RC Filter Calculator
Cutoff frequency fc = 1/(2πRC) and time constant for RC and RL low-pass or high-pass filters. Solve for R, C or L and check attenuation on a Bode plot.
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About this toolWhat is the RC Filter Calculator?
A resistor and a capacitor make the simplest filter there is. This RC filter calculator gives its cutoff frequency, fc = 1 ÷ (2πRC), and its time constant τ = RC, for both low-pass and high-pass versions. Switch to RL to do the same for a resistor and an inductor, where fc = R ÷ (2πL) and τ = L ÷ R.
Any one of the three values can be the unknown. Choose Solve for cutoff frequency, resistance or capacitance (or inductance), enter the other two, and the missing one is filled in. The calculator then rounds the result to standard E24 resistors and E12 capacitors and shows the cutoff those real parts give.
Enter a signal frequency to read the gain in decibels, the fraction of the amplitude that gets through and the phase shift at that frequency. A Bode magnitude plot spanning three decades either side of fc shows the flat passband, the −3 dB corner and the 20 dB per decade roll-off, with your check frequency marked on the curve.
- RC and RL, low-pass and high-pass
- Solve for fc, R, C or L
- Gain, amplitude and phase at any frequency
- Bode magnitude plot over six decades
- Time constant, rise time and settling time
- Nearest standard E24 resistor and E12 capacitor with resulting fc
How to use the RC Filter Calculator
- Choose RC or RL, and Low-pass or High-pass.
- Choose what to solve for: cutoff frequency, resistance, or capacitance (inductance for RL).
- Enter the two known values with their units. The solved field is greyed out and filled in for you.
- Optionally enter a signal frequency to read the gain, amplitude ratio and phase at that frequency.
- Read fc, τ, rise time and the −20 dB point, the standard-part version of the filter, and the Bode plot.
Worked examples
The textbook low-pass
Input
RC low-pass, R 1 kΩ, C 100 nF, check 10 kHz
Output
fc 1.5915 kHz, τ 100 µs; at 10 kHz −16.07 dB (15.72%), phase −81.0°
1 ÷ (2π × 1000 × 100 × 10⁻⁹) = 1,591.5 Hz. 10 kHz is about 6.3 times fc, so the output is about one sixth of the input.
Designing a 1 kHz filter with a 10 kΩ resistor
Input
Solve for capacitance: fc 1 kHz, R 10 kΩ
Output
C = 15.915 nF; with the E12 part 15 nF, fc = 1.061 kHz (+6.1%)
Calculated capacitances rarely land on stock values. If 6% matters, adjust R instead: keep 15 nF and solve for R to get 10.61 kΩ, then use 10.7 kΩ from E96.
PWM to analogue voltage
Input
RC low-pass, R 10 kΩ, C 1 µF, check 20 kHz
Output
fc 15.915 Hz; at 20 kHz −61.98 dB (0.08%)
A 20 kHz PWM ripple is reduced about 1,250 times, leaving a smooth DC level. The cost is a slow response: τ is 10 ms, so the output takes about 50 ms to settle after a change.
Audio coupling high-pass
Input
RC high-pass, R 10 kΩ, C 1 µF, check 50 Hz
Output
fc 15.915 Hz; at 50 Hz −0.42 dB, phase +17.7°
A coupling capacitor and the next stage's input resistance form a high-pass. With fc near 16 Hz, bass at 50 Hz loses less than half a decibel.
Background and insight
A capacitor's impedance falls with frequency, Xc = 1 ÷ (2πfC). In a low-pass RC filter the output is taken across the capacitor, so it forms a divider with R whose ratio drops as frequency rises. The transfer function is H(f) = 1 ÷ (1 + j f/fc), giving a magnitude of 1 ÷ √(1 + (f/fc)²) and a phase of −atan(f/fc). At f = fc the resistor and capacitor impedances are equal in size, the output is 1/√2 = 70.7% of the input (−3.01 dB, half power), and the phase is −45°. Swap the parts so the output is across the resistor and you have a high-pass filter with magnitude (f/fc) ÷ √(1 + (f/fc)²).
