Chapter 24: Filters, Snubbers, and EMI Mitigation — Power Electronics

Where this sits

Part 5 · Passive Components
Chapter 24 of 30 — the last of Part 5

You should already know

Leakage inductance and turn-off spikes from Chapter 14, loop stability from Chapter 15, and winding capacitance from Chapter 23.

By the end you can

Separate differential- from common-mode noise, design an input filter that does not destabilise the converter, size an RCD snubber, and read a conducted-emissions plot.

Time

≈ 55 min reading · ≈ 45 min problems

i What you'll learn

Why EMI is a legal requirement rather than a quality target, and what the limits actually say.
The distinction that organises everything: differential mode versus common mode, and why they need completely different filters.
How a common-mode choke can present 10 mH to noise and 5 μH to the load current, using the same winding.
Y-capacitors and the leakage-current limit that caps them — the one component a safety standard sizes for you.
The Middlebrook criterion: why adding an input filter can destabilise a perfectly good converter, and how to damp it.
Snubbers: RC, RCD and active clamp — what each is for, and how to size one from measurement rather than guesswork.
Why layout beats components, and which three loops decide whether a converter passes.

Section 24-1

Two Kinds of Noise

A switching converter is, by construction, a generator of electromagnetic interference. It exists to change voltages and currents as fast as possible, and \( dv/dt \) and \( di/dt \) are precisely what radiates and conducts.

Since the 1990s this has not been a matter of good practice. Conducted and radiated emission limits — CISPR 32, EN 55032, FCC Part 15 — are mandatory for sale in essentially every market. A converter that fails is not a poor product; it is an illegal one.

All of it organises around one distinction.

Differential noise circulates between live and neutral; common-mode noise flows out on both together and returns through the earth. They need different filters because they take different paths.

◇Intuition · two ways for water to escape a pipe

Picture a flow-and-return pipe pair carrying water to a machine.

Differential mode is a leak between the two pipes — water crosses from flow to return and circulates in a loop that never leaves the pair. The flow and return carry equal and opposite extra current.
Common mode is a leak from both pipes to the ground — water escapes downwards from each and finds its way back through the soil. Both pipes carry extra current in the same direction, and the return path is somewhere else entirely.

Where each comes from in a converter:

Differential noise is the switching current itself. A buck converter draws pulses from its input; those pulses flow out on live and back on neutral. The source is a current, and the frequencies are the switching frequency and its harmonics — typically 150 kHz to a few megahertz.
Common-mode noise comes from \( dv/dt \) acting on stray capacitance to earth. A switching node moving 400 V in 20 ns, with 50 pF to the heatsink, drives \( i = C\,dv/dt = 1 \) A into the chassis. The source is a voltage, and the frequencies extend to tens of megahertz.

Why the distinction organises the whole subject. A series inductor in each line blocks differential current beautifully and does nothing for common mode, because both lines carry it in the same direction and it returns outside the pair. The two need different components, in different places, and diagnosing which one you have is the first step in every EMI investigation.

And a rough rule that holds surprisingly often: below about 1 MHz, failures are usually differential; above about 5 MHz, almost always common mode. Between the two, look at both.

Section recap. EMI limits are mandatory, not aspirational. Differential noise is the switching current circulating between live and neutral; common-mode noise is \( dv/dt \) driving current through stray capacitance to earth. They take different paths, so they need different filters — and identifying which one dominates is the first diagnostic step.

Section 24-2

The Input Filter

A mains input filter is a small number of components arranged so that each addresses one mode. Once you know which is which, the schematic reads itself.

What each component in a mains filter does
Component	Where it sits	Blocks	Sized by
X-capacitor	Line to neutral	Differential	Attenuation needed; 0.1–2.2 μF typical. Needs a bleed resistor.
Y-capacitor	Line and neutral to earth	Common mode	The earth leakage current limit — a safety constraint, not an EMI one
Common-mode choke	Both lines, same core, same sense	Common mode	Impedance needed at the failing frequency; 1–30 mH typical
Differential inductor	One or both lines, separate cores	Differential	Often just the choke's leakage inductance
Damping network	Across the filter's output	Nothing — it damps the filter's own resonance	The Middlebrook criterion of Section 24-3

◈How a common-mode choke ignores the load current

A common-mode choke is the most elegant component in the filter, because it presents an enormous impedance to one thing and essentially none to another — with a single winding arrangement.

The construction. Both lines are wound on the same core, in the same direction, with equal turns.

For the load current — which flows out on live and back on neutral — the two windings' magnetomotive forces are equal and opposite. The net flux is zero. The core does nothing, and the choke presents only its small leakage inductance, typically 0.5–2% of the rated value.

For common-mode current — which flows the same way in both lines — the two MMFs add. The core sees the full flux, and the choke presents its full inductance, often 10 mH or more.

Two consequences that make the component possible at all:

The core never saturates on load current. A 10 A appliance's current produces no net flux, so a tiny high-permeability core can carry it — which is why a 10 mH choke rated for 10 A is the size of a large coin rather than a brick.
Very high permeability is usable. Nanocrystalline or high-\( \mu \) MnZn ferrite with \( \mu_r = 10{,}000 \) would saturate instantly in any ordinary inductor. Here it is ideal.

And the leakage inductance is not waste. The small imbalance flux that does escape the core gives a differential inductance of a few microhenries — which is often exactly the differential filtering needed, so many designs use it deliberately and fit no separate differential inductor.

The limitation worth knowing. A choke's own interwinding capacitance resonates with its inductance, typically between 1 and 10 MHz. Above that resonance it is a capacitor, and its attenuation falls. This is why high-performance filters use two chokes of different values in series — one for the low megahertz, one for the high.

