Solved Problems · Set 31

Cycloconverters and Matrix Converters

Part 5 · AC–AC Converters — changing frequency with no DC link at all. Six problems on the oldest large drive and the one that never quite arrived.

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

Set 31 — Cycloconverters and Matrix Converters

Set 30's controller could change an AC voltage's magnitude but never its frequency — two thyristors can only select which parts of the supply waveform reach the load. Select from three phases instead of one, and re-select many times per output cycle, and something else becomes possible: an output at a genuinely different frequency, assembled from pieces of the input, with no DC link anywhere.

Two converters do this. The cycloconverter is the older, built from thyristor groups running the phase control of Set 14 with a slowly varying firing angle — and it drives some of the largest machines ever built. The matrix converter replaces them with self-commutating bidirectional switches and is more elegant in every way except the two that matter: it cannot exceed 0.866 of its input voltage, and it has no freewheeling path at all.

Part 5 · Chapter 22 · 6 solved problems

i Method Recap
  • A cycloconverter is two phase-controlled converters whose firing angle is modulated at the output frequency:

    \[ v_o(t) = V_{dc0}\cos\alpha(t), \qquad \alpha(t) = \cos^{-1}\left(r\cos\omega_ot\right) \]

    Cosine modulation makes the output fundamental exactly \(rV_{dc0}\).

  • Its output frequency is limited by the number of input pulses available per output cycle:

    \[ f_o \lesssim \frac{f_i}{3} \quad\text{for acceptable waveform quality} \]
  • Device count scales with groups:

    \[ \text{3-}\phi\text{ to 3-}\phi,\ \text{6-pulse: } 6\times2\times3 = 36\ \text{thyristors} \]
  • A matrix converter uses nine bidirectional switches and has a hard voltage ceiling:

    \[ q = \frac{V_o}{V_{in}} \le \frac{\sqrt3}{2} = 0.866 \]
  • It has two rules that cannot both be broken:

    \[ \text{never short two input phases}; \qquad \text{never open an output phase} \]
  • Four-step commutation uses the current's direction to satisfy both:

    \[ t_{comm} = 4t_{step} \approx 4\ \mu\text{s} \]
Problem 1CoreBuilding a Slow Sine From a Fast One

A single-phase cycloconverter produces 16.7 Hz from a 230 V, 50 Hz supply using two anti-parallel bridges. Explain the cosine modulation law, find the maximum output fundamental, and explain why the output frequency is limited to about a third of the input.

vo 16.7 Hz envelope each step is one 50 Hz half cycle, phase-controlled
Six input half-cycles per output half-cycle — a coarse staircase, and that is the good case
Solution

The principle. A phase-controlled converter produces \(V_{dc0}\cos\alpha\) on average, and Set 12 established that the average can be made anything from \(+V_{dc0}\) to \(-V_{dc0}\). If \(\alpha\) is varied slowly and sinusoidally rather than held constant, the average output traces a sinusoid:

\[ \overline{v_o}(t) = V_{dc0}\cos\alpha(t) \]

The converter is doing nothing new — it is the same bridge, with a firing angle that moves. What has changed is that the output average is now a function of time rather than a constant.

The cosine modulation law. To make the output a clean sinusoid of amplitude \(rV_{dc0}\), the firing angle must satisfy:

\[ V_{dc0}\cos\alpha(t) = rV_{dc0}\cos\omega_ot \;\Longrightarrow\; \alpha(t) = \cos^{-1}\left(r\cos\omega_ot\right) \]

Which is why it is called cosine-wave-crossing control: the firing instant is found by intersecting a cosine timing wave, derived from the supply, with the output reference. The inverse cosine is performed by the crossing itself rather than by any computation — an elegance that mattered greatly in the analogue era.

The output magnitude:

\[ V_{dc0} = \frac{2V_m}{\pi} = \frac{2(325.3)}{\pi} = 207.1\ \text{V} \]
\[ \hat V_{o1} = rV_{dc0} \le 207.1\ \text{V} \;\Longrightarrow\; V_{o1} \le 146.4\ \text{V rms} \]

At \(r = 1\) the firing angle would have to reach 0 and 180° exactly, which the extinction-angle limit of Set 14 forbids. Practical designs use \(r \le 0.8\), giving about 117 V — roughly half the supply. That is the cycloconverter's characteristic penalty.

