Inverter Applications
•Motor Drives
• Power back-up systems
•Others:
Example HVDC Transmission systems
1
Single Phase Inverter
Square-wave “Modulation” (1)
+
q1(t)
Vdc
1-q1(t)
-
Q1
1-q2(t)
Q’2
+v out
q2(t)
Q’1
Vdc
vout(t)
Q2
-Vdc
sin(n / 2)
vout (t )  Vdc 
cos(nt  n0 )

n
n 1
4
Vout 1  Vdc  1.27Vdc
4


2
t
Single Phase Inverter
Square-wave “Modulation” (2)
THD = 0.48
Characteristics: - High harmonic content.
- Low switching frequency.
- Difficult filtering.
- Little control flexibility.
3
Single Phase Inverter
Square-wave “Modulation” (3)
+
q1(t)
Q1
1-q2(t)
+v out
q2(t)
Vdc
1-q1(t)
Q’1
-
Q’2
Vdc
vout(t)
Q2
t
-Vdc
sin(n / 2)
cosn S t   cosn S t  n 
vout (t ) 

 n1
n
4


Vout 1  Vdc cos 

2
2Vin
4

Single Phase Inverter
Square-wave “Modulation” (4)
Example with Vout-1=1.21Vdc
THD = 0.3
Characteristics: - High harmonic content.
- Low switching frequency.
- Difficult filtering.
- More control flexibility.
5
Single Phase Inverter
Pulse Width Modulation (1)
vout  DVdc  (1  D)( Vdc )  Vdc (2 D  1)
D is the duty cycle of switch Q1. D is the portion of
the switching cycle during which Q1 will remain
closed.
TON Q11
D
TS
In PWM D is made a function of time
D=D(t)
6
Single Phase Inverter
Pulse Width Modulation (2)
•Let’s
1 1
D   m(t )
2 2
m(t )  m[cos(0t )]
where
Modulation
index
Fundamental
Signal
Vout 1
m
Vdc
7
Modulation
function
Single Phase Inverter
Pulse Width Modulation (3)
1
D  [1  m cos(0t )]
2
n m(t ) 



 sin 
4Vdc
2
2

 cos(n t )
vout (t )  m(t )Vdc 

S
 n1
n
Moving average
vout  Vdc (2 D  1)  m(t )Vdc
8
t
Single Phase Inverter
Pulse Width Modulation (4)
Implementation issue: time variable “t” needs to be
sampled. Two basic sampling methods: UPWM
NSPWM
9
Single Phase Inverter
Pulse Width Modulation (5)
t
10
Single Phase Inverter
Pulse Width Modulation (6)
Considering that m(t )  m cos( 0t )
Fundamental
(n=1) and its
sidebands
NSPWM
mVdc
no sidebands
UPWM
 nm 
J


4Vdc  k n  2k  é n
 ù
sin ê  n  ú

 i 1 n
1
 2û
ë k
4Vdc  1  im   i 
 J 0   sin  
 i 1 i  2   2 
Carrier
4Vdc 
 im   i 
J

 sin  

0
component
 i 1  2   2 
(n = k)
Sidebands of
 im 
J
the carrier 4Vdc  n  2  é
 ù 4Vdc


sin
n

i
 i
êë
 i 1
2 úû 
11
 nm 
J



k n  2k  é n
 ù
sin ê  n  ú

1
 2û
ë k
i 1 n
Three Phase Inverter
Pulse Width Modulation (1)
Active States:
Zero States:
State Qa Qb Qc
S0
0 0 0
S7
1 1 1
12
State Qa Qb Qc
S1
0 0 1
S2
0 1 0
S3
0 1 1
S4
1 0 0
S5
1 0 1
S6
1 1 0
Three Phase Inverter
Pulse Width Modulation (2)
ma (t )  m[cos(0t )  ea 0 (t )]
Modulation m (t )  m[cos( t  2 / 3)  e (t )]
b
0
b0
Functions
mc (t )  m[cos(0t  2 / 3)  ec 0 (t )]
ea 0 (t )  e03 (t )  ea 0H (t )
“Zero-Sequence” e (t )  e (t )  e (t )
b0
03
b 0 H
Signals
ec 0 (t )  e03 (t )  ec 0H (t )
13
Three Phase Inverter
Pulse Width Modulation (3)
Triplen Harmonics (3, 9, 15, …)
ma (t )  m[cos(0t )  e03 (t )  ea 0H (t )]
Modulation Fundamental
Signal
index
V ph peak
m
Vdc / 2
14
Other
Harmonics
(5, 7, 11, 13, 17, ….)
Three Phase Inverter
Pulse Width Modulation (4)
m cos(0t )
t
e03 (t )  m(1/ 6) cos(30t )
ma (t )  m[cos( 0t )  e03 (t )]  m[cos( 0t )  (1 / 6) cos( 30t )]
15
Classic Approaches to PWM (1)
Time Domain
Use of modulation signal: Duty cycle computation:
 D  1  1 m (kT )
 a 2 2 a s

