Chapter 17: Three-Phase Voltage Source Inverters — Power Electronics

Where this sits

Part 4 · DC–AC Converters
Chapter 17 of 30

You should already know

The full-bridge inverter, quasi-square harmonics and feedback-diode conduction from Chapter 16, and three-phase phasor conventions from Chapter 7.

By the end you can

Derive the six-step phase and line voltages from the switching states, explain why triplens vanish between lines, and choose between 180° and 120° conduction with reasons.

Time

≈ 55 min reading · ≈ 50 min problems

i What you'll learn

Why three legs on one DC link is not three separate inverters — the legs interact through the load's neutral.
How the eight switching states map to phase and line voltages, and why only seven are distinct.
The six-step waveform: phase voltage a staircase of \( \pm V_s/3, \pm 2V_s/3 \); line voltage a clean 120° quasi-square.
Why triplen harmonics cancel between lines, so the line voltage is cleaner than the phase voltage for free.
The harmonic rule \( n = 6k \pm 1 \) — no 3rd, no 9th, and why the 5th and 7th matter so much in drives.
180° versus 120° conduction: 15% more output against immunity from shoot-through.
What the DC link current looks like, and why the link capacitor is sized by ripple current rather than capacitance.

Section 17-1

Three Legs, One Link

A three-phase inverter is three half-bridge legs sharing a DC link, with each leg driving one motor terminal. That is the entire circuit — six transistors, six diodes, one capacitor.

It is tempting to treat it as three independent single-phase inverters. It is not, and the reason is the load's neutral point.

Three half-bridge legs sharing one DC link. Because the star point is not connected to the inverter, each phase voltage depends on all three leg states, not just its own.

◇Intuition · the see-saw with three seats

Three people sit on a rigid platform balanced on a central pivot that is free to move up and down. Each person can raise or lower their own seat. But the platform's height — the average — floats to wherever the three positions put it.

So how high any one person actually is depends on what the other two did.

That is exactly the three-phase inverter's neutral. Each leg pulls its terminal to either \( +V_s \) or 0 relative to the negative rail. The load's star point is connected to nothing, so it settles at whatever voltage makes the three phase currents sum to zero — which for a balanced load is the average of the three terminal voltages:

\[ v_{nN} = \frac{v_{aN} + v_{bN} + v_{cN}}{3} \]

And so each phase voltage is its own terminal minus that average:

\[ v_{an} = v_{aN} - \frac{v_{aN} + v_{bN} + v_{cN}}{3} \]

Two consequences follow immediately, and both are important:

The phase voltage is never simply \( V_s \) or 0. It takes the values \( \pm V_s/3 \) and \( \pm 2V_s/3 \) — a four-level staircase produced by legs that only have two levels each.
Anything common to all three legs disappears. If all three terminals move up together, so does the neutral, and the phase voltages do not change at all. This is the mechanism behind triplen cancellation in Section 17-4 and behind third-harmonic injection in Chapter 18 — and it is a genuinely free gift.

🔑

From switching states to voltages

\[ v_{an} = V_s\left(S_a - \frac{S_a + S_b + S_c}{3}\right), \qquad v_{ab} = V_s\,(S_a - S_b) \]

\( S_a, S_b, S_c \in \{0, 1\} \), where 1 means the leg's upper device is on. Note that the line voltage depends on only two legs and takes just three values — while the phase voltage depends on all three and takes four.

Section recap. Three legs share one link, and because the load's neutral floats, each phase voltage is its own terminal minus the average of all three. That makes a four-level phase voltage out of two-level legs — and makes anything common to all three legs invisible to the load.

Section 17-2

Eight States, Seven Vectors

Each leg has two states, so three legs give \( 2^3 = 8 \) combinations. Enumerating them once is worth the effort, because this table is the foundation of both six-step operation and the space vector modulation of Chapter 19.

All eight switching states, in units of \(V_s\)
State \(S_aS_bS_c\)	Phase \(v_{an}\)	Phase \(v_{bn}\)	Phase \(v_{cn}\)	Line \(v_{ab}\)	Name	Angle
0 0 0	0	0	0	0	\(V_0\) — zero	—
1 0 0	\(+2/3\)	\(-1/3\)	\(-1/3\)	\(+1\)	\(V_1\)	0°
1 1 0	\(+1/3\)	\(+1/3\)	\(-2/3\)	\(0\)	\(V_2\)	60°
0 1 0	\(-1/3\)	\(+2/3\)	\(-1/3\)	\(-1\)	\(V_3\)	120°
0 1 1	\(-2/3\)	\(+1/3\)	\(+1/3\)	\(-1\)	\(V_4\)	180°
0 0 1	\(-1/3\)	\(-1/3\)	\(+2/3\)	\(0\)	\(V_5\)	240°
1 0 1	\(+1/3\)	\(-2/3\)	\(+1/3\)	\(+1\)	\(V_6\)	300°
1 1 1	0	0	0	0	\(V_7\) — zero	—

◈Why two of the eight states do nothing, and why that is useful

States 000 and 111 produce zero volts across every phase. In 000 all three terminals sit at the negative rail; in 111 all three sit at the positive rail. Either way all three are at the same potential, so no phase sees any voltage.

Eight states, six of which do something, two of which do nothing. Hence six active vectors and one zero vector, reachable two ways.

Zero states are not wasted, they are essential:

They set the output amplitude. Spending part of each interval in a zero state reduces the average voltage — exactly as the zero level did for the single-phase bridge in Chapter 16. Without them the inverter could only produce full output.
They give the load current somewhere to go. In 111 the current circulates through the three upper devices without touching the DC link at all — a genuine freewheeling path.
Having two of them is a free choice. 000 and 111 are identical at the load but use different devices, so alternating between them shares the conduction duty and equalises junction temperatures.

