Chapter 19: Space Vector Modulation — Power Electronics

Where this sits

Part 4 · DC–AC Converters
Chapter 19 of 30

You should already know

The eight switching states and their voltages from Chapter 17, and modulation index, carrier ratio and third-harmonic injection from Chapter 18.

By the end you can

Locate a reference in the hexagon, compute the dwell times, design a symmetric switching sequence, and explain why SVM reaches 0.707 \(V_s\) where sinusoidal PWM stops at 0.612.

Time

≈ 50 min reading · ≈ 45 min problems

i What you'll learn

How three phase voltages collapse into one rotating vector, and why that is a genuine simplification rather than a notational trick.
The hexagon: six active vectors of magnitude \( 2V_s/3 \), 60° apart, with the zero vectors at its centre.
How to synthesise any reference by time-sharing the two adjacent vectors and a zero vector.
The dwell-time equations \( T_1 = m T_s\sin(60^\circ - \theta) \), \( T_2 = m T_s\sin\theta \), and where they come from.
Why the inscribed circle sets the linear limit at \( 0.707\,V_s \) — the same 15.5% gain as third-harmonic injection, reached by a different route.
How to order the states so that only one leg switches at a time, and why that halves the switching loss.
How SVM handles over-modulation by riding the hexagon's edges.

Section 19-1

Three Voltages, One Vector

Chapter 18 treated the three legs as three independent modulators, each comparing its own sine against a shared carrier. That works, but it hides something: the three phase voltages of a balanced system are not three independent quantities. They always sum to zero, so only two of them are free.

Two independent quantities describe a point on a plane. So the entire instantaneous state of a three-phase system can be represented by a single vector — and for a balanced sinusoidal set, that vector rotates at constant speed with constant length.

◇Intuition · three shadows of one spinning arrow

Hold an arrow spinning at constant speed in a plane, and shine three lights on it from directions 120° apart. Each casts a shadow on its own wall, and each shadow moves back and forth sinusoidally, 120° out of step with the next.

Those three shadows are the three phase voltages. They look like three separate things, but they are three views of one rotating arrow — and the arrow is a far simpler object to think about.

Recovering the arrow from the shadows is the space vector transform:

\[ \vec{v} = \frac{2}{3}\left(v_a + a\,v_b + a^2 v_c\right), \qquad a = e^{j2\pi/3} \]

The \( a \) and \( a^2 \) factors point each phase's contribution along its own axis, 120° apart; the \( 2/3 \) is a scaling convention chosen so the vector's length equals the phase voltage's peak.

Why this reframing earns its keep:

The problem becomes geometric. "Produce three sinusoids of the right amplitude, frequency and phase" becomes "make a vector of the right length go round at the right speed". The second is far easier to reason about.
The inverter's options become a picture. Chapter 17's eight switching states become eight points on the plane — six on a hexagon, two at its centre.
The question becomes concrete. How do you make a vector that must be anywhere using only eight fixed points? Answer: spend a fraction of each interval at each of the nearest ones, so the average lands where you want.

And that last sentence, made precise, is the whole of space vector modulation.

🔑

The eight states as vectors

\[ |\vec V_k| = \frac{2}{3}V_s \ \ (k = 1\ldots6), \qquad \angle\vec V_k = (k-1)\times 60^\circ, \qquad \vec V_0 = \vec V_7 = 0 \]

Six active vectors of equal length, evenly spaced, forming a regular hexagon; two zero vectors at the origin. This is the complete list of what the inverter can produce at any instant.

Section recap. A balanced three-phase set has only two independent quantities, so it can be represented by one rotating vector. The inverter's eight switching states become six hexagon corners plus the origin — and modulation becomes the problem of averaging fixed points to reach an arbitrary one.

Section 19-2

Reaching a Point That Is Not a Corner

The inverter can only ever be at one of eight points. The reference is almost never at any of them. The resolution is the same volt-second argument that justified PWM in Chapter 18, applied to a vector instead of a scalar.

Over a switching period \( T_s \), short compared with anything the load can follow, spend \( T_1 \) at one vector, \( T_2 \) at another, and the remainder at zero. The average vector is then:

\[ \vec V_{ref}\,T_s = \vec V_1 T_1 + \vec V_2 T_2 + \vec V_0 T_0, \qquad T_1 + T_2 + T_0 = T_s \]

Choosing \( \vec V_1 \) and \( \vec V_2 \) to be the two vectors adjacent to the reference is not obligatory, but it is optimal: any other pair would require larger dwell times to reach the same point, and would therefore produce more ripple.

