Space Vector PWM

Anandarup Das
Asst. Professor
Room-402A, Department of Electrical Engineering, IIT Delhi.
anandarup@ee.iitd.ac.in
Space vectors
• The origin of space vectors lies in rotating mmf in machines.
• The resultant mmf for a three phase system is a rotating mmf having a
fixed magnitude and direction at every instant of time.
• Space vector is a mathematical concept which is useful for visualizing the
effect of three phase variables in space.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 13

Space vectors
• Resultant space vector for load phase voltage or current are defined as,
𝑗2𝜋 𝑗4𝜋
2
• 𝑽𝑹 𝒕 = 𝑣𝐴𝑛 𝑡 + 𝑣𝐵𝑛 𝑡 𝑒 3 + 𝑣𝐶𝑛 𝑡 𝑒 3
3
𝑗2𝜋 𝑗4𝜋
2
• 𝑰𝑹 𝒕 = 𝑖𝐴 𝑡 + 𝑖𝐵 𝑡 𝑒 3 + 𝑖𝐶 𝑡 𝑒 3
3
• The space vectors VR(t) or IR(t) have both magnitude and angle. Individual
voltages/currents can be balanced or unbalanced and need not be
sinusoidal.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 14

Current space Vector
• For the sinusoidal three phase
currents, the resultant current
space vector is shown.
• The resultant space vector
(pink) is rotating at a uniform
speed and having a constant
radius.

wt=0 wt=p/3 wt=2p/3 wt=p wt=4p/3 wt=5p/3

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 15

Space vectors
• The pole voltage of one phase of the
converter has two switching states: 1 (=VD)
and 0(=0).
• The converter has total eight switching states
(2*2*2=8). These are:
(000,111,100,110,010,011,001,101).
• There are six active vectors and two zero
vectors.
• What is the load phase voltage space vector
for 100 combination?

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 16

Space vector for 100 combination
• 𝑣𝐴𝑂 𝑡 = 𝑉𝐷 , 𝑣𝐵𝑂 𝑡 = 0, 𝑣𝐶𝑂 𝑡 = 0
2 1 1 2
• 𝑣𝐴𝑛 𝑡 = 𝑣𝐴𝑂 𝑡 − 𝑣𝐵𝑂 𝑡 − 𝑣𝐶𝑂 𝑡 = 𝑉𝐷
3 3 3 3
2 1 1 1
• 𝑣𝐵𝑛 𝑡 = 𝑣𝐵𝑂 𝑡 − 𝑣𝐶𝑂 𝑡 − 𝑣𝐴𝑂 𝑡 = − 𝑉𝐷
3 3 3 3
2 1 1 1
• 𝑣𝐶𝑛 𝑡 = 𝑣𝐶𝑂 𝑡 − 𝑣𝐴𝑂 𝑡 − 𝑣𝐵𝑂 𝑡 = − 𝑉𝐷
3 3 3 3

𝑗2𝜋 𝑗4𝜋
2 2
• 𝑽𝑹 𝒕 = 𝑣𝐴𝑛 𝑡 + 𝑣𝐵𝑛 𝑡 𝑒 3 + 𝑣𝐶𝑛 𝑡 𝑒 3 = 𝑉𝐷 𝑒 𝑗0
3 3

• Similarly we can deduce the resultant space vector for other combinations.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 17

Space vector for all combinations
Space Vector Switching States Resultant space
vector (VR(t) )
V0 000 V0 = 0 Zero Vector
2
V1 100 V1 = VD e j 0
3
2
V2 110 V2 = VD e jπ/3
3
2
V3 010 V3 = VD e j 2π/3
3
Active Vector
2
V4 011 V4 = VD e j 3π/3
3

2
V5 001 V5 = VD e j 4π/3
3
2
V6 = VD e j 5π/3
V6 101 3

V7 111 V7 = 0 Zero Vector

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 18
 Graphical way
100 Condition 110 Condition 010 Condition • The space
vectors can be
obtained also
from a graphical
method.

