LC Filter Calculator (Low-Pass Cutoff, Q & Attenuation)
By Santaji GadepatilinUpdated November 9, 2026
A series inductor followed by a shunt capacitor makes a second-order low-pass filter that falls at 40 dB per decade above its cutoff — the output filter of every buck converter, and a common supply and audio filter. This LC filter calculator gives the cutoff frequency, the characteristic impedance √(L/C), the Q factor set by the load and the attenuation at any frequency, so you can check both ripple rejection and resonant peaking.
How to use the LC filter calculator
Enter the series inductance, the shunt capacitance, the load resistance across the capacitor and a frequency to check (for a switching regulator, the switching frequency). The calculator gives the cutoff frequency, the characteristic impedance, the Q factor and the gain at your frequency as a ratio and in dB, plus the gain at f₀, which shows how much the filter peaks. To design for a target cutoff, enter f₀ and the inductance and press × on Capacitance.
LC low-pass filter formulas
f₀ = 1 ÷ (2π√(LC)) · Z₀ = √(L ÷ C) · Q = R ÷ Z₀
|H(f)| = 1 ÷ √[(1 − x²)² + (x ÷ Q)²], x = f ÷ f₀
At f = f₀ the gain equals Q. Far above f₀ it approaches 1 ÷ x², i.e. −40 dB per decade. The formula is for ideal parts — the ESR of the capacitor adds damping but also a zero that flattens the roll-off at high frequency.
Worked example
A buck converter output filter with L = 33 µH, C = 22 µF and a 1.2 Ω load, switching at 100 kHz:
- f₀ = 1 ÷ (2π√(33 µH × 22 µF)) = 5,907 Hz
- Z₀ = √(33 µH ÷ 22 µF) = 1.2247 Ω, so Q = 1.2 ÷ 1.2247 = 0.980 — well damped
- At 100 kHz, x = 16.930 and the gain is 0.003495 = −49.1 dB, so the ripple is reduced about 286 times
- Gain at f₀ = 20 log₁₀(0.980) = −0.18 dB — no significant peaking
Cutoff frequency and characteristic impedance
| L | C | f₀ | Z₀ = √(L/C) |
|---|---|---|---|
| 10 µH | 10 µF | 15.9 kHz | 1 Ω |
| 22 µH | 47 µF | 4.95 kHz | 0.684 Ω |
| 33 µH | 22 µF | 5.91 kHz | 1.22 Ω |
| 47 µH | 100 µF | 2.32 kHz | 0.686 Ω |
| 100 µH | 100 µF | 1.59 kHz | 1 Ω |
| 1 mH | 10 µF | 1.59 kHz | 10 Ω |
| 10 mH | 1 µF | 1.59 kHz | 100 Ω |
Gain versus frequency and Q
| Frequency | Q = 0.5 | Q = 0.707 | Q = 1 | Q = 2 | Q = 5 |
|---|---|---|---|---|---|
| 0.5 × f₀ | −1.9 dB | −0.3 dB | +0.9 dB | +2.0 dB | +2.4 dB |
| 1 × f₀ | −6.0 dB | −3.0 dB | +0.0 dB | +6.0 dB | +14.0 dB |
| 2 × f₀ | −14.0 dB | −12.3 dB | −11.1 dB | −10.0 dB | −9.6 dB |
| 5 × f₀ | −28.3 dB | −28.0 dB | −27.8 dB | −27.7 dB | −27.6 dB |
| 10 × f₀ | −40.1 dB | −40.0 dB | −40.0 dB | −39.9 dB | −39.9 dB |
| 100 × f₀ | −80.0 dB | −80.0 dB | −80.0 dB | −80.0 dB | −80.0 dB |
Practical notes
- Light load = high Q. A converter filter that is well damped at full load can ring at no load; check Q at the minimum load too.
- Place f₀ at least a decade below the switching frequency, and well away from the control-loop crossover frequency of the regulator.
- Use the capacitance at the working voltage — ceramic X5R/X7R capacitors can lose 50% or more of their value under DC bias.
- The inductor must not saturate at the peak load current, and its self-resonant frequency limits attenuation at very high frequencies.
- An LC input filter in front of a switching regulator can make it oscillate (negative input impedance); its output impedance Z₀ should be well below the regulator’s input impedance.
Frequently asked questions
What is the cutoff frequency of an LC filter?
f₀ = 1 ÷ (2π√(LC)), with L in henries and C in farads. Above f₀ the output falls at 40 dB per decade (12 dB per octave), twice as fast as an RC filter.
What does the load resistance do to an LC filter?
It sets the damping. Q = R ÷ √(L/C). A light load (high R) gives a high Q and a resonant peak at f₀; a load near √(L/C) gives Q ≈ 1 and a flat response. At Q = 0.707 the response is maximally flat (Butterworth).
How much does an LC filter attenuate ripple?
Well above f₀ the gain is about (f₀ ÷ f)². A filter with f₀ = 1 kHz reduces 100 kHz ripple by about 10,000 times (−80 dB), before capacitor ESR and ESL limit it.
How do I damp an LC filter?
Add resistance in series with an extra capacitor across C (an RC damper), use a capacitor with suitable ESR, or put a resistor in parallel with the inductor. The damper should give an effective Q of about 0.5–1.