Work in sector 1, where \(0 \le \theta < 60^\circ\) The two adjacent vectors are \( \vec V_1 \) at 0° and \( \vec V_2 \) at 60°, both of magnitude \( 2V_s/3 \). Resolve the volt-second balance into real and imaginary parts.
Imaginary part gives \(T_2\) immediately Only \( \vec V_2 \) has an imaginary component, \( (2V_s/3)\sin60^\circ \):

Working

\[ |\vec V_{ref}|\sin\theta\,T_s = \frac{2V_s}{3}\sin60^\circ\,T_2 \;\Longrightarrow\; T_2 = \frac{\sqrt3\,|\vec V_{ref}|}{V_s}T_s\sin\theta \]
Real part then gives \(T_1\)

Working

\[ |\vec V_{ref}|\cos\theta\,T_s = \frac{2V_s}{3}T_1 + \frac{2V_s}{3}\cos60^\circ\,T_2 \]

\[ \Longrightarrow\; T_1 = \frac{\sqrt3\,|\vec V_{ref}|}{V_s}T_s\left(\frac{\sqrt3}{2}\cos\theta - \frac{1}{2}\sin\theta\right) = \frac{\sqrt3\,|\vec V_{ref}|}{V_s}T_s\sin(60^\circ - \theta) \]
Define the modulation index and the result becomes clean Setting \( m = \sqrt3\,|\vec V_{ref}|/V_s \):

Result

\[ T_1 = m\,T_s\sin(60^\circ - \theta), \qquad T_2 = m\,T_s\sin\theta, \qquad T_0 = T_s - T_1 - T_2 \]

Two sines and a subtraction. In every other sector the same expressions apply, with \( \theta \) measured from the start of that sector.

Interactive · the hexagon and the dwell times

Move the reference vector round the hexagon and watch the two adjacent active vectors take turns. The inscribed circle is the limit of linear operation — outside it the reference cannot be reached without over-modulating.

Reference angle 30°

Modulation index m 0.85

Sector1

T₁ / T_s0.425

T₂ / T_s0.425

T₀ / T_s0.150

—

Section recap. Volt-second balance on a vector gives \( T_1 = mT_s\sin(60^\circ - \theta) \) and \( T_2 = mT_s\sin\theta \), with the remainder spent at zero. The same two expressions serve all six sectors once \( \theta \) is measured from the sector's start.

Section 19-3

Why the Inscribed Circle Is the Limit

The dwell times must be non-negative and must sum to no more than \( T_s \). That single arithmetic requirement has a clean geometric meaning, and it explains where SVM's 15.5% advantage comes from.

Write the constraint

Working

\[ T_1 + T_2 \le T_s \;\Longrightarrow\; m\bigl[\sin(60^\circ - \theta) + \sin\theta\bigr] \le 1 \]
Simplify the bracket Using \( \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \):

Working

\[ \sin(60^\circ - \theta) + \sin\theta = 2\sin30^\circ\cos(30^\circ - \theta) = \cos(30^\circ - \theta) \]
Find the worst angle \( \cos(30^\circ - \theta) \) is largest at \( \theta = 30^\circ \), where it equals 1. So the binding case is mid-sector, and the constraint there is simply \( m \le 1 \).
Translate back into voltage

Result

\[ m = \frac{\sqrt3\,|\vec V_{ref}|}{V_s} \le 1 \;\Longrightarrow\; |\vec V_{ref}|_{max} = \frac{V_s}{\sqrt3} = 0.577\,V_s \]

\[ V_{L1(rms)} = \frac{\sqrt3\,|\vec V_{ref}|}{\sqrt2} = \frac{V_s}{\sqrt2} = 0.707\,V_s \]

◈Why the answer is a circle, and why it is inscribed

The constraint came out as \( |\vec V_{ref}| \le V_s/\sqrt3 \) — a bound on length only, independent of angle. A set of points at constant distance from the origin is a circle, so the reachable region is a disc.

And that disc is exactly the hexagon's inscribed circle. The hexagon's circumradius is \( 2V_s/3 \) (the active vector magnitude), so its inradius is

\[ r_{in} = \frac{2V_s}{3}\cos30^\circ = \frac{2V_s}{3}\cdot\frac{\sqrt3}{2} = \frac{V_s}{\sqrt3} \]

which is the limit just derived. The geometry and the algebra agree, as they must.

Why the reachable region cannot be the whole hexagon. The hexagon's corners are reachable — they are the active vectors themselves. But a reference must rotate at constant magnitude to produce a balanced sinusoidal output, and a circle of radius greater than the inradius would pass outside the hexagon near the mid-sector angles. Those points require dwell times summing to more than \( T_s \), which is not physically available.

Now compare with sinusoidal PWM. SPWM constrains each leg independently to \( \pm V_s/2 \) about the mid-point. Translated into the vector plane, that confines the reference to a circle of radius \( V_s/2 \) — smaller than \( V_s/\sqrt3 = 0.577V_s \) by exactly:

\[ \frac{0.577}{0.500} = 1.155 = \frac{2}{\sqrt3} \]

The same 15.5% as third-harmonic injection, arrived at with no harmonics mentioned anywhere.