011 Condition 001 Condition 101 Condition

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 19

Eight space vectors

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 20

Boundary of space vector diagram

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 21

Sectors in space vector diagram

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 22

Space Vector PWM
• How to switch the eight vectors so that the correct voltage is impressed on
the load?
• Space vector PWM is an extension of sine triangle PWM. Here the PWM is
done by using space vectors.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 23

Space Vector PWM
• The space vectors are switched for certain duration of time in a cycle so as
to produce the resultant vector.
• 𝑽𝑹 𝑇𝑆 = 𝑽𝟏 𝑇1 + 𝑽𝟐 𝑇2 + 𝑽𝟎 𝑇0 = 𝑽𝟏 𝑇1 + 𝑽𝟐 𝑇2 + 𝑽𝟎 𝑇01 + 𝑽𝟎 𝑇07
• 𝑇𝑆 = 𝑇1 + 𝑇2 + 𝑇0
• In space vector PWM, T01= T07= T0/2

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 24

Mathematical expression of timings
𝑂𝐴 𝑂𝐵 𝑂𝐶
• 𝜋 = = 2𝜋
sin( 3 −𝜃) sin(𝜃) sin( 3 )
𝑉1 𝑇1 𝑉2 𝑇2 𝑉𝑅 𝑇𝑆
• 𝜋 = = 2𝜋
sin( 3 −𝜃) sin(𝜃) sin( 3 )
𝜋 𝑉𝑅 2 𝜋 𝑉𝑅
• 𝑇1 = sin( − 𝜃) 𝑇𝑆 = sin( − 𝜃) 3𝑇𝑆
3 𝑉1 3 3 𝑉𝐷
𝑉𝑅 2 𝑉𝑅
• 𝑇2 = sin 𝜃 𝑇 = sin 𝜃 3𝑇𝑆
𝑉1 3 𝑆 𝑉𝐷
• 𝑇0 = 𝑇𝑆 − 𝑇1 − 𝑇2
• What happens at 𝜃 = 0, and VR=2/3 VD?

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 25

Zero vector
• Usually the zero vectors are kept equal. This gives the best harmonic
performance.
• 𝑇0 = 𝑇𝑆 − 𝑇1 − 𝑇2 is divided into equal parts of T0/2 at the beginning and
𝑇0
end of the cycle i.e. 𝑇01 = 𝑇07 =
2
• For special switching sequences (e.g. discontinuous PWM), the division is
made not equal.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 26

Example of switching
• For example, we can switch in a switching cycle Ts: 111 for T01 time period,
110 for T1 time period, 100 for T2 time period and 000 for T07 time period.
This will realize the reference vector (𝑽𝑹 ) in the switching cycle Ts.
• 𝑽𝑹 𝑇𝑆 = 𝑽𝟏 𝑇1 + 𝑽𝟐 𝑇2 + 𝑽𝟎 𝑇0 = 𝑽𝟏 𝑇1 + 𝑽𝟐 𝑇2 + 𝑽𝟎 𝑇01 + 𝑽𝟎 𝑇07

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 27

Example of switching
• The instantaneous pole voltages can be seen from the diagram. The
switching sequence is 111-110-100-000-100-110-111 and so on in sector 1.
• The sequence ensures minimum switching.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 28

Example of switching
• Similarly, 111-110-010-000-010-110-111 and so on in sector 2.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 29

What is the maximum voltage?
• The maximum voltage is obtained in linear
modulation when the inscribed circle touches the
hexagon.
2 𝜋
• 𝑉𝑅𝑚𝑎𝑥 = 𝑉 𝑐𝑜𝑠 =0.577 VD
3 𝐷 6
• In sine-PWM the peak AC voltage that was
obtained was 0.5 VD.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 31

How to realize using carriers?
• The SVPWM technique discussed so far involves substantial calculation,
sector identification etc.
• It can be done very easily using carriers where no calculation, sector
identification or switching sequence design is required.
• In order to realize SVPWM through carriers, we can observe the sine PWM
more in details.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 32

Sine PWM for 3 phases Zoomed later

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 33

Zoomed view in sine PWM

• What is the pattern of switching?

