d q transormation
TRANSCRIPT
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dq Theory for Synchronous Machinewithout Damper Winding
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Relevance to Synchronous Machine
dq means direct and quadrature. Direct axis is aligned withthe rotors pole. Quadrature axis refers to the axis whoseelectrical angle is orthogonal to the electric angle of directaxis.
a axis
d axisq axis
b axis
c axis
m a
d
mq
22
2
memqr
mme
P
P min g
max,or r
min,or r max g
isr
d
a axis
m a
q axis
mq
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Parks Transformation
Stator quantities ( S abc) of current, voltage, or flux can beconverted to quantities ( S dq0) referenced to the rotor.
This conversion comes through the K matrix.
01
0
dqabc
abcdq
SK S
KSS
13/2sin3/2cos
13/2sin3/2cos1sincos
2/12/12/1
3/2sin3/2sinsin3/2cos3/2coscos
3
2
1
meme
meme
meme
mememe
mememe
K
K
13/2cos3/2sin13/2cos3/2sin
1cossin
2/12/12/13/2cos3/2coscos
3/2sin3/2sinsin
32
1
r r
r r
r r
r r r
r r r
K
K
where
or
(MITs notation)
(Purdues notation)
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Voltage Equations (1)
abcabcS abc dt d iR v
010101 dqdqS dq dt d
K iK R vK
010101 dqdqS dq dt d K K iK KR vK K
01
01
01
0 dqdqdqS dq dt d
dt d
K K KK iK KR v
01
000 dqdqdqS dq dt d
dt d
K K iR v
100
010001
sS RR
For stator windings
For field winding:
f f f f dt d
i Rv
Under motor reference convention for currents(i.e. the positive reference direction for currents is into the machine):
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Voltage Equations (2)
We derive the derivative of K-1:
Then, we get
000
00
001
r
r
dt d
K K
f f f
s
r d qq s
r qd d s
f
q
d
dt d i R
dt d
i R
dt d i R
dt d
i R
v
v
vv
000
03/2sin3/2cos
03/2sin3/2cos0sincos
03/2cos3/2sin03/2cos3/2sin0cossin
1
r r
r r
r r
r
meme
meme
meme
me
dt
d K
dt d me
me
And for voltage, we get
2 P
dt d
mmer
r
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Dynamical Equations for Flux Linkage
f f f
s
r d q sq
r qd sd
f
q
d
i Rvi Rv
i Rv
i Rv
dt
d
000
The derivations so far are valid for both linear and nonlinear models.
f
q
d
dqf
0
f f f
s
r d q sq
r qd sd
i Rv
i Rv
i Rv
i Rv
00
V
Let
we haveV
dt
d dqf
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Flux Linkage vs. Current (1)
The next step is to relate current to flux linkage throughinductances. For salient pole rotor, the inductances canbe approximately expressed as
cf
bf
af
sf
L
L
LL
32
2cos
322cos
2cos
me B Alscc
me B Alsbb
me B Alsaa
L L L L
L L L L
L L L L
or:
32
2cos
322cos
2cos
r B Alscc
r B Alsbb
r B Alsaa
L L L L
L L L L
L L L L
2
mer
f T sf
sf ssabcf LL
LLL
cccbca
bcbbba
acabaa
ss
L L L
L L L
L L LL
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Flux Linkage vs. Current (2)
32
2cos21
2cos21
32
2cos21
me B Acaac
me B Acbbc
me B Abaab
L L L L
L L L L
L L L L
Note: Higher order harmonics are neglected.
3
2cos
32
cos
cos
me sf fccf
me sf fbbf
me sf faaf
L L L
L L L
L L Lor:
32
2cos21
2cos21
32
2cos21
r B Acaac
r B Acbbc
r B Abaab
L L L L
L L L L
L L L L
32
sin
32
sin
sin
r sf fccf
r sf fbbf
r sf faaf
L L L
L L L
L L L
2
mer
Ref: 1. A. E. Fitzgerald, C. Kingsley, Jr., and S. D. Umans, Electric Machinery,6 th Edition, pages 660-661.
