Part 5 · Chapter 22

Cycloconverters and Matrix Converters

The standard way to change AC frequency is to rectify and then invert, with a capacitor in between. That capacitor is the converter's largest component and usually the first to fail, so it is worth asking whether it can be removed. Two families try — and both discover that without stored energy, the output can only ever be assembled from pieces of the input that happen to be available at that instant.

Power Electronics Prof. Mithun Mondal Reading time ≈ 50 min
Where this sits
Part 5 · AC–AC Converters
Chapter 22 of 30
You should already know
Phase control and \(V_{dc} = 1.35V_L\cos\alpha\) from Chapter 9, dual converters from Chapter 10, and PWM from Chapter 18.
By the end you can
Explain why a cycloconverter is a dual converter with a moving firing angle, why its output frequency is capped near a third of the supply, and what a matrix converter fixes and costs.
Time
≈ 50 min reading · ≈ 45 min problems
i What you'll learn
  • Why removing the DC link is worth attempting at all, and what it costs.
  • The cycloconverter: a dual converter whose firing angle is modulated sinusoidally instead of held constant.
  • Why the output frequency is limited to roughly \( f_s/3 \), and where that limit comes from.
  • Circulating-current and blocking modes, exactly as in Chapter 10 — and why the choice matters more here.
  • The matrix converter: nine bidirectional switches, no energy storage at all, and a hard gain limit of 0.866.
  • The commutation problem that kept matrix converters in the laboratory for thirty years, and the four-step solution.
  • Why, despite thirty years of research, the back-to-back inverter still wins nearly everywhere.
Section 22-1

Why Remove the DC Link

The standard way to change AC frequency is to do it in two stages: rectify to DC, then invert to AC at the new frequency. Chapters 7 to 20 built both halves, and the combination — a back-to-back converter — runs most variable-speed drives in the world.

The DC link between them is a capacitor, and it is worth asking what that capacitor costs.

What the DC link gives, and what it takes
The link providesThe link costs
Decoupling — input and output can be controlled independently Lifetime. An electrolytic capacitor is usually the first component to fail, and its life halves per 10 °C
Energy buffering — ride-through of brief supply dips Volume. Often the largest single component in the converter
A stiff voltage for the inverter to work against Temperature range. Electrolytics perform poorly below −25 °C and age fast above 85 °C
Freedom to use a simple diode rectifier Two conversions. Every watt passes through two stages, each with its own loss
Section recap. The DC link buys decoupling and ride-through at the price of the converter's least reliable and largest component. Removing it gives a single conversion stage and no electrolytics, but rigidly couples the input to the output — which limits what voltage can be produced.
Section 22-2

The Cycloconverter

Chapter 10 built a dual converter: two three-phase bridges in anti-parallel across one load, one supplying positive current and the other negative, with \( \alpha_1 + \alpha_2 = 180^\circ \). Holding \( \alpha \) constant gave an adjustable DC output.

A cycloconverter is that same circuit with one change: let \( \alpha \) vary sinusoidally with time.

  1. Start from the dual converter's output From Chapter 9, a three-phase bridge produces
    Working
    \[ V_{dc} = 1.35\,V_L\cos\alpha \]
    averaged over each 60° segment. Treat this as an instantaneous average that the firing angle can move from segment to segment.
  2. Ask what \(\alpha\) must do to make the output sinusoidal We want \( V_o(t) = \hat V_o\sin\omega_o t \). Setting the two equal:
    Working
    \[ 1.35\,V_L\cos\alpha(t) = \hat V_o\sin\omega_o t \;\Longrightarrow\; \alpha(t) = \cos^{-1}\!\left(r\sin\omega_o t\right) \]
    where \( r = \hat V_o/(1.35V_L) \) is the voltage ratio, \( 0 \le r \le 1 \).
  3. Read what that means physically \( \alpha \) sweeps from 0° when the wanted output is at its positive peak, through 90° at the zero crossing, to 180° at the negative peak. The converter moves smoothly from rectifying to inverting and back, once per output cycle.
  4. Note which bridge is in charge The positive bridge conducts while the output current is positive, and the negative bridge while it is negative — regardless of the voltage's sign. All four quadrants are available, exactly as in Chapter 10.
🔑
Cycloconverter
\[ \alpha(t) = \cos^{-1}\bigl(r\sin\omega_o t\bigr), \qquad V_{o1} = r\,\frac{1.35\,V_L}{\sqrt2}, \qquad f_o \lesssim \frac{f_s}{3} \]

A three-phase-to-single-phase cycloconverter needs two six-pulse bridges; a three-phase output needs six, or 36 thyristors. There is nothing subtle in the hardware — it is the firing schedule that carries the idea.