Which raises the obvious question, and the answer is yes. SVM and third-harmonic injection are the same thing seen from two directions. Compute the common-mode voltage that SVM's zero-state splitting produces, and it is a triangular wave at \( 3f_1 \) — a zero-sequence component the load cannot see. SVM does not avoid harmonic injection; it produces the optimal injection automatically, as a consequence of choosing the nearest vectors and splitting the zero time evenly.

Which of the two you implement is therefore a matter of convenience, not of performance. Digital drives generally use the min–max formulation of Chapter 18 because it is three comparisons and a subtraction, and gives waveforms identical to SVM.

Section recap. Requiring \( T_1 + T_2 \le T_s \) bounds the reference's length regardless of angle, so the reachable region is the hexagon's inscribed circle of radius \( V_s/\sqrt3 \). That gives \( 0.707V_s \) line-to-line — 15.5% more than SPWM, and identical to third-harmonic injection.

Section 19-4

Ordering the States

The dwell times say how long to spend at each vector. They say nothing about the order, and the order is where much of SVM's practical advantage lies.

Take sector 1, with vectors \( V_1 \) (100), \( V_2 \) (110), and the two zero states 000 and 111. There are many possible orderings; only a few are sensible.

Candidate switching sequences for sector 1
Sequence	Leg transitions per \(T_s\)	Symmetric?	Comment
000 → 100 → 110 → 111	3	No	Each step changes one leg. Good, but asymmetric within the period.
000 → 100 → 110 → 111 → 110 → 100 → 000	6 over two half-periods = 3 per leg per \(T_s\)	Yes	The standard. Symmetric, one leg per transition, zero time split evenly.
111 → 110 → 100 → 000 → 100 → 110 → 111	Same	Yes	Mirror image; equally valid, used alternately in some schemes.
100 → 000 → 110 → 111	5	No	Wasteful — two legs change between 000 and 110.
000 → 110 → 100 → 111	6	No	Worst case: every transition changes two legs.

◈Why "one leg at a time" is worth so much

Two adjacent states in the hexagon differ in exactly one bit — 100 and 110 differ only in leg b. Two non-adjacent states differ in two or three bits.

Every changed bit is a switching event, and switching loss is proportional to the number of events. A sequence in which every transition changes one leg therefore uses the minimum possible switching for the vectors chosen.

The standard sequence achieves this by construction:

\[ \underbrace{000}_{} \to \underbrace{100}_{\text{leg a}} \to \underbrace{110}_{\text{leg b}} \to \underbrace{111}_{\text{leg c}} \to \underbrace{110}_{\text{leg c}} \to \underbrace{100}_{\text{leg b}} \to \underbrace{000}_{\text{leg a}} \]

Seven states, six transitions, each changing exactly one leg — and each leg switches exactly twice per period, once on and once off.

Three separate benefits follow:

Minimum switching loss. Compared with a poorly ordered sequence needing 12 transitions, this halves it.
Symmetry, so a clean spectrum. Because the pattern is a mirror image about the centre of the period, the resulting waveform has quarter-wave symmetry — which suppresses a whole family of sideband harmonics. This is why symmetric SVM measurably outperforms asymmetric SVM at the same switching frequency.
Even zero-state sharing. Half the zero time is spent at 000 and half at 111, so the upper and lower devices share the freewheeling duty and run at similar temperatures.

And the sequence is easy to implement. The pattern is exactly what three comparators fed from a single triangular carrier produce, if each leg's threshold is set from the computed dwell times. So SVM needs no special hardware — it is a different way of computing the three thresholds that ordinary carrier PWM already uses.

One variant worth knowing. Discontinuous PWM uses only one zero state instead of two, clamping one leg to a rail for 60° at a time. That leg does not switch at all during those 60°, so switching loss falls by about a third — at the cost of higher current ripple and uneven device heating. It is used at high power, where switching loss dominates, and where the extra ripple is acceptable.

Section recap. Ordering the states so every transition changes one leg minimises switching events; making the sequence symmetric about the period's centre suppresses sideband harmonics; splitting the zero time between 000 and 111 balances device heating. The standard 000-100-110-111-110-100-000 does all three.

Section 19-5

Over-Modulation in the Hexagon

Beyond the inscribed circle the dwell times sum to more than \( T_s \), which cannot be delivered. SVM's response is geometric, and it makes the over-modulation of Chapter 18 much easier to picture.