2. P. C. Krause, O. Wasynczuk, and S. D. Sudhoff, Analysis of Electric Machinery and Drive Systems, 2 nd Edition, page 52.
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Flux Linkage vs. Current (3)
This matrix can be transformed into dq0 form and used tofind flux linkage.
abcf abcf abcf iL
f sf dq ssdq iLiK L K 01
01
dqf dqf dqf iL
f sf dq ssdq iKLiK KL 01
0
f T sf
sf ss
abcf LL
LLL
f
abc
abcf
f
abc
abcf i
ii
f sf abc ssabc iLiL
f f abcT sf f i L iL f f dq
T sf f i L 0
1iK L
f T sf
sf ssdqf
L1
1
K L
KLK KLL
f
dqdqf
0
f
dqdqf
i
0i
i
From with
where
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Inductance Matrix in dq0 Frame
f sf
q
sf d
dqf
L L
L
L
L L
0023
000
000
00
0L
where
)(23
)(
2
3
B Amq
B Amd
L L L
L L L
ls
mqlsq
md lsd
L L
L L L
L L L
0
and
dqf dqf dqf iL
From d sf f f f
ls
qqq
f sf d d d
i Li L
i L
i Li Li L
23
00
Through derivations, we have
f T sf
sf ssdqf L1
1
K L
KLK KL
L
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Dynamical Equation in terms of Current
V
dt
d dqf For linear model
from
VLi
1dqf
dqf
dt
d dynamical equationin terms of current
dqf dqf dqf iL
f f f
s
r d q sq
r qd sd
i Rv
i Rv
i Rv
i Rv
00
V
and
where
qqq
f sf d d d
i L
i Li L
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Power
Electrical instantaneous Input Power on Stator can also beexpressed through dq0 theory.
011
0 )( dqT T
dqabcT abcccbbaain iviviv p iK K viv
00223 iviviv p qqd d in
200010
001
23
)(11
K K T
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Torque
00223
iviviv p qqd d in
000
dt d
i Rdt
d i R
dt d
i R
v
v
v
s
r d qq s
r qd d s
q
d
From
we have
)(223
223
223 0
020
22d qqd m
q
qd
d qd sin ii P
dt d
idt
d idt
d iiii R p
Copper Loss Mechanical PowerMagnetic Power inWindings
Therefore, electromagnetic torque on rotor
)(223
d qqd m
meche ii P p
T
mech p
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Equivalent Circuits (1)
f f f
s
r d qq s
r qd d s
f
q
d
dt d
i R
dt d
i Rdt
d i R
dt d
i R
v
v
v
v
000
d sf f f f
ls
qqq
f sf d d d
i Li L
i L
i L
i Li L
23
00
d axisdt
di L
dt di
Li Rv f sf d
d r qd sd
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Equivalent Circuits (2)
q axisdt
di Li Rv qqr d q sq
0 axisdt di
Li Rv s0
000
This circuit is not necessaryfor Y connected windingssince i 0=0.
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Equivalent Circuits (3)
Field windingdt di
Ldt
di Li Rv d sf
f f f f f 2
3
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Combined Equivalent Circuit on d Axis (1)
dt
iid
Ldt
di
Li R
dt
di L
dt di
Li Rv
f d
md
d
lsr qd s
f sf
d d r qd sd
)( '
d axis equivalent circuit and field winding equivalent circuit can be combined:
md lsd L L L
mf lf f L L L
mf
sf
sf
md
f
a
L
L
L
L
N
N N
3
2
N
ii
L
Li f f
md
sf f
3
2'
f ad sf
f d mf
ad md
N N C L
N C L
N C L
23
2
2)
21(
82
0 g
avd P g
Dl C
From
(Details @ InductanceSM.ppt)
Let
a N f N and are effective number of turns of armature andfield windings.