▲Common mistake

"The common-mode emissions are 10 dB too high — fit bigger Y-capacitors."

Wrong Treating the Y-capacitor as an EMI component to be sized for attenuation.

Right Its value is fixed by a safety limit on earth leakage current, and that limit is usually already reached.

The constraint. A Y-capacitor connects a live conductor to the earthed chassis, so at mains frequency it passes a continuous current to earth:

\[ I_{leak} = 2\pi f C V \]

Safety standards cap that current because it flows through anyone who touches the chassis if the earth connection is broken:

IEC 62368-1, portable equipment: 0.25 mA
IEC 62368-1, stationary pluggable: 3.5 mA
IEC 60601-1, medical (patient-connected): 0.1 mA or less

Work out what that permits. At 230 V, 50 Hz with a 3.5 mA budget:

\[ C_{max} = \frac{I}{2\pi f V} = \frac{3.5\times10^{-3}}{2\pi(50)(230)} = 48\ \text{nF total} \]

and a portable product at 0.25 mA is limited to about 3.5 nF. Standard Y2 capacitors are 2.2 nF or 4.7 nF precisely because of this.

So the Y-capacitor is one of the few components a standard sizes for you, and there is normally no headroom to increase it.

What to do instead when common-mode emissions are too high:

Reduce the source. Slow the switching edges with a larger gate resistor — a few nanoseconds of \( dv/dt \) is worth many decibels, at the cost of some switching loss.
Reduce the coupling capacitance. Shrink the switching-node copper, and insulate the transistor from the heatsink with a thicker or lower-permittivity pad.
Add a transformer screen. A grounded copper foil between primary and secondary intercepts the displacement current before it reaches the secondary. This is frequently worth 10–20 dB for the cost of a strip of foil.
Add a second common-mode choke of a different value, to extend the attenuation above the first one's self-resonance.

And a caution about medical designs: at 0.1 mA the permitted Y-capacitance is so small that the filter must rely almost entirely on chokes and screening. This is why medical power supplies are physically larger than commercial ones of the same rating.

Section recap. X-capacitors block differential noise, Y-capacitors and the common-mode choke block common mode. The choke works because load current produces no net flux, so a tiny high-permeability core gives 10 mH. Y-capacitors are sized by a safety leakage limit, not by attenuation, and are usually already at maximum.

Section 24-3

When the Filter Destabilises the Converter

An input filter is a passive \( LC \) network placed in front of a converter, and it is easy to assume it can do no harm. It can, and the mechanism is one of the more surprising results in power electronics.

The reason is that a regulated converter is not a resistor. It draws constant power, so if its input voltage falls, its input current rises.

Write the converter's input characteristic

Working

\[ P = V_{in}I_{in} = \text{constant} \;\Longrightarrow\; I_{in} = \frac{P}{V_{in}} \]
Differentiate to get the small-signal input resistance

Result

\[ \frac{dI_{in}}{dV_{in}} = -\frac{P}{V_{in}^2} \;\Longrightarrow\; R_{in} = \frac{dV_{in}}{dI_{in}} = -\frac{V_{in}^2}{P} \]

Negative. A regulated converter presents a negative incremental resistance at its input, of magnitude \( V^2/P \).
See what that does to the filter An \( LC \) filter's resonance is damped by the resistance loading it. A negative resistance supplies energy instead of absorbing it. If it is large enough to overcome the filter's own losses, the resonance grows rather than decays — the filter oscillates.
State the condition for safety

The Middlebrook criterion

\[ \bigl|Z_{out,filter}\bigr| \ll \bigl|Z_{in,converter}\bigr| \quad\text{at all frequencies} \]

The filter's output impedance must stay well below the converter's input impedance — typically by 6 dB or more — everywhere, and especially at the filter's resonant peak.

Interactive · filter resonance and the Middlebrook margin

An input filter (\( C_f = 2\ \mu\text{F} \)) feeding a 200 W converter from 400 V. The filter's output impedance peaks at its resonance; the converter's input impedance is a flat −800 Ω. Add damping until the peak clears the limit by a comfortable margin.

Damping resistor R_d 10 kΩ (none)

Filter inductance L_f 100 µH

Resonance f₀11.3 kHz

Peak |Z_out|2.50 kΩ

Margin−9.9 dB

VerdictUnstable

—

▣In practice · damping without wasting power

The obvious fix — a resistor across the filter capacitor — would work and would dissipate continuously at full load. That is unacceptable, so real designs use one of three arrangements that damp only at the resonance.

Series RC damping. Put \( R_d \) in series with a blocking capacitor \( C_d \), and place the pair across the filter capacitor. \( C_d \) blocks DC and the mains frequency, so \( R_d \) dissipates nothing in normal operation. Choose \( C_d \approx 4 \) to \( 10 \) times \( C_f \) so the damping is effective near \( f_0 \), and \( R_d \approx \sqrt{L_f/C_f} \).
Parallel RL damping. \( R_d \) in parallel with a blocking inductor \( L_d \), in series with the filter inductor. The inductor carries the DC, so again the resistor dissipates only at high frequency. Better where the capacitance is fixed.
Lossy magnetics. Wind the filter inductor on a deliberately lossy core, or fit a ferrite bead. Free, and often enough on its own — the damping comes from core loss rather than a discrete resistor.

