EMI Input Filter Calculator
EMI Input Filter Calculator
The LC filter in front of a switching converter, and the stability question that comes with it. A regulated converter presents a negative incremental resistance to its input, so an undamped filter can make a working converter oscillate. This page gives the corner, the attenuation at the switching frequency, the filter’s output impedance peak, the Middlebrook verdict, and the parallel R–C damping branch that fixes it.
EMI input filter and its stability
a 400 kHz converter drawing 22 W from 24 V, needing 40 dB of attenuation at the switching frequency, built round a 2.2 µF capacitor with 100 mΩ of total series resistance, damped by a parallel R–C branch with Cd = 4 × Cf
Attenuation, and the impedance that has to stay under
Zin = −Vin² ÷ Pin (a regulated converter is a constant-power load)
Middlebrook: |Zo,filter(f)| << |Zin,converter(f)| at every frequency
Undamped peak: |Zo| ≈ R₀² ÷ Rs at f₀
R–C damping, n = Cd/Cf: Rd,opt = R₀·√[(2+n)(4+3n) ÷ (2n²(4+n))]
peak |Zo| = R₀·√(2(2+n)) ÷ n at f₀·√(2/(2+n))
- A
- attenuation wanted at the switching frequency. A second-order filter rolls off at 40 dB per decade, which is where the 40 in the exponent comes from
- R 0
- the filter’s characteristic impedance, √(L/C). Everything about the stability problem scales with it, so for a fixed corner, more C and less L is always the safer filter
- Z in
- the converter’s incremental input impedance. Negative, because holding the output power constant means the input current FALLS when the input voltage rises
- n
- the damping capacitor’s size relative to the filter capacitor. It buys a lower impedance peak for a bigger part
Worked example
a 400 kHz converter drawing 22 W from 24 V, needing 40 dB of attenuation at the switching frequency, built round a 2.2 µF capacitor with 100 mΩ of total series resistance, damped by a parallel R–C branch with Cd = 4 × Cf
f₀ = 400 kHz ÷ 10^(40/40) = 40 kHz, so L = 1 ÷ ((2π × 40 kHz)² × 2.2 µF) = 7.196 µH (8.2 µH in E12)
R₀ = √(L/C) = 1.809 Ω. The exact attenuation at 400 kHz is 39.91 dB — slightly under the 40 dB the asymptote promised, because a second-order filter is not yet on its asymptote one decade out
The converter is a constant-power load, so its incremental input resistance is −Vin²/Pin = −24² ÷ 22 = −26.18 Ω. That magnitude, 26.18 Ω, is the ceiling the filter's output impedance has to stay under
Undamped, the filter peaks at about R₀² ÷ Rs = 32.76 Ω at its corner — above the 26.18 Ω ceiling. Middlebrook margin -1.9 dB: it fails, and the converter can oscillate at 40 kHz
With Cd = 4Cf = 8.8 µF and the optimum Rd = R₀·√[(2+n)(4+3n)/(2n²(4+n))] = 1.108 Ω, the peak drops to 1.388 Ω at 23.09 kHz
Margin = 20·log₁₀(26.18 ÷ 1.388) = 25.5 dB — comfortably past the 6 dB usually asked for
Damping a filter whose corner is 40 kHz, R₀ = 1.809 Ω
| Damping | R_d | C_d | Peak |Z_o| | Middlebrook margin |
|---|---|---|---|---|
| None | — | — | 32.76 Ω | -1.9 dB |
| R–C branch, optimum R, n = 4 | 1.108 Ω | 8.8 µF | 1.388 Ω | 25.5 dB |
| R–C branch, rule of thumb R_d = R₀, n = 4 | 1.809 Ω | 8.8 µF | 1.962 Ω | 22.5 dB |
| R–C branch, optimum R, n = 10 | 0.690 Ω | 22.0 µF | 0.886 Ω | 29.4 dB |
What moves the margin, and in which direction
| Change | Corner f₀ | Peak |Z_o| | Margin |
|---|---|---|---|
| More filter capacitance, L adjusted to keep f₀ | same | falls as √(L/C) | improves |
| More filter inductance, C adjusted to keep f₀ | same | rises | worsens |
| Lower corner (more attenuation) at fixed C | falls | rises | worsens |
| Lower input voltage | same | same | worsens — |Z_in| goes as V_in² |
| Heavier load | same | same | worsens — |Z_in| goes as 1/P_in |
| Bigger damping capacitor ratio n | same | falls | improves |
| Electrolytic capacitor in place of ceramic | same | falls, its ESR damps | improves |
Why an input filter can stop a working converter working
Two things go in front of a switching converter’s input, and they are the same two components. A series inductor and a shunt capacitor keep the converter’s pulsed input current out of the supply wiring, which is what conducted-emissions limits are about. Above its corner the filter rolls off at 40 dB per decade, so the corner needed for A decibels at the switching frequency is fsw/10A/40 — and the lower that corner, the better the filtering. Every instinct says make the inductor bigger.
