RC Low-Pass Filter Calculator
From a resistor and a capacitor, get the cutoff frequency and the time constant that define a first-order RC low-pass filter.
Two inputs, two answers
Enter the resistance and the capacitance (in microfarads) and the calculator returns the cutoff frequency (1 / (2π·R·C)) and the time constant (R·C) at once.
Capacitance is in microfarads
Enter the capacitance in microfarads (µF), not farads — a 100 nF capacitor is 0.1 µF, a 1 nF capacitor is 0.001 µF. The calculator converts to farads for you.
What is an RC low-pass filter calculator?
Resistor and capacitor in, cutoff out
An RC low-pass filter calculator turns two component values — a resistor and a capacitor — into the numbers that describe how the simplest passive filter behaves: the cutoff frequency, above which signals are increasingly attenuated, and the time constant, which sets how quickly the filter settles after a change. A low-pass filter passes slow (low-frequency) signals and blocks fast (high-frequency) ones, so it is the go-to circuit for smoothing a noisy supply, debouncing a sensor, or rolling off the treble in audio. Just a resistor in series and a capacitor to ground is all it takes.
Enter the resistance and capacitance (in microfarads) to get the cutoff frequency and time constant instantly.
Two short formulas, both built from the resistance R (in ohms), the capacitance C (converted from microfarads to farads), and the constant π (about 3.14159).
f_c = 1 / (2π × R × C)The time constant τ = R × C is the product of the resistance and the capacitance in farads; it has units of seconds and tells you how fast the capacitor charges and discharges. The cutoff frequency is its inverse scaled by 2π: f_c = 1 / (2π × R × C). Because you enter the capacitance in microfarads, the calculator first multiplies by one-millionth to get farads before applying either formula.
Suppose you build a filter with a 1 kΩ resistor (1000 Ω) and a 1 µF capacitor.
Convert capacitance
1 µF × 0.000001 = 0.000001 F — the capacitance in farads.
Time constant
1000 × 0.000001 = 0.001 s — the filter settles in roughly 5 τ, about 5 ms.
Cutoff frequency
1 / (2π × 1000 × 0.000001) = 159.154943 Hz — signals above this are increasingly attenuated.
The two outputs answer two related questions. The cutoff frequency (about 159.15 Hz for 1 kΩ and 1 µF) is the −3 dB point: exactly there, the output amplitude drops to about 70.7 % of the input (and the power to half). Below it, signals pass almost untouched; above it, they roll off at 6 dB per octave, so a signal at ten times the cutoff is attenuated roughly tenfold. The time constant τ (here 0.001 s) is the same circuit seen in the time domain: after a step change the capacitor reaches about 63 % of the new level in one τ and is within 1 % after five τ. Cutoff and time constant are two views of one filter — f_c = 1 / (2πτ) — so making τ larger (a bigger R or C) both slows the response and lowers the cutoff. Pick R and C together: the same cutoff can come from a large resistor with a small capacitor or the reverse, and the choice affects loading and noise.
The formulas are exact for an ideal first-order filter, but a couple of practical points are worth keeping in mind.
Ideal components and gentle roll-off
This calculator models an ideal first-order RC low-pass filter — a single resistor and a single capacitor, with no source resistance, load, or parasitic effects. Real resistors and capacitors have tolerances (often ±5 % to ±20 %), so the measured cutoff will drift from the computed value. The roll-off is gentle at 6 dB per octave, so a single RC stage is no brick-wall filter; cascade stages or use an active filter for a sharper transition. The next stage's input impedance also loads the filter unless it is much larger than R.