Interactive · assembling an output from supply segments

A six-pulse cycloconverter building a low-frequency target from segments of a 50 Hz three-phase supply. The actual output is stitched together from whichever supply segment the firing angle selects. Raise the output frequency and watch the fit fall apart.

10 Hz
0.80
Cycloconverter output assembled from supply segments One cycle of the wanted low-frequency output, drawn as a dashed sine, with the actual cycloconverter output overlaid as a stepped trace assembled from six-pulse segments of the fifty hertz supply. At low output frequencies there are many supply segments per output cycle and the stepped trace follows the sine closely. As the output frequency rises the number of segments per output cycle falls, the steps become coarse, and the fit deteriorates sharply. +1 −1 t one output cycle
Segments/cycle30
Ratio fo/fs0.20
Fit error4.2%
VerdictGood

Section recap. A cycloconverter is a dual converter whose firing angle follows \( \alpha = \cos^{-1}(r\sin\omega_ot) \), sweeping from rectifying to inverting once per output cycle. A six-pulse bridge can only change six times per supply cycle, so the output frequency is capped near \( f_s/3 \) — which happens to suit the very large, very slow machines it drives.
Section 22-3

Circulating Current, Again

Chapter 10's dual converter faced a choice: let both bridges conduct at once with a reactor to limit the circulating current, or block one entirely and accept a dead time. A cycloconverter faces exactly the same choice, and it matters more here because the current reverses every output cycle rather than occasionally.

The two operating modes
PropertyCirculating-current modeBlocking (non-circulating) mode
Both bridges gatedYes, alwaysNo — only the one carrying current
Reactor requiredYes, and it is largeNone
Current reversalInstant and smoothRequires a dead time of 1–5 ms
Distortion near zero currentNoneA flat spot — the current sticks at zero
LossesHigher — circulating current is real currentLower
Control complexitySimple gating, more hardwareComplex logic, less hardware
Typical useFast reversing mill drivesMost modern installations
Section recap. Circulating-current mode gates both bridges and needs reactors, but reverses current instantly. Blocking mode needs no reactor but flattens the current around every zero crossing — twice per output cycle, producing third-harmonic torque distortion at frequencies a mill drive's mechanics respond to.
Section 22-4

The Matrix Converter

The cycloconverter's limitation traces to the thyristor: it can only be fired once per supply segment, so the output is sampled at \( 6f_s \). Replace the thyristors with fully controllable, bidirectional switches and that limit disappears entirely.

The result is startlingly simple to state: nine switches, connecting any input phase to any output phase, at any time.

3 × 3 SWITCH MATRIX ONE BIDIRECTIONAL SWITCH A B C in a b c out T₁ T₂ 4 devices per switch × 9 = 36 in total
Nine bidirectional switches connect any input to any output. Each needs four semiconductors, because it must block and conduct in both directions.
🔑
The matrix converter's hard gain limit
\[ \frac{\hat V_{out}}{\hat V_{in}} \le \frac{\sqrt3}{2} = 0.866 \]

Unlike the cycloconverter, the output frequency is unrestricted — but the output voltage can never exceed 86.6% of the input, and no modulation scheme changes that.

Section recap. Nine bidirectional switches connect any input to any output, with no storage anywhere. The output frequency is unrestricted, but the output voltage cannot exceed \( \sqrt3/2 \) of the input — a structural limit that requires a non-standard motor and has held the topology back commercially.
Section 22-5

The Commutation Problem

The matrix converter's gain limit is an inconvenience. Its commutation problem was, for three decades, the thing that actually stopped it working.

