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ABOUT

TIP JAR

SPACE VECTOR
PWM INTRO
M AY 1 , 2 0 1 7
BY YNGVE SOLBAKKEN

INTRODUCTION
Space Vector Pulse Width Modulation (SV-PWM) is a modulation
scheme used to apply a given voltage vector to a three-phased electric
motor (permanent magnet or induction machine).

The goal is to use a steady state DC-voltage and by the means of six
switches (e.g. transistors) emulate a three-phased sinusoidal
waveform where the frequency and amplitude is adjustable.

There are two challenges to this:

1. The only voltage level available is the DC-link voltage which can
be assumed constant (well, at least for sake of simplicity).

2. There are only six di erent voltage angles available. With no

middle ground. To rotate a motor, a smoothly rotating voltage
vector is required - not one that skips 60 degrees per step.

GRID SUPPLIED THREE PHASE

VOLTAGE
A grid supplied three phase voltage will look like the animation
below. This is what an induction machine is experiencing when
connected directly to the grid.

The three phased system is illustrated using two di erent, but equal
forms:

• A vector diagram showing all three phases and their vector sum
(space vector).

• An ordinary instantaneous sine wave representation, also

showing the resultant space vector.
THE TOPOLOGY OF A TWO-
LEVEL INVERTER
The two-level inverter is the most used topology today because it is
simple and cheap. Higher level converters are mostly used for high
voltage applications and will not be treated in this article.

The above schematic is the well-known and well-used inverter

topology. From left to right the following is shown:

• A three phase supply and a three-phased diode recti er

• A DC-link capacitor for energy storage and voltage stabilization

• An inverter bridge with six transistors

 • Three output terminals and a star-connected (ungrounded)
induction machine equivalent

• Not shown: snubber circuit, anti-parallel diodes over the transistors

and back-EMF in motor.

To better understand how space vector PWM is used, the above

schematic can be greatly simpli ed: 

First, the DC-link voltage can be simpli ed so that it is constant.

Normally it varies with load, but not that much during the time
frames given in this article. The supply and diode recti er will thus
not be shown for the rest of the article.

Second, each leg of the inverter bridge can be simpli ed by replacing

the two bridge leg-transistors by one single SPDT-switch which
shall indicate that either the top or the bottom transistor is closed.
The scenario of both transistors in one leg being open is not
interesting and will not be necessary. Further, the scenario of both
transistors being closed implies that the DC-link is short circuited
and we can all go home and order new parts.

The simpli cation from two transistors to one SPDT switch is

possible because only one transistor in each bridge leg can be closed
at any given time and that one transistor must be closed in each leg
in order to have a three-phased current owing. Additionally there
are always anti-parallel diodes across each transistor to allow bi-
directional current ow.

The simpli ed schematic is presented below:

The above example shows a transistor combination where the
leftmost leg (W) has its upper transistor closed and bottom
transistor open. The two other legs (V and U) has its upper transistor
open and lower transistor closed. This will correspond to a positive
voltage being applied to W-phase while the two other phases are
negative. The current arrows illustrate the current path from the
positive DC link to the negative DC link.

From the simpli ed

schematic, it is now seen that
there are three switches which
can be in two di erent positions
each. The total number of
possible switch con guration is
thus \(2^3 = 8\).

Six of these con gurations

correspond to di erent voltages
applied to the connected motor
and are called basic vectors from
now on. The last two are
referred to as zero vectors as
they represent zero volts on the
terminals.

The U-phase normally forms

the basis for the basic vector's
angles at 0°. The origin of the angles are the windings physical
location inside the stator; installed around the circumference at 120°
apart. Because each winding can have positive and negative voltage,
it occupies two angles at 180° separation, e.g. 240° and 60° is W-
phase in positive and negative state respectively.

In binary, these vector combinations can be represented as eight

di erent binary values, here named from \(v_0\) to \(v_7\):

000 - \(v_0\) (zero vector)

001 - \(v_1\) (Phase +U)
010 - \(v_2\) (Phase +V)
011 - \(v_3\) (Phase -W)
100 - \(v_4\) (Phase +W)
101 - \(v_5\) (Phase -V)
110 - \(v_6\) (Phase -U)
111 - \(v_7\) (zero vector)

Each of the three binary digits refers to one bridge leg where the
value 1 indicates that the top transistor is closed whereas the value 0
indicates that the bottom transistor is closed.

All eight combinations and the resulting motor voltage and current
directions are shown below.
Press any image to enlarge.
 v0 - 000 v1 - 001 v2 - 010 v3 - 011

v4 - 100 v5 - 101 v6 - 110 v7 - 111

SYNTHESIZING OUTPUT
VOLTAGE
So far it is only shown that the converter can output DC-link voltage
to the terminals and how each of the six usable vectors a ects
current- and voltage directions in the motor.

