/ωc) where ωc = 2πfc and A = 1 + Rf/R1.

Transfer function (2nd order): H(s) = ωc^2 / (s^2 + (ωc/Q)s + ωc^2) where Q = 0.707 for Butterworth flatness.

Gain-bandwidth check: the op-amp unity-gain bandwidth (GBW) must satisfy GBW >> A × fc. For TL071, GBW = 3 MHz, so at A = 10 and fc = 10 kHz, GBW must exceed 100 kHz. It does.

Given:
  R1 = R2 = 10 kΩ = 10,000 Ω
  C1 = C2 = 10 nF = 10 × 10^-9 F
  Op-amp: TL071 (unity gain Sallen-Key)

Why this formula:
  Equal-R, equal-C Sallen-Key: fc = 1/(2πRC)

Formula:
  fc = 1 / (2π × R × C)

Substitution:
  fc = 1 / (2π × 10,000 × 10×10^-9)
     = 1 / (2π × 10^-4)
     = 1 / (6.2832 × 10^-4)

Calculation:
  fc = 1 / 0.00062832
     = 1591.5 Hz

Final Answer:
  fc ≈ 1.59 kHz
  Roll-off = -40 dB/decade (2nd order)
  At 15.9 kHz (10× fc): attenuation = -40 dB

Exam Tip: GATE tests whether you know that the cut-off frequency of an active filter depends only on the RC network, not on the op-amp gain. A common wrong answer is to include Rf or R1 in the fc formula. Also watch for the unity-gain Sallen-Key: gain is 1 (Rf = 0, R1 = open), Q = 0.5, and the response is Butterworth only when the feedback capacitor ratio is set correctly. For equal R and equal C, the Butterworth condition requires op-amp gain = 1.586 (i.e., Rf/R1 = 0.586).

Key Properties

• First-order active LPF has -20 dB/decade roll-off; second-order gives -40 dB/decade using one op-amp.
• Cut-off frequency depends only on R and C: fc = 1/(2πRC). Op-amp gain does not shift fc.
• Output impedance is very low (near 0 Ω) because the op-amp buffers the output, unlike a passive RC filter.
• TL071 or LM741 are common choices; TL071 has higher GBW (3 MHz) and lower offset, making it better for audio-range filters.
• Q factor controls peaking near fc. Q = 0.707 gives maximally flat Butterworth. Q > 0.707 causes gain peaking before cut-off.
• For Sallen-Key with equal R and C, Butterworth response requires op-amp closed-loop gain = 1.586 (Rf = 0.586 × R1).
• The active filter can provide pass-band gain greater than 1, unlike a passive RC which always attenuates by at least some amount.

Quick Revision

• Active LPF uses op-amp to achieve gain, low output impedance, and sharp roll-off.
• fc = 1/(2πRC) for first-order and equal-component Sallen-Key second-order.
• First-order: -20 dB/decade. Second-order (Sallen-Key): -40 dB/decade.
• Gain A = 1 + Rf/R1 for non-inverting configuration. Gain does not change fc.
• Butterworth Sallen-Key with equal R and C needs gain = 1.586 for maximally flat response.
• Op-amp GBW must be much greater than A × fc to avoid bandwidth limitation.
• TL071 preferred over LM741 for active filter design due to FET input and 3 MHz GBW.
• Exam trap: Students include the gain-setting resistors Rf and R1 in the fc formula. They do not appear in fc = 1/(2πRC). Only the frequency-determining R and C set the cut-off point.