1 inverter applications motor drives power back-up systems others: example hvdc transmission systems
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1
Inverter ApplicationsInverter Applications
•Motor Drives
• Power back-up systems
•Others:
Example HVDC Transmission systems
2
Single Phase InverterSingle Phase InverterSquare-wave “Modulation” (1)Square-wave “Modulation” (1)
Vdc
Q1
Q2Q’1
Q’2
vout
+
+
-
-
q1(t)
q2(t)1-q1(t)
1-q2(t)
vout(t)
01
4 sin( / 2)( ) cos( )out dc
n
nv t V n t n
n
-Vdc
Vdc
dcdcout VVV 27.14
1
t
3
THD = 0.48
Characteristics: - High harmonic content. - Low switching frequency. - Difficult filtering. - Little control flexibility.
Single Phase InverterSingle Phase InverterSquare-wave “Modulation” (2)Square-wave “Modulation” (2)
4
1
coscos)2/sin(2
)(n
SSin
out ntntnn
nVtv
2
cos4
1
dcout VV
Single Phase InverterSingle Phase InverterSquare-wave “Modulation” (3)Square-wave “Modulation” (3)
vout(t)
-Vdc
Vdc
tVdc
Q1
Q2Q’1
Q’2
vout
+
+
-
-
q1(t)
q2(t)1-q1(t)
1-q2(t)
5
Characteristics: - High harmonic content. - Low switching frequency. - Difficult filtering. - More control flexibility.
THD = 0.3
Example with Vout-1=1.21Vdc
Single Phase InverterSingle Phase InverterSquare-wave “Modulation” (4)Square-wave “Modulation” (4)
6
Single Phase InverterSingle Phase InverterPulse Width Modulation (1)Pulse Width Modulation (1)
)12())(1( DVVDDVv dcdcdcout
D is the duty cycle of switch Q1. D is the portion ofthe switching cycle during which Q1 will remainclosed.
S
QON
T
TD 11
In PWM D is made a function of time
D=D(t)
7
Single Phase InverterSingle Phase InverterPulse Width Modulation (2)Pulse Width Modulation (2)
)(21
21
tmD •Let’s
where 0( ) [cos( )]m t m t
Modulationfunction
FundamentalSignal
Modulationindex
dc
out
VV
m 1
8
0
1[1 cos( )]
2D m t
1
)cos(2)(
2sin
4)()(
nS
dcdcout tn
n
tmnV
Vtmtv
Moving average
dcdcout VtmDVv )()12(
Single Phase InverterSingle Phase InverterPulse Width Modulation (3)Pulse Width Modulation (3)
t
9
Single Phase InverterSingle Phase InverterPulse Width Modulation (4)Pulse Width Modulation (4)
Implementation issue: time variable “t” needs to be sampled. Two basic sampling methods: UPWM
NSPWM
10
Single Phase InverterSingle Phase InverterPulse Width Modulation (5)Pulse Width Modulation (5)
t
11
NSPWM UPWM Fundamental (n=1) and its
sidebands
dcmV no sidebands
2sin
124
1
n
k
nk
nmJ
n
kV
i
ndc
Carrier component
(n = k)
2sin
2
4
10
iimJ
V
i
dc
2sin
2
14
10
iimJ
i
V
i
dc
Sidebands of the carrier
2sin
24
1
in
i
imJ
V
i
ndc
2sin
124
1
n
k
nk
nmJ
n
kV
i
ndc
Single Phase InverterSingle Phase InverterPulse Width Modulation (6)Pulse Width Modulation (6)
Considering that )cos()( 0tmtm
12
Three Phase InverterThree Phase InverterPulse Width Modulation (1)Pulse Width Modulation (1)
Active States:
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
Zero States:
State Qa Qb Qc
S0 0 0 0
S7 1 1 1
13
)]()[cos()( 00 tetmtm aa )]()3/2[cos()( 00 tetmtm bb )]()3/2[cos()( 00 tetmtm cc
Three Phase InverterThree Phase InverterPulse Width Modulation (2)Pulse Width Modulation (2)
ModulationFunctions
)()()( 0300 tetete Hcc
)()()( 0300 tetete Hbb
)()()( 0300 tetete Haa “Zero-Sequence”
Signals
14
Three Phase InverterThree Phase InverterPulse Width Modulation (3)Pulse Width Modulation (3)
)]()()[cos()( 0300 tetetmtm Haa
FundamentalSignal
Modulationindex
2/dc
peakph
V
Vm
Triplen Harmonics (3, 9, 15, …)
OtherHarmonics
(5, 7, 11, 13, 17, ….)
