How it works
In a first-order low-pass filter made of a series resistor and a shunt capacitor, the cutoff frequency is the point where the output is −3 dB below the input. The time constant τ describes how fast the filter responds to a step input.
fc = 1 / (2π × R × C)τ = R × C63.2 % is reached at 1τ, about 99 % at 5τReverse: R = 1 / (2π × fc × C)
Worked example
With R = 1 kΩ and C = 1 µF, fc = 159 Hz and τ = 1.00 ms. A step response reaches 99 % of its final value after about 5 ms. Conversely, for fc = 1 kHz with C = 0.1 µF, the required resistance is about 1.59 kΩ.
Design tips
- Ceramic capacitors lose effective capacitance with DC bias and temperature. If the real capacitance falls, fc rises above the calculated value, so consider the dielectric type such as X7R.
- A low downstream impedance loads the filter and shifts fc. In that case a voltage-follower buffer amplifier is the safer choice.
- Set an anti-aliasing filter in front of an ADC well below half the sampling frequency. Placing it just above the signal band can damage the usable bandwidth.
- A series resistor in a power line causes a voltage drop (I × R) under load. Check this drop along with the filter performance.
Limitations
- A first-order filter with no load is assumed. A downstream load changes the real cutoff frequency.
- Capacitor ESR, resistor tolerance and temperature coefficients are not included.
- At the cutoff frequency the phase lags by 45°. At high frequencies, parasitic elements of the components can change the response.
Frequently asked questions
How much attenuation is there at the cutoff frequency?
For a first-order RC low-pass filter, the cutoff frequency is the −3 dB point and the phase is −45°. Above it, the attenuation is about −20 dB per decade of frequency.
What if the real capacitance is smaller than the value I use?
fc is inversely proportional to capacitance, so if the effective capacitance drops, fc rises. MLCCs can lose a lot of capacitance with DC bias and temperature, so check the temperature characteristic in the datasheet.
Does a downstream load make the calculation wrong?
Yes. A downstream input impedance is in parallel with R, which lowers the effective resistance and raises fc. A voltage-follower buffer makes the calculation accurate again.