The frequency limit. Each output half cycle is assembled from a whole number of input half cycles:

\[ \text{segments per output half cycle} = \frac{f_i}{f_o} = \frac{50}{16.7} = 3 \]

Or, for a six-pulse bridge, six segments per output half cycle. As \(f_o\) rises the count falls:

\(f_o\)Segments per half cycleWaveform
5 Hz20excellent
16.7 Hz6acceptable
25 Hz4coarse
50 Hz2unusable

Below about six segments the staircase can no longer approximate a sinusoid, and the output carries large subharmonic and interharmonic content. Hence the practical rule \(f_o \lesssim f_i/3\) — and it is a hard structural limit, not a design choice.

The two operating modes, exactly as in the dual converter of Set 16:

ModeReactorBehaviour
Blocked (circulating-current-free)nonedead time at each current zero — distortion
Circulating currentrequiredcontinuous, linear, no dead zone

The same trade as Set 16, Problems 3 and 4, and the same conclusion: a drive that must cross zero current smoothly uses the circulating-current mode and pays for the reactor. Since a cycloconverter output crosses zero twice per output cycle — many times per second — the distortion of the blocked mode is continuous rather than occasional.

A cycloconverter is a dual converter whose firing angle never stops moving. Everything about it — the cosine law, the circulating-current modes, the extinction angle limit — comes directly from Sets 14 and 16. What is new is only that the output average is now a slow sinusoid instead of a constant, and that the input frequency sets a hard ceiling on how fast that sinusoid can be.
Answera\(\alpha(t) = \cos^{-1}(r\cos\omega_ot)\)   b\(\hat V_{o1} \le 207.1\ \text{V}\), practically \(r \le 0.8\)   c\(f_o \lesssim f_i/3\) for six segments per half cycle
Problem 2CoreThree Phases, Thirty-Six Thyristors

A three-phase to three-phase cycloconverter uses six-pulse bridges on a 415 V, 50 Hz supply. Find the maximum output voltage, the thyristor count, and the input power factor behaviour. Explain why the input power factor is poor regardless of the load.

Solution

The structure. Each output phase needs a full dual converter — two six-pulse bridges in anti-parallel:

\[ 6\ \text{thyristors}\times2\ \text{bridges}\times3\ \text{phases} = 36 \]

Thirty-six devices, each with its own gate drive, snubber and protection — against six for a voltage source inverter of comparable rating. Device count alone explains why cycloconverters are found only where their specific advantages are decisive.

The maximum output:

\[ V_{dc0} = \frac{3V_{m(line)}}{\pi} = \frac{3(586.9)}{\pi} = 560.4\ \text{V} \]
\[ \hat V_{o,ph} = rV_{dc0} \;\Longrightarrow\; V_{o,ph} \le \frac{560.4}{\sqrt2} = 396.3\ \text{V rms at } r = 1 \]
\[ \text{practically } r \le 0.8:\ V_{o,ph} \le 317\ \text{V},\ V_{o,line} \le 549\ \text{V} \]

So a 415 V supply yields at most about a 550 V line output — but at a third of the frequency. A machine designed for that combination needs many poles or a low base speed, which is exactly the application: slow, high-torque drives.

The input power factor problem. Each bridge is a phase-controlled converter, and Set 12 established that its displacement factor is \(\cos\alpha\):

\[ \alpha(t)\ \text{sweeps from near }0^\circ\ \text{to near }180^\circ\ \text{every output cycle} \]
\[ \overline{\text{DPF}} \ll 1 \quad\text{even at } r = 1 \]

Averaged over the output cycle, the firing angle spends most of its time far from zero — so the input displacement factor is poor even when the converter is delivering full output. It gets worse as \(r\) falls, exactly as a phase-controlled rectifier does at low output.

And the crucial point:

\[ \text{input PF is lagging } \textbf{regardless of the output power factor} \]

Even driving a load that returns power — a machine regenerating — the converter still draws lagging reactive current from the supply, because the reactive demand comes from the phase control itself rather than from the load. A cycloconverter can neither correct nor leading-compensate, and large installations always include substantial capacitor banks or a static VAr compensator.