How to size the series-RC network, in three steps:

\[ Z_0 = \sqrt{\frac{L_f}{C_f}}, \qquad R_d \approx Z_0, \qquad C_d \approx (4\text{–}10)\,C_f \]

Worked instance. \( L_f = 100\ \mu\text{H} \), \( C_f = 5\ \mu\text{F} \):

\[ Z_0 = \sqrt{\frac{10^{-4}}{5\times10^{-6}}} = 4.5\ \Omega, \qquad f_0 = 7.1\ \text{kHz} \]

So \( R_d = 4.7\ \Omega \) and \( C_d = 22\ \mu\text{F} \). The resistor's dissipation is only the ripple current at \( f_0 \) squared times 4.7 Ω — typically well under a watt.

What happens if this is skipped. The symptom is distinctive and often misdiagnosed: the converter's input voltage oscillates at a few kilohertz, the output has a matching low-frequency ripple, and the whole thing may audibly hum. Engineers often chase this in the control loop for days before realising the loop is fine and the input filter is oscillating. Measuring the input voltage with a scope, rather than the output, identifies it in seconds.

Section recap. A regulated converter presents a negative input resistance \( -V^2/P \), which can feed an input filter's resonance instead of damping it. The Middlebrook criterion requires the filter's output impedance to stay well below the converter's input impedance everywhere. Series-RC damping achieves this while dissipating nothing at DC.

Section 24-4

Snubbers

A filter deals with noise once it has left the converter. A snubber deals with it at the source, by limiting the \( dv/dt \) or \( di/dt \) that created it — and, just as importantly, by keeping the device inside its safe operating area.

The snubber families and what each is for
Type	Circuit	Limits	Energy goes to	Used for
RC across the switch	\(R\) and \(C\) in series, device to device	Ringing and \(dv/dt\)	The resistor	Damping the leakage-inductance ring in almost any converter
RCD turn-off	Diode into \(C\), \(R\) discharging it	Peak voltage	The resistor	Flyback primary clamp; IGBT turn-off
Turn-on (series \(L\))	Small inductor in series with the switch	\(di/dt\) and diode recovery	Its own snubber	Thyristors, hard-switched bridges
Active clamp	MOSFET and capacitor	Peak voltage	Recycled to the source	Forward converters; efficient high-power designs
TVS / avalanche	Transient-voltage-suppressor diode	Absolute peak	The TVS	Last-resort clamping; fault conditions only

◇Intuition · a door closer, not a doorstop

A door that slams shut makes a bang, damages the frame, and rattles the windows. Two responses are possible:

A doorstop — put something rigid in the way so the door cannot travel past a point. That is a clamp: it limits the peak, absorbing whatever energy arrives.
A door closer — a damper that slows the motion so the door arrives gently. That is a snubber in the strict sense: it does not limit the endpoint, it limits the rate.

Both are used, and the names are often muddled — an "RCD snubber" is really a clamp, and an "RC snubber" is really a damper. What matters is knowing which problem you have:

The peak voltage exceeds the device rating → you need a clamp. RCD, active clamp or TVS.
The peak is fine but the ringing radiates → you need a damper. RC across the switch or the diode.
The device is being destroyed at turn-off → you need to move the switching trajectory inside the safe operating area, which usually means a turn-off snubber that holds the voltage down while the current falls.

Why the distinction matters practically. A clamp sized to fix ringing will be far too large and will waste power continuously. A damper sized to fix an overvoltage will not clamp anything and the device will still fail. Diagnosing which you have takes one oscilloscope measurement across the device, and it is always worth doing before fitting anything.

Measure the ringing, do not guess it Probe across the device and record the ringing frequency \( f_r \). This is set by the parasitic inductance and the device capacitance, neither of which you know accurately.
Add a known capacitance and measure again Add \( C_{add} \) across the device and record the new frequency \( f_r' \). Two measurements, two unknowns:

Working

\[ C_p = \frac{C_{add}}{(f_r/f_r')^2 - 1}, \qquad L_p = \frac{1}{(2\pi f_r)^2 C_p} \]

A convenient choice is \( C_{add} \) large enough to halve the frequency, which gives \( C_p = C_{add}/3 \).
Choose the snubber components

Result

\[ R_s = \sqrt{\frac{L_p}{C_p}} = Z_0, \qquad C_s \approx (3\text{–}4)\,C_p \]

\( R_s = Z_0 \) gives critical damping; \( C_s \) large enough that the snubber dominates the parasitic capacitance.
Check the resistor's dissipation

The check that is usually forgotten

\[ P_{R} = C_s V^2 f_{sw} \]

Note the full \( CV^2 \), not \( \tfrac12CV^2 \): the capacitor is charged and discharged once each per cycle, and both events dissipate in the resistor.

▲Common mistake

"The ringing is bad — fit a bigger snubber capacitor until it goes away."

Wrong Increasing \( C_s \) until the waveform looks clean, without checking the resistor's dissipation.

Right \( P_R = C_sV^2f_{sw} \) rises linearly with \( C_s \), and it can reach tens of watts before you notice.

Work an example. A 400 V bridge at 100 kHz with a 10 nF snubber:

\[ P_R = (10\times10^{-9})(400)^2(10^5) = 160\ \text{W} \]

Per snubber. Four of them in a full bridge is 640 W — from components chosen to "clean up the waveform".

A sensible value would be 1 nF, giving 16 W per snubber, and even that needs a proper wirewound resistor with a heatsink.

Three consequences that follow, and are all commonly missed:

The resistor must be rated for real power, and for the pulse current \( V/R_s \) — which for a 4.7 Ω snubber on 400 V is 85 A peak. A small film resistor will fail open within minutes, at which point the snubber does nothing and nobody notices until the devices fail.
The capacitor must be a pulse type. Its ripple current is severe and its dielectric must tolerate the fast edges. C0G/NP0 ceramic or polypropylene film; never X7R, whose losses and DC bias behaviour are both unsuitable.
The snubber adds to the switch's turn-on current. Whatever charge sits on \( C_s \) is dumped through the device when it turns on — increasing turn-on loss even as turn-off loss falls. There is an optimum, and past it a snubber makes total loss worse.