The converter is a negative resistance. A converter with a closed feedback loop delivers a fixed output power whatever its input voltage does. Power in is power out over efficiency, so it is fixed too — which means that when the input voltage rises, the input current falls. The incremental input impedance dV/dI is therefore negative, and its magnitude is Vin²/Pin. This page derives that by differentiating the constant-power relation rather than quoting it. The negative sign is not a curiosity: it means the converter supplies energy to any disturbance at its input terminals, within the bandwidth over which its loop can hold the output constant.
Middlebrook’s criterion. Put a filter between a source and that load and the converter’s transfer functions get multiplied by a correction factor built from the ratio Zo,filter/Zin,converter — Middlebrook’s 1976 result. Keep that ratio small at every frequency and nothing changes: the converter behaves as though the filter were not there. Let the filter’s output impedance rise above the converter’s input impedance and the minor loop can encircle −1 and the system oscillates, typically as a sustained growl at the filter’s resonance rather than anything at the switching frequency. An undamped LC is exactly where that happens, because at resonance its output impedance peaks at roughly R₀²/Rs, and with ceramic capacitors Rs is almost nothing. Making the inductor bigger makes this worse, because R₀ = √(L/C) rises. For a fixed corner, more capacitance and less inductance is always the safer filter.
What the criterion is and is not. It is sufficient, not necessary. It compares magnitudes and ignores phase, so a design that violates it by a few decibels may still be perfectly stable — the overlap has to occur where the phase makes it dangerous. What that means in practice is that passing the criterion proves stability and failing it proves nothing except that you now have to do the full Nyquist analysis. Since passing it is usually cheap, most designers simply pass it. Six decibels of separation is the usual minimum, and defence and aerospace practice often specifies more.
The damping branch. The fix is a resistor that is present at the filter’s resonance and absent at DC: Rd in series with a blocking capacitor Cd = n·Cf, the pair connected across Cf. Too large a resistor and the branch does nothing; too small and it shorts Cf and a new, lower resonance appears against Cd. The optimum sits where those two limiting curves cross, and this page computes both the optimum resistor and the peak it leaves by searching the real impedance function. That search turned up a correction worth recording: the peak output impedance under optimum R–C damping is R₀·√(2(2+n))/n, not the R₀·√(2(2+n)/n) the result is frequently printed as. The two agree only at n = 1. For the common n = 4 the correct figure is 0.866·R₀ and the misprinted one is 1.73·R₀ — a factor of two, and in the unsafe direction if you trust it.
Two more things this page does not do. It assumes an ideal source, so it says nothing about the supply’s own impedance or the cable between them, which can add a resonance of its own. And it models one stage: where a single stage would need a corner low enough to be awkward, two cascaded stages of half the attenuation each are usually smaller, cheaper and better behaved. For the response shapes themselves see the RC and RLC filter calculator and the LC resonant frequency calculator; the converters this filter feeds are designed on the buck, boost and flyback pages.
Frequently asked questions
What is the Middlebrook criterion?
That the output impedance of the filter in front of a switching converter must stay well below the converter’s own input impedance at every frequency: |Z_o| << |Z_in|. Middlebrook showed in 1976 that the converter’s transfer functions are multiplied by a correction factor built from that ratio, so keeping the ratio small leaves the converter behaving as though the filter were not there. It is a sufficient condition, not a necessary one.
Why does a converter have negative input impedance?
