RC / RL Filter Calculator
Compute time constant, cutoff frequency and step response for a first-order RC or RL filter, with preferred-value design.
- Cutoff frequency
- 159 Hz
- Time constant τ
- 1 ms
- Angular ω
- 1 k rad/s
- Rise time (10–90%)
- 2.2 ms
- Time to 63.2%
- 1 ms
Design for a target cutoff
Enter a target cutoff — R above is held fixed while the capacitor is solved and snapped.
How RC / RL Filter Calculator works
A resistor and a capacitor in series form a first-order filter whose behaviour is governed by one number: the time constant τ = R × C. In the time domain, τ is how long the capacitor takes to reach about 63% of a step change, and five time constants brings it to over 99% — the usual definition of "settled".
In the frequency domain the same circuit rolls off at the cutoff frequency fc = 1 / (2πRC), where the output has fallen to 1/√2 of the input — that is −3 dB, or half the power. Beyond the cutoff, a first-order filter attenuates at 20 dB per decade: ten times the frequency gives ten times the attenuation.
Whether the circuit is a low-pass or a high-pass depends only on where you take the output. Across the capacitor it is a low-pass, since the capacitor’s impedance falls with frequency and shorts high frequencies to the output’s reference. Across the resistor it is a high-pass. The components and the cutoff are identical; only the tap moves.
An RL filter behaves the same way with τ = L / R and fc = R / (2πL). Inductors are bulkier, lossier, and more expensive than capacitors, so RC is the default in signal work — RL appears mainly in power circuits and where an inductor is already present for other reasons.
Reference
- RC time constant: τ = R × C
- RC cutoff: fc = 1 / (2π × R × C)
- RL time constant: τ = L / R
- RL cutoff: fc = R / (2π × L)
- Step response: V(t) = V_final × (1 − e^(−t/τ))
- 10–90% rise time = 2.2 × τ; settling to 99% ≈ 5τ; roll-off = 20 dB/decade
How to use this calculator
Pick the topology
Choose RC or RL, and whether you are building a low-pass or a high-pass.
Enter two of three
Give any two of resistance, capacitance (or inductance), and cutoff frequency — the third is solved.
Read the time-domain figures
Check the time constant and rise time if your concern is settling or debouncing rather than frequency response.
Choose standard parts
Take the suggested preferred values and check how far the resulting cutoff moves from your target.
Worked examples
Anti-aliasing before an ADC
- Given
- R = 1.6 kΩ, C = 100 nF, low-pass
- Result
- fc ≈ 995 Hz, τ = 160 µs
A first-order filter rolls off gently, so place the cutoff well below half your sampling rate — it gives only 20 dB per decade of rejection.
Sizing for a rise time
- Given
- A wanted 10–90% rise time of 1 ms
- Result
- τ = 455 µs, so fc ≈ 350 Hz
Rise time and bandwidth are two views of one constant. Specifying either fixes the other completely.
AC coupling a signal
- Given
- R = 100 kΩ, C = 1 µF, output across the resistor
- Result
- High-pass with fc ≈ 1.6 Hz
Blocks DC offset while passing audio intact. The same parts across the capacitor would give a 1.6 Hz low-pass instead.
When to use it
- Adding an anti-aliasing filter ahead of an analogue-to-digital converter.
- Smoothing a PWM output into a usable analogue voltage.
- Debouncing a mechanical switch in hardware rather than firmware.
- AC-coupling a signal to strip a DC offset without affecting the wanted band.
- Suppressing high-frequency noise on a sensor line.
Things to watch out for
- A first-order filter rolls off at only 20 dB per decade. Where you need sharp separation between passband and stopband, cascade stages or use an active filter topology.
- The source impedance driving the filter and the load impedance following it both shift the cutoff. The formula assumes an ideal source and no loading.
- Cascading two RC stages does not simply double the roll-off unless they are buffered — the second stage loads the first and the combined response is not the product of the two.
- Component tolerance moves the cutoff directly. A 10% capacitor and a 5% resistor give a cutoff that can sit 15% away from nominal, so do not design with a cutoff sitting exactly on a critical boundary.
Frequently asked questions
What is the RC cutoff frequency formula?
fc = 1 / (2π × R × C), with R in ohms and C in farads. At that frequency the output has fallen to 1/√2 — about 70.7% — of the input amplitude, which is half the power, the standard −3 dB definition.
What does the time constant τ actually mean?
It is R × C, the time for the capacitor to reach about 63.2% of a step change. Five time constants gets you past 99%, which is the usual working definition of settled. It is the same physical quantity as the cutoff frequency, viewed in the time domain.
How do I make a high-pass instead of a low-pass?
Take the output across the resistor rather than the capacitor. The components and the cutoff frequency are unchanged — only the tap point differs, which is why the same pair of parts serves either purpose.
How much attenuation does a first-order filter give?
20 dB per decade — a factor of ten in amplitude for every factor of ten in frequency past the cutoff. That is a gentle slope, so signals just above the cutoff are barely reduced. Cascade stages or use an active filter where you need a sharper transition.
How is rise time related to bandwidth?
For a first-order filter, the 10–90% rise time is 2.2 × τ, and bandwidth is 0.35 divided by the rise time. Specifying either one fixes the other, which is why a scope’s rise time and its bandwidth are two ways of stating the same specification.
When would I choose RL over RC?
Rarely in signal work. Inductors are larger, more expensive, lossier, and radiate and pick up magnetic fields. RL filters appear mainly in power supplies, motor drives, and EMI suppression, where an inductor is already present or where its current-smoothing behaviour is the point.
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