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Combined Equivalent Circuit on d Axis (2)
dt di
Ldt
di Li Rv d sf
f f f f f 2
3 '23
f f Nii
dt di
NLdt
di NLdt
di NLi NR Nv
d sf
f mf
f lf f f f 2
3
mf
sf
sf
md
L
L
L L
N 32
dt di
Ldt
di L
dt
di L N i R N Nv d md
f sf
f lf f f f
'2'2
23
23
dt
iid L
dt
di Li Rv f d md
f lf f f f
)( ''''''
lf lf
f f
f f
L N L
R N R
Nvv
2'
2'
'
2323
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Combined Equivalent Circuit on d Axis (3)
dt
iid L
dt di
Li Rv f d md d
lsr qd sd
)( '
dt iid
Ldt
di Li Rv f d md
f lf f f f
)( ''''''
From
f f Nvv'
'
23
f f Nii
we get
md md d ls
f sf d d d
i Li L
i Li L
' f d md iii
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dq Theory for Permanent MagnetSynchronous Machine (PMSM)
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Parks Transformation
Stator quantities ( S abc) of current, voltage, or flux can beconverted to quantities ( S dq0) referenced to the rotor.
This conversion comes through the K matrix.
01
0
dqabc
abcdq
SK S
KSS
13/2sin3/2cos
13/2sin3/2cos1sincos
2/12/12/1
3/2sin3/2sinsin3/2cos3/2coscos
32
1
meme
meme
meme
mememe
mememe
K
K
13/2cos3/2sin
13/2cos3/2sin1cossin
2/12/12/13/2cos3/2coscos3/2sin3/2sinsin
32
1
r r
r r
r r
r r r
r r r
K
K
where
or
(MITs notation)
(Purdues notation)
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Voltage Equations (1)
abcabcS abc dt d iR v
010101 dqdqS dq dt d
K iK R vK
01
01
01
dqdqS dq dt d
K K iK KR vK K
01
01
01
0 dqdqdqS dq dt d
dt d
K K KK iK KR v
01
000 dqdqdqS dq dt d
dt d
K K iR v
100
010001
sS RR
For stator winding
Under motor reference convention for currents(i.e. the positive reference direction for currents is into the machine):
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Voltage Equations (2)
We derive the derivative of K-1:
Then, we get
000
00
001
r
r
dt d
K K
000
dt
d i R
dt d
i Rdt
d i R
v
vv
s
r d qq s
r qd d s
q
d
03/2sin3/2cos03/2sin3/2cos0sincos
03/2cos3/2sin
03/2cos3/2sin0cossin
1
r r
r r
r r
r
meme
meme
meme
medt d
K
dt d me
me
And for voltage, we get
2 P
dt d
mmer
r
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Dynamical Equations for Flux Linkage
000 i Rv
i Rv
i Rv
dt
d
s
r d q sq
r qd sd
q
d
The derivations so far are valid for both linear and nonlinear models.
0
0
q
d
dq
00 i Rv
i Rvi Rv
s
r d q sq
r qd sd
V
Let
we haveV
dt
d dq 0
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Flux Linkage vs. Current (1)
The next step is to relate current to flux linkage throughinductances. For salient pole rotor, the inductances canbe approximately expressed as
cccbca
bcbbba
acabaa
abc
L L L
L L L
L L LL
32
2cos
322cos
2cos
me B Alscc
me B Alsbb
me B Alsaa
L L L L
L L L L
L L L L
Note: Higher order harmonics are neglected.
or:
32
2cos
322cos
2cos
r B Alscc
r B Alsbb
r B Alsaa
L L L L
L L L L
L L L L
2
mer
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Flux Linkage vs. Current (2)
32
2cos21
2cos21
32
2cos21
me B Acaac
me B Acbbc
me B Abaab
L L L L
L L L L
L L L L
Note: Higher order harmonics are neglected.
or:
32
2cos21
2cos21
32
2cos21
r B Acaac
r B Acbbc
r B Abaab
L L L L
L L L L
L L L L
2
mer
Ref: 1. A. E. Fitzgerald, C. Kingsley, Jr., and S. D. Umans, Electric Machinery,6 th Edition, pages 660-661.
