How this calculator works
Charge a capacitor through a resistor and the voltage across it doesn't jump — it climbs along an exponential curve. τ = R·C sets the pace of that curve: one τ after the switch closes, the capacitor has covered 63.2% of the gap to its final voltage, not 100%.
- R — resistance (Ω)
- C — capacitance (F) — this calculator takes µF
- τ — time constant (s)
- V(t) — capacitor voltage at time t
- V₀ — supply / final voltage
- f_c — -3 dB cutoff frequency (Hz)
Common values
| Resistance (Ω) | Time constant (τ) |
|---|---|
| 100 Ω | 0.100 ms |
| 250 Ω | 0.250 ms |
| 500 Ω | 0.500 ms |
| 1,000 Ω | 1.000 ms |
| 1,500 Ω | 1.500 ms |
| 2,000 Ω | 2.000 ms |
| 3,000 Ω | 3.000 ms |
| 5,000 Ω | 5.000 ms |
| 10,000 Ω | 10.00 ms |
More detail
Why 5τ counts as "fully charged"
The exponential curve never mathematically touches 100% — it only gets closer. At t = 5τ it has covered 1 − e⁻⁵ ≈ 99.3% of the gap, which is where engineers draw the line for "settled." A 1 kΩ resistor with a 1 µF capacitor has τ = 1 ms, so its output is close enough to final by 5 ms; waiting for a mathematically exact 100% would mean waiting forever.
Reading the cutoff frequency
The same R and C also define an RC low-pass filter: signals below f_c pass through close to full strength, and signals above it get progressively attenuated. f_c is exactly 1/τ divided by 2π, so a slow, high-τ pair (say 10 kΩ and 100 µF, τ = 1 s) makes a filter with a very low cutoff — around 0.16 Hz — useful for smoothing out anything faster than a slow drift.
Design tip. Debounce and power-up delay circuits want a large 5τ relative to the noise or ramp you're waiting out. Filter circuits instead care about f_c matching the frequency you want to pass or block — pick R and C to hit that number, not a round τ.
Frequently asked questions
1 kΩ and 1 µF — what's the time constant and the settle time?
τ = 1 kΩ × 1 µF = 1 ms. That's also the 63.2%-charged time. Full settle (5τ) is 5 ms, and the cutoff frequency if used as a filter is about 159.15 Hz.
10 kΩ and 1 µF — what cutoff frequency does that give?
τ = 10 kΩ × 1 µF = 10 ms, so f_c = 1/(2π × 10 ms) ≈ 15.92 Hz — useful for filtering out anything faster than roughly a 16 Hz signal.
I want a ~235 ms power-up delay with a 4.7 kΩ resistor — what capacitor?
4.7 kΩ and 10 µF gives τ = 47 ms, so 5τ (the practical settle point) is 235 ms — enter those two values above to see it live.
10 kΩ and 100 µF — how slow is that filter?
τ = 10 kΩ × 100 µF = 1 s, giving f_c = 1/(2π × 1 s) ≈ 0.16 Hz — this pair only passes changes slower than about one cycle every 6 seconds.