The order to attack ringing in, before reaching for a snubber at all:

Reduce the loop inductance. The ringing energy is \( \tfrac12L_pI^2 \); halving \( L_p \) halves it. Tighter layout is free and always the first move.
Slow the edges with a larger gate resistor. Costs switching loss but reduces the excitation.
Use a device with less recovery charge — a SiC diode has essentially none, and often removes the ringing entirely.
Only then fit a snubber, sized from measurement.

Section recap. A clamp limits the peak; a damper limits the rate. Size either from two ringing measurements rather than by guesswork, and always check \( P_R = C_sV^2f_{sw} \) — which can reach hundreds of watts. Reduce the loop inductance before fitting anything.

Section 24-5

Layout Beats Components

Filters and snubbers treat symptoms. The source of nearly all of it is the physical arrangement of three current loops and one voltage node — and getting those right is free, while getting them wrong cannot be fully corrected afterwards.

1 The three loops that matter

The high-\( di/dt \) power loop. In a buck converter this is the input capacitor, the switch and the diode. Current in it jumps from zero to full load in nanoseconds, so its inductance produces the turn-off spike and its area radiates. Make it as small as physically possible, with the input capacitor directly across the switch pair.
The gate drive loop. Driver output, gate, source, and back to the driver. Any inductance shared with the power loop couples switching current into the gate, causing false turn-on or oscillation. Use a separate source connection — a Kelvin source pin exists for exactly this.
The current-sense loop. Small signals in the noisiest place in the converter. Route differentially, keep it short, and filter at the destination rather than the source.

2 The node that matters

The switching node — the point that swings the full DC voltage at every transition — is the converter's antenna, and the source of essentially all common-mode current.
Keep its copper area minimal. It must carry the current, and nothing more. Every extra square millimetre is capacitance to everything nearby.
Watch what is underneath it. Copper on an adjacent layer forms a capacitor; a heatsink behind an insulating pad forms another. 50–200 pF is typical, and at 400 V in 20 ns that is amps of common-mode current.
Where the device must be heatsinked, consider an insulated package, a thicker pad, or a screened pad with a grounded foil layer that intercepts the displacement current.

✦Think about it

A converter fails conducted emissions by 12 dB at 20 MHz. The filter is already at the limit — the Y-capacitors cannot grow and the choke is as large as will fit. What now?

Attack the source, not the symptom

First, read the frequency. 20 MHz is well above the switching frequency and its low harmonics, so this is common mode — \( dv/dt \) driving current through stray capacitance to earth. That immediately rules out X-capacitors and differential inductors.

The source current is \( i = C_{stray}\,dv/dt \), so there are exactly two things to reduce.

Reduce \( dv/dt \):

Increase the gate resistor. Going from 5 Ω to 15 Ω might slow the edge from 20 ns to 50 ns — a 2.5× reduction, worth 8 dB. It costs switching loss, and that is the trade.
Add a small gate-to-drain capacitor to slow only the voltage transition, leaving the current edge alone. More surgical, and less costly in loss.

Reduce \( C_{stray} \):

Shrink the switching node's copper. Often worth several decibels for no cost at all.
Change the insulating pad. A thicker pad or a lower-permittivity material halves the capacitance to the heatsink, which is usually the dominant path.
Use a screened pad — an insulator with a grounded foil in the middle. The displacement current is intercepted and returned locally instead of travelling to the mains.

Redirect the current so it never reaches the mains:

Add a transformer screen. A grounded copper foil between primary and secondary catches the displacement current at the barrier. This is frequently worth 10–20 dB for the cost of a strip of foil, and it is the single highest-value change available in an isolated supply.
Add a small capacitor from the secondary return to the primary return — typically 1–2.2 nF. It gives the common-mode current a short local path home instead of a long one through the earth. Must be a Y-rated part, since it bridges the isolation barrier.

And spread the spectrum, if the standard's measurement bandwidth allows: dithering the switching frequency by a few percent smears each harmonic across a band, reducing the peak a quasi-peak detector reports by 6–10 dB. It does not reduce the total energy, and some argue it is gaming the measurement — but it is legal, widely used, and built into most modern controllers.

The order to try them: transformer screen first (large, cheap), then layout and pad changes (free), then gate resistance (costs efficiency), then spread spectrum (costs nothing but does not reduce real interference).

Section recap. Three loops — power, gate drive and current sense — and one node decide most EMI performance. Minimise the power loop's area, give the gate its own source connection, and keep the switching node's copper small. These cost nothing and cannot be recovered by filtering later.

Section 24-6

Worked Examples

1 Sizing the Y-capacitors

Problem. A 230 V, 50 Hz stationary pluggable appliance may leak 3.5 mA to earth. Find the maximum total Y-capacitance, and repeat for a 0.1 mA medical limit.

Working

\[ I_{leak} = 2\pi f C V \;\Longrightarrow\; C = \frac{I}{2\pi f V} \]

\[ \text{3.5 mA}: \quad C = \frac{3.5\times10^{-3}}{2\pi(50)(230)} = 48.4\ \text{nF} \]

\[ \text{0.1 mA}: \quad C = \frac{1.0\times10^{-4}}{2\pi(50)(230)} = 1.38\ \text{nF} \]

Read the practical consequence. For the appliance, two 22 nF Y2 capacitors (one line-to-earth, one neutral-to-earth) total 44 nF — comfortably inside the limit, and enough for good common-mode attenuation.