! The two rules, and why they conflict
  • Rule 1: never connect two input phases to the same output — that is a line-to-line short with nothing to limit it.
  • Rule 2: never leave an output phase unconnected — the load is inductive and its current must have a path, or the voltage rises without limit.
  • A VSI escapes this because dead time satisfies both: turn everything off, and the antiparallel diodes carry the current back into the DC link.
  • A matrix converter cannot. There is no DC link and no freewheeling path. Overlap violates Rule 1; a gap violates Rule 2. No simultaneous transition is safe.
4 Four-step current-direction commutation
  1. Measure the output current's direction. Everything depends on this.
  2. Turn off the outgoing switch's device that carries the opposite direction. Nothing changes — that device was not conducting.
  3. Turn on the incoming switch's device that carries the present direction. The current now transfers naturally to whichever source is more favourable. Crucially, the two remaining devices cannot short the supply, because each blocks in one direction only.
  4. Turn off the outgoing switch's remaining device, and finally turn on the incoming switch's other device to restore full bidirectional capability.

At no step are both rules broken, because the four-device structure of each bidirectional switch is used deliberately: each half is controlled separately, so a path always exists for the actual current direction and never for a short.

Section recap. With no DC link and no freewheeling path, a matrix converter can neither overlap nor gap its switches. Four-step commutation sequences each bidirectional switch's halves separately, but depends on knowing the current's direction — which fails near every zero crossing. Hybrid schemes and a clamp circuit are used in practice.
Section 22-6

Why Back-to-Back Still Wins

The three ways to change AC frequency
PropertyBack-to-back VSICycloconverterMatrix converter
Devices12 (6 + 6)36 (six bridges)36 (nine switches)
Energy storageDC link capacitorReactors (circulating mode)None
Output frequencyUnrestricted\(\lesssim f_s/3\)Unrestricted
Voltage ratioUp to 1.0 and beyond\(\approx 0.5\)0.866 maximum
Input power factorPoor with diodes; unity with an active front endPoor, and varies with loadControllable, near unity
Input current qualityPoor with diodes; excellent with AFEPoor — beat harmonicsGood, needs a small filter
RegenerationNeeds an active front endInherentInherent
Ride-throughGood — the link stores energyNoneNone
Control complexityLow — two independent halvesModerateHigh — commutation is coupled
Commercial statusUniversalInstalled base at very high powerNiche
Section recap. The matrix converter beats the back-to-back inverter on power factor, input current, regeneration and reliability, and loses on the one thing that matters commercially: it cannot drive a standard motor at rated voltage. It survives where weight, temperature or lifetime dominate.
Section 22-7

Worked Examples

1 Cycloconverter for a gearless mill drive

Problem. A 16 MW gearless ball-mill drive uses a 44-pole synchronous motor turning at 12 rpm, fed by a cycloconverter from a 50 Hz supply. Find the output frequency, the number of supply segments per output cycle, and comment on the choice of topology.

Output frequency
\[ f_o = \frac{p\,n}{120} = \frac{44 \times 12}{120} = 4.4\ \text{Hz} \]
Segments per output cycle, six-pulse bridges
\[ N = \frac{6f_s}{f_o} = \frac{6 \times 50}{4.4} = 68 \]

68 segments per output cycle is excellent — the staircase is very fine and the beat harmonics are small. The ratio \( f_o/f_s = 0.088 \) sits far below the \( 1/3 \) limit, with enormous margin for speed variation.

Why the topology fits so well here:

  • The frequency is low because the machine has many poles, which it has because a gearless drive must produce enormous torque at low speed. The mechanical design and the converter's limitation coincide rather than conflict.
  • 16 MW at perhaps 10 kV is beyond IGBT ratings without series connection, but comfortable for thyristors.
  • Natural commutation throughout, so switching loss is negligible and efficiency reaches 97–98%.
  • No DC link capacitors to maintain in a dusty, hot mineral-processing plant.

The penalty to design around. The input power factor of a cycloconverter is poor and varies with load — typically 0.7 or worse at partial load. A 16 MW drive at 0.7 draws 23 MVA, so the installation needs substantial reactive compensation and harmonic filtering. On a mine site this is a significant part of the electrical infrastructure, and it is budgeted for accordingly.