And the choice matters more than that. Moving from state 100 to 000 requires switching only leg a; moving from 100 to 111 requires switching legs b and c. A modulator that always picks the nearer zero state halves the number of switchings, and therefore the switching loss. Choosing well between them is one of the main things space vector modulation does better than naive carrier PWM — Chapter 19.

One more observation worth carrying forward. Look at the "Angle" column: the six active states are evenly spaced 60° apart. Plotted on a plane they form a regular hexagon, with the zero states at its centre. Any desired output can be synthesised by spending appropriate fractions of each interval on the two adjacent corners and the centre. That single picture is space vector modulation, and it is worth having in mind before Chapter 19 develops it.

Section recap. Eight switching states give six active voltage vectors, evenly spaced 60° apart, plus a zero vector reachable two ways. The zero states set the amplitude, freewheel the current, and offer a choice that determines how often the devices switch.

Section 17-3

Six-Step Operation

The simplest way to run the inverter is to visit the six active states in order, spending 60° in each. This is six-step or 180° conduction mode — so called because each device is on for a continuous 180° of the cycle.

Reading down the table of Section 17-2 in the order \( V_1 \to V_2 \to V_3 \to V_4 \to V_5 \to V_6 \) gives the two waveforms that define the mode.

Interactive · switching states to waveforms

Step round one electrical cycle. The bar shows which leg is high; the traces show the resulting phase voltage \( v_{an} \) and line voltage \( v_{ab} \). Notice that the phase voltage has four levels and the line voltage only three.

Electrical angle ωt 30°

State S_aS_bS_c100

VectorV₁

Phase v_an+2/3 V_s

Line v_ab+1 V_s

—

🔑

Six-step (180° conduction) output

\[ V_{ph(rms)} = \frac{\sqrt2}{3}V_s = 0.471\,V_s, \qquad V_{L(rms)} = \sqrt{\frac{2}{3}}\,V_s = 0.816\,V_s \] \[ V_{ph1(rms)} = \frac{\sqrt2}{\pi}V_s = 0.450\,V_s, \qquad V_{L1(rms)} = \frac{\sqrt6}{\pi}V_s = 0.780\,V_s \]

The line voltage is a quasi-square wave of exactly 120° conduction — which is the \( \alpha = 30^\circ \) case of Chapter 16, with its THD of 31.1% and no triplen harmonics at all. The inverter gets that for free, simply from being three-phase.

▲Common mistake

"180° conduction means all three upper devices could be on at once — that is a shoot-through waiting to happen."

Wrong Confusing "three upper devices on together" with "both devices of one leg on together".

Right State 111 — all three upper devices on — is a legitimate zero state. Shoot-through means \( S_1 \) and \( S_4 \) on together, which is a fault in a single leg.

The distinction. Each leg's two devices are complementary — exactly one should be on. There are three legs, so three independent complementary pairs. Whether the three legs agree with each other is a matter of what output you want; whether a leg agrees with itself is a matter of survival.

Why 180° conduction is nevertheless the riskier mode. In 180° operation a leg's upper device turns off at the same instant its lower device turns on, so there is a transition to protect at every commutation. Dead time is mandatory, and its cost is the distortion described in Chapter 16.

In 120° conduction there is a built-in 60° gap between one device turning off and its complement turning on. Shoot-through is impossible by construction, not by timing. That is genuinely valuable, and it is why the mode survived the era of slow, hard-to-turn-off devices.

What it costs. During the gap, that leg's terminal is connected to neither rail — its voltage is set by the load's back-EMF, not by the inverter. For a motor that is tolerable and even useful (it is how a sensorless BLDC drive detects rotor position). For a general load it means the output is no longer fully controlled.

Section recap. Visiting the six active vectors in turn for 60° each gives a four-level phase voltage and a three-level line voltage. The line voltage is a 120° quasi-square wave — 0.816 \( V_s \) RMS, 31.1% THD, and free of triplens by construction.

Section 17-4

Where the Triplen Harmonics Went

The phase voltage is a four-level staircase and the line voltage is a three-level quasi-square. The line voltage is visibly cleaner. The reason is one of the more elegant results in power electronics, and it is worth understanding properly rather than accepting as a rule.

Write the three phase voltages with their phase displacement Each phase's \( n \)th harmonic is displaced by \( n \times 120^\circ \) from the next, because the fundamental waveforms are displaced by 120°.

Working

\[ v_{an,n} = V_n\sin n\theta, \quad v_{bn,n} = V_n\sin n(\theta - 120^\circ), \quad v_{cn,n} = V_n\sin n(\theta - 240^\circ) \]
Ask what happens when \(n\) is a multiple of 3 For \( n = 3, 9, 15, \ldots \), the displacement \( n \times 120^\circ \) is a whole multiple of 360°:

Working

\[ n = 3: \quad 3 \times 120^\circ = 360^\circ \equiv 0^\circ \]

So the third harmonic is identical in all three phases — in phase, not displaced.
Take the difference to get the line voltage

Result

\[ v_{ab,3} = v_{an,3} - v_{bn,3} = V_3\sin3\theta - V_3\sin3\theta = 0 \]

The triplen harmonics subtract to exactly zero between any two lines.
Confirm what survives For \( n \) not a multiple of 3, the displacement is a genuine 120° or 240°, so the difference is non-zero — and scaled by \( \sqrt3 \), like the fundamental. Combined with the even-harmonic cancellation from half-wave symmetry, only \( n = 6k \pm 1 \) remains: 5, 7, 11, 13, 17, 19, …

🔑

The three-phase harmonic rule

\[ n = 6k \pm 1 \quad (k = 1, 2, 3, \ldots) \;\Longrightarrow\; n = 5, 7, 11, 13, 17, 19, \ldots \]

No even harmonics (half-wave symmetry), no triplens (three-phase cancellation). Whatever the modulation scheme, a three-phase inverter with an isolated neutral produces only these.