For the medical product, 1.38 nF is the total, so two 680 pF parts. That is 35 times less capacitance, and the common-mode attenuation it provides is correspondingly small.

How medical supplies compensate:

Much larger common-mode chokes, often two in series with different self-resonant frequencies.
Transformer screening, sometimes two or three screens, so the displacement current never crosses the barrier.
Slower switching edges, accepting lower efficiency.
Careful attention to the switching node's area, since every picofarad now matters.

This is why a medical power supply is physically larger and less efficient than a commercial one of the same rating — a direct and unavoidable consequence of one number in a safety standard.

2 Checking the input filter for stability

Problem. A 300 W converter runs from 380 V through an input filter of \( L_f = 220\ \mu\text{H} \), \( C_f = 4.7\ \mu\text{F} \) with 80 mΩ of series resistance. Check the Middlebrook criterion and design damping if needed.

Converter input impedance

\[ |Z_{in}| = \frac{V^2}{P} = \frac{380^2}{300} = 481\ \Omega \quad\text{(negative incremental)} \]

Filter resonance and peak impedance

\[ f_0 = \frac{1}{2\pi\sqrt{L_fC_f}} = \frac{1}{2\pi\sqrt{(2.2\times10^{-4})(4.7\times10^{-6})}} = 4.95\ \text{kHz} \]

\[ Z_0 = \sqrt{\frac{L_f}{C_f}} = \sqrt{\frac{2.2\times10^{-4}}{4.7\times10^{-6}}} = 6.84\ \Omega, \qquad Q = \frac{Z_0}{R_L} = \frac{6.84}{0.08} = 86 \]

\[ |Z_{out}|_{peak} \approx Q\,Z_0 = 86(6.84) = 585\ \Omega \]

585 Ω against 481 Ω — the criterion is violated. The filter's peak output impedance exceeds the converter's input impedance, so the negative resistance will feed the resonance and the input voltage will oscillate near 5 kHz.

Design the damping. Target a peak below 120 Ω, giving 12 dB of margin:

Working

\[ R_d \approx Z_0 = 6.8\ \Omega \;\Longrightarrow\; \text{use } 6.8\ \Omega \]

\[ C_d \approx 5\,C_f = 22\ \mu\text{F} \quad\text{(blocks DC; effective near } f_0) \]

Check the resistor's dissipation. At DC and 50 Hz, \( C_d \)'s impedance is 145 Ω at 50 Hz, so almost no mains current flows through the branch. The resistor sees only the switching-frequency ripple and any transient at \( f_0 \) — typically well under 0.5 W. A standard 1 W part is adequate.

The symptom this prevents. Without damping, the input voltage oscillates at about 5 kHz, the output carries a matching ripple, and the supply may audibly hum. It is easy to spend days looking for it in the control loop, because the loop is perfectly healthy.

3 Sizing a snubber from measurement

Problem. A 400 V bridge at 80 kHz rings at 14 MHz across the switch. Adding 470 pF drops the ringing to 6.5 MHz. Design an RC snubber and find the resistor's dissipation.

Extract the parasitics from the two measurements

\[ C_p = \frac{C_{add}}{(f_r/f_r')^2 - 1} = \frac{470\times10^{-12}}{(14/6.5)^2 - 1} = \frac{470\times10^{-12}}{3.64} = 129\ \text{pF} \]

\[ L_p = \frac{1}{(2\pi f_r)^2 C_p} = \frac{1}{(2\pi \times 14\times10^6)^2(129\times10^{-12})} = 1.00\ \mu\text{H} \]

Snubber values

\[ R_s = \sqrt{\frac{L_p}{C_p}} = \sqrt{\frac{10^{-6}}{129\times10^{-12}}} = 88\ \Omega \;\Longrightarrow\; \text{use } 82\ \Omega \]

\[ C_s \approx 3C_p = 390\ \text{pF} \]

Dissipation — the check that is usually skipped

\[ P_R = C_sV^2f_{sw} = (390\times10^{-12})(400)^2(8\times10^4) = 5.0\ \text{W} \]

5 W per snubber is significant, and in a full bridge with four of them that is 20 W. The resistor must be a genuine 10 W wirewound or thick-film power part, and it must also survive the peak pulse current of \( 400/82 = 4.9 \) A.

Before accepting that, reduce the source. The ringing energy is \( \tfrac12L_pI^2 \), and \( L_p = 1\ \mu\text{H} \) is a large loop inductance — typical of a layout with the input capacitor several centimetres from the switches.

Halve the loop inductance to 500 nH (a tighter layout, or ceramic capacitors placed directly across the devices) and:

The ringing frequency rises to 20 MHz, and its amplitude falls.
\( R_s \) becomes 62 Ω and \( C_s \) can be smaller, so the dissipation drops correspondingly.
The turn-off voltage spike, \( L_p\,di/dt \), also halves — which may remove the need for a snubber entirely.

The general rule this illustrates: a snubber sized to fix a bad layout is always more expensive than fixing the layout. Measure \( L_p \) first, and if it exceeds about 100 nH, improve the layout before designing anything.

4 Reading a conducted-emissions failure

Problem. A supply fails CISPR 32 Class B with peaks of +8 dB at 300 kHz and +14 dB at 18 MHz. Diagnose both and propose fixes.

The 300 kHz peak — differential mode.

The switching frequency is probably around 100 kHz, so 300 kHz is the third harmonic — squarely in the differential range.
Confirm it with a mode separator, or by observing that the noise on live and neutral is in antiphase.
Fixes: increase the X-capacitor (0.47 μF to 1 μF is a cheap 6 dB); add a differential inductor, or simply choose a common-mode choke with more leakage inductance; increase the converter's own input capacitance so less ripple reaches the filter.