1 1

 Db   mb (kTs )
2 2
t 
 D  1  1 m (kT )
 c 2 2 c s
16
Sector limits
Classic Approaches to PWM (2)
Classic SVM - Application
Park’s
Transformation
3
Park’s Transformation
ia* (t )
ib* (t )
2
1
1  ia 

1


 
M  2
2
2
 * 
 ib 
M  3
3
3
 0
 
 q

i

2
2  c 
ic* (t )
M (kTs ) M d* (kTs )
*
q
*
d
M d (kTs )
-
-
M de
d-PI
Controller
+
q-PI
Domain
Transformer
(Modulator)
Space
Time
Vector
M qe Controller M q (kTs )
+
M qref (kTs ) M dref ( kTs )

vref
(From Vector Controller)
M d*  cos(t )
Switch
Control
DC
Source
M q*  sin(t )
Inverter
Motor
ia  cos(t )
ib  cos(t  2 / 3)
ic  cos(t  2 / 3)
Disadvantages: Can’t see evolution in time
Loss of information about e0-3(t)
17
Classic Approaches to PWM (3)
Classic SVM
S2
S3
v̂q
S6
Space vector domain
SECTOR
v̂2
v̂6
II
 SECTOR
Bdq  vˆd vˆq 
vref I
SECTOR
v̂4 2 Bases
t
v̂3 III v̂0
BSVM  vˆi vˆ j 
v̂7 O
v̂
d
SECTOR
SECTOR S
4
IV
v̂1
S1
S
In O: 7
S0
18
VI
SECTOR
V
v̂5
S5
BSVM changes in
each sector
BSVM in sector I
vˆi  vˆ4
vˆ j  vˆ6
Classic Approaches to PWM (4)
Classic SVM
Transitions within
sector I
S6
S2
T7=T0
SVM Computation
S4
S7
S3

-Track the sector in which vref is in and based on it
select the appropriate set of basis Bij
S0
S1
S5

-Calculate the coordinates of vref in the basis Bij
- Change the coordinates of the reference voltage vector
from basis Bdq to basis Bij. The sector dependant transformation yields the period of
time Ti that the machine remains in each state in a given sampling period.
- When the time Ti is finished move to the next state following the sequence given by
the SVM state machine.
19
Mathematical Framework (1)
Complete representation
involves a 3-D space
Then, a 3-D vector 
can
be introduced: v
R
Space Vector
Domain
W
20
Control Time
Domain
Output Time
Domain
S
Mathematical Framework (2)
Control Time Domain
Functions of time are used as basis
tˆa  cos(t )  ea 0 (t )
ˆ
ˆ
ˆ
ˆ
Bt  ta , tb , tc   tb  cos(t  2 / 3)  eb 0 (t )
tˆ  cos(t  2 / 3)  e (t )
c0
c

v  m(tˆa  tˆb  tˆc )
21
Mathematical Framework (3)
Space Vector Domain

Bdq 0  dˆ , qˆ , nˆ0 t

dˆ  1 0 0T

T
 qˆ  0 1 0 
T

n


0
0
e
(
t
)

ˆ
0 3
 0t

v  vd (t )dˆ  vq (t )qˆ  v0 (t )nˆ0t

When e0-H(t)=0, v describes a circumference
with radius equal to m.
22
Mathematical Framework (4)
Output Time Domain
vˆa  1 0 0 

T
 vˆb  0 1 0 
T

vˆc  0 0 1
T
BS  vˆa , vˆb , vˆc 
v  va (t )vˆa  vb (t )vˆb  vc (t )vˆc
23
Mathematical Framework (5)
Output Time Domain
Length of sides equal to 2
Each corner represents one state
State sequence is obtained naturally
Bdq  va  vb  vc  0
Balanced system plane
Sectors: six pyramid-shaped volumes bounded by sides of
the cube and |vi|=|vj| planes (i, j = a,b,c; i  j).
24
Mathematical Framework (6)
Matrix R

When e0-H(t)=0, v describes a circumference
with radius equal to m.
1st Idea: Use Park’s transformation to a synchronous
rotating reference frame:
m(cos(t )  e03 (t ))
 vd 
 cos(t ) cos(t  2 / 3) cos(t  2 / 3) 

  2


v

sin(

t
)
sin(

t

2

/
3
)
cos(

t

2

/
3
)
m
(cos(

t

2

/
3
)

e
(
t
))
 q



0 3
3
v 
 1
 m(cos(t  2 / 3)  e (t )) 
1
1
 0



0 3
Instantaneous values of the voltages
25
Mathematical Framework (7)
Matrix R
1
 vd  