2 The firing schedule

Problem. A cycloconverter produces 5 Hz output from a 415 V, 50 Hz supply at a voltage ratio \( r = 0.7 \). Find the fundamental output voltage, and tabulate the firing angle at several instants of the output cycle.

Output magnitude
\[ \hat V_o = r \times 1.35\,V_L = 0.7 \times 1.35 \times 415 = 392\ \text{V (peak)} \]
\[ V_{o1(rms)} = \frac{392}{\sqrt2} = 277\ \text{V} \]

Firing angle through one output cycle, from \( \alpha = \cos^{-1}(0.7\sin\omega_ot) \):

Firing schedule at \(r = 0.7\)
Output angleSine \(\sin\omega_o t\)Scaled \(0.7\sin\omega_o t\)Firing angle \(\alpha\)Mode
0.000.0090.0°Zero output
30°0.500.3569.5°Rectifying
90°1.000.7045.6°Rectifying, peak
150°0.500.3569.5°Rectifying
210°−0.50−0.35110.5°Inverting
270°−1.00−0.70134.4°Inverting, peak

Read the pattern. The firing angle swings between 45.6° and 134.4°, symmetric about 90°. The converter spends half of each output cycle rectifying and half inverting — which is exactly what producing an alternating output from a DC-capable bridge requires.

Why \( r \) is limited well below 1 in practice. At \( r = 1 \) the firing angle would reach 0° and 180°. At 0° there is no margin for supply variation; at 180° there is no margin for the commutation overlap and recovery time of Chapter 10, and the converter would suffer commutation failure. Practical designs stop near \( r = 0.8 \), which is why a cycloconverter's realistic voltage ratio is about 0.5 of the supply line voltage rather than the 0.955 a rectifier achieves.

3 Matrix converter for a standard motor

Problem. A 400 V, 50 Hz, 22 kW induction motor is to be driven by a matrix converter from a 400 V supply. Assess whether this works, and propose remedies if not.

Available output
\[ V_{out(max)} = 0.866 \times 400 = 346\ \text{V} \]
Flux at rated frequency
\[ \frac{\Phi}{\Phi_{rated}} = \frac{V/f}{(V/f)_{rated}} = \frac{346/50}{400/50} = 0.866 \]

The motor runs at 86.6% of rated flux, and torque is proportional to flux, so:

Consequence
\[ T_{max} \approx 0.866\,T_{rated} \quad\text{— a 13.4\% derating} \]

Four possible responses, in order of practicality:

  1. Accept the derating and fit a 30 kW motor to do a 22 kW job. Simple, and often the cheapest answer — but the larger motor is less efficient at partial load and costs more.
  2. Specify a 346 V motor. Technically correct and available to order, but it is a non-standard item with long lead times and no local spares. For a plant that stocks standard motors, this is a serious operational objection.
  3. Feed the converter from a higher voltage. A 460 V supply gives \( 0.866 \times 460 = 398 \) V — enough. Free where 460 V exists; a transformer otherwise, which reintroduces the mass the topology was avoiding.
  4. Reduce the base frequency. Run 43 Hz with a proportionally reduced voltage, keeping \( V/f \) constant, and accept 14% less speed. Suits fans and pumps; unsuitable where speed is fixed by the process.

What this example is really showing. None of these is difficult, and all of them make the drive a system decision rather than a component swap. That is the commercial barrier, and it is worth more than the topology's genuine technical advantages.

4 Choosing between the three

Problem. Recommend a converter for each: (a) a 12 MW ore mill at 8 rpm; (b) a 30 kW conveyor drive in a factory; (c) a 50 kW actuator on an aircraft. Justify each.

(a) 12 MW ore mill — cycloconverter, or MMC for a new installation.

  • A many-pole ring motor at 8 rpm needs only a few hertz, comfortably inside the \( f_s/3 \) limit.
  • 12 MW at medium voltage is thyristor territory; IGBTs would need series connection.
  • Natural commutation gives 97–98% efficiency and negligible switching loss.
  • Four-quadrant operation is inherent, which the mill needs.
  • For a new plant, an MMC (Chapter 20) is now the stronger choice — unlimited frequency, near-unity input power factor and no beat harmonics — but the cycloconverter remains correct for the existing installed base.