The complete picture:

PropertyCycloconverter
Thyristors36 (6-pulse) or 72 (12-pulse)
Max output~0.8 of supply, at \(f_i/3\)
Input power factorpoor, always lagging
Input harmonicscomplex — interharmonics, not integer orders
Output frequency0 to \(f_i/3\)
Four-quadrantinherent
Devicesline-commutated thyristors — no forced commutation
Power rangeup to tens of MW

Rows six and seven are why it survives. Line-commutated thyristors are the most robust, highest-rated switching devices made — a single press-pack device handles thousands of amps at kilovolts — and four-quadrant operation comes free from the dual-converter structure. At 10 MW those two advantages outweigh everything above them.

The interharmonic problem deserves separate mention:

\[ f_{input\ components} = \left|kf_i \pm nf_o\right| \quad\text{— not integer multiples of } f_i \]

Because the modulation frequency is unrelated to the supply frequency, the input current contains components at frequencies that are not harmonics at all. These are far harder to filter than harmonics — a tuned filter has nothing fixed to tune to — and they can excite mechanical resonances in nearby turbo-generators. Large cycloconverter installations require careful system studies.

The cycloconverter's input current is not harmonic. Because the output frequency bears no relation to the supply, the input contains interharmonics at \(\left|kf_i \pm nf_o\right|\) — components that no tuned filter addresses and that can excite resonances elsewhere on the network. That, more than the device count, is what limits where one can be installed.
Answera\(V_{dc0} = 560.4\ \text{V}\), output \(\le 396\ \text{V}\) per phase   b 36 thyristors   c input PF poor and lagging whatever the load
Problem 3Exam levelNine Switches and a Hard Ceiling

A three-phase matrix converter operates from a 415 V supply. Find the maximum output voltage with and without zero-sequence injection, compare with a back-to-back voltage source converter on the same supply, and list what the matrix converter gains for that loss.

Solution

The structure. Nine bidirectional switches connect every input phase to every output phase:

\[ 3\times3 = 9\ \text{bidirectional switches} = 18\ \text{devices}+18\ \text{diodes} \]

Each output phase can be connected to any input phase at any instant, so the output is assembled from pieces of the three inputs at the switching frequency — typically several kilohertz, against the cycloconverter's line frequency. That alone removes the \(f_i/3\) limit: a matrix converter can produce any output frequency, above or below the input.

The voltage ceiling. The output must at every instant be constructible from the instantaneous input voltages, and those are themselves varying:

\[ q = \frac{\hat V_o}{\hat V_{in}} \le \frac{\sqrt3}{2} = 0.866 \]

Without zero-sequence injection the limit is only 0.5; adding a common component to all three outputs — the same freedom Sets 26 to 28 exploited — raises it to 0.866. And 0.866 is absolute: it is the ratio of the largest inscribed sinusoid to the input envelope, and no modulation scheme exceeds it.

Compare with the alternative:

\[ \text{matrix: } V_o = (0.866)(415) = 359\ \text{V} \]
\[ \text{back-to-back VSC: } V_{dc} = 415\sqrt2 = 587\ \text{V},\ V_o = (0.707)(587) = 415\ \text{V} \]
\[ \text{deficit} = 1-\frac{359}{415} = 13.4\% \]

The rectifier-plus-inverter chain reaches the full supply voltage because the DC link sits at the input peak — a capacitor stores energy from the peaks and releases it continuously. The matrix converter has no storage, so it can only ever draw what the input is providing at that instant.

What that 13.4% costs in practice. A standard 415 V motor cannot be driven at full voltage:

ConsequenceDetail
Motor derated13.4% less voltage → less torque at base speed
Or a custom motorwound for 359 V — not a stock item
Or a step-up transformerdefeats the compactness argument
Field weakening starts earlierreduced constant-torque range

This is the single largest reason the matrix converter has never displaced the back-to-back VSC despite thirty years of research. Every other advantage is real; none of them compensates for being unable to drive a standard motor at rated voltage.

