Low-Pass Filters

Last modified by Microchip on 2023/11/09 08:59

Low-pass filters pass low frequencies and attenuate signals at frequencies higher than the cut-off frequency. The actual amount of attenuation for each frequency differs from filter to filter depending on the values of the passive components as well as the configuration of the low-pass filters.

Passive Low-Pass Filters

A passive low-pass filter makes use of only passive components such as capacitors and inductors. The accompanying figure shows an example of the bode plot of a passive low-pass RC filter.

bode plot of a passive low-pass RC filter

Active Low-Pass Filter

An active filter uses op-amps as part of the filter network benefiting from high input and low output impedances. There are various kinds of active filters. The rest of this article demonstrates several popular filters including active low-pass, high-pass, and band-pass filters, in Butterworth, Chebyshev, and Bessel topologies.

Active Low-Pass FilterThe cut-off frequency (in Hertz) is defined as: fc =1/(2 π R1 C) or equivalently (in radians per second): ωc=1/(R1 X C), where f is in Hertz, R is in Ohms, and C is in Farads.

At low frequencies, where f « fc, the capacitor is open, so the gain of the amplifier is –R1/R2. At high frequencies, where f » fc, the capacitor is short and the gain of the circuit goes to zero.

The accompanying image shows the bode plot of the active low-pass filter.

Bode plot of the active low-pass filter

The gain in the pass band is –R1/R2 and the stop band drops off at -20 dB/decade (or −6 dB/octave). This is essentially a first-order filter. The closed-loop gain of the filter from the effect of the op-amp is Vout/Vin = fc = 1/2 π R1 C).

Design Example

Design an active op-amp LPF to achieve a 1.6 kHz cut-off frequency and closed-loop gain of 10.


If R1 = 10 kΩ, fp = fc = 1/(2 π R1 C) = 1.6 kHz, C ≈ 0.01 μF.
At low frequency, |Vout/Vin| = R1 / R2 = 10, hence R2 = 1 kΩ.