(b) 30 kW conveyor — back-to-back VSI, without hesitation.

  • Standard 400 V motor, full voltage available, no derating.
  • DC link ride-through survives the supply dips a factory sees weekly.
  • Commodity pricing, universal spares, and any electrician can replace it.
  • A matrix converter's advantages — power factor, no capacitor — are worth far less here than compatibility.

(c) 50 kW aircraft actuator — matrix converter, and this is where it belongs.

  • No DC link capacitor, which removes both mass and the component least able to survive −55 °C to +85 °C cycling.
  • Power density is the dominant metric, and the capacitor bank is what a back-to-back drive cannot shrink.
  • Regeneration is inherent, and an actuator regenerates constantly as it works against and with its load.
  • The motor is designed for the application anyway, so the 0.866 limit costs nothing — it is simply wound for 346 V.
  • Near-unity input power factor matters on an aircraft's limited generation capacity.

The pattern across all three. The correct converter is decided by which constraint is hardest — device ratings at 12 MW, compatibility at 30 kW, mass and temperature at altitude. No topology wins on merit alone.

Section 22-8

Summary & Formula Sheet

Chapter 22 in five sentences:

  1. Removing the DC link removes the converter's least reliable component and one conversion stage, at the cost of rigidly coupling input to output.
  2. A cycloconverter is a dual converter with \( \alpha = \cos^{-1}(r\sin\omega_ot) \), sweeping from rectifying to inverting once per output cycle.
  3. Its output is sampled at \( 6f_s \), so the output frequency is capped near \( f_s/3 \) — which suits the very large, very slow machines it drives.
  4. A matrix converter's nine bidirectional switches remove the frequency limit but impose a hard voltage ratio of \( \sqrt3/2 = 0.866 \).
  5. With no freewheeling path, its switches can neither overlap nor gap, so commutation must be sequenced in four steps using the current's direction.
Formula sheet · Chapter 22
Cycloconverter firing schedule\(r\) = voltage ratio
\( \alpha(t) = \cos^{-1}\bigl(r\sin\omega_o t\bigr) \)
Output fundamentalthree-phase bridge segments
\( \hat V_o = r\,(1.35\,V_L) \)
Segments per output cycle\(p\)-pulse bridge
\( N = \dfrac{p\,f_s}{f_o} \)
Output frequency limitpractical, not absolute
\( f_o \lesssim \dfrac{f_s}{3} \)
Beat frequenciessome fall below \(f_o\)
\( f = |m f_s \pm n f_o| \)
Synchronous machine frequency\(p\) poles, \(n\) rpm
\( f = \dfrac{p\,n}{120} \)
Matrix converter gain limitstructural; cannot be exceeded
\( q_{max} = \dfrac{\sqrt3}{2} = 0.866 \)
Naive modulation limitwithout harmonic injection
\( q = 0.5 \)
Matrix converter devices4 per bidirectional switch
\( 9 \times 4 = 36 \)
Motor flux deratingmatrix converter on a standard motor
\( \dfrac{\Phi}{\Phi_{rated}} = 0.866 \)

Key terms

Direct AC–AC converter
A converter that changes AC frequency without an intermediate DC stage. Cycloconverters and matrix converters.
Back-to-back converter
A rectifier and inverter sharing a DC link. The standard variable-frequency drive.
Cycloconverter
A dual converter whose firing angle is modulated sinusoidally to produce a low-frequency output.
Voltage ratio, \(r\)
Wanted output peak divided by the bridge's maximum DC output. Limited to about 0.8 in practice.
Beat frequency
A component at \(|mf_s \pm nf_o|\) produced because the input and output frequencies are unrelated. Some are sub-harmonic.
Matrix converter
Nine bidirectional switches connecting any input phase to any output phase, with no energy storage.
Bidirectional switch
A switch that blocks and conducts in both directions. Needs four semiconductors — two transistors and two diodes.
Four-step commutation
Sequencing each bidirectional switch's halves separately so no short and no open ever occurs. Requires the current's direction.
Clamp circuit
A small diode bridge and capacitor that catch the load's energy if a commutation fails. Present on every practical matrix converter.
Gearless mill drive
A very large many-pole synchronous machine driving a mill directly. The cycloconverter's principal application.
Check yourself

Test Yourself

Chapter 22 · six questions answers hidden until you ask
Why can a cycloconverter not produce an output frequency above about a third of the supply's?