◈Why the 5th and 7th matter so much more than their size suggests

In a six-step waveform the 5th is 20% of the fundamental and the 7th is 14% — not large. Yet drive engineers spend a great deal of effort on them, and the reason is not their amplitude but their sequence.

Work out the phase sequence of each harmonic. Phase b's \( n \)th harmonic lags phase a's by \( n \times 120^\circ \):

Fundamental (\( n = 1 \)): 120° — positive sequence. Rotates forwards.
5th: \( 5 \times 120^\circ = 600^\circ \equiv 240^\circ \) — negative sequence. Rotates backwards.
7th: \( 7 \times 120^\circ = 840^\circ \equiv 120^\circ \) — positive sequence. Rotates forwards.
11th: negative. 13th: positive. And so on, alternating.

What a backwards-rotating field does in a motor:

It produces a braking torque, opposing the useful one. Small, but it is pure loss.
It induces rotor currents at high slip, so the rotor heats up while producing nothing.
Interacting with the fundamental field, it makes a torque pulsation at \( 6f_1 \) — 300 Hz on a 50 Hz supply.

And the 7th does the same thing at the same frequency. The 5th (backwards) and 7th (forwards) both beat against the fundamental at \( 6f_1 \), so their torque ripples add. This is why a six-step drive has a characteristic 6th-harmonic torque ripple — audible as a hum, felt as vibration, and hard on gearboxes and couplings.

Why it is worst at low speed. At 5 Hz fundamental the ripple is at 30 Hz — right in the mechanical resonance range of many drivetrains, and slow enough that the rotor's inertia cannot smooth it. Six-step drives run visibly roughly below about 10% speed, which is the main reason PWM displaced them.

The one consolation. Because harmonic currents fall as \( 1/n^2 \) in an inductive load (Chapter 16), a 20% 5th-harmonic voltage produces only about a 4% 5th-harmonic current. The torque ripple is real but far smaller than the voltage spectrum alone would suggest.

Section recap. Triplen harmonics are in phase in all three legs, so they cancel between lines. Only \( n = 6k \pm 1 \) survives. The 5th is negative-sequence and the 7th positive, and both beat with the fundamental to produce a 6th-harmonic torque ripple that is worst at low speed.

Section 17-5

180° or 120° Conduction

Six-step operation runs each device for 180° of the cycle. The alternative is to run each for 120°, leaving a 60° gap before its complement turns on. Both modes are used, for quite different reasons.

The two conduction modes compared
Property	180° conduction	120° conduction
Devices on at once	3 (one per leg)	2
Fundamental phase voltage	\(0.450\,V_s\)	\(0.390\,V_s\)
Output relative to 180°	100%	86.6%
Phase voltage shape	Four-level staircase	Quasi-square, \(\pm V_s/2\)
Shoot-through risk	Present — needs dead time	Impossible by construction
Terminal voltage during the gap	Always defined	Set by the load's back-EMF
Device utilisation	Better	Each device idle a third of the time
Typical use	Nearly all AC drives, PWM inverters	BLDC drives, older thyristor inverters

▣In practice · why BLDC motors want 120° conduction

A brushless DC motor sounds like it should be driven with a sinusoid. Most are not — they are driven with 120° conduction, and the reason is a genuinely elegant piece of matching between the machine and the converter.

1. The back-EMF is trapezoidal, not sinusoidal. A BLDC's windings are concentrated rather than distributed, so each phase's back-EMF has a flat top lasting about 120° of the electrical cycle. Torque is \( T = \sum e_k i_k/\omega \), so the smoothest torque comes from driving constant current during the flat top — which is precisely a 120° conduction interval.

2. The idle phase is a free sensor. During its 60° gap, one phase is disconnected from both rails. Its terminal voltage is then pure back-EMF, and the instant that EMF crosses zero locates the rotor to within 30° electrical.

So a sensorless drive can find the rotor with no encoder, no Hall sensors and no extra wiring — just a voltage comparator on each terminal. This is why a computer fan, which cannot justify a position sensor, still commutates correctly.

3. Only two phases carry current at a time, so conduction loss is lower than in a three-phase-on scheme of the same torque, and the current sensing is simpler — a single shunt in the DC link sees the whole phase current.

The limitations, stated fairly:

Commutation torque ripple. Every 60° the current must transfer between phases, and it cannot do so instantly through the winding inductance. The result is a dip in torque six times per cycle — typically 5–15%, and the main reason high-performance servos use sinusoidal drive instead.
Zero-crossing detection fails at standstill, since there is no back-EMF. Starting requires open-loop alignment or a high-frequency injection scheme.
It suits trapezoidal machines only. Driving a genuinely sinusoidal PMSM this way produces significant torque ripple, which is why the two machine types are driven differently even though they look identical from outside.

Chapter 27 develops BLDC and PMSM drives properly; the point here is that the choice of conduction mode is dictated by the machine, not by the converter.

Section recap. 180° conduction gives 15.5% more fundamental and uses the devices better, at the price of needing dead time. 120° conduction cannot shoot through and leaves one phase free for back-EMF sensing — which is exactly what a trapezoidal BLDC machine wants.

Section 17-6

What the DC Link Sees

Everything so far has looked outward, at the load. Looking inward, at the DC link, reveals the component that most often determines an inverter's size, cost and lifetime.