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 34

Zoomed view in sine PWM
SECTOR 1 SECTOR 2

• 111-110-100-000-100-110-111 and so on in sector 1.

• 111-110-010-000-010-110-111 and so on in sector 2.
ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 35
Zoomed view in one carrier in sine PWM
• Let us zoom further into one
carrier.
• We observe that the two zero
vector periods are not equal.
• In Sine PWM, T01 time period and
T07 time period are not always
equal.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 36

Switching sequence in space vector PWM
• The switching sequence here in one
carrier period is 111-110-100-000-
100-110-111.
• This is same as sine PWM, however
the two zero periods are equal.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 37

Switching sequence in space vector PWM
• Minimum switching is ensured.
• Switching frequency is same as carrier
frequency.
• Thus space vector PWM is an
extension of sine PWM, and can also
be realized using carriers.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 38

Extension of sine PWM
• The space vector PWM is an extension
of sine PWM by addition of a common
mode voltage.
2𝜋
• 𝑣𝑎 = 𝑉𝑚 𝑐𝑜𝑠 𝜃,𝑣𝑏 = 𝑉𝑚 𝑐𝑜𝑠(𝜃 − ),
3
4𝜋
𝑣𝑐 = 𝑉𝑚 𝑐𝑜𝑠(𝜃 − )
3

• What are the line voltages 𝑣𝑎𝑏 and 𝑣𝑏𝑐 ?

𝜋
• 𝑣𝑎𝑏 = 3 𝑉𝑚 sin( − 𝜃)
3
• 𝑣𝑏𝑐 = 3 𝑉𝑚 sin 𝜃

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 39

Extension of sine PWM
• The line voltage expressions follow the
T1 and T2 expressions.
𝜋
• 𝑣𝑎𝑏 = 3 𝑉𝑚 sin( − 𝜃)
3
𝑉𝑅 𝜋
• 𝑇1 = 3 𝑇𝑆 sin( − 𝜃)
𝑉𝐷 3
• 𝑣𝑏𝑐 = 3 𝑉𝑚 sin 𝜃
𝑉𝑅
• 𝑇2 = 3 𝑇 sin 𝜃
𝑉𝐷 𝑆

• The active vectors are represented by

the line voltages.
• What about the zero vectors?

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 40

Common mode voltage
• To make the two zero vectors equal we
have to add a common mode voltage.
• What should be its value?
• 1 − 𝑣𝑎 + 𝑣𝑐𝑚 = 1 − 𝑣𝑐 + 𝑣𝑐𝑚
𝑣𝑎 +𝑣𝑐
• 𝑣𝑐𝑚 = −
2

𝑣𝑚𝑎𝑥 +𝑣𝑚𝑖𝑛
• In general, 𝑣𝑐𝑚 = −
2

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 42

Waveforms
m=1 Resultant Waveform

m=1.15

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 43

Simulation Waveforms
Pole Voltage

• VDC=600V, the fundamental pole

voltage is (1.154*600)*0.5*0.98=339.27
V
• mf =21, harmonics reside around mf ,
2mf , 3mf …

Triplen Harmonics

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 44

Simulation Waveforms
Line voltage

• VDC=600V, the fundamental line voltage

is (1.154*0.98*600)*0.5*1.732*=587.62 V
• mf =21, harmonics reside around mf ,
2mf , 3mf …

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 45

Simulation Waveforms
Pole voltage

• The phase voltage does not contain any

triplen harmonic, so the phase current will be
absent from it.

ANANDARUP DAS, INDIAN INSTITUTE OF TECHNOLOGY, DELHI, INDIA 46