What it does gain:

AdvantageAgainst back-to-back VSC
No DC link capacitorremoves the largest, shortest-lived component
Sinusoidal input currentvs the diode bridge's 31% THD
Controllable input power factorcan be set to unity or leading
Inherently bidirectionalregeneration with no extra hardware
Any output frequencyno \(f_i/3\) limit
Compact, high power densityno electrolytics, no bulk storage
Long lifecapacitor ageing is the usual failure mode

The first row is the strongest argument. Set 23's adapter and every drive in Part 4 relied on a DC link capacitor, and electrolytic capacitors are the shortest-lived component in any converter. Removing them entirely is worth a great deal in aerospace, downhole and other high-temperature or high-reliability applications — which is precisely where matrix converters are actually used.

A subtlety about the input current. Without storage, instantaneous input power must equal instantaneous output power:

\[ p_{in}(t) = p_{out}(t)\ \text{at every instant} \]

For balanced three-phase in and out, both are constant, so this is satisfiable — which is why a three-phase matrix converter works at all. But an unbalanced input or output makes \(p(t)\) pulsate, and with no capacitor to absorb the difference, the imbalance is transferred straight through to the other side. A matrix converter is unusually sensitive to supply unbalance.

No storage means no voltage boost. A DC link capacitor charges to the input peak and holds it, so the inverter works from a voltage the input only reaches momentarily. A matrix converter must build its output from whatever the input offers at that instant, and the best achievable is 0.866 — a limit imposed by geometry, not by the switches.
Answera\(q \le 0.866\) with injection, 0.5 without   b 359 V against 415 V — a 13.4% deficit   c gains: no DC capacitor, sinusoidal input, controllable input PF
Problem 4Exam levelNowhere for the Current to Go

Explain why a matrix converter cannot use a simple dead time, describe the four-step commutation sequence that solves the problem, and evaluate the timing overhead at a 10 kHz switching frequency.

Solution

The two rules, and why they conflict:

\[ \text{(1) never connect two input phases together — they are voltage sources} \]
\[ \text{(2) never open an output phase — the load is inductive} \]

Rule 1 forbids overlap. Rule 2 forbids a gap. Every other converter in this book escapes by having a third path: a VSI has anti-parallel diodes to freewheel into, a CSI has an overlap that is harmless because the link is a current source. A matrix converter has neither.

Why dead time fails. Insert a gap between turning one switch off and the next on, as in Set 7:

\[ \text{gap} \;\Longrightarrow\; \text{inductive load current interrupted} \;\Longrightarrow\; v = L\frac{di}{dt} \to \infty \]

The load inductance generates whatever voltage it needs to keep the current flowing, and with no path available that voltage appears across the switches and destroys them. Overlap instead, and two input phases are shorted through the switches with only the source inductance limiting the current. Neither option is survivable.

The resolution: exploit the current direction. A bidirectional switch is two devices with opposite orientations, and each conducts current in only one direction:

\[ i_o > 0 \;\Longrightarrow\; \text{only the forward device matters} \]

So if the current direction is known, the two halves of each bidirectional switch can be controlled independently — and a sequence exists that never opens the conducting path and never closes a shorting one. That is four-step commutation.

The sequence, commutating from input phase A to phase B with \(i_o > 0\):

StepActionWhy it is safe
1Turn off the reverse device of Ait carries no current
2Turn on the forward device of Bno short: reverse paths both open
3Turn off the forward device of Acurrent transfers naturally to B
4Turn on the reverse device of Brestores full bidirectionality

At no point is the load open, and at no point can current circulate between A and B — because whichever pair is on can only conduct one way, and the natural voltage difference determines which way that is. The commutation happens by itself between steps 2 and 3.

The timing overhead:

\[ t_{step} \approx 1\ \mu\text{s} \;\Longrightarrow\; t_{comm} = 4\ \mu\text{s} \]
\[ \text{at } f_s = 10\ \text{kHz}:\ \frac{4}{100} = 4\%\ \text{of the period} \]

Four per cent, against the 1–2% a VSI's dead time costs — and it produces the same kind of output voltage error as Set 25, Problem 5, requiring the same kind of compensation. Raising the switching frequency makes it proportionally worse, so a matrix converter cannot switch as fast as a VSI of the same rating.