This section will show how to

synthesize any voltage vector by
quickly alternating between
adjacent voltage vectors and
timing of the on- and o times.

To the right there is an example

of a reference vector which the
inverter has to synthesize using
the basic vectors available.
From the gure, it is seen that
the reference is located between
\(v_1\) and \(v_3\). This sector
is called sector 1. It is now possible to quickly alternate between
these two basic vectors to emulate a voltage vector at \(38.3^\circ\). 

The two zero vectors \(v_0\) and \(v_7\) are also used to add dead
time to the switching pattern. This dead time reduces the voltage
magnitude and is necessary when the voltage reference magnitude
is less then 100%.  

The actual synthesizing can be achieved using a triangular wave as

trigger. First is it assumed that the desired voltage reference is
already available, e.g calculated by the reverse Clarke/Park
transformation described in the Vector Control for Dummies post. 

The gures below illustrate how a desired voltage is transformed

into binary on/o signals for each of the three inverter legs by
means of the triangular wave. Six di erent examples are shown,
each with a di erent angle in di erent sectors. 
Sector 1 - 18° Sector 2 - 95° Sector 3 - 150° Sector 4 - 189°
 Sector 5 - 285° Sector 6 - 347°

CONCEPT EXPLANATION
Let us take a breathing pause there. This gure contains a lot of
information, and to better explain di erent parts of it, I have
clipped out four di erent regions of it that will be explained below.

PART 1 - INPUT AND PROJECTION TO

THE TRIG WAVE:

The left circle is the "input" or

"reference". Some also call it
the order signal. It is what we
want to achieve; three di erent
voltages at di erent angles -
together forming a space vector
with one magnitude and one
angle.

As seen, the U-phase (red) is close to its maximum positive value.

This level is also shown on the graph to the right of the circle
together with a triangle wave where it is referred to as a control
voltage. The frequency of this waveform is many times higher than
the frequency of the space vector. Typically between 1 and 20 kHz,
but applications with a few hundred or hertz or up to hundreds of
kilohertz are not unheard of.
The V-phase (green) is close to its maximum negative value and the
W-phase (blue) just stretching out from zero in positive direction.

How these three values are intersected with the triangular wave will
directly translate to which switches we want to close and how long
we are keeping them closed.

PART 2 - TRANSLATING INPUT TO

SWITCHING STATES:

Look at the right part of the

gure.

The main focus in this part are

the dashed lines protruding
downwards from the triangular
wave. These originate from
where the control voltages
intersect the trig wave. At all
times, four di erent sections
are visible below the W-axis
right above the \(\frac{1}{2}T\)
arrow. For other space vector
angles and magnitudes, the width of these four sections change, but
the total width will always be the same.

At the angle in this example (46°), the rst dashed line is the red
one. Between the start of the three axes and the red line, there are no
active voltage vectors. All three control voltages in this region are
below the trig wave. This means that all switches are o . 000. Zero
vector \(v_0\). 
Next section; between the red and the blue dashed line. The U-phase
control voltage is now above the trig line while the two others are
still below. This activates the U-switch. 001 there. Also called \
(v_1\). 

Third section: between the blue and the green line. Now both U and
V is activated; 011 - \(v_3\).

Last section: All three switches are now active. This is 111, \(v_7\),
the second zero vector.

At every angle and magnitude, the same pattern will be seen; Zero
vector - basic vector - basic vector - zero vector. Repeat. The only
thing that change is the width (i.e. the on-time) of these sectors and
which of the six basic vectors are used.

Note that the on-time percentage is also displayed to the left of the
W-axis. When the desired amplitude is low, the zero vectors will
have an increased on-time and opposite when the amplitude is high.
This is controlled by a scaling factor called modulation index. If the
desired amplitude is higher than the trig wave, something called
over modulation occurs. This is explained later.

PART 3 - ILLUSTRATING THE VECTORS 

Now we know that the two

vectors we have to alternate
between is \(v_1\) and \(v_3\).

By getting the on-times

correctly, this will emulate a
voltage vector at the same angle as the input voltage space vector.
You can see that the green \(v_3\)-vector is close to maximum both
in the circle vector diagram and in the "sector diagram" below the
W-axis. The same can be seen for the \(v_1\)-vector which is close
to its minimum.

 The thin line green line inside the space vector circle are visual aides
to illustrate that these two basic vectors do indeed form the desired
space vector as a vector sum.

