RC, LC & Active Filter Calculator
Cutoff from parts, or parts from a cutoff — with standard values and a Bode plot.
Parts
| Part | Calculated | Standard value | Difference |
|---|
Frequency response (Bode plot)
How it was calculated
About the RC, LC & Active Filter Calculator
Pick a filter — first-order RC or RL low- or high-pass, the passive RC band-pass, a second-order LC filter or a unity-gain Sallen-Key active filter — and work in either direction. Analyse gives the cutoff frequency, Q and time constant from the parts you have; Design calculates the parts for a target cutoff, rounds them to standard E-series values and shows the cutoff you will really get.
Second-order designs can be Butterworth (flattest pass band), Bessel (best step response) or Chebyshev with 0.5 to 3 dB of ripple. Every result comes with a magnitude and phase plot (Bode plot), the gain and phase at any frequency you type, the formulas with your numbers, and a CSV of the response.
How to use it
- Choose the filter type and Analyse (parts → cutoff) or Design (cutoff → parts).
- Analyse: type the part values (10k, 100n, 4.7m …). Design: type the cutoff frequency, choose the response for second-order filters, and enter the part you want to fix — usually the capacitor, because capacitors come in fewer values.
- Choose the E-series for the calculated parts (E24 or E96 for resistors, E6 or E12 for capacitors and inductors) and read the standard values and the cutoff they give.
- Type any frequency under Gain at a frequency to read the attenuation and phase there; download the response as CSV if you need it.
Examples
R 1 kΩ, C 100 nF
fc = 1/(2π × 1k × 100n) = 1.592 kHz, τ = 100 µs; −20 dB at 15.9 kHz
fc 1 kHz, C 100 nF, E24
R = 1.592 kΩ → 1.6 kΩ, fc = 994.7 Hz (−0.53%)
fc 1 kHz, C1 10 nF (to ground)
C2 = 22 nF (≥ 4Q² × C1), R1 ≈ 14.6 kΩ, R2 ≈ 7.9 kΩ before rounding
fc 10 kHz, Butterworth, R 8 Ω
L = 180 µH, C = 1.41 µF
100 Hz to 10 kHz, C1 1 µF, C2 1 nF
R1 = 1.592 kΩ, R2 = 15.92 kΩ
First-order filters
- RC: fc = 1 / (2πRC), time constant τ = RC. A low-pass takes the output across C, a high-pass across R.
- RL: fc = R / (2πL); R includes the source and load resistance in the loop.
- At fc the output is 3 dB down (70.7%) with a phase of −45° (low-pass) or +45° (high-pass); beyond it the response falls by 20 dB per decade.
- The 10–90% rise time of a first-order low-pass is 2.2τ (≈ 0.35 / fc).
Second-order filters, f0 and Q
A second-order filter is described by its natural frequency f0 and its quality factor Q: H(s) = 1 / (1 + s/(Q·ω0) + s²/ω0²) for a low-pass. Q sets the shape of the knee: Q = 0.707 is Butterworth (no peak, −3 dB at f0), Q = 0.577 is Bessel (gentle step response, −3 dB at 0.786 × f0), and Chebyshev filters trade a small pass-band ripple for a sharper knee (Q = 0.864, 0.957, 1.129 and 1.305 for 0.5, 1, 2 and 3 dB of ripple). The calculator always reports the true −3 dB frequency and, for Q above 0.707, the height and position of the peak. Beyond the cutoff the response falls by 40 dB per decade.
LC filters
The LC low-pass puts L in series and C across the load R; the high-pass swaps them. The load sets the damping: Q = R × √(C/L), and f0 = 1 / (2π√(LC)). To design, the calculator finds f0 for the chosen response and uses L = R / (2πf0 × Q) and C = Q / (2πf0 × R). The source driving the filter is assumed to have negligible resistance.
Sallen-Key active filters (unity gain)
The unity-gain Sallen-Key low-pass (TI SLOA024B, Figure 2) has R1 and R2 in series to the op-amp input, C1 from the input to ground and C2 from the junction of R1 and R2 to the output: f0 = 1 / (2π√(R1R2C1C2)) and Q = √(R1R2C1C2) / (C1(R1 + R2)). Real resistors need C2 ≥ 4Q² × C1, so the design picks the next standard C2 above that and solves R1 and R2 exactly before rounding. The high-pass swaps resistors and capacitors; with C1 = C2 = C it needs R1 (to ground) = 4Q² × R2 (feedback). SLOA024B recommends capacitors of at least 100 pF (C0G/NP0 preferred), resistors from a few hundred ohms to a few kilohms, and 1% parts.
Passive RC band-pass
A high-pass R1C1 followed by a low-pass R2C2. The design corners are 1/(2πR1C1) and 1/(2πR2C2), but the second stage loads the first, so the calculator works out the real −3 dB points from the full transfer function sR1C1 / (1 + s(R1C1 + R2C2 + R1C2) + s²R1C1R2C2). Keep R2 at least ten times R1 and the corners well apart; the peak gain is always below 0 dB.
Sources
- fc = 1 / (2πRC) — the first-order RC filter
- Texas Instruments SLOA024B, J. Karki, Analysis of the Sallen-Key Architecture
- Williams, A. B. and Taylor, F. J., Electronic Filter Design Handbook, 4th ed. (McGraw-Hill, 2006) — Butterworth, Bessel and Chebyshev responses
- IEC 60063:2015, Preferred number series for resistors and capacitors
Limitations
- Parts and op-amps are ideal. Real op-amps need a bandwidth far above the cutoff, and a Sallen-Key high-pass becomes a band-pass where the op-amp runs out of gain (TI SLOA024B).
- Source and load impedances are not included except where the filter defines them (R in the RL and LC filters). Connecting a passive filter to a low-impedance load changes its response.
- Inductors are taken as pure inductance; their winding resistance (DCR) lowers Q and adds loss.
- Higher-order filters (fourth order and above) are built from several second-order stages with different Qs; this calculator designs one stage at a time.
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Frequently asked questions
How do I calculate the cutoff frequency of an RC filter?
fc = 1 / (2πRC). With R = 1 kΩ and C = 100 nF: 1 / (2π × 1000 × 0.0000001) = 1,592 Hz. At that frequency the output is 3 dB (29%) lower than in the pass band.
What resistor do I need for a given cutoff?
R = 1 / (2π × fc × C). For 1 kHz with 100 nF that is 1.592 kΩ; the nearest E24 value, 1.6 kΩ, gives 995 Hz. Use Design mode to get this rounding and the resulting cutoff automatically.
How much does a filter attenuate above the cutoff?
A first-order filter falls by about 20 dB per decade (6 dB per octave), a second-order filter by 40 dB per decade. Type a frequency under Gain at a frequency to get the exact figure — a first-order low-pass is −20.04 dB at ten times fc.
Butterworth, Bessel or Chebyshev?
Butterworth for the flattest pass band, Bessel when pulses and steps must keep their shape (it has the most constant delay), Chebyshev when you want a sharper transition and can accept some ripple in the pass band.
Why does the Sallen-Key design need two different capacitors?
For a unity-gain low-pass with equal resistors, Q = ½√(C2/C1), so a Butterworth filter (Q = 0.707) needs C2 = 2 × C1. The calculator takes the next standard value at or above 4Q² × C1 and solves the resistors exactly for it.