What the method demands:

RequirementDifficulty
Reliable current direction sensingfails near the current zero crossing
18 independent gate signalsvs 6 for a VSI
Precise inter-step timingtypically a hardware state machine
Input voltage sensing (for the alternative method)fails when two phases are nearly equal
Clamp circuit for faultsa small diode bridge and capacitor, always required

The first row is the practical Achilles heel, and it is the same problem dead-time compensation had in Set 25: near the current zero the sign is uncertain, noise makes it flicker, and a wrong decision either shorts the supply or opens the load. Voltage-based commutation has the mirror problem near the input voltage crossings, so robust implementations use both and switch between them.

Every other converter has a freewheeling path; the matrix converter has none. That single structural absence turns commutation from a two-microsecond dead time into a four-step sequence conditional on the current's direction — and makes the converter fail catastrophically rather than merely distort when the direction is misjudged. It is the main reason a topologically superior converter is rarely built.
Answera dead time opens an inductive load; overlap shorts the supply   b four steps using the current direction   c\(4\ \mu\text{s}\) = 4% of the period at 10 kHz
Problem 5ChallengeDirect or Through a Link

Compare the four ways of converting AC to AC — AC voltage controller, cycloconverter, matrix converter and back-to-back voltage source converter — on frequency range, voltage ratio, component count, input quality and where each is actually used.

Solution

The fundamental division. Direct converters have no energy storage; indirect ones do:

\[ \text{direct: } p_{in}(t) = p_{out}(t)\ \text{always}; \qquad \text{indirect: the link absorbs the difference} \]

That single distinction explains most of what follows. Storage decouples the input from the output — the same argument that Set 17 made for the chopper drive — and buys voltage boost, independent control of both sides, and ride-through. Its cost is the capacitor.

The comparison:

PropertyAC controllerCycloconverterMatrixBack-to-back VSC
Output frequency\(f_i\) only\(0\text{–}f_i/3\)anyany
Voltage ratio\(\le1\)~0.80.8661.0
Switches6361812
Energy storagenonenonenoneDC capacitor
Input currentdistortedinterharmonicssinusoidaldepends on front end
Input PF controlnonenone — always laggingyesyes (active front end)
Four quadrantnoyesyesyes (active front end)
DevicesthyristorthyristorIGBTIGBT
Power ceilingMWtens of MW~1 MWMW

Read the two device rows together. The cycloconverter's power ceiling comes from what it is built from:

\[ \text{line-commutated thyristor: thousands of amps, kilovolts, per device} \]

No IGBT approaches that rating, and no forced commutation circuit is needed because the supply does the commutating. At 20 MW there is simply no self-commutated alternative of comparable robustness, which is why the cycloconverter survives in an application niche rather than being obsolete.

Where each is actually installed:

ApplicationConverterDeciding factor
Fan and pump soft startAC controllercost — bypassed after start
Furnace, heaterAC controllercost, thermal load
Gearless ore and cement millsCycloconvertertens of MW at a few Hz
Ship propulsion, rolling millsCycloconvertersame
Aerospace actuatorsMatrixno electrolytics, high temperature
Downhole and spaceMatrixreliability, capacitor lifetime
Everything elseBack-to-back VSCfull voltage, mature, cheap

The last row covers the overwhelming majority. A diode bridge, a capacitor and a six-switch inverter reach the full supply voltage, handle any frequency, and cost less than anything else — and adding an active front end buys sinusoidal input current and regeneration when needed.

Why the matrix converter never displaced the VSC. Three reasons, in order:

\[ \text{(i) } 0.866 < 1.0; \qquad \text{(ii) commutation complexity}; \qquad \text{(iii) no ride-through} \]

The third is easily overlooked. A DC link capacitor holds the converter up through a short supply dip — the hold-up requirement Set 19 sized a bulk capacitor for. A matrix converter has no stored energy at all, so a one-cycle supply interruption stops it dead. For industrial equipment that is often disqualifying on its own.