2. P. C. Krause, O. Wasynczuk, and S. D. Sudhoff, Analysis of Electric Machinery and Drive Systems, 2 nd Edition, page 52,
also pages 264-265.
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Flux Linkage vs. Current (4)
This matrix can be transformed into dq0 form and used tofind flux linkage.
PMabcabcabcabc iL
PMabcdqabcf dq iK L K 0101
PMabcdqabcdq K iK KL K K 0101
0000 PMdqdqdqdq iL
PMabcdqabcdq K iK KL 01
0
where
)3/2cos(
)3/2cos(
)cos(
me
me
me
PM PMabc
or:
2
mer )3/2sin(
)3/2sin(
)sin(
r
r
r
PM PMabc
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Inductance Matrix in dq0 Frame
Therefore, we get the following inductance matrix in dq0frame:
0
10
0000
00
L L
L
q
d
abcdq K KLL
where
)(23
)(23
B Amq
B Amd
L L L
L L L
ls
mqlsq
md lsd
L L
L L L
L L L
0
and
From00 i L
i L
i L
ls
qqq
PM d d d
0000 PMdqdqdqdq iL
000
PM
PMabc PMdq
K
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Dynamical Equation in terms of Current
V
dt
d dq 0For linear model from
VLi
1
0
0
dq
dq
dt
d dynamical equationin terms of current
00 i Rv
i Rv
i Rv
s
r d q sq
r qd sd
V
and
whereqqq
PM d d d
i L
i L
0000 PMdqdqdqdq iL
0
0
00
00
00
L
L
L
q
d
dqL
0000 /)(/)(
/)(
Li Rv Li Li Rv
Li Li Rv
ii
i
dt d
s
q PM r d d r q sq
d qqr d sd
q
d
For Y connected winding, since , only need to considerthe first two equations for i d and i q .
0)(3
10 cba iiii
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Power
Electrical instantaneous Input Power on Stator can also beexpressed through dq0 theory.
011
0 )( dqT T
dqabcT abcccbbaain iviviv p iK K viv
00223 iviviv p qqd d in
200010
001
23
)(11 K K T
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Torque
00223
iviviv p qqd d in
000
dt d i R
dt d
i R
dt d
i R
v
v
v
s
r d qq s
r qd d s
q
d
From
we have
)(223
22
3
22
3 00
2
0
22
d qqd m
q
q
d
d qd sin ii
P
dt
d
idt
d
idt
d
iiii R p
Copper LossMechanical PowerMagnetic Power in
WindingsTherefore, electromagnetic torque on rotor
)(22
3d qqd
m
meche ii
P pT
mech p
qqq
PM d d d
i L
i L
qd qd q PM e ii L Li P T )(
22
3
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Dynamical Equations of Motion
mm
damp Lem
dt d
T T T dt
d J
where
qT qd qd q PM e i K ii L Li P T )(22
3
For round rotor machine, qd L L q PM e i P T 43
mmdamp DT D m is combined damping coefficient of rotorand load.
d qd PM q
eT i L L
P iT
K )(4
3 torque constant
PM T P K 43
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dq Theory for Synchronous Machinewith Damper Winding
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Relevance to Synchronous Machine
dq means direct and quadrature. Direct axis is aligned withthe rotors pole. Quadrature axis refers to the axis whoseelectrical angle is orthogonal to the electric angle of direct
axis.
a axis
d axisq axis
b axis
c axis
m a
d
mq
22
2
memqr
mme
P
P min g
max,or r
min,or r max g
isr
d
a axis
m a
q axis
mq
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Parks Transformation
Stator quantities ( S abc) of current, voltage, or flux can beconverted to quantities ( S dq0) referenced to the rotor.
