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Full changelogEngine 1.18.3

LC Resonance & Damping

Signal & Timing

Resonant frequency, Q and characteristic impedance of an LC pair, and the resistance that damps it.

Inputs

H
10 µH
F
22 µF
Inductor DCR + capacitor ESR + trace
Ω
50 mΩ
A resistive load across C; gives the loaded Q
Ω
The target when solving for L or C
Hz

Results

1 to check
Resonant frequency f0

Not judged: a resonant frequency has no pass or fail. Damping decides whether the network behaves: see the quality factor below.
10.73kHz
Characteristic impedance Z0

√(L/C)
674.2mΩ
Quality factor Q

damping ratio ζ = 0.037
13.48
All results (4)
Step overshoot
89.0%
Peak gain at f₀
22.6dB
Series R for Q = 1
674.2mΩ
Critical damping R

ζ = 1, no overshoot
1.348Ω
Q = 13.5: this network rings (about 89 % overshoot, 23 dB of peaking at 10.73 kHz). Raise the series resistance to 674.2 mΩ for Q = 1, or add an R–C damping leg. An R–C leg across the capacitor damps the peak without the DC loss of a series resistor.
|Vout/Vin|

Accuracy

Verified against

3 independent cases. See the working.

This is a design aid. The engineer remains responsible for the design and for checking the standard itself.

Parameter sweep

Vary one input over a range and see the answer and verdict at each step, as a table and a curve.Pro

Worst-case corners

Put a tolerance on each input and get the worst-case band around the answer.Pro

Circuit

The principle

An inductor and a capacitor exchange energy at a frequency set by √(LC). How hard they ring depends on the loop resistance compared with the characteristic impedance √(L/C). That ratio decides whether a filter behaves.

Resonant frequency
f0=12πLCf_0 = \frac{1}{2\pi\sqrt{LC}}
Characteristic impedance
Z0=LCZ_0 = \sqrt{\frac{L}{C}}
The reactance of L, and of C, at f0. Size damping resistance against this, not against the load.
Quality factor and damping (series RLC)
Q=Z0Rs=1RsLCζ=12Q=Rs2Z0Q = \frac{Z_0}{R_s} = \frac{1}{R_s}\sqrt{\frac{L}{C}} \qquad \zeta = \frac{1}{2Q} = \frac{R_s}{2 Z_0}
Step overshoot
Mp=e−πζ/1−ζ2×100  %M_p = e^{-\pi\zeta/\sqrt{1-\zeta^{2}}}\times 100\;\%
Bandwidth
BW=f0QBW = \frac{f_0}{Q}
Critical damping
Rs=2Z0  ⇒  ζ=1  (no overshoot)R_s = 2 Z_0 \;\Rightarrow\; \zeta = 1 \;\t{(no overshoot)}
  • Z0Z_0characteristic impedance of the LC pair
  • RsR_stotal series resistance: inductor DCR + capacitor ESR + copper
  • ζ\zetadamping ratio; ζ < 1 rings, ζ = 1 is critical, ζ > 1 is overdamped
More detail

Input filters and negative resistance

A switching converter draws constant power, so its input looks like a negative resistance, −V²/P. If the filter's output impedance at resonance exceeds that magnitude, the supply oscillates at the filter's resonant frequency. This is Middlebrook's criterion. The cure is damping, not more filtering.

Damp without wasting power. Put the damping resistor in series with a separate electrolytic or polymer capacitor, in parallel with the main ceramic. It carries no DC current but dominates at resonance. Typical values: Cdamp ≈ 3–5 × Cmain, Rdamp ≈ Z₀.

The same maths elsewhere

  • Snubbers. Switch-node ringing is the layout inductance against the FET's Coss. Measure the ring frequency, add a known capacitor and measure again. The shift gives L and C, and the snubber resistor is R = √(L/C).
  • Crystals and tank circuits, where you want Q high rather than low.
Further reading: R. D. Middlebrook, "Input filter considerations in design and application of switching regulators" (1976).

Engine version ⁨1.18.3⁩