The DC link is not overhead — it is a feature that costs a capacitor. Storage buys the full input voltage, independent control of both sides, ride-through and a trivial commutation problem. Direct conversion saves the capacitor and gives all four back, which is why every AC–AC converter in general use has a link and the ones that do not occupy narrow niches.
Answera ratios 1.0 / 0.8 / 0.866 / 1.0   b switches 6 / 36 / 18 / 12   c cycloconverter above ~10 MW; matrix where capacitors cannot survive; VSC everywhere else
Problem 6ChallengeA 5 MW Mill Drive

A 5 MW gearless ore mill uses a low-speed synchronous motor rated 3 kV at a maximum of 10 Hz, driven by a six-pulse cycloconverter. Determine the required transformer secondary voltage, the motor current, the thyristor duty, and explain why this application defeats every alternative.

Solution

The application. A gearless mill motor is a very large, many-poled synchronous machine wrapped directly around the mill shell:

\[ \text{mill speed} \approx 12\ \text{rpm}, \qquad \text{poles} \approx 100 \;\Longrightarrow\; f = \frac{12\times50}{60} \approx 10\ \text{Hz} \]

No gearbox, no pinion, no alignment problem — and an electrical requirement that is unusual in every respect: five megawatts at ten hertz, with full torque needed from standstill.

Check the frequency limit first:

\[ \frac{f_o}{f_i} = \frac{10}{50} = 0.20 \;<\; 0.33 \quad\checkmark \]

Comfortably inside the cycloconverter's range, with about ten input segments per output half cycle at a six-pulse configuration. The waveform will be good, which is the opposite of the usual situation — this application sits in the cycloconverter's sweet spot rather than at its limit.

Size the transformer secondary. The peak phase voltage required:

\[ \hat V_{ph} = \frac{3000}{\sqrt3}\sqrt2 = 2449\ \text{V} \]

With a modulation ratio limited to \(r = 0.8\) for commutation margin:

\[ V_{dc0} \ge \frac{2449}{0.8} = 3062\ \text{V} \]
\[ V_{dc0} = 1.35V_{LL} \;\Longrightarrow\; V_{LL} = \frac{3062}{1.35} = 2268\ \text{V} \]

So a 2.3 kV secondary — and since each of the six bridges needs its own isolated supply to avoid shorting the phases together, the transformer has six secondaries. That transformer is a major item, and it is also where the 12-pulse arrangement is implemented for harmonic cancellation.

The motor current:

\[ I = \frac{5\times10^{6}}{\sqrt3(3000)(0.9)} = 1069\ \text{A} \]
\[ I_{thyristor,avg} = \frac{1069}{3} = 356\ \text{A}, \qquad I_{rms} = \frac{1069}{\sqrt3} = 617\ \text{A} \]

Six hundred amps RMS per device, at 2.3 kV — well within a single press-pack thyristor's capability, with no series or parallel connection needed. That is the point: the device count is 36, but every one of them is a single, standard, extremely rugged component.

Why every alternative fails:

AlternativeWhy it fails here
Geared drive + standard motorno gearbox handles 4 MNm reliably
Two-level VSI5 MW at 3 kV needs series devices — fragile
Cascaded H-bridgeviable, but needs 27+ isolated cells and forced cooling
Matrix converterpower ceiling ~1 MW; no ride-through
Current source inverterviable — the modern competitor
Cycloconverter36 standard thyristors, no forced commutation

The fifth row is where the technology is moving. A load-commutated current source inverter using the same thyristors handles the same power with fewer devices and a better input power factor — and modern installations increasingly choose it. But the cycloconverter's four-quadrant operation and its excellent low-frequency torque remain hard to beat for a mill that must be inched into position and reversed.

What the installation must also include:

ItemReason
Reactive power compensationinput PF is poor at all outputs — Problem 2
Harmonic and interharmonic filterscomponents at \(\left|kf_i\pm nf_o\right|\)
12-pulse transformer arrangementcancels the 5th and 7th input harmonics
Circulating current reactorsfor smooth zero crossing — Set 16
Network resonance studyinterharmonics can excite nearby generators

The first row is typically 40–60% of the drive rating in capacitors — two or three megavars of compensation for a five-megawatt drive. That auxiliary plant is a substantial fraction of the installation cost and is the price of line commutation.