This conversion comes through the K matrix.
01
0
dqabc
abcdq
SK S
KSS
13/2sin3/2cos
13/2sin3/2cos1sincos
2/12/12/1
3/2sin3/2sinsin3/2cos3/2coscos
32
1
meme
meme
meme
mememe
mememe
K
K
13/2cos3/2sin
13/2cos3/2sin1cossin
2/12/12/13/2cos3/2coscos3/2sin3/2sinsin
32
1
r r
r r
r r
r r r
r r r
K
K
where
or
(MITs notation)
(Purdues notation)
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Voltage Equations (1)
abcabcS abc dt d iR v
010101 dqdqS dq dt d
K iK R vK
01
01
01
dqdqS dq dt d
K K iK KR vK K
01
01
01
0 dqdqdqS dq dt d
dt d
K K KK iK KR v
01
000 dqdqdqS dq dt d
dt d
K K iR v
100
010
001
sS RR
For stator windings
Under motor reference convention for currents(i.e. the positive reference direction for currents is into the machine):
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Voltage Equations (2)
We derive the derivative of K-1:
Then, we get
00000
001
r
r
dt d
K K
000
dt d
i R
dt d
i Rdt
d i R
v
vv
s
r d qq s
r qd d s
q
d
03/2sin3/2cos
03/2sin3/2cos0sincos
03/2cos3/2sin
03/2cos3/2sin0cossin
1
r r
r r
r r
r
meme
meme
meme
medt d
K
dt d me
me
And for stator voltage, weget
2 P
dt d
mmer
r
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Voltage Equations (3)
For rotor windings:
We assume the rotor has field winding (magnetic field along d axis),one damper with magnetic field along d axis and one damper
with magnetic field along q axis.
qd qd qd k fk k fk r k fk dt d
iR v
q
d
k
k
f
r
R
R
R
00
00
00R
0
0 f
k fk
v
qd v
q
d qd
k
k
f
k fk
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Voltage Equations (4)
qqq
d d d
k k k
k k k
f f f
s
qr d q s
d r qd s
f
q
d
dt
d i R
dt d
i Rdt
d i R
dt d
i Rdt
d i R
dt d
i R
v
v
v
v
000
00
In summary:
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Dynamical Equations for Flux Linkage
qq
d d
q
d
k k
k k
f f f
s
r d q sq
r qd sd
k
k
f
q
d
i R
i Ri Rv
i Rv
i Rv
i Rv
dt
d 000
The derivations so far are valid for both linear and nonlinear models.
Let
we have V
dt
d dqf
q
d
k
k
f
q
d
dqf
0
qq
d d
k k
k k
f f f
s
r d q sq
r qd sd
i R
i R
i Rv
i Rv
i Rv
i Rv
00
V
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Flux Linkage vs. Current (1)
The next step is to relate current to flux linkage throughinductances. For salient pole rotor, the inductances canbe approximately expressed as
32
2cos
322cos
2cos
me B Alscc
me B Alsbb
me B Alsaa
L L L L
L L L L
L L L L
or:
32
2cos
322cos
2cos
r B Alscc
r B Alsbb
r B Alsaa
L L L L
L L L L
L L L L
2
mer
rr T sr
sr ssabcf LL
LLL
cccbca
bcbbba
acabaa
ss
L L L
L L L
L L LL
qd
qd
qd
ck ck cf
bk bk bf
ak ak af
sr
L L L
L L L
L L LL
qd qq
qd d d
qd
k k k f k
k k k f k
fk fk f
rr
L L L
L L L
L L LL
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Flux Linkage vs. Current (2)
32
2cos21
2cos21
32
2cos21
me B Acaac
me B Acbbc
me B Abaab
L L L L
L L L L
L L L L
32
cos
3
2cos
cos
me sf fccf
me sf fbbf
me sf faaf
L L L
L L L
L L Lor:
32
2cos21
2cos21
32
2cos21
r B Acaac
r B Acbbc
r B Abaab
L L L L
L L L L
L L L L
32
sin3
2sin
sin
r sf fccf
r sf fbbf
r sf faaf
L L L
L L L
L L L
2
mer
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Flux Linkage vs. Current (3)
32
cos
32
cos
cos
me sk ck ck
me sk bk bk
me sk ak ak
d d d
d d d
d d d
L L L
L L L
L L L
or:
32
sin
32
sin
sin
r sk ck ck
r sk bk bk
r sk ak ak
d d d
d d d
d d d
L L L
L L L
L L L
2 mer
32
sin
32
sin
sin
me sk ck ck
me sk bk bk
me sk ak ak
qqq
qqq
qqq
L L L
L L L
L L L
32
cos
32
cos
cos
r sk ck ck
r sk bk bk
r sk ak ak
qqq
qqq
qqq
L L L
L L L
L L L
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Flux Linkage vs. Current (4)
Note: Higher order harmonics are neglected in the above expressions.