The 18 MHz peak — common mode.

Far above any switching harmonic, so this is \( dv/dt \) into stray capacitance.
Confirm it by observing that live and neutral noise are in phase, or by clamping a current probe around both together — common-mode current shows, differential cancels.
Fixes, in order of value: a transformer screen (often 10–20 dB, very cheap); check the Y-capacitors are as large as the leakage limit allows; a second common-mode choke resonant higher than the first; slow the switching edges; reduce the switching node's copper and improve the heatsink insulation.

The order to work in. Fix the common-mode problem first: it is the larger failure, its remedies are cheap, and some of them (slower edges) also help the differential result. Then re-measure — EMI fixes interact, and a change made for one mode frequently shifts the other.

And measure with the product in its final enclosure. A metal case, its earthing arrangement and the cable dress all change the common-mode result substantially. A supply that passes on the bench routinely fails in the box.

Section 24-7

Summary & Formula Sheet

Chapter 24, and Part 5, in five sentences:

Differential noise circulates between live and neutral; common-mode noise flows out on both and returns through earth. They take different paths and need different filters.
A common-mode choke works because load current produces no net flux, so a tiny high-permeability core gives tens of millihenries.
Y-capacitors are sized by a safety leakage limit rather than by attenuation, and are usually already at maximum.
A regulated converter presents a negative input resistance \( -V^2/P \), which can drive an input filter into oscillation unless the Middlebrook criterion is satisfied.
Snubbers should be sized from two ringing measurements, and \( P_R = C_sV^2f_{sw} \) checked — but reducing the loop inductance is always the cheaper fix.

Formula sheet · Chapter 24

Common-mode source currentthe origin of most CM noise

\( i_{cm} = C_{stray}\dfrac{dv}{dt} \)

Earth leakage currentsizes the Y-capacitors

\( I_{leak} = 2\pi f C_Y V \)

Converter input resistancenegative; magnitude only

\( |Z_{in}| = \dfrac{V_{in}^2}{P} \)

Middlebrook criterionat every frequency

\( |Z_{out,filt}| \ll |Z_{in,conv}| \)

Filter resonancewhere the peak sits

\( f_0 = \dfrac{1}{2\pi\sqrt{L_fC_f}} \)

Filter characteristic impedancealso the damping value

\( Z_0 = \sqrt{L_f/C_f} \)

Undamped peak impedance\(Q = Z_0/R_L\)

\( |Z_{out}|_{pk} \approx Q\,Z_0 \)

Series-RC damping\(C_d\) blocks DC, so \(R_d\) is lossless there

\( R_d \approx Z_0, \quad C_d \approx (4\text{–}10)C_f \)

Parasitic capacitancefrom two ringing measurements

\( C_p = \dfrac{C_{add}}{(f_r/f_r')^2 - 1} \)

Parasitic inductancethen from \(f_r\)

\( L_p = \dfrac{1}{(2\pi f_r)^2 C_p} \)

Snubber valuescritical damping

\( R_s = \sqrt{L_p/C_p}, \quad C_s \approx 3C_p \)

Snubber dissipationfull \(CV^2\), not half — check this

\( P_R = C_sV^2f_{sw} \)

Key terms

Differential-mode noise: Noise current circulating between live and neutral. Blocked by X-capacitors and series inductors.
Common-mode noise: Noise current flowing the same way in both lines and returning through earth. Caused by \(dv/dt\) into stray capacitance.
X-capacitor: Line-to-neutral capacitor. Fails open by design, so it cannot cause a short. Needs a bleed resistor.
Y-capacitor: Line-to-earth capacitor. Fails open by design. Sized by the earth leakage current limit.
Common-mode choke: Both lines on one core in the same sense. Load current cancels; common-mode current sees the full inductance.
Negative input resistance: A regulated converter's small-signal input resistance, \(-V^2/P\). Can destabilise an input filter.
Middlebrook criterion: The requirement that a filter's output impedance stay well below the converter's input impedance at every frequency.
Snubber: A network limiting \(dv/dt\) or \(di/dt\) at a switching device. Strictly a damper, as distinct from a clamp.
Clamp: A network limiting the peak voltage. RCD, active clamp or TVS.
Switching node: The point swinging the full DC voltage at every transition. The converter's antenna and the main common-mode source.
Transformer screen: A grounded foil between primary and secondary that intercepts displacement current. Often worth 10–20 dB.
Spread spectrum: Dithering the switching frequency so each harmonic is smeared across a band, reducing the measured peak.

Check yourself

Test Yourself

Chapter 24 · six questions answers hidden until you ask

How can a 10 mH common-mode choke carry 10 A without a core the size of a brick?

Because the load current produces no net flux in the core at all, so saturation is never approached.

The construction. Both lines are wound on the same core, in the same sense, with equal turns.

Load current flows out on live and back on neutral, so the two windings' magnetomotive forces are equal and opposite. Net MMF is zero, net flux is zero, and the core is magnetically idle.
Common-mode current flows the same way in both lines, so the MMFs add. The core sees the full flux and the choke presents its full inductance.

Two things follow, and together they make the component possible:

The core can be tiny. It never carries load flux, so its cross-section is set only by the common-mode flux — which is small, since common-mode currents are milliamps.
Very high permeability is usable. Nanocrystalline or high-\( \mu \) MnZn ferrite at \( \mu_r = 10{,}000 \) would saturate instantly in any ordinary inductor, but here it is ideal, and it is what gives 10 mH from a coin-sized core.

