Solved Problems · Set 32

Filters, Snubbers and EMI Mitigation

Part 5 · AC–AC Converters — every converter is now built. Six problems on what surrounds them: the overshoot, the emissions, the resonance and the cable.

Prof. Mithun Mondal 6 solved problems GATE · ESE · University

Set 32 — Filters, Snubbers and EMI Mitigation

Every converter in this book is now built. What remains is everything around them — and it has been accumulating throughout. The snubber of Set 4, the notches of Set 15, the capacitor resonance of Set 21, the common-mode current of Set 26, the bearing damage that followed: each was raised where it arose and left unresolved.

They share a cause. A converter works by changing voltages and currents as fast as possible, and every fast edge finds the nearest parasitic inductance or capacitance and does something unwanted with it — overshoot across a device, current into a heatsink, a reflection at the end of a cable, a resonance in the supply. The remedies form a layered defence, and this set builds it.

Part 5 · Chapter 24 · 6 solved problems

i Method Recap
  • Stray inductance converts current slope into voltage:

    \[ \Delta V = L_s\frac{di}{dt}, \qquad E_{stored} = \tfrac12L_sI^2 \]
  • An RCD clamp absorbs that energy into a capacitor and burns it in a resistor:

    \[ C = \frac{L_sI^2}{\Delta V^2}, \qquad P_R = \tfrac12L_sI^2f_s, \qquad RC \ll T_s \]
  • Common-mode current comes from \(dv/dt\) into stray capacitance:

    \[ i_{cm} = C_{stray}\frac{dv}{dt}, \qquad V_{LISN} = i_{cm}\times25\ \Omega, \qquad \text{dB}\mu\text{V} = 20\log_{10}\frac{V}{1\ \mu\text{V}} \]
  • A second-order filter attenuates by 40 dB per decade:

    \[ A_{dB} = 40\log_{10}\frac{f}{f_c} \quad\text{per stage} \]
  • Y capacitance is capped by the leakage current limit:

    \[ I_{leak} = 2\pi fC_YV \le 3.5\ \text{mA} \]
  • A cable reflects when the edge is faster than the round trip:

    \[ t_{r} < \frac{2\ell}{v} \;\Longrightarrow\; V_{motor} \to 2V_{dc}, \qquad \ell_{crit} = \frac{vt_r}{2} \]
Problem 1CoreThe Volts That Stray Inductance Makes

An IGBT switching 50 A off a 400 V link in 100 ns sits in a loop with 100 nH of stray inductance. Find the overshoot, the effect of doubling the stray inductance, and design an RCD clamp limiting the excursion to 30 V at a 10 kHz switching frequency.

Solution

The overshoot:

\[ \frac{di}{dt} = \frac{50}{100\times10^{-9}} = 5\times10^{8}\ \text{A/s} = 500\ \text{A}/\mu\text{s} \]
\[ \Delta V = L_s\frac{di}{dt} = \left(100\times10^{-9}\right)\left(5\times10^{8}\right) = 50\ \text{V} \]
\[ V_{peak} = 400+50 = 450\ \text{V} \]

Fifty volts from a hundred nanohenries — which is a few centimetres of busbar. This is why converter layout is a design activity rather than a packaging afterthought.

Double the stray inductance and the overshoot doubles:

\[ L_s = 200\ \text{nH} \;\Longrightarrow\; \Delta V = 100\ \text{V} \;\Longrightarrow\; V_{peak} = 500\ \text{V} \]
\(L_s\)Typical sourceOvershoot
20 nHlaminated busbar, capacitor on the module10 V
50 nHshort, wide PCB planes25 V
100 nHreasonable layout50 V
200 nHcapacitor a few cm away100 V
500 nHwired connection250 V — device fails

The last row is a real failure mode: a converter that works on the bench with a short lead destroys devices when the capacitor is moved. The energy involved is trivial; the voltage is not.

Why faster devices make it worse. The overshoot scales with the switching speed:

\[ \Delta V = L_s\frac{I}{t_{fall}} \;\propto\; \frac{1}{t_{fall}} \]

A SiC MOSFET switching the same current in 20 ns produces five times the overshoot — 250 V instead of 50 — from the same layout. Wide-bandgap devices therefore demand low-inductance packaging, and much of their packaging development has been about reducing \(L_s\) rather than improving the die.