Ref: 1. A. E. Fitzgerald, C. Kingsley, Jr., and S. D. Umans, Electric Machinery,6 th Edition, pages 660-661.
2. P. C. Krause, O. Wasynczuk, and S. D. Sudhoff, Analysis of Electric Machinery and Drive Systems, 2 nd Edition, pages 52and 195.
qqq
d d d
mk lk k
mk lk k
mf lf f
L L L
L L L
L L L
0
0
d qqd
qq
d d
k k k k
f k fk
f k fk
L L
L L
L L
l i k C ( )
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Flux Linkage vs. Current (5)
This matrix can be transformed into dq form and used tofind flux linkage.
abcf abcf abcf iL
qd k fk sr dq ssdqiLiK L K
01
01
dqf dqf dqf iL
qd k fk sr dq ssdqiKLiK KL
0
1
0
r T sr
sr ssabcf
LL
LLL
qd k fk
abc
abcf
qd k fk
abc
abcf i
ii
qd k fk sr abc ssabciLiL
qd qd k fk rr abcT sr k fk
iLiL qd qd k fk rr dq
T sr k fk
iLiK L 0
1
rr
T
sr
sr ssdqf LK L
KLK KLL
1
1
qd k fk
dqdqf
0
qd k fk
dqdqf i
ii
0
From with
where
I d M i i d F
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Inductance Matrix in dq Frame
qq
d d d
d
q
d
k sk
k f k sk
fk f sf
sk q
sk sf d
dqf
L L
L L L
L L L L
L L
L L L
000230
00023
00023 00000
0000
000
0
L
where
)(23
)(23
B Amq
B Amd
L L L
L L L
mqlsq
md lsd
L L L
L L L
and
dqf dqf dqf iL From q sk k k k
f fk d sk k k k
k fk d sf f f f
k sk qqq
k sk f sf d d d
i Li L
i Li Li L
i Li Li L
i L
i Li L
i Li Li L
qqqq
d d d d d
d d
qq
d d
2323
23
000
Through derivations, we have
rr T sr
sr ss
dqf LK L
KLK KLL
1
1
ls L L0
D i l E ti i t f C t
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Dynamical Equation in terms of Current
V
dt
d dqf For linear model
from
VLi 1dqf
dqf
dt d dynamical equation
in terms of current
dqf dqf dqf iL
qq
d d
k k
k k
f f f
s
r d q sq
r qd sd
i R
i Ri Rv
i Rv
i Rvi Rv
00
V
and
where
qq
d d
k sk qqq
k sk f sf d d d
i Li L
i Li Li L
P
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Power
Electrical instantaneous Input Power on Stator can also beexpressed through dq0 theory.