The values to the left in the gure shows the angle, which sector we
are in and the weighing factors, \(x\) and \(y\), for each of the two
basic vectors. We will get back to those later.

PART 4 - ILLUSTRATION OF THE TWO

ALTERNATING SWITCHING
CONFIGURATIONS

The two di erent switching

combinations (basic vectors)
which the drive will alternate
between is shown here. The
progress bar above each con guration indicates how long it is on
compared to the total on-time for both basic vectors. I.e, how long
they are on relative to each other.

The zero-vectors come in addition and will increase for lower

voltage levels.

Pay attention to that they are in opposite "phase" to each other, just
as two sine waves 90 degrees apart. When the left con guration has
completed its cycle, the right con guration is halfway.

This switching scheme also employs a nifty feature; only one switch
is changing state at the time. This reduce harmonics in the machine
and also reduces switching losses, a very important aspect to
consider when designing any type of power electronics.

SUMMARY
The by far most important part of the gure is the area below the
lowest axis (W). It shows the portion each basic vector is active for
each period \(T\). The percentage number is presented to the left of
that axis. 

It can also be seen that every pattern starts and ends with a zero
vector, and that between every transition, only one switch is being
changed. This greatly reduces harmonics in the motor which cause
overheating and torque uctuations. Also, any current that is not
contributing to torque, or is necessary to maintain the required
magnetic eld, is unwanted. This is because the losses in the power
electronic components are current dependent and they translate to
heat which will stress the cooling system. Over-heating of
transistors and diodes are one of the main challenges when
designing power electronics today. The alternative is to design a
converter with higher current ratings, but this increases the price.

Further, at the bottom of the above gure two schematics are

shown. These represent the state of the switches for the two basic
vectors which are in use in the current section. The percentage
indication above indicate their on-time relative to the other. For
example, if the angle is exactly between to basic vectors (i.e. 30°),
both basic vector would be turned on 50% of the time each. Please
note that this percentage does not indicate their total on time with
respect to the period \(T\). If the voltage magnitude is less than
maximum, there is added zero-vectors to averange down the
voltage. So at 70% voltage magnitude and still at 30°, the ON-time
of each basic vector is still 50% with respect to each other, but in
total there would be 30% zero vectoring within this period. The
actual on-time for each transistor thus become 35%;

\begin{equation}
T = T_{v_1} + T_{v_3} + T_{v_0} + T_{v_7} = 35\% + 35\% +
15\% + 15\% = 100\%
\end{equation}

A full animation is shown below.

DETAILED EXAMPLE
PROBLEM:
• The drive has estimated that it needs to modulate a voltage
vector of 630 V at an angle of 38.3°:
\begin{equation}
\vec{v}_s^u=630\angle38.3^\circ
\end{equation}
where the subscript \(s\) denotes a stator frame reference (i.e., not
rotor) and the superscript \(u\) denotes that the angle reference (0°)
is at phase U (0°) .

• 38.3° lies between basic vector 1 (0°) and 3 (60°), so these are
the two we are going to utilize.

• The equation to determine how much we are using each basic

vector can be written as

\begin{equation}
\vec{v}_s^u = \hat{V}_s \angle \theta_{v_s} = x \cdot v_3 + y
\cdot v_1
\end{equation}

where \(x\) and \(y\)  are the unknown weighting factors for each
vector and \(v_1\) and \(v_3\)  are the two basic vectors in use.

• The DC link voltage is 975 V and the motor nominal terminal

voltage is 690 V (typical values in the marine industry)

First, the real component is established:

\begin{equation}
\mathfrak{R} \{\vec{v}_s^u\}
= \mathfrak{R}\{630\angle 38.3^\circ \}
= \mathfrak{R}\{x \cdot 975 \angle 60^\circ + y \cdot 975 \angle
0^\circ \}
\end{equation}

\begin{equation}
630\angle38.3^\circ = x \cdot 975 \cos{60^\circ} + y \cdot 975
\cos{0^\circ}
\end{equation}

\begin{equation}
630 \cdot 0.7848 = x \cdot 975 \cdot 0.5 + y \cdot 975 \cdot 1
\end{equation}

\begin{equation}
\label{eq:x}
494.409 = x \cdot 487.50 + y \cdot 975
\end{equation}