The RCD clamp. Rather than slowing the device, divert the inductive energy into a capacitor:

\[ E = \tfrac12L_sI^2 = \tfrac12\left(100\times10^{-9}\right)(50)^2 = 125\ \mu\text{J} \]
\[ \tfrac12C\Delta V^2 = E \;\Longrightarrow\; C = \frac{2E}{\Delta V^2} = \frac{2\left(125\times10^{-6}\right)}{30^2} = 278\ \text{nF} \]

The clamp diode conducts as soon as the device voltage exceeds the capacitor's, dumping the inductive energy into it and limiting the excursion. Note that a smaller \(\Delta V\) needs a quadratically larger capacitor — halving it to 15 V would need four times the capacitance.

The resistor must discharge the capacitor before the next switching event:

\[ RC \ll T_s = 100\ \mu\text{s}; \qquad \text{choose } RC = 20\ \mu\text{s} \]
\[ R = \frac{20\times10^{-6}}{278\times10^{-9}} = 72\ \Omega \]
\[ P_R = Ef_s = \left(125\times10^{-6}\right)\left(10\times10^{3}\right) = 1.25\ \text{W} \]

The stored energy is dissipated once per switching cycle, so the resistor's rating is proportional to frequency — a 100 kHz converter would need 12.5 W. That is the clamp's real cost, and it is why reducing \(L_s\) is always preferable to clamping the consequences.

The hierarchy of remedies, best first:

ApproachEffectCost
Reduce \(L_s\) by layoutattacks the causefree — design effort
Decoupling capacitor at the moduleshortens the commutation loopa small film capacitor
Slow the gate drivereduces \(di/dt\)more switching loss
RCD clamplimits the excursion1.25 W plus components
Higher-voltage devicetolerates the overshootworse conduction, more cost

The first two are what a good design does; the last three are what a compromised one needs. Set 4 designed an RC snubber for a thyristor's \(dv/dt\) withstand; this is the same idea applied to a different parasitic, and the same conclusion applies — a snubber is insurance, not a substitute for layout.

A few centimetres of conductor cost fifty volts. The energy in 100 nH at 50 A is only 125 microjoules — nothing — but delivered in a hundred nanoseconds it is fifty volts of overshoot on top of the link. Layout is the only remedy that costs nothing, and it is the only one that scales to faster devices.
Answera\(\Delta V = 50\ \text{V}\), peak 450 V   b 200 nH gives 100 V   c\(C = 278\ \text{nF},\ R = 72\ \Omega,\ P_R = 1.25\ \text{W}\)
Problem 2CoreSixty-Seven Decibels Over

A 24 V buck converter switches at 200 kHz with 20 ns edges. Its switching node has 50 pF of stray capacitance to an earthed heatsink. Estimate the conducted emission at the LISN and compare with the CISPR Class B limit of 56 dBµV.

Solution

The mechanism. The switching node swings the full input voltage at every transition, and its capacitance to the earthed heatsink carries a current spike:

\[ \frac{dv}{dt} = \frac{24}{20\times10^{-9}} = 1.2\times10^{9}\ \text{V/s} = 1.2\ \text{V/ns} \]
\[ i_{cm} = C_{stray}\frac{dv}{dt} = \left(50\times10^{-12}\right)\left(1.2\times10^{9}\right) = 60\ \text{mA} \]

Sixty milliamps into the earth conductor from a converter that may only be delivering a few amps. The current is brief, but a spectrum analyser does not care about duration — it cares about the amplitude in each frequency bin.

What the LISN measures. The line impedance stabilisation network presents a defined 50 Ω to each line, which for common-mode current is two in parallel:

\[ V = i_{cm}\times25\ \Omega = (0.060)(25) = 1.5\ \text{V} \]
\[ \text{dB}\mu\text{V} = 20\log_{10}\frac{1.5}{1\times10^{-6}} = 123.5\ \text{dB}\mu\text{V} \]

The LISN exists so that measurements are repeatable — without it, the emission depends on whatever mains impedance happens to be present. Its 50 Ω is a standard, not a physical property of the supply.

Against the limit:

\[ 123.5-56 = 67.5\ \text{dB of attenuation required} \]
\[ 67.5\ \text{dB} = \text{a factor of } 2400\ \text{in voltage} \]

This is the number that surprises people meeting EMC for the first time. The converter is not marginally over — it is over by a factor of two thousand, and no amount of tidying will close that gap. A filter is mandatory, not optional.