The link current is whatever the three legs demand: \( i_{dc} = S_a i_a + S_b i_b + S_c i_c \). Because the \( S \) values are switching functions, this is a chopped combination of three sinusoids — and its shape is nothing like DC.

◇Intuition · three taps on one tank

A water tank feeds three taps that open and close in rotation. The tank's outflow at any instant is the sum of whichever taps are open — steady on average, but jumping every time a tap changes state.

The DC link is that tank, and the capacitor's job is to absorb the jumps so the source upstream sees something smooth.

Two distinct ripple components have to be absorbed, and they live at different frequencies:

Switching-frequency ripple. Every commutation redirects current between legs, producing a step in the link current. In a PWM inverter this is at the carrier frequency — kilohertz — where a small capacitor is effective.
Six-times-fundamental ripple. Even in ideal six-step operation the link current has a genuine 6th-harmonic component, because the six sectors are not identical from the link's point of view. At 50 Hz that is 300 Hz — much harder to filter, and it is what sizes the capacitor.

And in a real drive there is a third. If the inverter is fed by a diode rectifier, the link also carries the rectifier's own 300 Hz ripple. The two 300 Hz components beat with each other, so the link voltage can develop a low-frequency oscillation that no amount of capacitance fully removes.

Why capacitance is not what you specify. The link capacitor's RMS ripple current is roughly \( 0.5\,I_{phase} \) for a typical drive — for a 100 A machine, about 50 A of continuous ripple. That demands a physically large capacitor bank whose ripple rating, not its capacitance, sets its size. This is why the capacitor is usually the largest single component in a drive cabinet.

✦Think about it

A drive's DC link capacitor fails after two years in a hot factory, well short of its rated life. The inverter is correctly sized and the machine is not overloaded. What is likely to be going on?

Four candidates, and how to tell them apart

1. Ripple-current heating — the most likely. An electrolytic capacitor's life halves for roughly every 10 °C rise in core temperature. The core is heated by \( I_{ripple}^2 R_{ESR} \) from the inside, where it cannot easily escape. A capacitor rated for 5,000 hours at 105 °C might give 20,000 hours at 85 °C — or 2,500 at 115 °C. Check the actual ripple current against the rating at the operating frequency, since ratings are quoted at 100 or 120 Hz and must be derated.

2. Ambient temperature. A cabinet at 55 °C with a capacitor dissipating 15 W internally can easily reach 90 °C at the core. Measure the can's surface temperature under load, not the cabinet air.

3. Beating between the rectifier and inverter ripple. If the mains is 50 Hz and the motor runs at 50 Hz, both produce 300 Hz link ripple. At other speeds the two frequencies differ slightly and beat, producing a large low-frequency envelope on the link voltage — and a much higher RMS ripple current than the nameplate calculation predicts. Look for a slow amplitude modulation on the link voltage.

4. Light-load operation. Counter-intuitively, a lightly loaded drive can be worse: the power factor collapses, so more current circulates without doing work, and the ripple does not fall in proportion to the useful load.

What to do:

Parallel more capacitors. The ripple current divides, and \( I^2R \) heating divides as the square — so doubling the count quarters the heating per unit.
Add a DC link choke. It reduces the rectifier-side ripple substantially and improves the input power factor at the same time.
Move to film capacitors. Much better ripple rating and a far longer life, but larger and more expensive per microfarad. Standard in traction and increasingly in industrial drives.
Improve the cooling. Often the cheapest fix, and frequently overlooked because the capacitor is not the component anyone expects to be running hot.

The general lesson. In a converter, the component that fails first is rarely the one carrying the most power. It is the one whose lifetime is most sensitive to a stress that was estimated rather than measured.

Section recap. The DC link current is a chopped sum of the three phase currents, with ripple at the switching frequency and at six times the fundamental. The link capacitor is selected on RMS ripple current, not capacitance, and its temperature governs the drive's service life.

Section 17-7

Worked Examples

1 Six-step inverter feeding a motor

Problem. A three-phase six-step inverter runs from a 540 V DC link into a star-connected load of \( R = 8\ \Omega \), \( L = 30 \) mH per phase at 50 Hz. Find the line and phase voltages, the fundamental current, and the 5th and 7th harmonic currents.

Voltages

\[ V_{L(rms)} = 0.816(540) = 441\ \text{V}, \qquad V_{ph(rms)} = 0.471(540) = 255\ \text{V} \]

\[ V_{ph1} = 0.450(540) = 243\ \text{V}, \qquad V_{ph5} = \frac{243}{5} = 48.6\ \text{V}, \qquad V_{ph7} = \frac{243}{7} = 34.7\ \text{V} \]

Impedances

\[ \omega L = 2\pi(50)(0.03) = 9.42\ \Omega \]

\[ Z_1 = \sqrt{64 + 88.8} = 12.4\ \Omega, \quad Z_5 = \sqrt{64 + 2220} = 47.8\ \Omega, \quad Z_7 = \sqrt{64 + 4349} = 66.4\ \Omega \]

Currents

\[ I_1 = \frac{243}{12.4} = 19.6\ \text{A}, \qquad I_5 = \frac{48.6}{47.8} = 1.02\ \text{A}, \qquad I_7 = \frac{34.7}{66.4} = 0.52\ \text{A} \]

Compare the two distortion figures. The phase voltage THD is about 31% for the line and higher for the phase, but the current THD is

\[ \text{THD}_i \approx \frac{\sqrt{1.02^2 + 0.52^2 + \cdots}}{19.6} \approx 6\% \]

The load has done most of the filtering, exactly as in Chapter 16. A six-step drive therefore produces reasonably smooth current despite a visibly poor voltage waveform — which is why the mode was viable for decades before PWM became affordable.

