Active Low-Pass Filter Calculator

Example 1: Unity-Gain Active Low-Pass Filter

Mode

Unity-Gain Buffer

R

10 kΩ

C

10 nF

fc = 1 / (2 × π × R × C); τ = R × C

fc = 1 / (2 × π × 10,000 × 10 × 10^-9) ≈ 1.5915 kHz; τ = 100 µs

Passband Gain = 1 V/V, Gain = 0 dB, Magnitude at fc ≈ 0.707 V/V

The cutoff is the ideal -3 dB point, not a hard blocking frequency.

Example 2: Non-Inverting Active Low-Pass Filter

Mode

Non-Inverting Gain

R

4.7 kΩ

C

100 nF

Rf

10 kΩ

Rg

10 kΩ

fc = 1 / (2 × π × R × C); Av = 1 + Rf/Rg

fc ≈ 338.63 Hz; τ = 470 µs; Av = 1 + 10 kΩ / 10 kΩ = 2 V/V

Gain ≈ 6.02 dB, Magnitude at fc ≈ 1.414 V/V

The op-amp gain raises the passband magnitude, while the RC values set the ideal first-order cutoff.

Treating fc as a complete blocking frequency

A first-order low-pass filter attenuates gradually above fc; it does not abruptly remove all higher-frequency content.

Forgetting unit conversion

Convert kΩ, nF, µF, and pF correctly before checking manual calculations.

Confusing nF and µF

A 100 nF capacitor is 0.1 µF, not 100 µF. This mistake can shift cutoff by orders of magnitude.

Ignoring the -3 dB definition

At cutoff, the output is Av/√2, not zero and not equal to the full passband gain.

Assuming a first-order filter behaves like a high-order filter

A first-order filter has a gentle -20 dB/decade roll-off and may not provide enough attenuation for steep requirements.

Ignoring gain-bandwidth product

The op-amp must support the selected gain and frequency response. Verify the device datasheet instead of relying only on the RC equation.

Ignoring R and C tolerance

Real cutoff frequency can move because resistor and capacitor values are not exact.

Mixing up passband gain and cutoff magnitude

The magnitude at fc is lower than the passband gain by about 3 dB.