0
11
0 )( dqT T
dqabc
T
abcccbbaain iviviv piK K viv
00223 iviviv p qqd d in
200
010
001
2
3)( 11 K K T
Torque
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Torque
00223
iviviv p qqd d in
000
dt d i R
dt d
i R
dt d
i R
v
v
v
s
r d qq s
r qd d s
q
d
From
we have
)(22
32
2
32
2
3 00
20
22d qqd m
qq
d d qd sin ii
P
dt
d i
dt
d i
dt
d iiii R p
Copper LossMechanical PowerMagnetic Power inWindings
Therefore, electromagnetic torque on rotor
)(22
3d qqd
m
mech
e ii
P p
T
mech p
Equivalent Circuit on d Axis (1)
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Equivalent Circuit on d Axis (1)
d axis of stator, field winding and d axis damper of rotor can form an equivalentcircuit.
Let md lsd L L L
mf lf f L L L
d d d mk lk k L L L
d d
d d
d d
k f d fk
k ad sk
k d mk
f ad sf
f d mf
ad md
N N C L
N N C L
N C L
N N C L
N C L
N C L
2
3
2
2
2
)21(8
20 g
avd P g
Dl C
From
(Details @ Inductance for SM.ppt)
, a N f N and are effective number of turns of armature, field and d axis
damper windings, respectively.
d k N
Equivalent Circuit on d Axis (2)
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Equivalent Circuit on d Axis (2)
dt
iiid L
dt di
Li R
dt
di L
dt
di L
dt
di Li Rv
d
d
d
k f d md
d lsr qd s
k sk
f sf
d d r qd sd
)( ''
f a
f f
md
sf f i
N
N i
L
Li
32'Define
d
d
d
d
d k a
k k
md
sk k i
N
N i
L
Li
32'and
dt
di L
dt di
Ldt
di Li Rv d
d
k fk
d sf
f f f f f 2
3
dt di NL
dt di NL
dt di NL
dt di NLi NR Nv d d k fk d sf f mf f lf f f f 2
3
dt
di L
N
N
N
N
dt
di L
dt
di L
dt
di L N i R N Nv d
d
d
k fk
k
a
f
ad md
f sf
f lf f f f
''2'2
2
3
2
3
2
3
Define
f f Nvv'
a
f
N N
N and
Equivalent Circuit on d Axis (3)
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Equivalent Circuit on d Axis (3)
dt
iiid L
dt
di Li Rv d k f d md
f lf f f f
)( '''''''
lf f
alf
f f
a
f
L N
N L
R N
N
R2
'
2
'
23
2
3
where
2
32
a
k f
md
fk
N N N
L L d d
dt
di
Ldt di
Ldt
di
Ldt
di
Li R
dt
di L
dt di
Ldt
di Li R
f
fk
d
sk
k
mk
k
lk k k
f fk
d sk
k k k k
d d
d
d
d
d d d
d d
d
d d d
23
23
0From
above
3
2
d d k
a
sk
md
N
N
L
L
next page
Equivalent Circuit on d Axis (4)
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Equivalent Circuit on d Axis (4)
dt
di L
N N
N dt di
Ldt
di L
N
N dt
di L
N
N i R
N
N f fk
k f
ad md
k mk
k
ak lk
k
ak k
k
ad
d
d
d
d
d
d
d
d d
d
'2'2
'2
'
2
23
23
23
23
0
dt
iiid Ldt
di Li R
d d
d d d
k f d md
k lk k k
)(0
''''''
where
d
d
d
d
d
d
lk k
alk
k k
ak
L N
N L
R N
N R
2
'
2
'
23
2
3
d d
d d
k f
a
fk
md
k
a
mk
md
N N
N L L
N
N L L
23
23
2
2
Equivalent Circuit on d Axis (5)
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Equivalent Circuit on d Axis (5)
From
we get
md md d ls
k sk f sf d d d
i Li L
i Li Li Ld d
''d k f d md iiii
dt iiid L
dt di Li Rv d k f d md
d lsr qd sd
)(''
dt
iiid L
dt
di Li Rv d k f d md
f lf f f f
)( '''''''
dt iiid L
dt di Li R d d
d d d
k f d md
k lk k k )(0
'''
'''
'd k
i
Equivalent Circuit on q Axis (1)
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Equivalent Circuit on q Axis (1)
q axis equivalent circuit and q axis damper equivalent circuitcan be combined:
Letmqlsq L L L
qqq mk lk k L L L
qq
qq
k aq sk
k qmk
aqmq
N N C L
N C L
N C L
23
2
2)
21(8 2
0
P g Dl C av
qFrom
(Details @ Inductance for SM.ppt)
s N and are effective number of turns of
stator and q axis damperwindings, respectively.