The two limits that do apply:

Winding resistance sets the current rating — it is a thermal limit, not a magnetic one.
Imbalance. If the two windings are not exactly matched, or if the load current is not exactly balanced, some net flux does appear. Choke datasheets quote a rated current for this reason, and a partly-rectified or asymmetric load can saturate one that looks adequately rated.

And the leakage inductance is not a defect. The small imbalance flux escaping the core gives a differential inductance of a few microhenries — often exactly the differential filtering needed, so many designs use it deliberately and fit no separate differential inductor.

Common-mode emissions are 10 dB too high. Why can you not simply fit larger Y-capacitors?

Because their value is set by a safety limit that is normally already reached, not by an EMI target.

The constraint. A Y-capacitor connects a live conductor to the earthed chassis, so it passes a continuous mains-frequency current to earth:

\[ I_{leak} = 2\pi f C V \]

If the earth connection is ever broken, that current flows through whoever touches the chassis. Standards therefore cap it:

Portable equipment (IEC 62368-1): 0.25 mA
Stationary pluggable: 3.5 mA
Medical, patient-connected (IEC 60601-1): 0.1 mA

At 230 V, 50 Hz that permits:

3.5 mA → 48 nF total
0.25 mA → 3.5 nF total
0.1 mA → 1.4 nF total

which is why standard Y2 parts are 2.2 nF and 4.7 nF. There is normally no headroom.

What to do instead, in order of value:

Add a transformer screen — a grounded foil between primary and secondary. Often 10–20 dB for the cost of a strip of copper, and it stops the current at the barrier instead of filtering it afterwards.
Reduce \( dv/dt \). A larger gate resistor slowing the edge from 20 ns to 50 ns is worth about 8 dB, at the cost of switching loss.
Reduce \( C_{stray} \). Shrink the switching node's copper, use a thicker or lower-permittivity insulating pad, or a screened pad.
Add a second common-mode choke of a different value, extending attenuation above the first one's self-resonance.

And note the design consequence for medical equipment: at 1.4 nF the Y-capacitors provide very little attenuation, so the filter must rely on chokes and screening. That is precisely why medical supplies are physically larger than commercial ones of the same rating.

How can a purely passive input filter make a stable converter oscillate?

Because a regulated converter is not a resistor — it is a constant-power load, which presents a negative incremental resistance.

The derivation is three lines. A regulated converter draws constant power, so \( I_{in} = P/V_{in} \). Then:

\[ \frac{dI_{in}}{dV_{in}} = -\frac{P}{V_{in}^2} \;\Longrightarrow\; R_{in} = -\frac{V_{in}^2}{P} \]

Read what negative resistance means physically. If the input voltage dips, an ordinary load would draw less current. This converter draws more, which pulls the voltage down further. That is positive feedback.

What it does to an \( LC \) filter. The filter's resonance is normally damped by the resistance loading it, which absorbs the oscillation's energy. A negative resistance supplies energy instead. If it overcomes the filter's own losses, the resonance grows rather than decays — and the input voltage oscillates at the filter's resonant frequency.

The condition for safety is Middlebrook's:

\[ |Z_{out,filter}| \ll |Z_{in,converter}| \quad\text{at every frequency} \]

with at least 6 dB of margin, and especially at the filter's resonant peak — where an undamped filter's impedance is \( Q \) times its characteristic impedance and can easily reach hundreds of ohms.

Why this is such a common field failure:

The converter is stable on the bench without the filter.
The filter is passive and obviously cannot oscillate by itself.
The control loop measures perfectly healthy.
The oscillation is at a few kilohertz, nothing like the switching frequency, so it looks like a control problem.

The diagnostic that settles it in seconds: put a scope on the converter's input voltage, not its output. If it is oscillating near \( 1/(2\pi\sqrt{L_fC_f}) \), the filter is the cause, and a series-RC damping network across the filter capacitor fixes it — dissipating nothing at DC because the capacitor blocks it.

A snubber capacitor is increased from 1 nF to 10 nF and the ringing disappears. What else happened?

The resistor's dissipation went up tenfold, and it is probably now far beyond its rating.

The dissipation. The snubber capacitor is charged and discharged once each per switching cycle, and both events dissipate in the resistor. So the loss is the full \( CV^2 \), not half of it:

\[ P_R = C_sV^2f_{sw} \]

At 400 V and 100 kHz:

1 nF: \( P_R = 16 \) W
10 nF: \( P_R = 160 \) W

Per snubber. A full bridge with four is 640 W — from a component change made to tidy up a waveform.

Three further consequences, all easy to miss:

Peak current. At turn-on the capacitor discharges through the resistor at \( V/R_s \) — for 400 V and 82 Ω, nearly 5 A in a pulse. A small film resistor fails open within minutes, and once it does the snubber is inert and nobody notices until the devices fail.
Turn-on loss rises. The charge on \( C_s \) is dumped through the switch when it turns on, so turn-off loss falls while turn-on loss rises. There is an optimum, and past it a larger snubber makes total loss worse.
The capacitor must be a pulse type. C0G/NP0 ceramic or polypropylene film — never X7R, whose dielectric loss and DC bias behaviour are both unsuitable at these currents.

What should have been done instead, in order:

Reduce the loop inductance. The ringing energy is \( \tfrac12L_pI^2 \); a tighter layout halves it for free, and often removes the need for a snubber entirely.
Slow the edges with a larger gate resistor.
Use a device with less recovery charge — a SiC diode has essentially none.
Then size a snubber from two ringing measurements, and check \( P_R \) before fitting it.

A supply passes EMI on the bench and fails in its enclosure. Why?