2 Sizing the DC link from an AC requirement

Problem. A 400 V, 50 Hz induction motor must be driven at rated voltage by a six-step inverter. What DC link voltage is needed, and what does a standard three-phase rectifier from a 400 V supply actually provide?

Required link, six-step

\[ V_{L1} = 0.780\,V_s \;\Longrightarrow\; V_s = \frac{400}{0.780} = 513\ \text{V} \]

Available from a diode bridge (Chapter 7)

\[ V_{dc} = 1.35\,V_L = 1.35(400) = 540\ \text{V} \]

So six-step operation works comfortably — 540 V available against 513 V needed, a 5% margin.

Now ask the same question for a PWM inverter, which is what would actually be built:

Sinusoidal PWM, linear region (Chapter 18)

\[ V_{L1(max)} = \frac{\sqrt3}{2\sqrt2}V_s = 0.612\,V_s \;\Longrightarrow\; V_s = \frac{400}{0.612} = 653\ \text{V} \]

653 V needed against 540 V available — a 21% shortfall. Plain sinusoidal PWM cannot produce rated motor voltage from a standard rectifier, which is a genuinely awkward result and one that catches people out.

How real drives resolve it, and both methods are standard:

Third-harmonic injection or space vector modulation, which raise the limit to \( 0.707V_s \) — giving 382 V from a 540 V link, close enough that a small rectifier margin covers it. This is the usual answer, and Chapters 18 and 19 explain why adding a harmonic increases the usable output.
Over-modulation at the top of the speed range, accepting some low-order harmonics in exchange for amplitude — converging on six-step at maximum speed.

This example is worth remembering, because it explains why every serious drive uses SVM rather than plain sinusoidal PWM: not for elegance, but because plain SPWM simply cannot reach rated voltage.

3 Device ratings for a three-phase inverter

Problem. A 30 kW drive runs from a 540 V link at 0.85 power factor. Find the phase current, the device peak current, and the ratings for the transistors and diodes. Compare with a single-phase inverter of the same power.

Load current

\[ S = \frac{30\,000}{0.85} = 35.3\ \text{kVA}, \qquad I_{ph} = \frac{S}{\sqrt3\,V_L} = \frac{35\,300}{\sqrt3(441)} = 46.2\ \text{A} \]

\[ I_{pk} = \sqrt2(46.2) = 65.4\ \text{A} \]

Conduction split

\[ \varphi = \cos^{-1}(0.85) = 31.8^\circ \;\Longrightarrow\; \text{diodes conduct } \frac{31.8}{180} = 17.7\% \text{ of each cycle} \]

Ratings. Blocking voltage is the full link plus overshoot — specify 1200 V devices for a 540 V link, which is standard practice. Peak current 65.4 A, so a 100 A module gives sensible margin.

Now the comparison that matters. A single-phase inverter delivering the same 30 kW from the same link:

Single-phase equivalent

\[ V_1 = 0.900(540) = 486\ \text{V}, \qquad I = \frac{35\,300}{486} = 72.6\ \text{A}, \qquad I_{pk} = 103\ \text{A} \]

The three-phase inverter needs six devices at 65 A; the single-phase needs four at 103 A. Total device current rating is 392 A against 412 A — almost identical. The three-phase version is not cheaper in silicon.

Where it wins is everywhere else:

Constant instantaneous power. A balanced three-phase load draws steady power, so the DC link sees no 2nd-harmonic ripple. A single-phase inverter's power pulsates at \( 2f_1 \), forcing a link capacitor several times larger.
Smaller devices individually, so cheaper packages, easier gate drive and simpler thermal management.
Better waveform for free — no triplens, as Section 17-4 showed.
It matches the machine. Three-phase motors dominate above a few kilowatts, so the alternative barely exists.

4 Reading a state sequence back into a waveform

Problem. An inverter is observed cycling through the states 100, 110, 010, 011, 001, 101 with equal dwell times. Identify the mode, find \( v_{an} \) and \( v_{ab} \) in each interval, and determine the phase sequence.

Step 1 — identify the mode. Six active states in sequence with equal dwell and no zero states is six-step, 180° conduction. Each state lasts 60°.

Applying \(v_{an} = V_s(S_a - \bar S)\) and \(v_{ab} = V_s(S_a - S_b)\)

\[ \begin{array}{lccc} \text{State} & \bar S & v_{an}/V_s & v_{ab}/V_s \\ 100 & 1/3 & +2/3 & +1 \\ 110 & 2/3 & +1/3 & \phantom{+}0 \\ 010 & 1/3 & -1/3 & -1 \\ 011 & 2/3 & -2/3 & -1 \\ 001 & 1/3 & -1/3 & \phantom{+}0 \\ 101 & 2/3 & +1/3 & +1 \end{array} \]

Step 2 — read the phase voltage. \( +2/3, +1/3, -1/3, -2/3, -1/3, +1/3 \) — the classic six-step staircase, positive for 180° and negative for 180°.

Step 3 — read the line voltage. \( +1, 0, -1, -1, 0, +1 \). Grouping across the cycle boundary, that is \( +1 \) for 120°, \( 0 \) for 60°, \( -1 \) for 120°, \( 0 \) for 60° — a 120° quasi-square wave, as Section 17-3 predicted.

Step 4 — the phase sequence. Leg \( a \) goes high at 0°, leg \( b \) at 60°, leg \( c \) at 180°... but the sequence to check is when each phase voltage peaks. \( v_{an} \) peaks in state 100 (0°–60°), \( v_{bn} \) peaks in 010 (120°–180°), \( v_{cn} \) peaks in 001 (240°–300°). So the order is a, b, c — positive sequence, and the motor turns forwards.