Where the emission sits in the spectrum. The conducted band runs from 150 kHz to 30 MHz:

ComponentFrequencyIn band?
Fundamental switching200 kHzyes
Harmonics400 kHz – several MHzyes
Edge ringing10–100 MHzpartly — then radiated

A converter switching below 150 kHz keeps its fundamental out of the measured band — which is exactly why so many designs sit at 100–140 kHz. The harmonics still fall inside, but they are smaller, and the filter's job becomes much easier. Choosing the switching frequency is an EMC decision as much as a magnetics one.

Common mode versus differential mode. Two distinct mechanisms need two distinct filters:

Differential modeCommon mode
Pathline to lineboth lines to earth
Sourceinput current ripple\(dv/dt\) into stray capacitance
Dominant atlow frequencyhigh frequency
Filter elementX capacitor, DM inductorY capacitor, CM choke
Easier to fix?yesno

Differential-mode noise is a circuit problem with a circuit solution — more input capacitance, as Set 18 sized. Common-mode noise depends on stray capacitances that are hard to measure and harder to change, which is why it dominates most EMC failures and why the 67.5 dB above is a common-mode figure.

Conducted emission failures are measured in tens of decibels, not decibels. A perfectly reasonable buck converter exceeds Class B by 67 dB — a factor of 2400 — because 50 pF and a nanosecond edge are enough. EMC is therefore designed in from the start; it cannot be added to a finished converter.
Answera\(i_{cm} = 60\ \text{mA}\)   b\(123.5\ \text{dB}\mu\text{V}\) at the LISN   c 67.5 dB of attenuation needed
Problem 3Exam levelDesigning the Filter

Design a common-mode filter delivering the 67.5 dB of Problem 2 at 200 kHz. Show that a single stage is impractical given the Y-capacitor leakage limit, then design a two-stage filter.

Solution

The Y capacitance is capped first, because it sets a safety limit:

\[ I_{leak} = 2\pi fC_YV \le 3.5\ \text{mA} \]
\[ C_{Y,max} = \frac{3.5\times10^{-3}}{2\pi(50)(230)} = 48\ \text{nF} \]

That is the absolute limit for a protectively earthed appliance, and prudent designs stay far below it — typically 2.2 nF per line, giving 4.4 nF total. The leakage flows through the earth conductor and is what an RCD sees, so several such devices on one circuit can cause nuisance tripping.

Try a single stage. Working back from the required attenuation:

\[ 67.5 = 40\log_{10}\frac{200\times10^{3}}{f_c} \;\Longrightarrow\; f_c = 4.1\ \text{kHz} \]
\[ L_{cm} = \frac{1}{\left(2\pi f_c\right)^2C_Y} = \frac{1}{\left(2\pi\times4100\right)^2\left(4.4\times10^{-9}\right)} = 342\ \text{mH} \]

Three hundred and forty millihenries. A common-mode choke of that inductance would be enormous, and its self-resonant frequency — set by its own winding capacitance — would fall well below 200 kHz, at which point it stops behaving as an inductor at all. The single stage is not just impractical, it is self-defeating.

Two stages share the work, and because attenuation is logarithmic, sharing is very effective:

\[ \frac{67.5}{2} = 33.75\ \text{dB per stage} \]
\[ 33.75 = 40\log_{10}\frac{200\times10^{3}}{f_c} \;\Longrightarrow\; f_c = 28.6\ \text{kHz} \]
\[ L_{cm} = \frac{1}{\left(2\pi\times28{,}600\right)^2\left(4.4\times10^{-9}\right)} = 7.0\ \text{mH} \]

Seven millihenries per stage — an entirely ordinary common-mode choke, available as a standard part. Splitting the filter reduced the required inductance by a factor of nearly fifty, because the corner frequency only had to rise sevenfold and inductance goes as \(1/f_c^2\).

The complete filter:

ElementValueFunction
CM choke × 27 mH eachcommon-mode attenuation
Y capacitors × 42.2 nFCM return path to earth
X capacitor0.47 µFdifferential-mode
DM inductancethe choke's leakagefree — usually sufficient
Bleeder resistor1 MΩdischarges X within 1 s

The fourth row is a useful economy: a common-mode choke's windings are never perfectly coupled, and the resulting leakage inductance — typically 1–3% of the CM value, so 70 to 200 µH here — acts as a differential-mode inductor. With the X capacitor it forms the DM filter at no extra cost.