Because its output is a sampled version of the wanted waveform, and the sampling rate is fixed by the supply.

The sampling rate. A six-pulse bridge can change its output value only when the next thyristor becomes able to conduct — once per 60° of supply, so six times per supply cycle. On 50 Hz that is 300 opportunities per second, and no control strategy increases it.

Segments per output cycle:

\[ N = \frac{6f_s}{f_o} \]

  • \( f_o = 5 \) Hz: 60 segments — an excellent staircase.
  • \( f_o = 16.7 \) Hz: 18 segments — the conventional limit.
  • \( f_o = 50 \) Hz: 6 segments — barely a sine.

But the visible steps are not the real problem. Because \( f_s \) and \( f_o \) are unrelated, the output contains beat components at \( |mf_s \pm nf_o| \), and some of these land below the output frequency. A machine's inductance cannot filter sub-harmonics, and its inertia cannot smooth torque pulsations at a few hertz — so they appear directly as mechanical disturbance.

As \( f_o \) rises, more of these products move into the damaging band, which is why the quoted limit varies from \( f_s/2 \) to \( f_s/3 \) depending on how much distortion the application tolerates.

How the limit is raised in practice. Use more pulses: a 12-pulse cycloconverter samples at \( 12f_s \), doubling the usable output frequency. Large mill drives are built this way, and the phase-shifting input transformer that provides the 12-pulse operation also serves as harmonic filtering.

And why the limit rarely bites. Cycloconverters drive many-pole machines at low speed, where a few hertz is exactly what is wanted. The topology and its applications co-evolved.

Where does the matrix converter's 0.866 limit come from, and could better modulation beat it?

It comes from having no energy storage, and no modulation scheme can beat it.

The constraint. With nothing to store energy, the output at any instant must be a weighted average of the input voltages at that same instant. The output is therefore confined inside the envelope the input offers.

Why the envelope dips. The three input voltages are sinusoids 120° apart. The largest available difference at any instant varies through the cycle, falling to \( \cos30^\circ = \sqrt3/2 \) of its peak six times per cycle, where two phases cross.

To produce a constant-amplitude output, the modulator must succeed at the worst instant, so the limit is set by the envelope's minimum:

\[ q_{max} = \cos30^\circ = 0.866 \]

Could better modulation help? It already has, twice:

  • Naive modulation reaches only 0.5. The obvious scheme wastes a great deal.
  • Injecting a third harmonic of the input frequency and another of the output frequency — both zero-sequence, both invisible to a floating-neutral load — raises the limit to 0.866. This is the same trick as Chapter 18's, applied on both sides, and it is now standard.

Beyond that, no. Exceeding 0.866 would require producing an output larger than the instantaneous input can supply, which needs stored energy by definition. The limit is structural.

The only ways round it involve adding storage or a ratio: a step-up transformer at the input, or a hybrid topology with a small storage element — both of which forfeit the property the matrix converter existed to have.

Note the parallel with Chapter 19. A two-level inverter's linear region was the largest circle inside a hexagon. A matrix converter's is the largest circle inside the input voltage envelope. The same geometry, in a different space.

Why can a matrix converter not use dead time, when every inverter in Part 4 relies on it?

Because dead time works only if there is somewhere for the load current to go — and a matrix converter has nowhere.

How dead time works in a VSI. Turn both devices of a leg off, and the load current freewheels through one of the antiparallel diodes back into the DC link. The output voltage is briefly not commanded, but the current is never interrupted. Both safety rules are satisfied at once.

Why a matrix converter cannot do this. There is no DC link and no freewheeling path. If every switch feeding an output phase is turned off, the inductive load current has nowhere to flow, and \( v = L\,di/dt \) produces an unbounded voltage that destroys the switches within microseconds.

