LC Filter Calculator

Cutoff frequency and characteristic impedance of an L-C reactive filter, or design L and C for a target fc and impedance.
L & C → Cutoff & Zo
Design for Target fc & Zo

Cutoff Frequency & Characteristic Impedance

fc = 1 / (π√(LC))   •   Zo = √(L/C)
10mH, 100nF
2.546mH, 39.79µF (8Ω crossover)
mH
nF
Enter values and press Calculate.
Not the same as simple LC resonance: this constant-k filter cutoff fc=1/(π√LC) is exactly the plain resonance formula f0=1/(2π√LC) used for tank/resonant circuits — see the note below for why.

Magnitude Response (2nd-Order Roll-off)

Design L & C for a Target fc and Zo

L = Zo / (πfc)   •   C = 1 / (πfcZo)
1kHz, 8Ω speaker crossover
100Hz, 50Ω RF/line filter
Hz
Ω
Enter values and press Calculate.

How an LC Filter Differs From an RC Filter — and From LC Resonance

An LC filter section uses one inductor and one capacitor (instead of a resistor and a capacitor) to build a second-order filter: a series inductor followed by a shunt capacitor (low-pass), or a series capacitor followed by a shunt inductor (high-pass). Because it has two reactive elements instead of one, its roll-off past cutoff is −40 dB per decade — twice as steep as the single-element RC filter's −20 dB/decade — while also (ideally) dissipating no power in the passband, since ideal inductors and capacitors don't resist current the way a resistor does.

Why this fc formula looks different from the resonance formula

You may have seen f0=1/(2π√(LC)) before — that is the resonant frequency of an LC tank circuit (see the RLC Resonant Frequency & Q calculator), used for band-pass/band-stop and oscillator design. This calculator answers a different question: given an L and a C wired as a matched filter section (the classic "constant-k image-parameter" filter design from filter theory), where does the passband end? That cutoff works out to fc=1/(π√(LC)) — exactly twice the plain resonance frequency for the same L and C, because a filter section's cutoff condition (image impedance becoming imaginary) is mathematically distinct from a tank circuit's resonance condition (reactances cancelling). Both are valid, well-known formulas — they simply answer two different circuit questions.

QuantityFormula
Cutoff frequency (constant-k filter section)fc = 1/(π√(LC))
Characteristic (nominal) impedanceZo = √(L/C)
Design: inductor for target fc, ZoL = Zo/(πfc)
Design: capacitor for target fc, ZoC = 1/(πfcZo)
Roll-off rate−40 dB/decade (−12 dB/octave), second-order

The characteristic impedance Zo matters because this style of filter is designed to be driven from, and loaded by, a source/load resistance equal to Zo — exactly like an audio speaker crossover is designed around the speaker's nominal impedance (commonly 4 Ω or 8 Ω), or an RF filter is designed around a 50 Ω system.

Real-World Applications & Fully-Explained Examples

Worked examples — explained in full

1. L=10 mH, C=100 nF. fc=1/(π√(0.01×100×10−9))=1/(π√(10−9))=1/(π×3.162×10−5)≈10,065.8 Hz (≈10.07 kHz). Zo=√(0.01/10−7)=√100000≈316.2 Ω.
2. Designing an 8 Ω speaker crossover at 1 kHz. L=8/(π×1000)≈2.546 mH, C=1/(π×1000×8)≈39.79 µF — a textbook first-order-per-driver LC crossover pairing for an 8 Ω woofer/tweeter split at 1 kHz.
3. Designing a 50 Ω filter at 100 Hz. L=50/(π×100)≈159.2 mH, C=1/(π×100×50)≈63.66 µF — a much higher inductance than the speaker example, because a 50 Ω system impedance is over 6× higher than 8 Ω and the cutoff is 10× lower.
4. Doubling the inductor (keeping example 1's 100 nF capacitor). fc drops from 10,065.8 Hz to 7,117.6 Hz — a factor of 1/√2≈0.707, exactly as predicted by fc∝1/√(LC): doubling either reactive element alone lowers fc by √2, not by 2.
5. Doubling the capacitor instead (keeping example 1's 10 mH inductor). fc falls to the identical 7,117.6 Hz as example 4 — doubling L or doubling C has exactly the same effect on fc, since the formula only depends on their product LC.
6. Comparing to plain LC resonance. For the same 10 mH/100 nF pair as example 1, the simple tank-circuit resonance formula f0=1/(2π√(LC))≈5,032.9 Hz — exactly half example 1's filter cutoff of 10,065.8 Hz, concretely demonstrating the 2× relationship between the two formulas for identical components.