SVM operating regions
Region	Index \(m\)	What the reference does	Output \(V_{L1}\)
Linear	\(\le 1\)	Traces a circle inside the hexagon. \(T_0 > 0\) always.	up to \(0.707\,V_s\)
Over-modulation I	\(1\)–\(1.05\)	Circle intersects the hexagon; the excursions outside are clipped onto the edges.	\(0.707\)–\(0.74\,V_s\)
Over-modulation II	\(1.05\)–\(1.10\)	Reference held at the corners for part of each sector, sliding along edges between.	\(0.74\)–\(0.78\,V_s\)
Six-step	\(1.10\)	Reference jumps from corner to corner. \(T_0 = 0\) everywhere.	\(0.780\,V_s\)

◇Intuition · a circle that will not fit in a hexagon

Draw a circle inside a regular hexagon and grow it. At first it fits comfortably. At the inradius it touches all six edges. Grow it further and it pokes out through the middle of each edge while the corners still stand clear.

That is the entire geometry of over-modulation. The parts of the circle that lie outside the hexagon are simply unreachable — the inverter cannot produce them. The modulator's only choice is what to do instead.

Scale the dwell times down proportionally, which is what the explorer above does:

\[ T_1' = \frac{T_1}{T_1 + T_2}T_s, \qquad T_2' = \frac{T_2}{T_1 + T_2}T_s, \qquad T_0 = 0 \]

Geometrically this projects the reference radially onto the hexagon edge. The angle is preserved and the magnitude is reduced to whatever the hexagon allows there.

What the load experiences. The output no longer traces a circle but a rounded hexagon — full magnitude near the corners, reduced in between. A hexagonal path is exactly a sinusoid plus its 5th and 7th harmonics, which is why those particular harmonics reappear. The 6-fold symmetry of the hexagon and the \( 6k \pm 1 \) harmonic rule are the same fact.

Why the last step is discontinuous. Between \( m = 1.05 \) and \( m = 1.10 \) the reference must spend increasing time held at the corners to gain any more amplitude, because the corners are the only points further out than the edges. At \( m = 1.10 \) it jumps directly from corner to corner, dwelling 60° at each — which is precisely six-step operation. The hexagon picture makes the convergence obvious in a way the carrier picture does not.

Section recap. Beyond \( m = 1 \) the circular reference no longer fits inside the hexagon, and the modulator projects it onto the edges. The output path becomes hexagonal, which is a sinusoid plus 5th and 7th harmonics — and at \( m = 1.10 \) the reference sits only at the corners, which is six-step.

Section 19-6

Worked Examples

1 Computing the dwell times

Problem. A three-phase inverter with \( V_s = 600 \) V runs SVM at 5 kHz. The reference is 300 V peak per phase at an angle of 100° electrical. Find the sector, the dwell times, and the switching sequence.

Modulation index and sector

\[ m = \frac{\sqrt3\,|\vec V_{ref}|}{V_s} = \frac{\sqrt3(300)}{600} = 0.866 \]

\[ 100^\circ \text{ lies in } 60^\circ\text{–}120^\circ \;\Longrightarrow\; \textbf{sector 2}, \quad \theta = 100^\circ - 60^\circ = 40^\circ \]

Dwell times, with \(T_s = 1/5000 = 200\ \mu\text{s}\)

\[ T_1 = mT_s\sin(60^\circ - \theta) = 0.866(200)\sin20^\circ = 59.2\ \mu\text{s} \]

\[ T_2 = mT_s\sin\theta = 0.866(200)\sin40^\circ = 111.3\ \mu\text{s} \]

\[ T_0 = 200 - 59.2 - 111.3 = 29.5\ \mu\text{s} \]

The vectors. In sector 2 the adjacent active vectors are \( \vec V_2 \) (110) and \( \vec V_3 \) (010), so:

Symmetric sequence, one leg per transition

\[ 000 \to 010 \to 110 \to 111 \to 110 \to 010 \to 000 \]

Check it. 000→010 changes leg b; 010→110 changes leg a; 110→111 changes leg c. Each transition changes exactly one leg, as required.

Sanity check on the answer. The reference is at 100°, closer to \( \vec V_3 \) at 120° than to \( \vec V_2 \) at 60° — and indeed \( T_2 \) (the dwell at \( \vec V_3 \)) is nearly twice \( T_1 \). Dwell times should always be largest for the nearest vector, and it is worth checking that before proceeding.

2 Comparing SVM with sinusoidal PWM

Problem. A 415 V, 50 Hz motor is driven from a 540 V link. Find the maximum line voltage achievable with SPWM and with SVM, and state the modulation index each needs to reach 415 V.

Sinusoidal PWM

\[ V_{L1(max)} = 0.612(540) = 330\ \text{V} \quad\text{— 20\% short of 415 V} \]

Space vector modulation

\[ V_{L1(max)} = 0.707(540) = 382\ \text{V} \quad\text{— 8\% short} \]

Neither reaches 415 V from this link, but the difference between them is decisive.

With SPWM the machine could only reach \( 330/415 = 80\% \) of rated flux. Torque falls in proportion, so the drive would be substantially derated.
With SVM it reaches 92% of rated flux, and the remaining 8% comes from mild over-modulation at the top of the speed range — where, as Chapter 18 discussed, the harmonics cost little.