Far beyond fc a first-order filter falls by 20 dB per decade (6 dB per octave): at ten times fc a low-pass passes only about 10% of the amplitude, at a hundred times about 1%. In the time domain the same circuit charges with time constant τ = RC, reaching 63.2% of a step in one τ, 99.3% in five τ, and rising from 10% to 90% in 2.2τ, which is where the rule of thumb rise time ≈ 0.35 ÷ fc comes from. An RL filter behaves identically with τ = L ÷ R.
These formulas assume an ideal source (zero output impedance) and an ideal load (infinite input impedance). A real source resistance adds to R and lowers fc; a real load resistance in parallel with the output forms another divider, lowering the passband gain and raising fc. Keep the load at least ten times R, or buffer the output with an op-amp. Capacitor tolerance is often ±10% or worse, which moves fc by the same proportion. For steeper roll-off, cascade buffered stages or use an active filter design.
Common use cases
Anti-aliasing before an ADC
Pick R and C so frequencies above half the sample rate are attenuated before sampling.
Smoothing PWM into a DC voltage
Choose a cutoff far below the PWM frequency and check the ripple attenuation.
Audio tone and coupling networks
Set the bass cutoff of a coupling capacitor or a simple treble cut.
Debouncing and timing
Use the time constant to set how quickly a switch input or reset line settles.
Tips and limitations
- The −3 dB point is not where the filter "stops" anything. A first-order filter only reaches −20 dB a decade beyond fc, so put fc well away from the frequencies you want to remove.
- Include the source impedance in R and keep the load at least 10 × R, or the real cutoff will differ from the calculation.
- Ceramic capacitors such as X7R and Y5V change value with voltage and temperature. Use C0G or film capacitors when fc must be accurate.
- In RL filters, the inductor's winding resistance adds to R and its core can saturate; check both in the datasheet.
- Two RC stages in a row do not give a clean second-order response unless the second stage is buffered or has a much higher impedance than the first.
Frequently asked questions
What is the formula for the cutoff frequency of an RC filter?
fc = 1 ÷ (2π × R × C), with R in ohms, C in farads and fc in hertz. For 1 kΩ and 100 nF that is 1,591.5 Hz. The same formula applies to low-pass and high-pass versions.
Why is the cutoff frequency called the −3 dB point?
At fc the output amplitude is 1/√2, or 70.7%, of the input. In decibels that is 20 × log10(0.707) = −3.01 dB, which is also the point where the output power has halved.
How do I calculate the capacitor for a given cutoff frequency?
Rearrange the formula: C = 1 ÷ (2π × R × fc). For a 1 kHz cutoff with a 10 kΩ resistor, C = 15.9 nF; the nearest stock value, 15 nF, gives 1.06 kHz.
What is the relationship between time constant and cutoff frequency?
They are two views of the same circuit: fc = 1 ÷ (2πτ), where τ = RC. A 100 µs time constant corresponds to a 1.59 kHz cutoff.
How much does an RC filter attenuate above the cutoff?
About 20 dB per decade once you are well past fc. A low-pass passes roughly 10% of the amplitude at ten times fc and 1% at a hundred times fc. Exactly at 2 × fc it is −6.99 dB.
What is the difference between an RC low-pass and high-pass filter?
The same two parts in different places. Taking the output across the capacitor gives a low-pass filter; taking it across the resistor gives a high-pass filter. Both have the same cutoff frequency.
When should I use an RL filter instead of RC?
Mostly in power circuits, where inductors carry large currents with little loss, such as the LC output filters of switching regulators. At signal level, RC filters are smaller, cheaper and closer to ideal.
Is there an in-depth guide to the RC Filter Calculator?
Yes. “RC Filter Cutoff Frequency: Where 1/(2πRC) Comes From” covers the background, how it works, common pitfalls and worked examples. Read the full guide.
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