Time constant and cutoff frequency
An RC network is defined by one number, the time constant τ = R × C, in seconds. It sets both how fast the capacitor charges and where the filter rolls off. The −3 dB cutoff frequency is:
f = 1 / (2π R C)
10 kΩ with 10 nF gives τ = 100 µs and a cutoff of about 1.6 kHz. A low-pass filter passes frequencies below the cutoff; the same R and C wired as a high-pass passes those above it — the cutoff is the same either way.
Charging and discharging
A capacitor charges toward the supply on an exponential curve, reaching 63% of the way in one τ and settling within 1% after about 5τ. Discharging is the mirror image. This is why "5 time constants" is the usual rule for "fully charged/discharged":
| Time | Charged to |
|---|---|
| 1τ | 63.2% |
| 2τ | 86.5% |
| 3τ | 95.0% |
| 4τ | 98.2% |
| 5τ | 99.3% |
Worked example: a debounce filter
To debounce a switch with a ~1 ms bounce, pick τ a few times longer — say 10 kΩ and 1 µF for τ = 10 ms. The RC smooths the contact chatter, and a Schmitt-trigger input cleans up the slow edge.
Frequently asked questions
- Low-pass or high-pass — does the formula change?
- No. The cutoff f = 1/(2πRC) is identical; what changes is which element the output is taken across. Output across the capacitor → low-pass; across the resistor → high-pass.
- What exactly is "cutoff"?
- The frequency where the output has dropped to −3 dB (70.7% of the input voltage, half the power). It's a gentle knee, not a brick wall — a single-pole RC rolls off at 20 dB/decade.
- Why 63% at one time constant?
- It's 1 − 1/e. The charging curve is 1 − e^(−t/τ), and at t = τ that's 1 − 1/e ≈ 0.632.
- How does this relate to decibels?
- The cutoff is the −3 dB point — see the decibel calculator for what that ratio means.