The modulation index needed for 382 V:

Working

\[ m = \frac{V_{L1}}{0.707\,V_s} = \frac{382}{382} = 1.00 \quad\text{— exactly at the inscribed circle} \]

This is the practical reason SVM displaced carrier PWM in drives. Not elegance, not spectral purity — the 15.5% of DC link voltage that plain SPWM cannot reach, and without which a standard motor cannot be run at rated flux from a standard rectifier.

3 Switching count and loss

Problem. An inverter runs SVM at 8 kHz with the standard symmetric sequence. Find the number of switching events per second per device, and compare with a badly ordered sequence needing 12 transitions per period.

Standard sequence. Six transitions per \( T_s \), each changing one leg, so each leg changes twice — once on, once off.

Working

\[ \text{Per leg: } 2 \text{ transitions} \times 8000 = 16\,000 \text{ per second} \]

\[ \text{Per device: } 8000 \text{ turn-ons and } 8000 \text{ turn-offs} \]

So each device switches at exactly the carrier frequency, which is the expected and desirable result.

Badly ordered sequence with 12 transitions per period would give each device 16,000 turn-ons per second — double the switching loss for exactly the same output waveform.

Putting a number on it

\[ P_{sw} = (E_{on} + E_{off})\,f_{sw}, \quad E_{on}+E_{off} \approx 12\ \text{mJ at 540 V, 100 A} \]

\[ \text{Good: } 12\times10^{-3}(8000) = 96\ \text{W per device} \]

\[ \text{Bad: } 192\ \text{W per device} \;\Longrightarrow\; 576\ \text{W extra across six devices} \]

Over half a kilowatt, entirely from state ordering. No component changes, no waveform changes — only the sequence in which the same states are visited. This is why the ordering is treated as part of the modulator design rather than an afterthought.

And the further saving available. Discontinuous PWM, using only one zero state, clamps each leg to a rail for 60° per half cycle — cutting switching events by a further third, to about 64 W per device. The cost is roughly 30% more current ripple, which at high power is often a good bargain.

4 Reading a fault from the dwell times

Problem. A drive's modulator reports \( T_1 = 130\ \mu\text{s} \), \( T_2 = 95\ \mu\text{s} \) with \( T_s = 200\ \mu\text{s} \). What is wrong, and what will the inverter actually do?

Check the constraint first.

Working

\[ T_1 + T_2 = 225\ \mu\text{s} > T_s = 200\ \mu\text{s} \;\Longrightarrow\; T_0 = -25\ \mu\text{s} \]

A negative zero-time is physically meaningless — it means the reference lies outside the hexagon and cannot be produced.

Find the modulation index it implies. Working backwards from \( T_1 + T_2 = mT_s\cos(30^\circ - \theta) \), and noting that \( T_1 > T_2 \) puts \( \theta \) below 30°:

Working

\[ \frac{T_2}{T_1} = \frac{\sin\theta}{\sin(60^\circ-\theta)} = \frac{95}{130} = 0.731 \;\Longrightarrow\; \theta \approx 25.3^\circ \]

\[ m = \frac{T_1 + T_2}{T_s\cos(30^\circ - \theta)} = \frac{225}{200\cos(4.7^\circ)} = 1.13 \]

So \( m = 1.13 \) — well into over-modulation, and close to the six-step limit of 1.10... which it exceeds, meaning the command is beyond what the inverter can do at all.

What a correctly written modulator does:

Detects \( T_0 < 0 \) — a one-line check that must never be omitted.
Scales both dwell times by \( T_s/(T_1+T_2) \), giving \( T_1' = 116\ \mu\text{s} \), \( T_2' = 84\ \mu\text{s} \). This projects the reference onto the hexagon edge, preserving its angle.
Reports the saturation upward so the current regulator knows its command was not honoured, and its integrator can be prevented from winding up.

Why step 3 is the one that matters. Without anti-windup, the regulator sees a persistent error, integrates it, demands even more voltage, and saturates harder. When the load finally lightens, the accumulated integral produces a large overshoot. Saturation feedback from the modulator to the regulator is a small piece of code that prevents a well-known and destructive failure.

Section 19-7

Summary & Formula Sheet

Chapter 19 in five sentences:

Three phase voltages with a floating neutral have only two degrees of freedom, so they collapse into one rotating vector.
The eight switching states become six hexagon corners of magnitude \( 2V_s/3 \) plus the origin.
Any reference is synthesised by spending \( T_1 = mT_s\sin(60^\circ-\theta) \) and \( T_2 = mT_s\sin\theta \) at the two adjacent corners and the remainder at zero.
The requirement \( T_1 + T_2 \le T_s \) confines the reference to the inscribed circle, giving \( 0.707V_s \) — the same limit as third-harmonic injection, because SVM is optimal zero-sequence injection.
Ordering the states so one leg changes per transition, symmetrically, and splitting the zero time between 000 and 111, minimises switching loss and cleans the spectrum.