What makes real filters underperform:

EffectConsequence
Choke self-resonancebecomes capacitive above it — no attenuation
Capacitor ESLimpedance rises again above a few MHz
Layout coupling input to outputbypasses the filter entirely
Core saturation from DM currentinductance collapses at load
Filter resonance with the sourceamplifies at the corner — needs damping

The third row is the most common failure in practice. A filter that measures perfectly on a network analyser can achieve nothing in a product if its input and output tracks run near each other, because the noise couples around it. Physical separation of the filter's two sides matters as much as its component values.

And the last row echoes Set 19. The filter is an \(LC\) network on the converter's input:

\[ Z_{filter,out} \ \text{must stay below}\ \left|Z_{in,converter}\right| \ \text{at all frequencies} \]

A switching converter presents a negative input resistance — drawing constant power means drawing more current as the voltage falls — so an undamped input filter can oscillate. The Middlebrook criterion above is checked in the same way Set 23 checked a loop's phase margin, and the cure is a damping network across the X capacitor.

Two small filters beat one large one by a wide margin. Because attenuation is logarithmic and inductance goes as \(1/f_c^2\), splitting 67.5 dB into two stages cut the required choke from 342 mH to 7 — a factor of fifty. Any single-stage EMI filter design that produces an absurd component value is a signal to cascade instead.
Answera\(C_Y \le 48\ \text{nF}\); use 4.4 nF   b single stage needs 342 mH — impractical   c two stages of 7.0 mH at \(f_c = 28.6\ \text{kHz}\)
Problem 4Exam levelTuning to a Harmonic, Deliberately Off

Design a shunt filter for the fifth harmonic of a 230 V, 50 Hz supply, providing 5 kVAr of fundamental reactive compensation. Then explain why it is tuned to the 4.7th rather than the 5th, and what happens if it is not.

Solution

The principle. A series \(LC\) branch placed in shunt presents near-zero impedance at its resonance, so harmonic current at that frequency flows into it rather than into the supply:

\[ \text{at } f_{tune}:\ Z = R\ \text{only} \;\Longrightarrow\; \text{harmonic current diverted} \]

And below resonance the branch is capacitive, so it also supplies reactive power at the fundamental. One component does two jobs, which is why passive filters remain common despite active alternatives.

Size the capacitor from the reactive requirement:

\[ Q_C = \omega CV^2 \;\Longrightarrow\; C = \frac{5000}{(314.16)(230)^2} = 301\ \mu\text{F} \]

Then the inductor from the tuning frequency:

\[ L = \frac{1}{\left(2\pi f_{tune}\right)^2C} \]
\[ f = 250\ \text{Hz}:\ L = \frac{1}{\left(2\pi\times250\right)^2\left(301\times10^{-6}\right)} = 1.35\ \text{mH} \]
\[ f = 235\ \text{Hz (4.7th)}:\ L = 1.52\ \text{mH} \]

Why detune. Consider what happens to the tuning frequency as components drift:

\[ f_{tune} = \frac{1}{2\pi\sqrt{LC}} \;\propto\; \frac{1}{\sqrt{C}} \]
CauseEffect on \(C\)Effect on \(f_{tune}\)
Capacitor ageingfallsrises
A failed internal elementfalls in a steprises in a step
Temperaturevariesvaries
Manufacturing tolerance\(\pm5\%\)\(\mp2.5\%\)

Every mechanism moves the tuning frequency upward. A filter tuned exactly to 250 Hz will therefore drift above the fifth harmonic in service — and above resonance the branch is inductive, so it no longer absorbs the harmonic. Tuning to 235 Hz means the drift moves it towards 250 Hz rather than away.

The danger of tuning above. An inductive branch in parallel with the supply's capacitance forms a parallel resonance:

\[ f_{tune} > f_{harmonic} \;\Longrightarrow\; \text{parallel resonance appears near } f_{harmonic} \]

And a parallel resonance amplifies rather than absorbs. This is exactly the mechanism Set 21 found at 269 Hz — a capacitor bank resonating with the supply inductance close to the fifth harmonic, magnifying it instead of removing it. A mistuned filter can make a harmonic problem substantially worse than having no filter at all.