To reverse it, run the state sequence backwards: 100, 101, 001, 011, 010, 110. No hardware change, no contactor — a software decision. This is one of the quiet advantages of an inverter drive over a direct-on-line starter, and it costs nothing.

Section 17-8

Summary & Formula Sheet

Chapter 17 in five sentences:

Three legs share one DC link, and because the load's neutral floats, each phase voltage is its own terminal minus the average of all three.
Eight switching states give six active vectors 60° apart plus a zero vector reachable two ways — the picture Chapter 19 builds on.
Six-step operation makes a four-level phase voltage and a three-level line voltage of 0.816 \( V_s \) RMS with 31.1% THD.
Triplen harmonics are in phase in all three legs and cancel between lines, leaving only \( n = 6k \pm 1 \).
180° conduction gives 15.5% more output; 120° conduction cannot shoot through and leaves a phase free for back-EMF sensing.

Formula sheet · Chapter 17

Phase voltage from states\(S \in \{0,1\}\), floating neutral

\( v_{an} = V_s\Bigl(S_a - \dfrac{S_a+S_b+S_c}{3}\Bigr) \)

Line voltage from statesonly two legs involved

\( v_{ab} = V_s(S_a - S_b) \)

Six-step phase RMSfour-level staircase

\( V_{ph} = \dfrac{\sqrt2}{3}V_s = 0.471\,V_s \)

Six-step line RMS120° quasi-square

\( V_L = \sqrt{\dfrac{2}{3}}\,V_s = 0.816\,V_s \)

Fundamental phasesix-step

\( V_{ph1} = \dfrac{\sqrt2}{\pi}V_s = 0.450\,V_s \)

Fundamental linesix-step; sizes the DC link

\( V_{L1} = \dfrac{\sqrt6}{\pi}V_s = 0.780\,V_s \)

Harmonic amplitudesrelative to the fundamental

\( \dfrac{V_n}{V_1} = \dfrac{1}{n}, \quad n = 6k \pm 1 \)

Line voltage THDsix-step, before filtering

\( \text{THD} = 31.1\% \)

120° conduction, fundamental86.6% of 180° mode

\( V_{ph1} = \dfrac{\sqrt3}{\pi\sqrt2}V_s = 0.390\,V_s \)

Harmonic current, RL loadfalls as \(1/n^2\)

\( I_n = \dfrac{V_n}{\sqrt{R^2 + (n\omega L)^2}} \)

DC link currentchopped sum; 6th-harmonic ripple

\( i_{dc} = S_a i_a + S_b i_b + S_c i_c \)

Diode conduction fractionper device, per cycle

\( \dfrac{\varphi}{\pi} \)

Key terms

Six-step operation: Visiting the six active switching vectors in turn for 60° each. Also called 180° conduction.
Switching state: The triple \(S_aS_bS_c\) recording which leg has its upper device on. Eight exist; six are active.
Zero state / zero vector: States 000 and 111, in which all three terminals share a rail so no phase sees voltage.
Floating neutral: A star point not connected to the inverter. Its potential is the average of the three terminal voltages.
Triplen harmonics: Multiples of three. In phase in all three legs, so they cancel between lines and never reach the load.
Positive / negative sequence: Whether a harmonic's field rotates with or against the fundamental. 5th is negative, 7th positive, alternating thereafter.
Sixth-harmonic torque ripple: The pulsation produced when the 5th and 7th harmonics beat against the fundamental. Worst at low speed.
120° conduction: Each device on for 120° with a 60° gap before its complement. Shoot-through impossible; suits trapezoidal BLDC machines.
DC link capacitor: The energy buffer between rectifier and inverter. Selected on RMS ripple current; usually the life-limiting component.
DC link utilisation: Fundamental output per volt of link. Six-step 0.780; SVM 0.707; sinusoidal PWM 0.612.

Check yourself

Test Yourself

Chapter 17 · six questions answers hidden until you ask

Each leg produces only two voltage levels. Where do the phase voltage's four levels come from?

From the floating neutral, which makes each phase voltage depend on all three legs rather than one.

The mechanism. With no connection between the load's star point and the inverter, the star point settles wherever the three phase currents sum to zero — for a balanced load, at the average of the three terminal voltages. So

\[ v_{an} = V_s\left(S_a - \frac{S_a + S_b + S_c}{3}\right) \]

Enumerate what that expression can produce. \( S_a \) is 0 or 1, and the sum \( S_a + S_b + S_c \) is 0, 1, 2 or 3:

\( S_a = 1 \), sum = 1 (only phase a high): \( 1 - 1/3 = +2/3 \)
\( S_a = 1 \), sum = 2 (a and one other high): \( 1 - 2/3 = +1/3 \)
\( S_a = 0 \), sum = 1 (a low, one other high): \( 0 - 1/3 = -1/3 \)
\( S_a = 0 \), sum = 2 (a low, two others high): \( 0 - 2/3 = -2/3 \)
Sum = 0 or 3 (all agree): \( 0 \) — the zero states

Four non-zero levels plus zero, from legs that have two. The extra levels come from the relationships between legs, not from any leg's own capability.

Test the claim by connecting the neutral. Tie the star point to the DC link's mid-point and \( v_{an} \) becomes simply \( \pm V_s/2 \) — back to two levels, and each phase is now an independent single-phase half-bridge. The four-level behaviour vanishes, and with it the triplen cancellation.

Which is why drives never connect the neutral. Doing so would forfeit the extra levels, forfeit the triplen cancellation, forfeit the ability to inject a third harmonic for extra DC link utilisation (Chapter 18), and add a wire that would carry substantial triplen current. It is a rare case where leaving something unconnected is the whole design.