Formula sheet · Chapter 19

Space vector transform\(a = e^{j2\pi/3}\)

\( \vec v = \tfrac23\bigl(v_a + a\,v_b + a^2v_c\bigr) \)

Active vector magnitudehexagon circumradius

\( |\vec V_k| = \tfrac23 V_s \)

Modulation index1 at the inscribed circle

\( m = \dfrac{\sqrt3\,|\vec V_{ref}|}{V_s} \)

First dwell time\(\theta\) from the sector start

\( T_1 = m\,T_s\sin(60^\circ - \theta) \)

Second dwell timelargest for the nearer vector

\( T_2 = m\,T_s\sin\theta \)

Zero timemust not go negative

\( T_0 = T_s - T_1 - T_2 \)

Dwell-time sumworst at \(\theta = 30^\circ\)

\( T_1 + T_2 = m\,T_s\cos(30^\circ - \theta) \)

Linear limitthe inscribed radius

\( |\vec V_{ref}|_{max} = \dfrac{V_s}{\sqrt3} = 0.577\,V_s \)

Maximum line voltageagainst 0.612 for SPWM

\( V_{L1} = \dfrac{V_s}{\sqrt2} = 0.707\,V_s \)

Gain over SPWMsame as 3rd-harmonic injection

\( \dfrac{2}{\sqrt3} = 1.155 \)

Over-modulation scalingproject onto the hexagon edge

\( T_1' = \dfrac{T_1}{T_1+T_2}T_s \)

Symmetric sequencesector 1; one leg per transition

\( 000\!-\!100\!-\!110\!-\!111\!-\!110\!-\!100\!-\!000 \)

Key terms

Space vector: The single complex quantity representing three instantaneous phase voltages. Rotates at constant speed for a balanced sinusoidal set.
Voltage hexagon: The regular hexagon whose corners are the six active switching vectors, with the zero vectors at its centre.
Sector: One of six 60° regions between adjacent active vectors. Determines which two vectors are used.
Dwell time: How long within each switching period a particular vector is applied.
Inscribed circle: The largest circle fitting inside the hexagon, radius \(V_s/\sqrt3\). The limit of linear modulation.
Symmetric sequence: A state order that mirrors about the centre of the switching period, giving quarter-wave symmetry and a cleaner spectrum.
Zero-state splitting: Dividing the zero time between 000 and 111 so upper and lower devices share the freewheeling duty.
Discontinuous PWM: Using only one zero state, clamping one leg to a rail for 60° at a time. Cuts switching loss by about a third.
Zero-sequence injection: Adding a common-mode component the load cannot see. SVM produces the optimal one automatically.
Anti-windup: Preventing a saturated regulator's integrator from accumulating an error the modulator cannot act on.

Check yourself

Test Yourself

Chapter 19 · six questions answers hidden until you ask

SVM and third-harmonic injection both reach 0.707 \(V_s\). Is that a coincidence?

No — they are the same technique described in two languages, and the shared limit is the proof.

The connection, made explicit. Compute the common-mode voltage that SVM produces. The star point's potential relative to the DC link's negative rail is \( (v_{aN}+v_{bN}+v_{cN})/3 \), and evaluating that through a full cycle of SVM's state sequence gives a triangular wave at three times the fundamental frequency.

That is a zero-sequence component — identical in all three legs, invisible to a floating-neutral load, exactly like the injected third harmonic of Chapter 18. SVM does not avoid injection; it produces it as a by-product of choosing the nearest vectors and splitting the zero time evenly.

Why the limit must therefore be identical. Both techniques face the same constraint: each leg's output must fit between the rails. Both improve on plain SPWM by adding a common-mode component that flattens the leg references. The best any such method can do is set by geometry — the largest circle that fits in the hexagon — and both reach it.

Where the small difference lies. SVM's triangular injection and Chapter 18's sinusoidal \( -\tfrac16\sin3\omega t \) are not identical functions, and they produce marginally different harmonic spectra. SVM's is slightly better in ripple terms. But the fundamental limit is the same for both.

The practical consequence for implementation. Because they are equivalent, you can implement SVM in whichever form is cheapest:

Geometric: identify the sector, compute two sines, order the states. Conceptually clear, computationally heavier.
Min–max: compute three sinusoidal references, subtract \( \tfrac12(\max + \min) \) from all three, compare against a carrier. Three comparisons and a subtraction, and it produces waveforms identical to geometric SVM.

Almost every digital drive uses the second, and calls it SVM — correctly, because the output is the same.

Why must the two adjacent vectors be used? Could a reference in sector 1 be made from \(\vec V_1\) and \(\vec V_3\)?

It could, and the average would be correct. It would be a bad choice for three separate reasons.