The full design in practice:

ItemValue or note
Tuning order4.7th, not 5th
\(C\)301 µF
\(L\)1.52 mH
Quality factor30–100 typical — a damping resistor sets it
Capacitor voltage ratingabove supply — the inductor raises it
7th harmonic branchusually added, tuned to 6.7th
High-pass branchfor 11th and above

The capacitor rating deserves care. At the fundamental the branch is capacitive, and the series inductor's voltage adds to the capacitor's rather than subtracting — so the capacitor sees more than the line voltage, typically 5–10% more. Rating it at the line voltage is a common and expensive error.

A harmonic filter is tuned below its target on purpose. Every ageing and failure mechanism reduces capacitance and raises the tuning frequency, and a filter that drifts above a harmonic becomes inductive and creates a parallel resonance that amplifies it. Detuning to the 4.7th means the drift moves it towards correctness rather than into that failure.
Answera\(C = 301\ \mu\text{F}\)   b\(L = 1.35\ \text{mH}\) at the 5th, \(1.52\ \text{mH}\) at the 4.7th   c ageing raises \(f_{tune}\); drifting above creates an amplifying parallel resonance
Problem 5ChallengeThe Cable That Doubles the Voltage

A drive with a 587 V link and 100 ns switching edges feeds a motor through 50 m of cable with a propagation velocity of 150 m/µs. Determine the critical cable length, the voltage at the motor terminals, and the available remedies.

Solution

The cable is a transmission line. At a 100 ns edge the relevant wavelength is comparable with the cable length, so lumped analysis fails:

\[ t_{travel} = \frac{\ell}{v} = \frac{50}{150} = 0.333\ \mu\text{s} \]
\[ t_{round\ trip} = 2t_{travel} = 0.667\ \mu\text{s} \]

The impedance mismatch. A motor's surge impedance is far higher than a cable's:

\[ Z_{cable} \approx 50\text{–}100\ \Omega; \qquad Z_{motor} \approx \text{several k}\Omega \]
\[ \Gamma = \frac{Z_m-Z_c}{Z_m+Z_c} \approx +1 \quad\text{— near-total reflection} \]

An open circuit reflects a voltage wave with the same sign, so the incident and reflected waves add at the motor terminals. The machine sees twice what the inverter sent.

When it happens. The reflection returns to the inverter after the round trip, and only matters if the edge is still rising:

\[ t_{rise} < 2t_{travel} \;\Longrightarrow\; \text{full doubling} \]
\[ 100\ \text{ns} \;\ll\; 667\ \text{ns} \;\Longrightarrow\; \textbf{full doubling} \]
\[ V_{motor} = 2(587) = 1174\ \text{V} \]

Almost twelve hundred volts on the winding of a 415 V motor, thousands of times per second. Standard motor insulation is not qualified for that, and the failure mode is progressive partial discharge in the first turn of the first coil — where the whole step appears, because the winding is also a transmission line.

The critical cable length:

\[ \ell_{crit} = \frac{vt_{rise}}{2} = \frac{(150)(0.1)}{2} = 7.5\ \text{m} \]
Cable lengthMotor voltage
< 7.5 mno significant overshoot
7.5–15 mpartial — rising towards 2×
> 15 mfull doubling

Seven and a half metres is shorter than almost any real installation, so this is the normal case rather than an edge case. And faster devices shorten it further: a SiC drive with 20 ns edges has a critical length of 1.5 m.

The remedies:

RemedyMechanismCost
Output \(dv/dt\) filterslows the edge past \(2t_{travel}\)small \(LC\), some loss
Terminating network at the motormatches \(Z_{cable}\) — no reflectiona box at the machine
Sine filterremoves the PWM entirelylarge, lossy, costly
Inverter-duty motorinsulation rated for 1600 Vmotor premium
Shorter cableattacks the causeoften impossible

The first row is standard practice. A \(dv/dt\) filter that stretches the edge to about 1 µs makes the round trip short by comparison, and the reflection arrives while the edge is still rising — so the two waves never fully add. It costs a small inductor and capacitor per phase and a few tenths of a per cent in efficiency.

