15 dq theory
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Synchronous Machine
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Relevance to Synchronous Machine
dq means direct and quadrature. Direct axis is aligned with
the rotors pole. Quadrature axis refers to the axis whose
electrical angle is orthogonal to the electric angle of direct
axis.
a axis
daxisqaxis
baxis
caxis
ma
d
mq
22
2
memqr
mme
P
Pming
max,orr
min,orrmaxg
isr
d
a axis
m
a
q axis
mq
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Parks Transformation
Stator quantities (Sabc) of current, voltage, or flux can be
converted to quantities (Sdq0) referenced to the rotor.
This conversion comes through the K matrix.
0
1
0
dqabc
abcdq
SKS
KSS
13/2sin3/2cos
13/2sin3/2cos
1sincos
2/12/12/1
3/2sin3/2sinsin
3/2cos3/2coscos
3
2
1
meme
meme
meme
mememe
mememe
K
K
13/2cos3/2sin13/2cos3/2sin
1cossin
2/12/12/1
3/2cos3/2coscos
3/2sin3/2sinsin
3
2
1
rr
rr
rr
rrr
rrr
K
K
where
or
(MITs notation)
(Purdues notation)
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Voltage Equations (1)
abcabcSabcdtd iRv
010101 dqdqSdqdt
dKiKRvK
010101 dqdqSdq dtd KKiKKRvKK
0
1
0
1
0
1
0 dqdqdqSdqdt
d
dt
dKKKKiKKRv
0
1
000 dqdqdqSdqdt
d
dt
dKKiRv
100
010001
sS RR
For stator windings
For field winding:
ffff dt
diRv
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
KK
fff
s
rdqqs
rqdds
f
q
d
dtdiR
dt
diR
dt
diR
dt
diR
v
v
vv
000
03/2sin3/2cos
03/2sin3/2cos
0sincos
03/2cos3/2sin
03/2cos3/2sin
0cossin1
rr
rr
rr
r
meme
meme
meme
me
dt
dK
dt
d meme
And for voltage, we get
2
P
dt
dmme
rr
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Dynamical Equations for Flux Linkage
fff
s
rdqsq
rqdsd
f
q
d
iRviRv
iRv
iRv
dt
d
000
The derivations so far are valid for both linear and nonlinear models.
f
q
d
dqf
0
fff
s
rdqsq
rqdsd
iRv
iRv
iRv
iRv
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 through
inductances. For salient pole rotor, the inductances can
be approximately expressed as
cf
bf
af
sf
L
L
L
L
3
22cos
322cos
2cos
meBAlscc
meBAlsbb
meBAlsaa
LLLL
LLLL
LLLL
or:
3
22cos
3
22cos
2cos
rBAlscc
rBAlsbb
rBAlsaa
LLLL
LLLL
LLLL
2
mer
f
T
sf
sfss
abcf LL
LLL
cccbca
bcbbba
acabaa
ss
LLL
LLL
LLL
L
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Flux Linkage vs. Current (3)
This matrix can be transformed into dq0 form and used to
find flux linkage.
abcfabcfabcf iL
fsfdqssdq iLiKLK
0
1
0
1
dqfdqfdqf iL
fsfdqssdq iKLiKKL
0
1
0
fTsf
sfss
abcf LL
LLL
f
abc
abcf
f
abc
abcf i
ii
fsfabcssabc iLiL
ffabc
T
sff iL iL ffdqT
sff iL
0
1iKL
f
T
sf
sfss
dqf
L1
1
KL
KLKKLL
f
dq
dqf
0
f
dq
dqf
i
0ii
From with
where
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Dynamical Equation in terms of Current
V
dt
d dqf
For linear model
from
VLi
1 dqfdqf
dt
d dynamical equationin terms of current
dqfdqfdqf iL
fff
s
rdqsq
rqdsd
iRv
iRv
iRv
iRv
00
V
and
where
qqq
fsfddd
iL
iLiL
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Power
Electrical instantaneous Input Power on Stator can also be
expressed through dq0 theory.