Because a regulated converter draws constant power. If the input voltage rises and the output power is held fixed, the input current must fall — so dV/dI is negative, with magnitude V_in² ÷ P_in. It only behaves this way within its control loop’s bandwidth; above the loop’s crossover the input looks like the passive components that are actually there.
Why does my converter oscillate after I added an input filter?
Almost certainly because the filter is undamped and its output impedance peak at resonance has risen above the converter’s input impedance. The symptom is a low-frequency oscillation on the input rail at the filter’s corner — a few kilohertz to a few tens of kilohertz — not anything at the switching frequency. Add a parallel R–C damping branch, or swap some inductance for capacitance at the same corner.
How do you damp a switching converter’s input filter?
A resistor in series with a blocking capacitor, the pair across the filter capacitor. The blocking capacitor keeps the resistor out of the DC path so it dissipates nothing in normal operation; it is usually 2 to 10 times the filter capacitor. For a ratio n the optimum resistor is R₀·√[(2+n)(4+3n)/(2n²(4+n))] with R₀ = √(L/C), and it leaves a peak of R₀·√(2(2+n))/n.
Can I use an electrolytic capacitor instead of a damping resistor?
Often, yes — that is the same trick with the resistor built in, since an aluminium electrolytic’s ESR is in series with its capacitance by construction. Texas Instruments’ own note on the problem fixes an oscillating LMR14050 with a 47 µF electrolytic of 100 mΩ ESR across the input. The catch is that ESR rises sharply at low temperature and falls over life, so the damping is not a controlled quantity the way a resistor is.
Should I make the input filter inductor bigger for more attenuation?
Be careful: it improves attenuation and worsens stability at the same time, because the filter’s characteristic impedance √(L/C) rises and the impedance peak rises with it. For the same corner frequency, more capacitance and less inductance gives the same attenuation with a lower peak. If one stage cannot give the attenuation you need at a safe √(L/C), use two.
Related calculators
References
- Middlebrook RD. Input Filter Considerations in Design and Application of Switching Regulators. IEEE Industry Applications Society Annual Meeting, Chicago, October 1976, pp. 366–382. The original: the correction factor built from Z_o/Z_in, and the criterion that the filter’s output impedance must stay well below the converter’s input impedance. Everything on this page’s stability section descends from it.
- Erickson RW, Maksimović D. Fundamentals of Power Electronics, 3rd ed. Springer, 2020, Ch. 10 Input Filter Design. States the criterion as ‖Z_o‖ << ‖Z_N‖ and ‖Z_o‖ << ‖Z_D‖, gives the regulated converter’s incremental input resistance as −R/M², and works the R–C parallel damping optimum in §10.4.1 with n = C_b/C. The closed forms this page uses are checked against a numerical search rather than taken from the text.
- Hegarty T. The Engineer’s Guide to EMI in DC-DC Converters, Part 10: Input Filter Impact on Stability. Texas Instruments / How2Power Today, November 2019. Gives the criterion as |Z_out,F| << |Z_in,C| with a gain-margin factor, the constant-power input resistance −V_in/I_in, and the practical R_d ≈ √(L_f/C_f), C_d ≈ 4·C_f damping rule this page compares against the optimum.
- Texas Instruments, Simple Solution for Input Filter Stability Issue in DC/DC Converters, application report SLUA929. The negative incremental input impedance explained from ΔVin/ΔIin, and a measured case: an LMR14050 at 9 V in, 5 V/5 A out, oscillating until a 47 µF electrolytic of 100 mΩ ESR is added across the input.
- Würth Elektronik, 1-Phase Line Filter Design, application note ANP015 (2024-06-03). Gives the single-stage design step used here, f_CO = f_sw ÷ 10^(A/40 dB) for a second-order filter at 40 dB/decade, and the case for splitting a large attenuation requirement across two stages.
- Sclocchi M. Input Filter Design for Switching Power Supplies. National Semiconductor / Texas Instruments, literature number SNVA538. A worked single-stage example with a parallel damped filter — 33 µH and 47 µF, damping resistor 0.838 Ω, blocking capacitor 188 µF — in which the chosen resistor is exactly √(L/C), the rule of thumb this page quantifies against the optimum.