Because the enclosure, its earthing and the cable dress are all part of the common-mode current path, and none of them existed on the bench.

What changes when the supply goes into a box:

New stray capacitance. A metal enclosure a few millimetres from the switching node adds tens of picofarads to earth — directly increasing \( i = C\,dv/dt \). On the bench that capacitance was to distant, poorly-coupled objects.
A new return path. An earthed chassis gives common-mode current a low-impedance route back to the mains earth. On the bench it had to find a much worse one, so less flowed.
Cables become antennas. Output leads, sense wires and fan cables all carry common-mode current and radiate. Their length, routing and proximity to the switching node all matter, and a bench test with short leads sees none of it.
The heatsink is now bonded. A floating heatsink on the bench becomes an earthed one in the enclosure, completing the path from the transistor's tab through the insulating pad to earth — usually the dominant common-mode source.

Why the failure is nearly always common mode and above a few megahertz: differential noise circulates within the supply and is largely unaffected by what surrounds it, while common-mode noise depends entirely on the paths to earth.

What to do about it:

Test in the final enclosure, with the final cables, dressed as they will be. This is not a formality — it is the only meaningful measurement.
Control the earthing deliberately. A single, defined chassis bonding point beats several accidental ones, which form loops.
Fit the filter at the mains entry, not in the middle of the board, so noise cannot bypass it by coupling to the wiring downstream.
Add a screened insulating pad under the transistor, so its displacement current is intercepted and returned locally rather than through the chassis.

And a practical point about the schedule: because these effects are mechanical, EMI cannot be fully signed off until the mechanical design is fixed. Projects that leave the enclosure until last routinely discover this at the worst moment.

Part 5 has covered AC controllers, cycloconverters, magnetics and filters. What connects them?

They are the four places where a converter meets the physical world rather than the circuit diagram — and in each, the limiting quantity is not electrical.

Chapter 21, AC voltage controllers. The circuit is two thyristors and could not be simpler. What limits it is the power factor and the harmonics it imposes on the supply — a consequence for other people's equipment, enforced by standards rather than by physics.

Chapter 22, direct AC–AC converters. Removing the DC link removes the least reliable component, and immediately imposes limits that come from having no stored energy — a frequency cap for one topology, a voltage cap for the other. What you cannot store, you cannot exceed.

Chapter 23, magnetics. The only component that must be designed rather than bought, and the one that sets the converter's size. Its limit is temperature — not volts, not amps.

Chapter 24, filters and EMI. The converter's effect on everything around it, governed by legal limits, and dominated by parasitics that appear on no schematic.

The connecting theme: Parts 1 to 4 were about making a converter work. Part 5 is about making it acceptable — to the supply it draws from, to the equipment beside it, to the enclosure it must fit in, and to the standards it must satisfy.

And the practical lesson that recurs in all four:

A phase-controlled heater works perfectly and may still be unusable on its network.
A matrix converter is technically superior and cannot drive a standard motor.
A magnetic design can satisfy every equation and still run too hot.
A supply can meet every specification and be illegal to sell.

Which is why these chapters are not an appendix to the converter chapters. A great deal of real project time goes here, and it is where designs that looked finished turn out not to be.

Practice

Problems

Three habits for filter and EMI work:

Identify the mode before choosing a component. Below 1 MHz suspect differential; above 5 MHz suspect common mode.
Check the Middlebrook criterion whenever an input filter is added to a regulated converter.
Size snubbers from measurement, and always compute \( P_R = C_sV^2f_{sw} \).

Problems 1–5 are direct application; 6–9 need judgement; 10–12 are design questions worth discussing in a tutorial.

A 120 V, 60 Hz portable product is limited to 0.25 mA of earth leakage. Find the maximum total Y-capacitance, and compare with the 230 V, 50 Hz case.
A switching node swings 320 V in 15 ns with 80 pF to an earthed heatsink. Find the peak common-mode current, and the value if the edge is slowed to 40 ns.
An input filter has \( L_f = 150\ \mu\text{H} \), \( C_f = 10\ \mu\text{F} \), series resistance 60 mΩ. Find \( f_0 \), \( Z_0 \), \( Q \) and the peak output impedance.
The filter of Problem 3 feeds a 500 W converter from 350 V. Check the Middlebrook criterion and design a series-RC damping network if required.
A device rings at 22 MHz. Adding 220 pF reduces this to 11 MHz. Find \( C_p \) and \( L_p \), and design an RC snubber.
For the snubber of Problem 5 on a 600 V bus at 50 kHz, find the resistor's average dissipation and its peak pulse current, and specify a suitable resistor.
Explain why a common-mode choke's core does not saturate at rated load current, and state two conditions under which it would.
A converter passes conducted emissions with a resistive load and fails with a long output cable. Diagnose the mechanism and propose three remedies.
A designer proposes fitting 100 nF Y-capacitors to fix a common-mode failure on a 230 V portable product. Analyse the proposal and give an alternative plan.
Design the complete input filter for a 300 W, 230 V supply switching at 130 kHz: choose X- and Y-capacitor values, specify the common-mode choke, check the leakage current, verify the Middlebrook criterion against the converter's input impedance, and design any damping needed.
A 5 kW converter fails CISPR 32 Class A by 6 dB at 500 kHz and by 11 dB at 25 MHz. Identify the mode of each, propose a remedy for each, state the order in which you would apply them, and explain why the order matters.
Compare reducing \( dv/dt \) with adding filtering as a route to EMI compliance. Quantify the efficiency cost of slowing a 400 V, 20 A edge from 20 ns to 60 ns at 100 kHz, and state the circumstances under which each approach is preferable.