RC time constant calculator

Work out the time constant of a resistor and capacitor, the −3 dB corner it puts on a single-pole low-pass filter, and how long a step takes to settle to the accuracy you need.

Time constant

15.92 µs

Corner frequency (−3 dB)

9.997 kHz

Settles within 0.01 %

146.6 µs, or 9.21 time constants

After one time constant

63.2 % of the way there

τ = R × C · f₋₃dB = 1 ÷ (2π × R × C) · t = −ln(accuracy ÷ 100) × R × C

A resistor feeding a capacitor has one number that governs everything it does: the time constant, the resistance times the capacitance. It sets how fast the capacitor charges, where the circuit stops passing signal, and how long you have to wait before a reading means anything. This works out all three from the two component values.

How it is calculated

τ = R × C · f₋₃dB = 1 ÷ (2π × R × C) · t = −ln(accuracy ÷ 100) × R × C

The step response is 1 − e^(−t/RC), so the capacitor covers the same fraction of the remaining distance in every time constant: 63.2 % after one, 99.3 % after five. It never quite arrives, which is why settling is quoted to an accuracy rather than as a finish line.

Source: Burr-Brown — Fast Settling Low-Pass Filter, application bulletin AB-022 (SBOA011, January 1991): f-3dB = 1/(2 • π • R1 • C1) and tS = –ln(%/100) • R1 • C1

Questions people ask

How long until the capacitor is charged?
It never finishes, so the question has to be asked with a tolerance attached. Each time constant closes 63.2 % of whatever distance is left: one gets you to 63.2 %, five to 99.3 %, and 9.2 of them to within 0.01 % of the final value. Pick the accuracy your circuit actually needs and read the time off above.
Why is the corner not where the signal stops?
Because a single RC is a gentle filter. The corner frequency is where the output has fallen to −3 dB, about 71 % of the input, and beyond it the response only falls by a factor of ten per decade. Signal well above the corner is attenuated, not removed.
Does swapping the resistor and capacitor change anything?
Not the time constant, and not the corner. Doubling the resistance and halving the capacitance leaves both untouched. What it changes is everything around them: the current drawn, the impedance the circuit presents, and how much the following stage loads it.
Why does my 10 kHz filter not come out at 10 kHz?
Because the parts are rounded before the maths is. The bulletin behind this page pairs 10 kΩ with 1592 pF and calls it a 10 kHz filter; work it through exactly and the corner lands at 9.997 kHz. Nothing is wrong — capacitors come in the values they come in, and a corner three parts in ten thousand off nominal is far inside the tolerance of the capacitor itself.
Is filtering worth the wait?
That is the trade the bulletin behind this page was written about. Narrowing the bandwidth by a factor of a hundred cuts broadband noise by ten, because noise adds as the square root of the bandwidth — but the settling time grows in step. In fast measurement work that penalty is often the binding constraint.

Found a problem, or want more?

A number that disagrees with its source is a defect, not a rounding preference.

What did you enter, what did the tool show, and what did you expect instead? If you have a source that disagrees with ours, a link to it is the most useful thing you can send.

Write to us

Opens your mail app with the page and tool already filled in.

Related tools