15
Three Phase InverterThree Phase InverterPulse Width Modulation (4)Pulse Width Modulation (4)
)]3cos()6/1()[cos()]()[cos()( 00300 ttmtetmtma
)3cos()6/1()( 030 tmte
)cos( 0tm
t
16
Classic Approaches to PWM (1)Classic Approaches to PWM (1)Time DomainTime Domain
Use of modulation signal: Duty cycle computation:
)(21
21
)(21
21
)(21
21
scc
sbb
saa
kTmD
kTmD
kTmD
t
Sector limits
17
Disadvantages: Can’t see evolution in time Loss of information about e0-3(t)
Classic Approaches to PWM (2)Classic Approaches to PWM (2)Classic SVM - ApplicationClassic SVM - Application
InverterDC
SourceMotor
SwitchControl
d-PIController
q-PIController
DomainTransformer(Modulator)
SpaceVector Time
+
+
-
-
3 2
Park’sTransformation
* ( )d sM kT* ( )q sM kT
( )refd sM kT( )ref
q sM kT
( )d sM kT
( )q sM kTeqM
edM
refv
(From Vector Controller)
* ( )ai t* ( )bi t
* ( )ci t
Park’s Transformation
c
b
a
q
d
i
i
i
M
M
23
23
0
21
21
1
32
*
*
)cos( tia
)3/2cos( tib
)3/2cos( tic
)cos(* tM d
)sin(* tM q
18
SECTORI
4v̂trefv
SECTORII
SECTORIII
SECTORIV
SECTORV
SECTORVI
6v̂2v̂
3v̂
1̂v 5v̂
0S
O
7S
6S2S
3S
1S5S
In O:
4Sdv̂
qv̂
0v̂7v̂
Space vector domain
Classic Approaches to PWM (3)Classic Approaches to PWM (3)Classic SVMClassic SVM
2 Bases
qddq vvB ˆˆ
jiSVM vvB ˆˆ
BSVM in sector I4ˆˆ vvi
6ˆˆ vv j
BSVM changes ineach sector
19
-Track the sector in which is in and based on it
select the appropriate set of basis Bij
-Calculate the coordinates of in the basis Bij
- Change the coordinates of the reference voltage vector
S4
S6
S2
S5S1
S0
S7
S3
Transitions withinsector I
Classic Approaches to PWM (4)Classic Approaches to PWM (4)Classic SVMClassic SVM
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.
SVM Computation
refv
refv
T7=T0
20
Then, a 3-D vector canbe introduced:
Mathematical Framework (1)Mathematical Framework (1)
Control TimeDomainSpace Vector
DomainOutput Time
Domain
Complete representationinvolves a 3-D space
v
R
W
S
21
Mathematical Framework (2)Mathematical Framework (2)Control Time DomainControl Time Domain
Functions of time are used as basis
)()3/2cos(ˆ
)()3/2cos(ˆ
)()cos(ˆ
ˆ,ˆ,ˆ
0
0
0
tett
tett
tett
tttB
cc
bb
aa
cbat
)ˆˆˆ( cba tttmv
22
Mathematical Framework (3)Mathematical Framework (3)Space Vector DomainSpace Vector Domain
T
t
T
T
tdq
ten
q
d
nqdB
)(00ˆ
010ˆ
001ˆ
ˆ,ˆ,ˆ
300
00
tqd ntvqtvdtvv 00 ˆ)(ˆ)(ˆ)(
When e0-H(t)=0, describes a circumferencewith radius equal to m.
v
23
Mathematical Framework (4)Mathematical Framework (4)Output Time DomainOutput Time Domain
T
c
T
b
T
a
cbaS
v
v
v
vvvB
100ˆ
010ˆ
001ˆ
ˆ,ˆ,ˆ
ˆ ˆ ˆ( ) ( ) ( )a a b b c cv v t v v t v v t v
24
Mathematical Framework (5)Mathematical Framework (5)Output Time DomainOutput Time Domain
Length of sides equal to 2
Each corner represents one state
State sequence is obtained naturally
Sectors: six pyramid-shaped volumes bounded by sides of the cube and |vi|=|vj| planes (i, j = a,b,c; i j).