1. Larger dwell times mean larger ripple. Vectors further apart require longer dwells to reach the same average, so the instantaneous output spends longer far from the reference. Ripple current is proportional to how far the applied vector strays from the wanted one and for how long — so a wider pair produces substantially more ripple for the same average.

2. More switching. Adjacent states differ in one bit; \( \vec V_1 \) (100) and \( \vec V_3 \) (010) differ in two. Every transition between them switches two legs, roughly doubling the switching loss.

3. Reduced reach. The set of points reachable from two vectors 120° apart is a smaller triangle than that from two 60° apart. The linear region would shrink, throwing away the DC link utilisation that SVM exists to provide.

The general principle: among all vector pairs that can produce a given average, the nearest pair minimises the deviation, minimises the switching, and maximises the reachable region. There is no trade-off here — adjacency wins on every count.

The one legitimate exception. In a fault-tolerant drive with one leg failed, some vectors are unavailable and the modulator must use whatever remains — accepting worse ripple in exchange for continued operation. Similarly, in a three-level NPC inverter (Chapter 20) there are redundant vectors, and the choice between them is used to balance the neutral-point voltage rather than to minimise ripple. Both cases prove the rule: the nearest pair is chosen unless something more important is at stake.

Why does the linear limit occur at \(\theta = 30^\circ\) rather than at a sector boundary?

Because mid-sector is where the hexagon's boundary is nearest the origin.

The algebra says so. The dwell-time sum simplifies to

\[ T_1 + T_2 = m\,T_s\bigl[\sin(60^\circ - \theta) + \sin\theta\bigr] = m\,T_s\cos(30^\circ - \theta) \]

which is maximised when \( \cos(30^\circ - \theta) = 1 \), that is at \( \theta = 30^\circ \). There the constraint is \( m \le 1 \); at \( \theta = 0 \) it relaxes to \( m \le 1/\cos30^\circ = 1.155 \).

The geometry says the same thing. Distance from the origin to the hexagon boundary:

At a corner (\( \theta = 0^\circ \) or \( 60^\circ \)): the circumradius, \( 2V_s/3 = 0.667V_s \).
At an edge midpoint (\( \theta = 30^\circ \)): the inradius, \( V_s/\sqrt3 = 0.577V_s \).

So the boundary is closest mid-sector, and a rotating reference of constant magnitude is constrained by the closest point it must pass.

Why constant magnitude is the binding requirement. A balanced sinusoidal output demands a vector of fixed length rotating at constant speed. If the magnitude were allowed to vary with angle, the reference could ride the hexagon boundary and reach 0.667 \( V_s \) on average — but that is exactly over-modulation, and it introduces the 5th and 7th harmonics.

The intuition worth carrying: a circle inscribed in a hexagon touches the edges, not the corners. The corners are wasted capability from the point of view of undistorted operation — and getting at them is precisely what over-modulation does, at precisely the cost you would expect.

Two sequences apply the same vectors for the same dwell times but in different orders. Does the load notice?

The average is identical, so the fundamental output is identical. Everything else differs.

1. Switching loss — the largest effect. A sequence in which every transition changes one leg needs six transitions per period. A poorly ordered one can need twelve. Since switching loss is proportional to the number of events, ordering alone can double it.

2. Ripple current. Although the average over \( T_s \) is the same, the path differs. The load current integrates the instantaneous voltage, so an order that spends longer far from the reference before returning produces larger excursions. Symmetric sequences keep the deviation small and balanced.

3. Harmonic spectrum. A sequence symmetric about the centre of the switching period gives the waveform quarter-wave symmetry, which suppresses a whole family of sideband harmonics. This is measurable: symmetric SVM has noticeably lower WTHD than asymmetric SVM at the same switching frequency, for no extra cost.

4. Device temperature balance. Splitting the zero time between 000 and 111 shares the freewheeling duty between the upper and lower devices. Using only one zero state loads one half of the bridge and leaves the other cool — which is fine if deliberate (discontinuous PWM) and a problem if accidental.

5. Common-mode voltage. Every transition between 000 and 111 is a full-amplitude common-mode step, which drives current through the motor's stray capacitance to its frame and thence through the bearings. Some schemes deliberately restrict which zero states they use to reduce this, trading switching count against bearing life.

So the order is a genuine design variable, not an implementation detail — and the standard symmetric sequence is standard because it is a good compromise across all five, not because it is the only one that works.

A modulator computes \(T_0 = -18\ \mu\text{s}\). What should it do?

Scale both dwell times proportionally, and — just as importantly — tell the controller above it that the command was not honoured.

What the negative value means. \( T_1 + T_2 > T_s \), so the reference lies outside the hexagon. The inverter physically cannot produce it.

The standard remedy:

\[ T_1' = \frac{T_1}{T_1 + T_2}\,T_s, \qquad T_2' = \frac{T_2}{T_1 + T_2}\,T_s, \qquad T_0 = 0 \]

Geometrically this projects the reference radially onto the hexagon edge: the angle is preserved and the magnitude reduced to what is available. Preserving the angle matters — in a field-oriented drive the angle carries the torque/flux decomposition, and distorting it would misalign the current vector with the rotor flux.