What this joins. The cable is now doing three unwanted things at once:

EffectWhere it appeared
Voltage doubling at the motorthis problem
Common-mode current into bearingsSet 26, Problem 5
Radiated emission from the cableProblem 6

All three come from the same source — a fast edge on a long conductor — and all three are reduced by the same remedy. A \(dv/dt\) filter at the inverter output simultaneously protects the insulation, reduces the bearing current and lowers the radiated emission.

Any cable longer than a few metres doubles the voltage at the motor. The critical length is 7.5 m for a 100 ns edge and 1.5 m for a wide-bandgap drive — both far shorter than a real installation. Faster switching, which improves every efficiency figure in this book, makes this problem strictly worse.
Answera\(\ell_{crit} = 7.5\ \text{m}\)   b\(V_{motor} = 1174\ \text{V}\) — full doubling   c\(dv/dt\) filter, terminating network or inverter-duty motor
Problem 6ChallengeA Layered Defence

Specify a complete EMC and protection strategy for the 7.5 kW drive of Set 26, addressing every parasitic effect identified across this book, and order the measures by where they act.

Solution

Collect the problems. Each was identified in its own set and left open:

ProblemSourceMagnitude
Turn-off overshootProblem 150 V on 587
Conducted emissionProblem 267 dB over limit
Input current distortionSet 10THD ~100%, PF 0.44
Supply resonanceSet 21near the 5th harmonic
Common-mode currentSet 264 A spikes into bearings
Cable reflectionProblem 51174 V at the motor
Dead-time distortionSet 2512.7% at low speed

Seven distinct problems from one drive, and no single measure addresses more than two.

Layer 1 — the device and its loop. Fix what can be fixed for free:

MeasureAddresses
Laminated busbar, \(L_s < 50\) nHovershoot — halved
Film capacitor on the module terminalsovershoot, ringing, emission
Gate resistor chosen for \(dv/dt\)emission and CM current — costs some loss
RCD clamp if neededresidual overshoot

The third row is the layer's characteristic trade. Slowing the gate drive reduces every emission problem in this list at the cost of switching loss — a knob that trades Set 2's efficiency directly against this set's EMC.

Layer 2 — the converter's own terminals:

MeasureAddresses
Two-stage CM filter, 7 mH per stageconducted emission — 67 dB
X and Y capacitors within limitsDM and CM, leakage < 3.5 mA
DC link choke or active front endinput THD — Set 10
Damping across the X capacitorfilter–converter interaction
\(dv/dt\) filter on the outputcable reflection and CM current

The third row deserves emphasis: a DC link choke takes the input power factor from 0.44 to about 0.9 and roughly halves the input THD, for one inductor. It is the single most cost-effective power quality measure available to a drive.

Layer 3 — the installation:

MeasureAddresses
Symmetric shielded motor cable, 360° glandsradiated emission, CM containment
Cable as short as practicalreflection, CM current
Separate power and signal routingcoupling into the control
Single-point earthing per the drive manualCM current paths
Insulated non-drive-end bearingbearing EDM — Set 26
Shaft grounding brushdiverts residual shaft current

The first row is worth more than its cost suggests. A symmetric shielded cable bonded 360° at both ends confines the common-mode current to a loop that stays inside the shield — so it never enters the building's earth system and never radiates. A cable terminated by a pigtail achieves almost none of this.

Layer 4 — the system:

MeasureAddresses
Detuned harmonic filters at the switchboardsupply resonance — Set 21
12-pulse or 18-pulse supply for large drivesinput harmonics at source
Harmonic study before installationresonance with existing capacitors
Active filter if several drives share a busaggregate distortion

The third row is often skipped and is where Set 21's 269 Hz resonance came from — an existing power factor correction bank that becomes a resonant circuit as soon as a harmonic source is connected. That interaction cannot be found by testing the drive alone.

What each layer costs and buys:

LayerCostEffectiveness
1 — device and loopnear zerohighest per unit cost
2 — terminalsmoderateessential — nothing else gives 67 dB
3 — installationlow if plannedvery high — and free at design time
4 — systemhighonly for large or multiple drives

The pattern is the same one that ran through Set 24 and Set 23: the measures taken earliest and closest to the source are the cheapest and most effective. Layer 3 in particular costs almost nothing if specified before installation and is extremely expensive to retrofit — replacing a buried cable is not a filter change.