Why is the line voltage cleaner than the phase voltage, when it is made from the same switching?

Because subtraction removes everything the two phases have in common — and the triplen harmonics are entirely common.

The phase voltage is a four-level staircase containing the fundamental, the 5th, 7th, 11th, 13th and the triplens (3rd, 9th, 15th). It looks stepped and irregular.

The line voltage is \( v_{an} - v_{bn} \). For any harmonic displaced by \( n \times 120^\circ \):

If \( n \) is a multiple of 3, the displacement is a whole number of full turns, so the two are identical and subtract to zero.
Otherwise the displacement is a real 120° or 240°, and the difference is \( \sqrt3 \) times either — the same factor as the fundamental.

So the line voltage keeps the useful part and discards the triplens, arriving at a clean three-level 120° quasi-square wave with 31.1% THD against the phase voltage's higher figure.

An important consequence for measurement. If you probe a drive's output phase-to-neutral using the inverter's negative rail as your reference, you will see a two-level square wave with an enormous common-mode component — nothing like what the motor experiences. The motor sees phase-to-its own floating star point, which no oscilloscope terminal is connected to. Always measure line-to-line on a drive output; phase measurements referenced to the DC link are misleading.

And where the triplens actually go. They do not disappear — they appear between the load's star point and the DC link, as a common-mode voltage stepping at the switching frequency. That common-mode voltage drives current through the motor's parasitic capacitance to its frame, and thence through the bearings. Bearing currents are a well-known cause of premature motor failure in inverter drives, and they are exactly the triplens that the line voltage was so pleased to be rid of.

A standard 400 V drive uses a 540 V DC link. Why can plain sinusoidal PWM not produce 400 V from it, when six-step operation can?

Because sinusoidal PWM confines each leg to a sine wave that must fit inside the link voltage, while six-step lets each leg swing fully square.

The two limits.

Six-step: \( V_{L1} = (\sqrt6/\pi)V_s = 0.780V_s \). From 540 V, that is 421 V — more than enough.
Sinusoidal PWM, linear region: each leg's reference is a sine of peak at most \( V_s/2 \) about the mid-point, giving \( V_{L1} = (\sqrt3/2\sqrt2)V_s = 0.612V_s \). From 540 V, that is 331 V — a 17% shortfall.

Why square wave beats sine here. A square wave's fundamental is \( 4/\pi = 1.27 \) times its own amplitude, because Fourier packs more fundamental into a square than into a sine of the same peak. SPWM deliberately throws that away in exchange for a clean spectrum.

The two standard resolutions:

Third-harmonic injection or SVM. Add a triplen component common to all three legs. It flattens each leg's reference so a larger fundamental fits inside the same link — and because it is common to all three, the load never sees it (Section 17-4). The limit rises to \( 0.707V_s \), giving 382 V from 540 V. Essentially free, which is why every commercial drive does it. Chapters 18 and 19.
Over-modulation. Push the reference beyond the linear region, accepting some low-order harmonics for amplitude, converging on six-step at the extreme. Used at the top of the speed range, where the motor is field-weakened and waveform quality matters less.

Why this is worth remembering. It is the practical reason SVM displaced carrier PWM in drives — not elegance, but the 15% of output voltage that plain SPWM cannot reach. A drive built with naive SPWM would be unable to run a standard motor at rated speed.

Why do the 5th and 7th harmonics both produce torque ripple at six times the fundamental, when they are different frequencies?

Because torque comes from the interaction of two fields, and both harmonics are six times the fundamental away from it — one below, one above.

Set up the rotation. The fundamental field rotates forwards at \( \omega_1 \). Working out each harmonic's sequence from its \( n \times 120^\circ \) displacement:

5th: \( 5 \times 120^\circ = 600^\circ \equiv 240^\circ \) — negative sequence, so it rotates backwards at \( 5\omega_1 \).
7th: \( 7 \times 120^\circ = 840^\circ \equiv 120^\circ \) — positive sequence, rotating forwards at \( 7\omega_1 \).

Now take the relative speeds seen from the fundamental field:

5th relative to fundamental: \( -5\omega_1 - \omega_1 = -6\omega_1 \).
7th relative to fundamental: \( +7\omega_1 - \omega_1 = +6\omega_1 \).

Both are \( 6\omega_1 \) — one leading, one lagging. Torque is proportional to the product of the two fields, and a product of components separated by \( 6\omega_1 \) pulsates at exactly that frequency. So both contribute a 6th-harmonic torque ripple, and they add.

The pattern continues. The 11th (negative) and 13th (positive) both sit \( 12\omega_1 \) from the fundamental, giving a 12th-harmonic ripple; the 17th and 19th give an 18th. This is why the ripple spectrum of any three-phase drive is at multiples of six — a direct consequence of the \( 6k \pm 1 \) rule.

Why it matters at low speed. At 5 Hz fundamental the ripple is at 30 Hz, which is slow enough that the rotor's inertia cannot smooth it and squarely inside the mechanical resonance band of many drivetrains. At 50 Hz the ripple is at 300 Hz, where inertia filters it almost completely. Six-step drives are therefore acceptable at speed and unacceptable near standstill — which is precisely the operating region most modern applications care about.

State 000 and state 111 both produce zero output. Does the choice between them matter?

Electrically at the load, no. In every other respect, yes — and modulators exploit the choice deliberately.

1. Switching count, which is the biggest effect. Getting to a zero state from an active state requires switching however many legs disagree:

From 100 to 000: only leg a changes. One switching.
From 100 to 111: legs b and c change. Two switchings.

