RC time constant calculator
A capacitor charges 63.2% of the remaining gap in each time constant. After one tau it is at 63.2%, after two 86.5%, after three 95%, after five 99.3%. Nothing ever technically reaches full charge, so "five tau" became the universal engineering shorthand for settled. Beyond that the remaining error is smaller than the component tolerances.
The RC time constant is resistance times capacitance. A 10 kΩ resistor with a 10 µF capacitor gives τ = 100 ms, so it reaches 63.2% in 100 ms and effectively settles after five time constants. Half a second. The same pair is a low-pass filter with a 1.59 Hz cutoff.
How to use the RC time constant
The same RC pair is two different things depending on what you care about. In the time domain it is a delay or a debounce, characterised by tau. In the frequency domain it is a first-order filter with a −3 dB cutoff at 1/(2πRC) and a 20 dB per decade roll-off. They are the same physics viewed from two angles, and the relationship between them explains a trade nobody escapes: the 1.59 Hz filter these defaults describe knocks 50 Hz mains hum down by a factor of 31, about 30 dB, and it does so by making the circuit take half a second to respond to anything at all.
τ = RC is a two-component model of a circuit that has more than two components in it, and three of the omissions matter. The first is the source. The formula assumes whatever drives the network has zero output impedance, so the real time constant is (R_source + R) × C; a microcontroller pin at roughly 25 Ω disappears against 10 kΩ, and a high-impedance sensor does not. The second is the load. Anything drawing current from the output sits in parallel with the capacitor’s charging path, so both the final voltage and the time constant fall: work with R in parallel with the load resistance rather than with R alone. The third is that nothing ever quite reaches the supply, because a real capacitor leaks. With a 10 kΩ resistor the leakage is invisible; with a 10 MΩ one an ordinary electrolytic can settle a visible fraction below the rail and stay there.
The component that decides how close reality comes to the number on screen is almost always the capacitor. A 1 per cent resistor next to a ±20 per cent aluminium electrolytic gives a time constant good to ±20 per cent, so a design that depends on tau to better than that needs a film or class 1 ceramic part. Class 2 ceramics are worse than their tolerance suggests: an X7R loses capacitance as DC voltage is applied and keeps losing it while the voltage stays there. Vishay measured four manufacturers’ 0603 X7R 100 nF parts at 40 per cent of rated voltage and found every competing part more than 20 per cent down after 1,000 hours. A timing circuit that was right on the bench can be measurably wrong a month later for that reason alone.
One last practical point, because it changes which row of the panel you should be reading. A capacitor feeding a logic input does not do anything at 63.2 per cent; the input switches when the voltage crosses its own threshold. An ordinary CMOS gate switches near half the supply, and the time to reach half is 0.693τ, which is exactly the half-life row here. For a switch debounce, size that figure against the bounce you are suppressing, a few milliseconds on most tactile switches, and feed the result into a Schmitt-trigger input rather than a plain gate. A slow edge into a plain gate spends milliseconds crossing the undefined region between logic levels, and the output oscillates the whole way through.
What people use it for
- Designing a switch debounce
- Setting a low-pass filter cutoff
- Timing a 555 or a microcontroller RC input
- Estimating how long a capacitor takes to charge
- Sizing an RC snubber or a soft-start delay
Questions
τ = R × C. It is the time to reach 63.2% of the final voltage, and the natural time-scale of the circuit.
Five time constants gets to 99.3%, which is the accepted engineering definition of settled. Beyond that the remaining gap is smaller than the component tolerances.
f = 1 ÷ (2πRC). For 10 kΩ and 10 µF that is 1.59 Hz.
Not for tau. Only the product matters. It matters a great deal for the current drawn and the impedance the circuit presents.
Same time constant, mirrored curve. Discharge falls to 36.8% in one tau and to 0.7% in five.
Source resistance adds to R. The formula assumes an ideal driver, so the true constant is (R_source + R) × C. A logic pin adds tens of ohms and is invisible here; a sensor output can add tens of kilohms and is not.
The load sits in parallel with the capacitor’s charging path. Both the final voltage and the time constant drop, so recalculate with R in parallel with the load resistance.
As accurate as the capacitor, which is usually the loosest part in the circuit. Aluminium electrolytics are commonly ±20 per cent, class 2 ceramics ±10 to ±15, and class 1 ceramics and film parts ±5 or better.
Class 2 ones do, badly. Capacitance falls as soon as DC bias is applied and keeps falling while it stays applied: in Vishay’s comparison of 0603 X7R 100 nF parts at 40 per cent of rated voltage, every competing part was more than 20 per cent down after 1,000 hours. Use C0G or film where the value has to hold.
Leakage. A real capacitor passes a small current, and the voltage settles where that current matches the current through R. It is unmeasurable with a 10 kΩ resistor and obvious with a 10 MΩ one.
At its own threshold, not at 63.2 per cent. An ordinary CMOS input flips near half the supply, and the time to reach half is 0.693τ, which the panel shows as the half-life row.
Aim the half-life figure at a few milliseconds, comfortably longer than the switch bounces. A 10 kΩ resistor with a 100 nF capacitor gives τ = 1 ms and a half-life of 0.69 ms, which suits most tactile switches.
No. The slow edge it produces sits in the undefined region between logic levels for milliseconds, and a plain gate oscillates the whole way across. Feed it into a Schmitt-trigger input, or into a pin whose datasheet says it has hysteresis.
Divide the frequency by the cutoff and read it off the roll-off. At 50 Hz with a 1.59 Hz cutoff the ratio is 31, so the hum comes out about 30 dB down, near a thirtieth of its amplitude.
45 degrees, lagging. It approaches 90 degrees well above the cutoff, which is what matters if the filter sits inside a control loop.
Yes: swap the resistor and the capacitor. Same τ, same cutoff frequency, and the roll-off applies below the cutoff instead of above it.
Half C times V squared, so 125 microjoules for 10 µF at 5 V. Exactly the same amount is dissipated in the resistor while charging, whatever value the resistor is.
Dielectric absorption. Some of the charge sits in slow-relaxing parts of the dielectric and returns after the short circuit is removed, a per cent or two on an aluminium electrolytic and far less on a film part.