Space vector modulation theory

AN2154

6/16

Equation 4

Equation 5

Moreover, referring to Figure 2 it is easy to demonstrate that:

Equation 6

Equation 7

from which it is possible to get:

Equation 8

Equation 9

If we now substitute Equation 8 and 9 into Equation 3 and 4, it is possible to compute times

Equation 10

Equation 11

Equation 12

Please note that when the modulation index exceeds

negative for certain values of

α. Since this does not have a physical meaning, it can be

modulation in linear region is exactly

overmodulation, where Equation 10 - 12 are no longer valid.

Graphically, this means that for space vector modulation to work properly, the magnitude of

the vector V must be small enough to ensure that the vector is totally contained inside the

hexagon shown in Figure 2.

t

b

V

b

V

001

------------

T

0

⋅

=

t

0

T

0

t

a

–

t

b

–

=

V

π

3

---

α

–

⎝⎠

⎛⎞ V

a

π

3

---

sin

⋅

=

sin

⋅

V

α V

b

π

3

---

sin

⋅

=

sin

⋅

v

a

2

3

-------

V

π

3

---

α

–

⎝⎠

⎛⎞

sin

⋅⋅

=

V

b

2

3

-------

V

α

sin

⋅⋅

=

t

a

2

3

-------

T

0

m

i

π

3

---

α

–

⎝⎠

⎛⎞

sin

⋅⋅

⋅

=

0

α π

3

---

≤≤

t

b

2

3

-------

T

0

m

i

α

sin

⋅⋅

⋅

=

0

α π

3

---

≤≤

t

0

T

0

t

a

–

t

b

–

=