0 cbadq vvvB
Balanced system plane
25
Mathematical Framework (6)Mathematical Framework (6)Matrix RMatrix R
When e0-H(t)=0, describes a circumferencewith radius equal to m.
v
1st Idea: Use Park’s transformation to a synchronousrotating reference frame:
))()3/2(cos(
))()3/2(cos(
))()(cos(
111
)3/2cos()3/2sin()sin(
)3/2cos()3/2cos()cos(
32
30
30
30
0 tetm
tetm
tetm
ttt
ttt
v
v
v
q
d
Instantaneous values of the voltages
26
Mathematical Framework (7)Mathematical Framework (7)Matrix RMatrix R
)(2
1
1
300 tmev
v
v
q
d Problem: components in and are constant values
d̂ q̂
I am interested in having a constant value in tn0ˆ
2nd Idea: Freeze the rotational reference frame at t=0
)(2
)sin(
)cos(
))()3/2(cos(
))()3/2(cos(
))()(cos(
11123
23
0
21
21
1
32
3030
30
30
0 tme
tm
tm
tetm
tetm
tetm
v
v
v
q
d
27
Mathematical Framework (8)Mathematical Framework (8)Matrix RMatrix R
3rd Idea: Rearrange the product.
m
tet
tet
tet
v
v
v
q
d
)()3/2cos(
)()3/2cos(
)()cos(
11123
23
0
21
21
1
32
30
30
30
0
m
m
m
tetete
tt
ttt
)()()(
)3/2cos()2/3()3/2cos()2/3(0
)3/2cos()2/1()3/2cos()2/1()cos(
32
303030
0dqBv
tBv
28
Mathematical Framework (9)Mathematical Framework (9)Matrix RMatrix R
111
)3/2cos()2/3()3/2cos()2/3(0
)3/2cos()2/1()3/2cos()2/1()cos(
32
tt
ttt
R
2
)sin(
)cos(
0
t
t
m
m
m
m
R
v
v
v
q
d
4th Idea: Eliminate the dependency on e0-3(t) inorder to have a constant coordinate in tn0ˆ
29
111
)3/2cos()2/3()3/2cos()2/3(0
)3/2cos()2/1()3/2cos()2/1()cos(
32
tt
ttt
R
000
)()2/3()()2/3(0
)()2/1()()2/1()(
32
00
000
tete
tetete
HcHb
HcHbHa
111
))()3/2)(cos(2/3())()3/2)(cos(2/3(0
))()3/2)(cos(2/1())()3/2)(cos(2/1()()cos(
32
00
000
tettet
tettettet
R HcHb
HcHbHa
Mathematical Framework (10)Mathematical Framework (10)Matrix RMatrix R
In order to include e0-H(t) we need to follow thesame steps and apply superposition.