Secondly, we establish the imaginary component:

\begin{equation}
\mathfrak{I} \{\vec{v}_s^u\}
= \mathfrak{I} \{630\angle 38.3^\circ \}
= \mathfrak{I}\{x \cdot 975 \angle 60^\circ + y \cdot 975 \angle
0^\circ \}
\end{equation}

\begin{equation}
630\angle38.3^\circ = x \cdot 975 \sin{60^\circ} + y \cdot 975
\sin{0^\circ}
\end{equation}

\begin{equation}
630 \cdot 0.6198 = x \cdot 975 \cdot \frac{\sqrt{3}}{2} + y \cdot 975
\cdot 0
\end{equation}

\begin{equation}
\label{eq:y}
390.461 = x \cdot 844.37 + 0
\end{equation}

We can now nd \(x\) and \(y\)  by setting up equation \ref{eq:x} and

\ref{eq:y} towards each other:

\begin{equation}
390.461 = x \cdot 844.37 + 0
\end{equation}

\begin{equation}
x=\frac{390.461}{844.37} = 0.4624
\end{equation}

Inserting the \(x\)-value into the equation for \(y\):

\begin{equation}
494.409 = x \cdot 487.50 + y \cdot 975
\end{equation}

\begin{equation}
494.409 = 0.4624 \cdot 487.50 + y \cdot 975
\end{equation}

\begin{equation}
y=\frac{494.409-(0.4624 \cdot 487.50)}{975} = 0.2759
\end{equation}

Let us double-check the result by inserting the values we have

found:

\begin{equation}
\vec{v}_s^u
= \hat{V}_s \angle \theta_{v_s}
= x \cdot v_3 + y \cdot v_1
= 0.4624 \cdot 975 \angle 60^\circ + 0.2759 \cdot 975 \angle 0^\circ
= 629.999\angle38.297^\circ
\end{equation}

This result is pretty close to the original target, which was 630 V at
an angle of 38.3°

SO WHAT WILL THE OUTPUT

VOLTAGE BE?
The available voltage at the inverter terminals are not readily
apparent by looking at the schematics alone due to the averaging of
di erent vectors and the use of a modulation index.

Let us start at the beginning and see where we end up:

The grid voltage which the recti er is connected to:

\begin{equation}
V_{grid, RMS} = 690 V_{RMS}
\end{equation}

The resulting DC-link voltage (assumed low load):

\begin{equation}
V_{DC} = V_{grid, RMS} \times \sqrt{2} = 975.8 V
\end{equation}

Alternative formula DC-link voltage (average for three phase / six

pulse supply):

\begin{equation}
\cfrac{3\sqrt{2}V_{RMS}}{\pi} = 931 V
\end{equation}
(for now we will stick with the rst one, and get back to the last one
later)

This voltage level is what we have to play with to synthesize an

output voltage.

As shown earlier, there are six

basic vectors, but their
magnitude has not been stated
yet — only their angles. So here
goes:

\begin{equation}
\left|v_{1-6}\right| = V_{DC}
\end{equation}

Tada. That wasn't so di cult?

With the above information an

updated gure of the six basic
vectors can be presented. When
lines are drawn between each vector tip, it will form the well known
space vector hexagon. Note that in the literature, di erent notation
for the space vector hexagon are used; phase voltage, line voltage,
RMS voltage and some include over modulation as well (we'll get to
that, don't worry).

The observant reader might notice a circle located inside the

hexagon. Its radius is given by the maximum size permissible by the
boundaries of the hexagon. This radius is given at \(30^{\circ}\)
which is exactly between two adjacent basic vectors.

The maximum RMS output voltage then becomes:

\begin{equation}
V_{max,RMS} = V_{DC} \times \cos{30^{\circ}} \times \frac{1}
{\sqrt{2}} = 597 V
\end{equation}

This is lower than the input voltage, but the reason for this is
apparent when looking at the hexagon; when alternating between to
adjacent basic vectors, we are averaging and in the process some of
the voltage is "lost".

This low voltage might come to some as a slight

disappointment. When your grid voltage is 690 V, you would want to
buy a 690 V motor to go with it. 

What do we do now?

THIRD HARMONIC INJECTION

Fasten your seat belts and whip out your wand, because we are now
heading towards the dark arts of electrical engineering.

As just shown, we are faced with a challenge with too low output
voltage compared to what we really want. But we are also faced with
physical limitations. Which we are going to bend. Slightly.

First, let us illustrate the problem below:

The maximum voltage output of the DC-link has to be the DC-link
voltage. Above it is shown as a red sine wave where the peak-to-
peak voltage equals \(V_{DC} = 975 V\). The RMS phase-to-phase
voltage then becomes:

\begin{equation}
V_{max,RMS} = \frac{V_{DC}}{2} \times \sqrt{3} \times \frac{1}
{\sqrt{2}}  = 597 V
\end{equation}

which is exactly the same as we just stated a few paragraphs up, just
presented from a di erent perspective. And we still haven't touched
our problem besides illustrating that we need to go through the DC-
link ceiling somehow by approximately 15% to reach 690 V. This
might seem impossible; The ceiling represent a switching state
where the switches are on 100% of the time, while the oor
represent a zero on-time. So how can a switch be on more than 100%
or less than 0% ?