Always choosing the nearer zero state roughly halves the switching loss, for no cost. This is the single largest advantage of space vector modulation over naive carrier PWM.

2. Device thermal balance. In 111 the load current freewheels through the three upper devices; in 000 through the three lower ones. Always choosing one would leave half the module hotter than the other. Alternating shares the duty.

3. Common-mode voltage. The star point relative to the negative rail sits at 0 in state 000 and at \( V_s \) in state 111. Each transition between them is a full-amplitude common-mode step — which drives current through the motor's stray capacitance to the frame and thence through the bearings. Some modulation schemes deliberately restrict which zero states they use to reduce common-mode stepping, at the cost of more switchings. Trading bearing life against efficiency is a real decision in large drives.

4. Neutral-point balance in multilevel converters. In a three-level NPC inverter (Chapter 20), the redundant states charge and discharge the mid-point capacitors differently, so the choice becomes an active control input rather than an optimisation.

The general point. Redundancy in a switching table is not waste. It is a free degree of freedom, and good modulators spend it on switching loss, thermal balance or common-mode voltage — whichever the application is short of.

A drive's DC link capacitor runs hot even though the inverter and motor are correctly sized. Why is the capacitor the component under stress?

Because it absorbs the difference between two currents that neither the rectifier nor the inverter wants to smooth — and it does so continuously, from the inside.

What it is absorbing. The rectifier supplies current in six pulses per mains cycle; the inverter draws current in a pattern set by the switching states and the motor current. The capacitor takes the difference, which contains:

Switching-frequency ripple from the inverter's commutations — kilohertz, and largely handled by a small film capacitor in parallel.
Six-times-fundamental ripple from the inverter — 300 Hz at 50 Hz output, and much harder to filter.
Six-times-mains ripple from the rectifier — also 300 Hz at 50 Hz mains.

The two 300 Hz components beat whenever the motor is not running at exactly mains frequency, producing a low-frequency envelope that raises the RMS ripple well above a naive calculation.

Why heating is the failure mode. The loss is \( I_{ripple}^2 R_{ESR} \), dissipated inside a sealed aluminium can with poor internal conduction. The core runs far hotter than the case, the electrolyte dries out, ESR rises, and the heating accelerates. Life halves for roughly every 10 °C.

Typical magnitude. For a drive delivering \( I_{phase} \), the link ripple current is roughly \( 0.5\,I_{phase} \) — about 25 A RMS for a 50 A machine, continuously.

The remedies, and what each actually buys:

Parallel more capacitors. Ripple divides between them and \( I^2R \) divides as the square, so doubling the count quarters the heating per unit.
Add a DC link choke. Reduces the rectifier-side contribution substantially and improves the input power factor at the same time — often the best value change.
Move to film capacitors. Far better ripple rating and life; larger and dearer per microfarad. Standard in traction.
Improve cooling. Frequently the cheapest fix, and often overlooked because nobody expects the capacitor to be the hot component.

The lesson worth generalising: the component that fails first is rarely the one handling the most power. It is the one whose life depends most sharply on a stress that was estimated rather than measured.

Practice

Problems

Three habits for three-phase inverter problems:

Start from the switching states, not from a memorised waveform. \( v_{an} = V_s(S_a - \bar S) \) and \( v_{ab} = V_s(S_a - S_b) \) give everything.
Work line-to-line where you can. The triplens are already gone and the arithmetic is simpler.
Check the DC link requirement against what a rectifier provides before assuming a modulation scheme will fit.

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

A six-step inverter runs from a 600 V link. Find the RMS and fundamental values of the phase and line voltages, and the amplitudes of the 5th, 7th and 11th harmonics relative to the fundamental.
For the inverter of Problem 1 feeding a star load of \( R = 10\ \Omega \), \( L = 25 \) mH at 50 Hz, find the fundamental, 5th and 7th harmonic currents and the current THD.
Using \( v_{an} = V_s(S_a - \bar S) \), tabulate all eight switching states and verify that exactly two produce zero output.
A 415 V, 50 Hz motor is to be driven at rated voltage. Find the DC link required for (a) six-step, (b) sinusoidal PWM, (c) space vector modulation, and state which are achievable from a 415 V diode bridge.
A 45 kW drive runs from a 650 V link at 0.88 power factor. Find the phase current, the device peak current, and the fraction of each cycle the feedback diodes conduct.
Show that the 5th harmonic is negative sequence and the 7th positive, and hence explain why both produce torque pulsation at six times the fundamental frequency.
Explain why a three-phase inverter's DC link needs no second-harmonic filtering while a single-phase inverter of the same power does. Quantify the difference for a 10 kW converter.
A drive is switched from 180° to 120° conduction with the DC link unchanged. Find the change in fundamental output voltage and in the current each device must carry, and state one application in which the change would be worthwhile.
The common-mode voltage between a motor's star point and the DC link's negative rail is \( (v_{aN}+v_{bN}+v_{cN})/3 \). Evaluate it for all eight switching states, and explain why transitions between 000 and 111 are the worst case for bearing currents.
Design the DC link for a 55 kW, 400 V drive: find the link voltage, estimate the capacitor's RMS ripple current, and specify a capacitor bank. State what you would measure to confirm the design in service.
A six-step drive produces audible 300 Hz noise and measurable torque ripple at 50 Hz output, but runs smoothly at 5 Hz output when a PWM inverter is substituted. Explain both observations from the harmonic analysis of this chapter, and estimate the ripple frequency in each case.
An engineer proposes connecting the motor's star point to the DC link mid-point "to give a stable reference". Analyse the consequences for phase voltage levels, harmonic content, neutral current, DC link capacitor stress and the achievable output voltage. Recommend for or against with reasons.