RC Time Constant and Filter Calculator

Find the time constant and cutoff frequency of a resistor and capacitor, the gain of a simple low-pass or high-pass filter at a frequency you choose, how long the capacitor takes to charge, or the part value for a cutoff you want.

Cutoff frequency fc
Time constant (tau = R x C)
Low-pass gain at the test frequency
High-pass gain at the test frequency
Charging: 63 percent, 95 percent, 99 percent
10 to 90 percent rise time

How it works

A resistor and a capacitor together have a time constant τ = R × C. A capacitor charging through R reaches 63 percent of the final voltage after one τ, 95 percent after 3 τ and 99 percent after 4.6 τ, and the 10 to 90 percent rise time is 2.2 τ. As a filter the same pair has a cutoff frequency fc = 1 ÷ (2π R C), where the output is 0.707 times the input (−3 dB). A first-order low-pass has a gain of 1 ÷ √(1 + (f ÷ fc)²) and a phase lag of arctan(f ÷ fc); a high-pass has f ÷ fc times that. See Capacitors for the theory.

Worked example

R = 10 kΩ and C = 100 nF give τ = 10 000 × 100 × 10⁻⁹ = 1 ms and fc = 1 ÷ (2π × 1 ms) = 159.2 Hz. At 1 kHz a low-pass passes 1 ÷ √(1 + (1000 ÷ 159.2)²) = 0.157, which is −16.1 dB, with a phase lag of 81°. The capacitor reaches 63 percent after 1 ms, 95 percent after 3 ms and 99 percent after 4.6 ms. For a 1 kHz cutoff with a 4.7 kΩ resistor you need C = 1 ÷ (2π × 4700 × 1000) = 33.9 nF, so choose a 33 nF or 39 nF part.

Things to know

This is a first-order passive filter with no load. A load, a source resistance or the capacitor’s tolerance (often 10 to 20 percent for ceramics) moves the cutoff.

Gain drops by 20 dB per decade above the cutoff for a first-order filter. For a steeper slope use two or more stages or an active filter.

Choose the nearest standard value and check the result: use the calculator again with the real R and C.

Capacitors have leakage, ESR and voltage coefficients that matter in precise timing circuits.

Common questions

What is the cutoff frequency?

The frequency where the output has fallen to 70.7 percent of the input voltage (−3 dB, half the power). Below it a low-pass passes the signal; above it, it attenuates.

Why 63 percent?

After one time constant a charging capacitor has covered 1 − e⁻¹ = 63.2 percent of the way to the final voltage.

How do I make a debounce or delay with this?

Choose τ longer than the contact bounce (a few milliseconds for a switch) or the delay you want, then check the threshold of the input you are driving.

Sources

  • Horowitz, P. and Hill, W., The Art of Electronics, 3rd ed., Cambridge University Press: RC circuits and filters.
  • IEC 60063, Preferred number series for resistors and capacitors.

Last reviewed: October 22, 2026