Every parasitic in this book comes from a fast edge meeting a stray element. Overshoot, emission, bearing current and cable reflection are four faces of the same thing, and the defence is layered — loop, terminals, installation, system — with the cheapest and most effective measures closest to the source. Faster devices improve every efficiency figure in this book and make every problem in this set worse.
Answera layer 1: busbar and decoupling   b layer 2: two-stage CM filter, DC choke, \(dv/dt\) filter   c layer 3: shielded cable, insulated bearing   d layer 4: detuned filters and a harmonic study
Formulas

Key Formulas

QuantityRelationNotes
Turn-off overshoot\(\Delta V = L_s\,di/dt\)50 V from 100 nH
Stored stray energy\(\tfrac12L_sI^2\)125 µJ here
Clamp capacitor\(C = L_sI^2/\Delta V^2\)Quadratic in \(\Delta V\)
Clamp resistor\(RC \ll T_s\)Typically \(T_s/5\)
Clamp dissipation\(P = \tfrac12L_sI^2f_s\)Scales with frequency
CM current\(i_{cm} = C_{stray}\,dv/dt\)60 mA from 50 pF
LISN voltage\(V = i_{cm}\times25\ \Omega\)CM: two 50 Ω in parallel
Decibel microvolts\(20\log_{10}\left(V/1\ \mu\text{V}\right)\)1.5 V = 123.5 dBµV
Filter attenuation\(40\log_{10}(f/f_c)\) per stageSecond order
Y capacitor limit\(I_{leak} = 2\pi fC_YV \le 3.5\ \text{mA}\)48 nF max at 230 V
CM choke\(L = 1/\left[\left(2\pi f_c\right)^2C_Y\right]\)7 mH for two stages
Harmonic filter \(C\)\(C = Q_C/(\omega V^2)\)From the kVAr requirement
Harmonic filter \(L\)\(L = 1/\left[\left(2\pi f_{tune}\right)^2C\right]\)Tune to 4.7th, not 5th
Cable travel time\(t = \ell/v\), \(v \approx 150\ \text{m}/\mu\text{s}\)
Critical length\(\ell_{crit} = vt_{rise}/2\)7.5 m at 100 ns
Motor overvoltage\(V_{motor} \to 2V_{dc}\)For \(\ell > 2\ell_{crit}\)
Pitfalls

Common Mistakes

  1. Treating stray inductance as negligible because its energy is small. The energy is 125 µJ and the voltage is 50 V — Problem 1.

  2. Moving the link capacitor away from the module. A working design fails when the loop inductance quadruples — Problem 1.

  3. Assuming faster devices only bring benefits. Overshoot, CM current and cable reflection all scale inversely with the edge time — Problems 1, 2 and 5.

  4. Expecting to fix EMC by tidying. The failure was 67 dB — a factor of 2400 — Problem 2.

  5. Designing a single-stage EMI filter. The required choke was 342 mH; two stages needed 7 — Problem 3.

  6. Exceeding the Y capacitor leakage limit. 3.5 mA is a safety limit, and RCDs see it — Problem 3.

  7. Letting filter input and output tracks run together. The noise couples around the filter entirely — Problem 3.

  8. Tuning a harmonic filter exactly to its harmonic. Ageing drives it above, where it becomes inductive and creates an amplifying resonance — Problem 4.

  9. Rating a filter capacitor at line voltage. The series inductor raises the voltage across it — Problem 4.

  10. Ignoring cable length in a drive installation. Anything over 7.5 m doubles the motor terminal voltage — Problem 5.

Looking Ahead

That completes Part 5. Three sets covered AC–AC conversion and the parasitics that surround every converter in this book. The recurring theme of this last set was that a converter's virtues and its problems have the same cause: switching fast is what makes it efficient, and it is also what makes it overshoot, emit, reflect and destroy bearings. Every improvement in device speed since Set 2 has sharpened both edges of that trade.

The converters are now complete. What has been missing throughout is the load. A duty ratio was chosen to give a certain output voltage, but not because a machine needed it; a modulation index reached 415 V because a motor was rated for it, without asking what the motor would do with it. Every design decision in Parts 3 to 5 was made against an assumed requirement, and those requirements come from the mechanical system.

Next: Part 6 begins with Set 33 — DC Motor Drives, where the chopper of Set 17 and the controlled rectifier of Set 12 finally meet the machine they were built for, four-quadrant operation is worked through in torque and speed rather than in voltage and current, and the speed loop is closed around the current loop of Set 23.