RC low-pass and high-pass filters: the cut-off frequency
One resistor and one capacitor make the simplest filter there is. Put the resistor in series and the capacitor to ground and high frequencies are shorted away: a low-pass. Swap them and DC and bass are blocked: a high-pass. Both turn over at the same corner, fc = 1 / (2π R C), where the output is 70.7% of the input (−3 dB) and shifted by 45°. This calculator gives the corner, the gain and phase at any frequency, the part values for a corner you want, and draws the Bode plot.
How to use the RC filter calculator
- Pick low-pass or high-pass; the drawing swaps the parts.
- Type R and C as 10k, 4k7, 100n or 4.7u. The corner, the time constant and the Bode plot appear at once.
- Optional: a signal frequency shows the gain (as a ratio and in dB) and the phase shift the filter gives it, marked on the plot.
- Optional: a target cut-off gives the C for your R and the R for your C, with the nearest E12 capacitor and E24 resistor, one click to use.
- Click R, C or the plot for its own calculation; the Response table lists the gain and phase from a hundredth to a hundred times the corner.
The formula
fc = 1 / (2π R C) low-pass |A| = 1 / √(1 + (f/fc)²) high-pass |A| = (f/fc) / √(1 + (f/fc)²)
The RC pair is a voltage divider whose lower leg changes with frequency. A capacitor's reactance is XC = 1 / (2π f C): large at low frequencies, small at high ones. The corner is simply the frequency where XC equals R. Below it the capacitor is the bigger impedance and takes most of the input; above it the resistor does. Which one you take the output from decides whether low or high frequencies come through.
What −3 dB means
At the corner R and XC are equal but 90° apart, so the series impedance is √2 × R and either part has 1/√2 = 0.707 of the input across it. In decibels that is 20 log₁₀(0.707) = −3.01 dB, and because power goes as the square of voltage it is exactly half the power. The "cut-off" is therefore not a wall: a signal at fc still comes through at 71%. The filter only really bites a decade away.
Roll-off and phase
Past the corner the gain of a single RC section falls 20 dB per decade (6 dB per octave): a tenth at 10 fc, a hundredth at 100 fc. Two sections give 40 dB per decade, but only if the second does not load the first (put a buffer between them, or make the second R ten times the first). The phase shifts too: a low-pass lags, from 0° far below the corner through −45° at it to −90° far above; a high-pass leads by the mirror image. That shift is why RC networks appear in oscillators and why a filter in a feedback loop can make it unstable.
Choosing R and C
Only the product sets the corner, so there is freedom in the split. Make R at least ten times the source impedance (or the source adds to it and moves the corner down) and at most a tenth of the load impedance (or the load appears in parallel with the capacitor leg). For op-amp and microcontroller circuits 1 kΩ to 100 kΩ with 1 nF to 1 µF covers almost everything; use C0G/NP0 ceramic or film capacitors where the corner matters, since X7R parts lose a third of their value with DC bias and temperature.
| R | C | fc | τ |
|---|---|---|---|
| 1 kΩ | 100 nF | 1.59 kHz | 100 µs |
| 10 kΩ | 100 nF | 159 Hz | 1 ms |
| 10 kΩ | 10 nF | 1.59 kHz | 100 µs |
| 10 kΩ | 1 nF | 15.9 kHz | 10 µs |
| 100 kΩ | 100 nF | 15.9 Hz | 10 ms |
| 4.7 kΩ | 22 nF | 1.54 kHz | 103 µs |
| 1 kΩ | 1 µF | 159 Hz | 1 ms |
| 10 kΩ | 10 µF | 1.59 Hz | 100 ms |
Typical uses
Low-pass: turning PWM into a DC level (corner a hundred times below the PWM frequency), anti-aliasing before an ADC (corner below half the sample rate), smoothing a noisy sensor, de-bouncing a switch, limiting the bandwidth of an audio line. High-pass: blocking DC between amplifier stages (coupling capacitor and the next stage's input resistor), removing hum and rumble below 20 Hz, pre-emphasis, and the edge detector that turns a step into a pulse.
Your filter, step by step
- Corner: 1 / (2π × 10 kΩ × 10 nF) = 1.592 kHz, where XC = R = 10 kΩ.
- Time constant: R C = 100 µs; a step settles in about 5τ = 500 µs.
- At 5 kHz: f/fc = 3.142, gain 0.303 (−10.4 dB), phase −72.3°: in the transition.
- A decade past the corner: −20.0 dB; two decades: −40.0 dB.
Worked example: 10 kΩ and 10 nF
A low-pass with R = 10 kΩ and C = 10 nF has fc = 1 / (2π × 10 000 × 10⁻⁸) = 1.592 kHz; its time constant is 100 µs. A 5 kHz signal is 3.14 times the corner, so it comes through at 1 / √(1 + 3.14²) = 0.303 of its amplitude (−10.4 dB), lagging by −72.3°. At 100 Hz the same filter is flat to within −0.02 dB; at 50 kHz it is down to 0.0318 (−29.9 dB).
Questions
Is the cut-off where the signal stops?
No. At fc the signal is still at 71%; it is a tenth a decade later and a hundredth two decades later. If you need a frequency really gone, put the corner at least a decade away from it, or use more sections.
Does it matter which way round R and C go?
Entirely: the output is always taken across the part nearest ground. Resistor in series and capacitor to ground is a low-pass; capacitor in series and resistor to ground is a high-pass. The corner frequency is the same for both.
Why is my corner lower than calculated?
Source resistance adds to R, load resistance sits across the capacitor leg, and the capacitor itself may be worth less than it says (X7R ceramics under DC bias). Measure with a buffer each side, or choose R between those two impedances with a factor of ten to spare.
Can I get a sharper filter with one more capacitor?
Two RC sections give 40 dB per decade, but with a soft knee; a Sallen-Key active filter with one op-amp gives the same slope with a sharp, Butterworth-flat corner. For anything over second order use an active or LC design.