The cycloconverter's disadvantages all concern the supply, and its advantages all concern the load. Poor power factor, interharmonics and thirty-six devices are tolerable when the alternative is a gearbox transmitting four meganewton-metres — and no other converter delivers full torque at 0.2 Hz using single standard thyristors with no forced commutation at all.
Answera secondary 2.27 kV, six windings   b\(I = 1069\ \text{A}\), 356/617 A per thyristor   c\(f_o/f_i = 0.20\) — well inside the limit
Formulas

Key Formulas

QuantityRelationNotes
Cycloconverter output\(\overline{v_o} = V_{dc0}\cos\alpha(t)\)Set 12's law, modulated
Cosine modulation\(\alpha(t) = \cos^{-1}\left(r\cos\omega_ot\right)\)Gives \(\hat V_{o1} = rV_{dc0}\)
1-φ \(V_{dc0}\)\(2V_m/\pi\)207.1 V at 230 V
3-φ \(V_{dc0}\)\(3V_{m(line)}/\pi = 1.35V_{LL}\)560.4 V at 415 V
Practical \(r\)\(\le 0.8\)Commutation margin
Frequency limit\(f_o \lesssim f_i/3\)Six segments per half cycle
Thyristor count\(6\times2\times3 = 36\)6-pulse, 3-φ to 3-φ
Input spectrum\(\left|kf_i \pm nf_o\right|\)Interharmonics
Matrix switches9 bidirectional = 18 devices
Matrix voltage ratio\(q \le \sqrt3/2 = 0.866\)0.5 without injection
Matrix rulesno input short, no output openNo freewheeling path
Four-step commutation\(t_{comm} = 4t_{step}\)~4% of \(T_s\) at 10 kHz
Direct converter constraint\(p_{in}(t) = p_{out}(t)\)No storage
Back-to-back VSC ratio1.0Link sits at the input peak
Pitfalls

Common Mistakes

  1. Treating a cycloconverter as a new topology. It is the dual converter of Set 16 with a moving firing angle — Problem 1.

  2. Assuming \(r = 1\) is achievable. The extinction angle limits it to about 0.8 — Problems 1 and 6.

  3. Exceeding \(f_i/3\). Below six input segments per output half cycle the waveform collapses — Problem 1.

  4. Expecting good input power factor at high output. The firing angle sweeps the full range every output cycle regardless — Problem 2.

  5. Calling the input current harmonics. They are interharmonics at \(\left|kf_i\pm nf_o\right|\), which tuned filters cannot address — Problem 2.

  6. Quoting the matrix converter's ratio as 0.5. That is without zero-sequence injection; with it, 0.866 — Problem 3.

  7. Assuming a matrix converter can drive a standard motor at rated voltage. It is 13.4% short — Problem 3.

  8. Using dead time in a matrix converter. There is no freewheeling path; the load cannot be opened — Problem 4.

  9. Forgetting the matrix converter's lack of ride-through. No stored energy means a one-cycle dip stops it — Problem 5.

  10. Omitting reactive compensation from a cycloconverter installation. It typically needs 40–60% of the drive rating in capacitors — Problem 6.

Looking Ahead

Six problems on converting AC to AC without a link. The cycloconverter turned out to be Set 16's dual converter with a firing angle that never stops moving — inheriting its four-quadrant capability, its circulating-current modes and its poor power factor, and adding a hard \(f_i/3\) frequency ceiling. The matrix converter was better in almost every respect and lost on the two that decide: 0.866 against 1.0, and no freewheeling path at all.

Every converter in this book is now built. What remains is what surrounds them. A converter does not sit alone: its switching edges radiate, its input current distorts the supply, its output cable reflects, and its devices need protection from the transients they themselves create. Those problems have appeared in every part — the snubber of Set 4, the notches of Set 15, the resonance of Set 21, the common-mode current of Set 26 — and they share a common set of remedies.

Next: Set 32 — Filters, Snubbers and EMI Mitigation, where turn-off overshoot is computed from stray inductance and clamped, conducted emissions are estimated against CISPR limits and found 67 dB over, a two-stage EMI filter is designed against the Y-capacitor leakage limit, and a motor cable is shown to double the voltage at the machine terminals.