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LC filter

LC filter: the circuit drawn with the example's values10µH10µF5 Ω
Low-pass f0 = 15.9 kHz, -31.7 dB at 100 kHz, Q 5 on 5 Ω

Twice the slope of an RC, but a resonance at the corner that a light load lets ring.

Opens in the app on 1 December, with Circuit Notebook 1.0 See it worked out

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The example, worked out

Given

Filter
Low-pass: L in series, C across
Inductance L
10 µH
Capacitance C
10 µF
Load R
5 Ω
Frequency to reject
100 kHz

Working

  1. where L and C meet
  2. the load sets the peak
  3. -31.7 dB at 100 kHz

Results

Corner f015.9 kHz
At 100 kHz-31.7 dB
Q on this load5
Peak, near f0+14 dB
−3 dB point24.6 kHz
Z0 = √(L/C)1 Ω
Load for a flat response707 mΩ

The formulas

  1. f0 = 1 / (2π √(L C))where L and C meet
  2. Z0 = √(L / C)the impedance of the pair at f0
  3. Q = R / Z0the load sets the peak: light load, high Q

The usual traps

  • Unloaded, an LC filter peaks at f0: a buck output filter rings when the load drops.
  • For a flat response the load must be about 0.7 × Z0, not any value.
  • The slope is 40 dB per decade past f0, not 20.

Questions

How do I calculate the cutoff of an LC filter?

f0 = 1 / (2π √(L C)): where L and C meet. Z0 = √(L / C): the impedance of the pair at f0. Q = R / Z0: the load sets the peak: light load, high Q.

What does it give for typical values?

Low-pass f0 = 15.9 kHz, -31.7 dB at 100 kHz, Q 5 on 5 Ω. Corner f0: 15.9 kHz; At 100 kHz: -31.7 dB.

What are the usual mistakes?

Unloaded, an LC filter peaks at f0: a buck output filter rings when the load drops. For a flat response the load must be about 0.7 × Z0, not any value. The slope is 40 dB per decade past f0, not 20.

In the app

Your circuit, drawn live

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The part you can buy

Results rounded to standard values (E12, E24, E96) and power ratings.

The working, shown

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Into your notebook

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Opens in the app on 1 December, with Circuit Notebook 1.0

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