What a poor implementation does instead:

Clipping \( T_1 \) and \( T_2 \) independently at \( T_s \) — this distorts the angle, not just the magnitude, and misaligns the current vector.
Ignoring the sign and letting a negative time propagate into the timer registers — which typically produces a very large unsigned value, and a shoot-through.

The step that is most often forgotten. The modulator must report the saturation back to the current regulator. Without it:

The regulator sees a persistent error and integrates it.
It demands even more voltage, which is also unavailable.
The integrator winds up over many cycles.
When the load lightens, the accumulated integral produces a large overshoot — sometimes an over-current trip, sometimes worse.

Anti-windup is a few lines of code that prevent a well-known and destructive failure, and the saturation flag from the modulator is what makes it possible.

Why does over-modulation produce specifically the 5th and 7th harmonics, and not some other pair?

Because the reference path becomes a hexagon, and a hexagon's Fourier series contains exactly those harmonics.

What the path becomes. In the linear region the reference traces a circle. In over-modulation the parts outside the hexagon are projected onto its edges, so the path becomes a rounded hexagon — and in the limit, the hexagon itself.

Why a hexagon gives \( 6k \pm 1 \). A regular hexagon has six-fold rotational symmetry: rotating it by 60° reproduces it exactly. A waveform whose vector path has \( N \)-fold symmetry can only contain harmonics of order \( Nk \pm 1 \). With \( N = 6 \):

\[ n = 6k \pm 1 = 5, 7, 11, 13, 17, 19, \ldots \]

This is the same rule as Chapter 17's, and now the reason is visible: it comes from the six-fold symmetry of the switching hexagon, which is itself a consequence of having three phases and two levels.

Why the 5th and 7th dominate. They are the lowest members of the family, and Fourier coefficients of a piecewise-linear path fall as \( 1/n^2 \). At full six-step operation the 5th is 20% and the 7th 14% of the fundamental — the figures of Chapter 17, arrived at from a different direction.

Why there is no way around it. Any two-level three-phase inverter has this hexagon; it is not a property of the modulation scheme. Reaching beyond the inscribed circle necessarily means following a six-fold-symmetric path, and that necessarily produces \( 6k \pm 1 \) harmonics.

Which points at the only real solution: more levels. A three-level inverter's vector diagram has 19 points rather than 7, with a larger inscribed circle relative to the switching step. More levels means a rounder reachable region and less distortion at any given utilisation — and that is exactly the argument for the multilevel converters of Chapter 20.

Practice

Problems

Three habits for SVM problems:

Find the sector first, then measure \( \theta \) from that sector's start. The same two formulas then work everywhere.
Check \( T_0 \ge 0 \) before believing any dwell-time answer.
Sanity-check by proximity: the larger dwell must belong to the nearer vector.

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

An inverter with \( V_s = 540 \) V runs SVM at 4 kHz. For a reference of 260 V peak at 25°, find \( m \), the sector, and the three dwell times.
Repeat Problem 1 for a reference at 195°, and state the switching sequence you would use.
Find the maximum reference magnitude and the maximum line-to-line RMS voltage for a 700 V link under SVM, and compare with sinusoidal PWM.
Show that \( \sin(60^\circ-\theta) + \sin\theta = \cos(30^\circ-\theta) \), and hence that the linear limit binds at \( \theta = 30^\circ \).
A modulator reports \( T_1 = 88\ \mu\text{s} \), \( T_2 = 140\ \mu\text{s} \), \( T_s = 200\ \mu\text{s} \). Find \( \theta \), \( m \), and the corrected dwell times.
Verify that the sequence 000-100-110-111-110-100-000 changes exactly one leg at each transition, and count the switching events per device per period. Repeat for 100-000-110-111 and comment.
Explain why the space vector hexagon's corners are reachable but a circle through them is not, and relate your answer to over-modulation.
Show that SVM's zero-state splitting produces a triangular common-mode voltage at \(3f_1\), and explain why the load does not see it.
A drive uses discontinuous PWM, clamping one leg to a rail for 60° at a time. Estimate the reduction in switching loss and state two disadvantages.
Design the modulator for a 90 kW, 400 V drive from a 540 V link: choose a switching frequency, find the modulation index at rated speed, compute the dwell times at \( \theta = 30^\circ \), and state what happens above rated speed.
A three-level NPC inverter has 27 switching states and 19 distinct vectors, several of which are redundant. Explain what the redundancy could be used for, and why a two-level inverter has only one such choice.
An engineer proposes implementing SVM by the min–max method rather than geometrically. Explain why the two give identical output, compare their computational cost, and state which you would choose for (a) a 16-bit microcontroller and (b) an FPGA.