RC Filter Cutoff Frequency & Time Constant Calculator
A first-order RC low-pass or high-pass filter reaches its half-power (-3.01 dB) corner frequency f_c where capacitive reactance X_C equals resistance R: f_c = 1 / (2 × π × R × C).
Half-power corner frequency in Hertz
Series or shunt resistance in Ohms
Capacitance in microfarads (switch inline unit for nF or pF)
Step-by-Step Calculation
A first-order RC low-pass or high-pass filter reaches its half-power (-3.01 dB) corner frequency f_c where capacitive reactance X_C equals resistance R: f_c = 1 / (2 × π × R × C).
Real-World Reference Values
63.2% step-response charge time (5τ = 99.3% settled)
First-order single-pole attenuation rate
Output lags by 45° in low-pass, leads by 45° in high-pass
20-Row Comparison Table
Cutoff Frequency (-3 dB) (f_c) · Filter Resistance (R)
| # | Filter Resistance (Ω) | Cutoff Frequency (-3 dB) (Hz) | Δ % |
|---|---|---|---|
| 01 | 1,000.00 ΩCurrent | 159.1549 Hz | 0.00% |
| 02 | 72,421.99 Ω | 2.1976 Hz | -98.62% |
| 03 | 144,842.99 Ω | 1.0988 Hz | -99.31% |
| 04 | 217,263.98 Ω | 0.732542 Hz | -99.54% |
| 05 | 289,684.97 Ω | 0.549407 Hz | -99.65% |
| 06 | 362,105.96 Ω | 0.439526 Hz | -99.72% |
| 07 | 434,526.96 Ω | 0.366272 Hz | -99.77% |
| 08 | 506,947.95 Ω | 0.313947 Hz | -99.80% |
| 09 | 579,368.94 Ω | 0.274704 Hz | -99.83% |
| 10 | 651,789.93 Ω | 0.244181 Hz | -99.85% |
| 11 | 724,210.93 Ω | 0.219763 Hz | -99.86% |
| 12 | 796,631.92 Ω | 0.199785 Hz | -99.87% |
| 13 | 869,052.91 Ω | 0.183136 Hz | -99.88% |
| 14 | 941,473.91 Ω | 0.169049 Hz | -99.89% |
| 15 | 1,013,894.90 Ω | 0.156974 Hz | -99.90% |
| 16 | 1,086,315.89 Ω | 0.146509 Hz | -99.91% |
| 17 | 1,158,736.88 Ω | 0.137352 Hz | -99.91% |
| 18 | 1,231,157.88 Ω | 0.129273 Hz | -99.92% |
| 19 | 1,303,578.87 Ω | 0.122091 Hz | -99.92% |
| 20 | 1,375,999.86 Ω | 0.115665 Hz | -99.93% |
Frequently Asked Questions
How do I design an anti-aliasing RC filter for a 10 kHz ADC?+
Click SOLVE FOR on R (or C), set f_c to your desired corner frequency (e.g., 4 kHz), and choose a standard E24 capacitor value in nF or μF.
Why is the cutoff point called the -3 dB frequency?+
At f = f_c, output voltage drops to 1/√2 (≈ 0.7071) of input voltage, which equals 20 × log₁₀(1/√2) = -3.0103 dB (half power).
How is the RC time constant τ related to f_c?+
Since τ = R × C, the cutoff frequency is simply f_c = 1 / (2πτ) ≈ 0.159155 / τ.