And overlap is equally forbidden. Turning the incoming switch on before the outgoing one is off connects two supply phases together through two switches — a line-to-line short limited only by the small input filter inductance.

So neither a gap nor an overlap is safe, and there is no simultaneous transition that avoids both.

The resolution: four-step commutation. Each bidirectional switch is two independently-controlled halves, each conducting one direction. Sequencing them separately means that at every step:

  • A path exists for the actual current direction — no open circuit.
  • No path exists that would let current flow between two inputs — no short.

The dependency this creates. The sequence depends on knowing the current's direction, and near a zero crossing that is exactly what noise and sensor offset make uncertain. Practical converters use hybrid schemes — current direction where the current is large, voltage direction where it is small — plus a clamp circuit to catch the energy if a commutation does fail.

A slight irony worth noting: that clamp contains a capacitor. The topology whose selling point is having no capacitor cannot be built without a small one.

A cycloconverter in blocking mode drives a mill at 5 Hz with a 4 ms dead time at each current reversal. Is that acceptable?

Probably not, and the arithmetic shows why quickly.

Put the dead time in context. At 5 Hz the output period is 200 ms, and the current passes through zero twice per cycle. So:

\[ \frac{2 \times 4\ \text{ms}}{200\ \text{ms}} = 4\% \text{ of the waveform held at zero} \]

What that does to the current waveform. A flat spot at each zero crossing turns a sine into something closer to a trapezoid with a notch. Symmetric flat spots produce predominantly third-harmonic distortion — at 15 Hz for a 5 Hz fundamental.

Why 15 Hz is the wrong frequency to have. A gearless mill drive's mechanical system — the ring gear, the mill shell, the foundations — has resonances in the single- and low-double-digit hertz range. Exciting one produces audible drumming, measurable vibration and fatigue in couplings.

And the effect is worst exactly where it matters. The current reversal happens when the torque changes direction, which on a reversing mill is during the controlled part of the cycle — so the disturbance lands where torque quality is most needed.

Three responses:

  1. Switch to circulating-current mode. Both bridges gated, current reverses instantly through the reactors, no dead time at all. This is why fast reversing mills use it despite the reactors and the extra loss.
  2. Reduce the dead time — but only so far. It is set by the certainty required that the outgoing bridge has stopped, plus \( t_q \), plus margin. Cutting it risks a line-to-line short through both bridges.
  3. Accept it if the drive is unidirectional and reversals are rare. A ball mill turning steadily one way crosses zero only at start and stop, so 4 ms twice a day is irrelevant.

The decision is entirely about how often the current reverses, which is a property of the application rather than the converter.

The matrix converter has been researched since the 1980s and is better on several counts. Why has it not displaced the standard drive?

Because of one incompatibility that outweighs several advantages.

The decisive problem: 0.866. A 400 V motor fed from a 400 V supply receives 346 V, so it runs at 86.6% flux and 86.6% torque. Fixing this requires a larger motor, a special 346 V machine, a higher supply, or a transformer — and every one of them turns a drive replacement into a system project. Industrial customers replace drives; they do not replace motors.

The founding argument has weakened. The topology was invented because electrolytic capacitors fail. Modern film capacitors have essentially unlimited life, and modern electrolytics are far better than 1980s parts. The problem has substantially solved itself.

No ride-through, which factories mind about. A DC link stores enough energy to survive supply dips of tens of milliseconds — common on real industrial networks. A matrix converter has no stored energy, so it trips. An unplanned process stop costs far more than a year of improved power factor.

Volume economics. Back-to-back drives are made in millions, so their devices and controllers are commodities. A matrix converter needs 36 devices, custom modules and bespoke FPGA commutation logic. That cost gap is a manufacturing-volume gap and does not close by itself.

Where it genuinely wins, and is sold:

  • Aerospace — mass and volume dominate, and −55 °C to +85 °C cycling destroys electrolytics.
  • Downhole and high-temperature applications where no capacitor survives at all.
  • 20-year maintenance-free requirements.
  • Regenerative drives that would need an active front end anyway, which erases much of the back-to-back cost advantage.