  

v

1
 q 

 v   2me (t ) 
 0 

0 3
Problem: components in d̂ and q̂
are constant values
I am interested in having a constant value in nˆ0 t
2nd Idea: Freeze the rotational reference frame at t=0
1
1

1




2
2
v
m(cos(t )  e03 (t ))
 d
  m cos(t ) 


  2
 

3
3 
 vq    0 
 m(cos(t  2 / 3)  e03 (t ))    m sin(t ) 

2
2 
v  3
  2me (t ) 
m
(cos(

t

2

/
3
)

e
(
t
))


1
1
1
 0

 

0 3
0 3




26
Mathematical Framework (8)
Matrix R
3rd Idea: Rearrange the product.

vB
dq 0
1
1

1




2
2
cos(t )  e03 (t )
 vd 



  2

3
3 
v

0

cos(

t

2

/
3
)

e
(
t
)
m
 q


0 3
2
2 
v  3

1
1
1  cos(t  2 / 3)  e03 (t ) 
 0




 cos(t )  (1 / 2) cos(t  2 / 3)  (1 / 2) cos(t  2 / 3)  m 
 
2
  0
 ( 3 / 2) cos(t  2 / 3) ( 3 / 2) cos(t  2 / 3)  m 
3
 m 
e
(
t
)
e
(
t
)
e
(
t
)
 0 3
 
0 3
0 3

vB
t
27
Mathematical Framework (9)
Matrix R
4th Idea: Eliminate the dependency on e0-3(t) in
order to have a constant coordinate in nˆ0 t
 cos(t )  (1 / 2) cos(t  2 / 3)  (1 / 2) cos(t  2 / 3) 

2
R  0
 ( 3 / 2) cos(t  2 / 3) ( 3 / 2) cos(t  2 / 3) 
3

1
1
1


 vd 
m
 cos(t ) 
 
 


 vq   R m   m sin(t ) 
v 
m
 2 
 0
 


28
Mathematical Framework (10)
Matrix R
In order to include e0-H(t) we need to follow the
same steps and apply superposition.
 cos(t )  (1 / 2) cos(t  2 / 3)  (1 / 2) cos(t  2 / 3) 

2
R  0
 ( 3 / 2) cos(t  2 / 3) ( 3 / 2) cos(t  2 / 3)  
3

1
1
 1

 ea 0 H (t )  (1 / 2)eb 0 H (t )  (1 / 2)ec 0 H (t ) 

2
  0
 ( 3 / 2)eb 0 H (t ) ( 3 / 2)ec 0 H (t ) 
3

0
0
0


 cos(t )  ea 0 H (t )  (1/ 2)(cos(t  2 / 3)  eb 0 H (t ))  (1 / 2)(cos(t  2 / 3)  ec 0 H (t )) 

2
R 
0
 ( 3 / 2)(cos(t  2 / 3)  eb 0 H (t )) ( 3 / 2)(cos(t  2 / 3)  ec 0 H (t )) 
3

1
1
1


29
Mathematical Framework (11)
Matrix W
1
1
ˆ
d '  vˆa  vˆb  vˆc
2
2
qˆ '  vˆb  vˆc
nˆ '0  qˆ  dˆ  vˆa  vˆb  vˆc
 dˆ '   1  1 / 2  1 / 2  vˆa 
  
 
1  vˆb 
 qˆ '    0  1
 nˆ '   1
 vˆ 
1
1
 c 
 0 
W’
30
Mathematical Framework (12)
Matrix W
 1  1/ 2  1/ 2 


W '  0 1
1 
1
1
1 

0
1
2

1
W ' '    1  3 1
2
3 1
 1
0
e03 (t ) 
2


1
W    1  3 e03 (t ) 
2
3 e03 (t ) 
 1
31
1) Take the transpose
and apply scaling
factor
2) Include e0-3(t) in
order to have it as
a component
Mathematical Framework (13)
Matrix S
S=WR
 2 cos(t )  e03 (t )  cost  2 / 3  e03 (t )  cos(t  2 / 3)  e03 (t ) 

1
S    cos(t )  e03 (t ) 2 cos(t  2 / 3)  e03 (t )  cos(t  2 / 3)  e03 (t )  
3

  cos(t )  e03 (t )  cos(t  2 / 3)  e03 (t ) 2 cos(t  2 / 3)  e03 (t ) 
 2ea 0 H (t )  eb 0H (t )  ec 0 H (t ) 