The trick we need to pull was invented by an engineer working for

GE in the 1980s and is called (fanfare); Third Harmonic Injection.

That's right folks; Adding a second sine wave with a frequency three

times the fundamental frequency will e ectively solve our problem
in a very elegant way as shown below:
Wait, you say? Did some pay for a smooth sine wave? 

Well, the trick is that it's all hidden inside. By expanding the blue
curve above to its constituent sine waves (a.k.a Fourier expansion),
the following will appear:

Now that is more like it! We

have now successfully
circumvented physical
limitations in terms of
maximum voltage output and
packed it all neatly with the use
of third harmonic injection.

Next we will look at what the

motor thinks of this. Harmonics
is almost always an unwanted phenomenon, so let's check the
consequences.

The gure to the right illustrates all three phase voltages with third
harmonic injection.

Some might notice something odd with those curves; The black
third harmonic curve is common to all three phases. All three are
super positioned and are totally identical which means that the third
harmonic is now a common mode wave form. Now all parts of the
motor circuit will oscillate at three times the frequency with respect
to the DC-link. 

The neutral point of the motor

is now of particular
interest. The question now
becomes: 

What is the phase-to-neutral

voltage across each winding?

Recall that the phase voltage

consist of a big sine wave and a
third harmonic wave added together. The scope will now display the
phase voltage (the left probe in the gure) minus the neutral voltage
(the right probe).

Also remember that every point in the motor is oscillating at the

third harmonic frequency and amplitude, including the neutral
point. The neutral-connected probe will therefore also see the same
third harmonic which is measurable at the phase-connected probe.

The mathematics then becomes something like this:

(Big sinewave + third harmonic sinewave) — (third harmonic sine
wave) = (Only big sinewave)

Oh yeah. The whole third harmonic component goes away and the
winding doesn't even know about it. 

Bazinga.

MODULATION INDEX, M
This is quite simply a factor used to scale the voltage. In the last
section it was shown that the output voltage can be scaled up to
115.4% by applying third harmonic injections. I.e. the modulation
index can span from 0 to 1.154

The formula for output phase-to-phase voltage then becomes

\begin{equation}
V_{max,RMS} = \frac{V_{DC}}{2} \times \sqrt{3} \times \frac{1}
{\sqrt{2}} \times m
\end{equation}

where \(m\) represent the modulation index factor.

The below animation illustrates the concept:

OVER MODULATION
For modulation index beyond 1.154 strange things will start to
happen. The control voltages used in conjunction with the triangular
wave will move above the trig wave. This will e ectively remove zero
vectors and depending on how far above the control voltage moves,
it will also dominate the other control voltages for longer and longer
periods within each own sector.

The inverter will cut pulses and distort (roughen up) the sinusoidal
wave more and more until a situation called six step occurs. In six
step operation, the inverter will stay at one basic vector at the time
and jump directly to the next basic vector and then to the next and
so forth. With 60 degree steps and square wave operation, the motor
will produce a lot for harmonics and probably feel really terrible.
Luckily, the motor inductance will lter (smoothen) the current to
avoid damage due to exessive cogging / torque ripple (remember, it
is the current that produces the torque - not the voltage).

This mode of operation is exactly what we tried to avoid initially by

doing our averging of two di erent and adjacent basic vectors.

Even though over modulation seems like motor torture (which it

is...), it is sometimes necessary. Up till now it has been assumed that
the DC-link always stays at a factor of \(\sqrt{2}\) above the supply
voltage, but that is not always the case.

When high currents are needed, such as when the motor is running
at full speed and full power, the current drawn from the grid is
substantial, causing the small DC resistance in the recti er diodes to
become huge voltage drops (Ohm's law there, folks). 
Suddenly the DC-link voltage is no longer 975 V, but 900 V and at
full speed the motor requires full voltage in order to keep the motor
ux at rated value. At this voltage, a modulation index of 1.25 is
required, and this will create an ugly sine wave. Also, note that over-
modulation is no longer linear. It actually looks more like the
saturation region of a magnetizing curve, meaning that the
controller will have to amplify the modulation index in order to
achieve the desired (linearized) modulation level.

Due to the ugly sine wave, harmonics other than our designed third
harmonic start to appear. These can cause over heating of both rotor
and main windings along with acoustic noise (whining) and torque
ripples.  Note that this applies to induction motors. Brushless DC-
machines are actually designed for this mode of operation

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