The general lesson. Superiority in several dimensions loses to a single hard constraint in another — especially when that constraint is compatibility with an installed base. This pattern recurs throughout power electronics, and recognising it is worth as much as any circuit analysis.

Both direct converters remove the DC link. Why is one used at 20 MW and the other at 50 kW, with nothing in between?

Because each survives only where a constraint that would defeat the standard drive happens to dominate — and those constraints sit at opposite ends of the scale.

At 20 MW, the constraint is device ratings.

  • No IGBT blocks 10 kV, so a back-to-back drive needs devices in series with all the timing difficulty that brings.
  • Thyristors handle 8 kV and 5 kA in one device, and a cycloconverter uses them naturally-commutated, so switching loss is negligible.
  • The load is a many-pole machine at 10 rpm, needing only a few hertz — so the \( f_s/3 \) limit is not a limit at all.
  • Four-quadrant operation comes free from the dual-converter structure.

At 50 kW in an aircraft, the constraint is mass and temperature.

  • A DC link capacitor bank is the largest and heaviest part of a back-to-back drive, and the least able to survive the thermal cycling.
  • The motor is purpose-designed, so the 0.866 limit costs nothing — it is simply wound for 346 V.
  • Regeneration is inherent, and an actuator regenerates constantly.
  • Near-unity input power factor matters on limited aircraft generation.

In between — from a few kilowatts to a few megawatts — no constraint dominates. Devices are available, standard motors must be driven at rated voltage, ride-through is valued, and volume pricing rules. The back-to-back drive wins every one of those, so it takes the entire middle of the market.

The pattern generalises, and it is worth carrying beyond this chapter: a specialised topology survives where a single constraint is extreme enough to override general-purpose economics. Ask which constraint is hardest before comparing converters on their merits.

Practice

Problems

Three habits for direct converter problems:

  1. Count the segments per output cycle before judging whether a cycloconverter suits an application.
  2. Check the voltage ratio against the motor's requirement before anything else in a matrix converter design.
  3. Ask what the load current does at a zero crossing. Both topologies have their hardest problem there.

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

  1. A cycloconverter produces 12 Hz from a 50 Hz supply using six-pulse bridges. Find the number of segments per output cycle and comment on the waveform quality.
  2. A 415 V, 50 Hz supply feeds a cycloconverter at \( r = 0.75 \). Find the peak and RMS fundamental output voltage, and the firing angle at output angles of 0°, 45°, 90° and 225°.
  3. A 36-pole synchronous motor is to run at 15 rpm. Find the required output frequency and state whether a six-pulse cycloconverter from a 50 Hz supply is suitable.
  4. A matrix converter is fed from 480 V. Find the maximum output voltage, and the largest standard motor voltage it can drive at rated flux.
  5. Show that a matrix converter needs 36 semiconductor devices, and explain why a bidirectional switch cannot be made from fewer than four.
  6. Explain why a cycloconverter produces components at \( |mf_s \pm nf_o| \), and identify which of these are most damaging to a mechanical drivetrain.
  7. A cycloconverter drive operates in blocking mode at 8 Hz with a 3 ms dead time. Estimate the fraction of the cycle flattened, name the dominant distortion harmonic and its frequency, and recommend whether to change mode.
  8. Compare a 12-pulse and a six-pulse cycloconverter on maximum output frequency, input harmonics and hardware cost. State when the extra cost is justified.
  9. An engineer proposes a matrix converter to replace a failed 55 kW drive on a standard 400 V motor. Analyse the proposal and recommend a course of action.
  10. Design the converter for a 20 MW gearless SAG mill drive: choose a topology, find the output frequency for a 48-pole machine at 9 rpm, count the segments per output cycle, and state what input compensation the installation will need.
  11. A matrix converter's commutation fails intermittently at low speed. Explain the likely mechanism, describe two commutation strategies that would reduce it, and state what protective hardware must be present regardless.
  12. Compare a cycloconverter and a modular multilevel converter for a new 15 MW mill drive on output frequency range, input power factor, harmonics, efficiency, part count and maintenance. Recommend one and identify the circumstances under which the other would win.