Frequently Asked Questions

What is the cutoff frequency of an LC filter?

For a constant-k (image-parameter) LC filter section, fc = 1/(π√(LC)). This is the frequency at which the filter's passband ends and its stopband begins, with a steep −40 dB/decade roll-off past that point.

Why is the LC filter cutoff formula different from the LC resonance formula?

They describe different circuit behaviours. f0=1/(2π√(LC)) is where a series or parallel LC tank circuit resonates (reactances cancel). fc=1/(π√(LC)) is where a constant-k filter section's image impedance becomes imaginary, marking the edge of its passband. For the same L and C, the filter cutoff is exactly twice the resonance frequency.

What is characteristic impedance (Zo) and why does it matter?

Zo=√(L/C) is the impedance the filter is designed to be driven from and loaded into for its response to match the ideal design equations — for example, a speaker crossover is designed around the speaker's nominal impedance (often 4 Ω or 8 Ω), and an RF filter is designed around 50 Ω.

How much steeper is an LC filter than an RC filter?

An LC filter section is second-order (two reactive elements) with a −40 dB/decade roll-off, twice as steep as a first-order RC filter's −20 dB/decade, because it has two reactive elements instead of one.

How do I design L and C for a target cutoff frequency and impedance?

Use L = Zo/(πfc) and C = 1/(πfcZo). Enter your target cutoff and system impedance in the "Design" tab above to get both values directly.

What is the difference between a low-pass and high-pass LC filter?

A low-pass section uses a series inductor followed by a shunt capacitor; a high-pass section swaps them — a series capacitor followed by a shunt inductor. Both use the same fc and Zo formulas for the same L and C values.

Are ideal LC filters lossless?

In the ideal case, yes — inductors and capacitors don't dissipate power the way resistors do, so an ideal LC filter passes in-band signal with no resistive loss. Real inductors have winding resistance and core losses that introduce some practical loss, especially at high current or high frequency.

What impedance should I design a speaker crossover around?

Use the driver's nominal impedance rating (commonly 4 Ω, 6 Ω, or 8 Ω for most consumer speakers) as Zo — check the speaker's datasheet, since using the wrong impedance shifts both the actual cutoff frequency and the crossover's intended response shape.

Can I cascade multiple LC sections for an even steeper filter?

Yes, this is standard practice in filter design (called a multi-section or "ladder" filter) and yields roll-offs of −80 dB/decade or steeper, at the cost of more components and more complex design equations than a single L-section.

How does this differ from the RLC Resonant Frequency calculator on this site?

That calculator finds a series or parallel RLC tank circuit's resonant frequency and Q/bandwidth — useful for oscillators, tuned amplifiers, and band-pass/notch applications. This calculator is for filter design: sizing a simple L-C section to hit a target low-pass or high-pass cutoff at a specified system impedance.

Does source or load resistance change the actual cutoff frequency?

Yes — the ideal formulas assume the filter is driven and loaded exactly at its designed characteristic impedance Zo. A significant mismatch between the actual source/load impedance and Zo will shift the real-world response away from the ideal calculated curve.

What inductor core type should I use for an LC filter?

It depends on frequency and current: air-core or powdered-iron toroids are common for RF and moderate-power filters, while ferrite cores suit higher-frequency, lower-current signal filtering; audio crossover inductors are often air-core or laminated-iron-core coils sized for the expected speaker current.

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