d k N
Equivalent Circuit on q Axis (2)
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Equivalent Circuit on q Axis (2)
dt
iid L
dt
di Li R
dt
di L
dt
di L
dt
di Li R
dt
di L
dt
di Li Rv
q
q
q
q
q
k q
mqq
lqr d q s
k
sk q
mqq
lqr d q s
k
sk q
qr d q sq
)(
'
q
q
q
q
d k
a
k
k
md
sk
k i N
N i
L
Li
3
2'where
dt
di
Ldt
di
Ldt
di
Li R
dt
di L
dt
di Li R
q
sk
k
mk
k
lk k k
q sk
k
k k k
q
q
q
q
qqq
q
q
qqq
23
23
0From
above
3
2
qq k
a
sk
mq
N
N
L
L
next page
Equivalent Circuit on q Axis (3)
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q q ( )
dt
di L
dt
di L
N
N dt
di L
N
N i R
N
N qmq
k
mk k
ak
lk k
ak k
k
a q
q
q
q
q
q
qq
q
'2'2'
2
23
23
23
0
dt
iid Ldt
di Li R
qq
qqq
k q
mq
k
lk k k
)(0
'''''
where
q
q
q
q
q
q
lk k
alk
k k
ak
L N
N L
R N
N R
2
'
2
'
23
2
3
qq
qq
k f
a
fk
mq
k
a
mk
mq
N N
N L
L
N
N
L
L
23
2
3
2
2
Equivalent Circuit on q Axis (4)
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q q ( )
From
we get
mqmqqls
k sk qqq
i Li L
i Li Lqq
'qk qmq
iii
dt iid L
dt di Li Rv q
k qmq
qlqr d q sq
)('
dt
iid L
dt
di Li R qq
qqq
k q
mq
k
lk k k
)(0
'''''
'qk
i
Equivalent Circuit on 0 Axis
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q
0 axisdt di
Li Rv s0
000
This circuit is not necessaryfor Y connected windingssince i 0=0.
Dynamical Equations fromEquivalent Circuits (1)
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Equivalent Circuits (1)
dt di
Li Rv
dt di L
dt di Li Rv
dt
di L
dt
di Li R
dt di L
dt di Li R
dt
di L
dt
di Li Rv
dt di L
dt di Li Rv
ls s
md md
f lf f f f
mqmq
k
lk k k
md md
k lk k k
mqmq
qlsqd r q sqq
md md
d lsd qr d sd d
q
qqq
d
d d d
00000
'''''
''''
''''
0
0
where
'
''
q
d
k qmq
k f d md
iii
iiii
Dynamical Equations fromEquivalent Circuits (2)
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Equivalent Circuits (2)The equations can be written in matrix form as:
VI
Ldt d
where
mqlk mq
md lk md md
md md lf md
ls
mqmqlsq
md md md lsd
L L L
L L L L
L L L L
L
L L L
L L L L
q
d '
'
'0
0000000
000
000000000
000
L
'
'
'0
q
d
k
k
f
q
d
i
i
ii
i
i
I
''
''
'''000
qq
d d
k k
k k
f f f
s
d r q sqq
qr d sd d
i R
i R
i Rvi Rv
i Rv
i Rv
V
VLI 1
dt d or