1
   ea 0 H (t ) 2eb 0 H (t )  ec 0 H (t ) 
3

  ea 0 H (t )  eb 0H (t )  2ec 0H (t ) 
 2 cos(t )  e03 (t )  2ea 0 H (t )  cost  2 / 3  e03 (t )  eb 0 H (t )  cos(t  2 / 3)  e03 (t )  ec 0 H (t ) 

1
S    cos(t )  e03 (t )  ea 0 H (t ) 2 cos(t  2 / 3)  e03 (t )  2eb 0 H (t )  cos(t  2 / 3)  e03 (t )  ec 0 H (t ) 
3

  cos(t )  e03 (t )  ea 0 H (t )  cos(t  2 / 3)  e03 (t )  eb 0 H (t ) 2 cos(t  2 / 3)  e03 (t )  2ec 0 H (t ) 
32
3D Analysis and Representation


v B  (W )v B
S
dq 0

vB
dq 0

 ( R )v B
t


v B  ( S )v B
S
t
cos(t S )  ea 0 (t S )
 va 


 


 vb   m cos(t S  2 / 3)  eb 0 (t S ) 
v 
 cos(t  2 / 3)  e (t ) 
 c

S
c0 S 
33
3D Analysis and Representation
1
ei 0 (t )  e03 (t )   cos(3t )
4
UPWM
34
3D Analysis and Representation
SVM
UPWM
35
3D Analysis and Representation
3D Representation: Plot

evolution of v in output
time domain during a
complete fundamental
period.
The resulting curve always
lays within the cube defined
by the switching states.
1
ei 0 (t )  e03 (t )   cos(3t )
6
36
3D Analysis and Representation
Triplen harmonic distortion:
Evolution away from the
plane va+vb+vc=0
Other harmonic distortion:
non circular projection
of the curve over the
plane va+vb+vc=0
Sharp corners indicate the presence of higher
order harmonics.
37
Commonly used schemes
1
ei 0 (t )  e03 (t )   cos(3t )
6
ei 0 (t )  0
SVM
Square wave
38
3D Analysis and Representation
ei 0 (t )  0
39
1
ei 0 (t )  e03 (t )   cos(3t )
6
Square wave
3D Analysis and Representation
Phase and Line Voltages (1)
1

vab (t )  Vdc (v .vab )
2
1

voa (t )  Vdc (v .voa )
2

 
vab  va  vb  vˆa  vˆb
1 
1


voa  (vab  vca )  (2vˆa  vˆb  vˆc )
3
3
Phase Leg
Phase
voltages
Line Voltages
40
13 2
1
a
 1  1
T
 1 0
T
13  1
0
b
2  1
T
1  1
T
13  1
 1
c
 1 2
T
0 1
T
3D Analysis and Representation
Phase and Line Voltages (2)
voa (t) va (t)
t
41
Direction of vab
Direction of voa
3D Analysis and Representation
Phase and Line Voltages (3)
e03 (t )  (1/ 4) cos(30t )
e0 H (t )  (1/ 10) cos(50t )
voa (t) va (t)
t
42
3D Analysis and Representation
Maximum non distorting range
Radius of circle is m if it is
measured in the Space Vector
Domain.
There is a scaling factor in W

vd

vd
1
ei 0 (t )  e03 (t )   cos(3t )
6
43
Bdq 0
BS

vd
 vˆd
 vˆa  0.5vˆb  0.5vˆc
mmax
ND

Bdq 0
vd
1
BS

2
2
2
 1.15
3
3
3
2
3D Analysis and Representation
vi  Di  (1  Di ) 1  2 Di  1
If fswitch>>ffund then
44
, i  a, b, c
vi  vi
3D Analysis and Representation
t
45
3D Analysis and Representation
Sector II
Sector I
T7 T6
T4 T0
T7 T6
Sector III
T 2 T0
T7 T 3
Qa
Qb
Qb
Qb
Qa
Qc
Qc
Qc
Qa
T2 T0
T7  DminTs
T  (1  D )T
 0
max
s

Ti  ( Dmax  Dmed )Ts , (i  1, 2, 4)
T j  ( Dmed  Dmin )Ts , (j  3, 5, 6)
46
Analysis of SVM (1)
T7=T0
Dmin=1-Dmax
vmax +vmin=0
S is sector e (t )  e (t )  1 v (t )
0
0 3
med
2
dependent
(considering e0-H(t)=0)
SVM tends to approximate the trajectory of a
square wave, but adds 3rd harmonic and
higher order triplen harmonics
No difference in sequence compared to other schemes
47
Analysis of SVM (2)
t
Fundamental and
zero-sequence
signal
48
t
Modulating signals
Analysis of SVM (3)
Control Time
Domain
Space Vector
Domain
Output Time
Domain
Digital implementation related to sampling method
selected, not to the modulation function used.
1
ei 0 (t )   cos(3t ) vs. SVM
6
mmax
49
2

 1.15
3