30
Mathematical Framework (11)Mathematical Framework (11)Matrix WMatrix W
cba vvvd ˆ21
ˆ21
ˆ'ˆ
cb vvq ˆˆ'ˆ
cba vvvdqn ˆˆˆˆˆ'ˆ 0
c
b
a
v
v
v
n
q
d
ˆ
ˆ
ˆ
111
110
2/12/11
'ˆ
'ˆ
'ˆ
0
W’
31
131
131
102
21
''W
Mathematical Framework (12)Mathematical Framework (12)Matrix WMatrix W
111
110
2/12/11
'W
)(31
)(31
)(02
21
30
30
30
te
te
te
W
1) Take the transpose and apply scaling factor
2) Include e0-3(t) in order to have it as a component
32
Mathematical Framework (13)Mathematical Framework (13)Matrix SMatrix S
)(2)()3/2cos(2)()()3/2cos()()()cos(
)()()3/2cos()(2)()3/2cos(2)()()cos(
)()()3/2cos()()(3/2cos)(2)()cos(2
31
030030030
030030030
030030030
tetettetettetet
tetettetettetet
tetettetettetet
S
HcHbHa
HcHbHa
HcHbHa
)()3/2cos(2)()3/2cos()()cos(
)()3/2cos()()3/2cos(2)()cos(
)()3/2cos()(3/2cos)()cos(2
31
303030
303030
303030
tettettet
tettettet
tettettet
S
)(2)()(
)()(2)(
)()()(2
31
000
000
000
tetete
tetete
tetete
HcHbHa
HcHbHa
HcHbHa
S=WR
33
3D Analysis and Representation3D Analysis and Representation
tdq BB vRv
)(0
0)(
dqS BB vWv
tS BB vSv
)(
)()3/2cos(
)()3/2cos(
)()cos(
0
0
0
ScS
SbS
SaS
c
b
a
tet
tet
tet
m
v
v
v
34
3D Analysis and Representation3D Analysis and Representation)3cos(
41
)()( 300 ttetei
UPWM
35
3D Analysis and Representation3D Analysis and Representation
SVM
UPWM
36
3D Analysis and Representation3D Analysis and Representation
3D Representation: Plotevolution of in outputtime domain during acomplete fundamentalperiod.
The resulting curve alwayslays within the cube definedby the switching states.
v
)3cos(61
)()( 300 ttetei
37
3D Analysis and Representation3D 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.
38
Commonly used schemesCommonly used schemes
)3cos(61
)()( 300 ttetei 0)(0 tei
SVM
Square wave
39
3D Analysis and Representation3D Analysis and Representation
)3cos(61
)()( 300 ttetei Square wave0)(0 tei
40
3D Analysis and Representation3D Analysis and Representation
Phase and Line Voltages (1)
babaab vvvvv ˆˆ ).(21
)( abdcab vvVtv
)ˆˆˆ2(31
)(31
cbacaaboa vvvvvv ).(
21
)( oadcoa vvVtv
Phase Leg a b c Phase
voltages T11231 T1213
1 T21131
Line Voltages T011 T110 T101
41
3D Analysis and Representation3D Analysis and Representation
Phase and Line Voltages (2)
Direction of vab
Direction of voa
voa (t)
t
va (t)
42
3D Analysis and Representation3D Analysis and Representation
Phase and Line Voltages (3)
)3cos()4/1()( 030 tte
)5cos()10/1()( 00 tte H
voa (t)
t
va (t)
43
3D Analysis and Representation3D Analysis and RepresentationMaximum non distorting range
Radius of circle is m if it ismeasured in the Space VectorDomain.
)3cos(61
)()( 300 ttetei
There is a scaling factor in W
dBd vvdq
ˆ0
cbaBd vvvvS
ˆ5.0ˆ5.0ˆ 3
2Sd B
v
10
dqBdv
15.13
22
32
max ND
m
44
3D Analysis and Representation3D Analysis and Representation
ii vv If fswitch>>ffund then
cbaiDDDv iiii , ,, 121)1(
45
3D Analysis and Representation3D Analysis and Representation
t
46
3D Analysis and Representation3D Analysis and Representation
Qa
Qb
Qc
Qb
Qa
Qc
Qb
Qc
Qa
Sector I Sector II Sector III
T7 T6 T4 T0 T7 T6 T2 T0 T7 T3 T2 T0
6) 5, 3, (j , )(
4) 2, 1, (i , )(
)1(
min
max
max0
min7
smedj
smedi
s
s
TDDT
TDDT
TDT
TDT
47
Analysis of SVM (1)Analysis of SVM (1)
T7=T0 Dmin=1-Dmax
vmax +vmin=0
S is sectordependent
)(21
)()( 300 tvtete med
(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
48
Analysis of SVM (2)Analysis of SVM (2)
Fundamental andzero-sequence
signal
Modulating signals
t t
49
Analysis of SVM (3)Analysis of SVM (3)
Control TimeDomain
Space VectorDomain
Output TimeDomain
Digital implementation related to sampling method selected, not to the modulation function used.
SVM vs.)3cos(61
)(0 ttei
15.13
2max m
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