0
11
0 )( dqTT
dqabc
T
abcccbbaain ivivivp iKKviv
00223 ivivivp qqddin
200010
001
2
3
)( 11
KK
T
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Torque
0022
3ivivivp qqddin
00
0
dt
diRdt
diR
dt
diR
v
v
v
s
rdqqs
rqdds
q
d
From
we have
)(223
22
3
22
3 00
2
0
22
dqqdm
q
q
d
dqdsin ii
P
dt
d
idt
d
idt
d
iiiiRp
Copper Loss Mechanical PowerMagnetic Power inWindings
Therefore, electromagnetic torque on rotor
)(22
3dqqd
m
meche iiPp
T
mechp
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Equivalent Circuits (1)
fff
s
rdqqs
rqdds
f
q
d
dt
diR
dt
diRdt
diR
dt
diR
v
vv
v
000
dsffff
ls
qqq
fsfddd
iLiL
iL
iL
iLiL
2
3
00
d axisdt
diL
dt
diLiRv
f
sfd
drqdsd
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Equivalent Circuits (2)
q axisdt
diLiRv
q
qrdqsq
0 axisdt
diLiRv s
0000
This circuit is not necessaryfor Y connected windingssince i0=0.
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Combined Equivalent Circuit on d Axis (1)
dt
iid
Ldt
di
LiR
dt
diL
dt
diLiRv
fd
md
d
lsrqds
f
sfd
drqdsd
)( '
d axis equivalent circuit and field winding equivalent circuit can be combined:
mdlsd LLL
mflff LLL
mf
sf
sf
md
f
a
L
L
L
L
N
NN
3
2
N
ii
L
Li
f
f
md
sf
f
3
2'
fadsf
fdmf
admd
NNCL
NCL
NCL
2
3
2
2
)2
1(8
2
0 g
av
dPg
DlC
From
(Details @ InductanceSM.ppt)
Let
aN fNand are effective number ofturns of armature andfield windings.
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Combined Equivalent Circuit on d Axis (2)
dt
diL
dt
diLiRv dsf
f
ffff2
3 '
2
3ff Nii
dt
di
NLdt
di
NLdt
di
NLiNRNv d
sf
f
mf
f
lffff2
3
mf
sf
sf
md
L
L
L
L
N 3
2
dt
diL
dt
diL
dt
diLNiRNNv dmd
f
sf
f
lffff
'
2'2
2
3
2
3
dt
iidL
dt
diLiRv
fd
md
f
lffff
)( ''''''
lflf
ff
ff
LNL
RNR
Nvv
2'
2'
'
2
3
2
3
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Combined Equivalent Circuit on d Axis (3)
dt
iidL
dt
diLiRv
fd
mdd
lsrqdsd
)( '
dt
iidL
dt
diLiRv
fd
md
f
lffff
)( ''''''
From
ff Nvv '
'
2
3ff Nii
we get
mdmddls
fsfddd
iLiL
iLiL
'fdmd iii
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dq Theory for Permanent Magnet
Synchronous Machine (PMSM)
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Parks Transformation
Stator quantities (Sabc) of current, voltage, or flux can be
converted to quantities (Sdq0) referenced to the rotor.
This conversion comes through the K matrix.
0
1
0
dqabc
abcdq
SKS
KSS
13/2sin3/2cos
13/2sin3/2cos
1sincos
2/12/12/1
3/2sin3/2sinsin
3/2cos3/2coscos
3
2
1
meme
meme
meme
mememe
mememe
K
K
13/2cos3/2sin
13/2cos3/2sin
1cossin
2/12/12/1
3/2cos3/2coscos
3/2sin3/2sinsin
3
2
1
rr
rr
rr
rrr
rrr
K
K
where
or
(MITs notation)
(Purdues notation)
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Voltage Equations (1)
abcabcSabcdtd iRv
010101 dqdqSdqdt
dKiKRvK
01
0
1
0
1
dqdqSdqdt
dKKiKKRvKK
0
1
0
1
0
1
0 dqdqdqSdqdt
d
dt
dKKKKiKKRv
0
1
000 dqdqdqSdqdt
d
dt
dKKiRv
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
dtd
KK
00
0
dt
diR
dt
diR
dt
diR
v
v
v
s
rdqqs
rqdds
q
d
03/2sin3/2cos
03/2sin3/2cos
0sincos
03/2cos3/2sin
03/2cos3/2sin
0cossin1
rr
rr
rr
r
meme
meme
meme
medt
dK
dt
d meme
And for voltage, we get
2
P
dt
dmme
rr
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Dynamical Equations for Flux Linkage
000 iRv
iRv
iRv
dt
d
s
rdqsq
rqdsd
q
d
The derivations so far are valid for both linear and nonlinear models.
0
0
q
d
dq
00 iRv
iRviRv
s
rdqsq
rqdsd
V
Let
we haveV
dt
d dq0
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Flux Linkage vs. Current (1)
The next step is to relate current to flux linkage through
inductances. For salient pole rotor, the inductances can
be approximately expressed as
cccbca
bcbbba
acabaa
abc
LLL
LLL
LLL
L
3
22cos
322cos
2cos
meBAlscc
meBAlsbb
meBAlsaa
LLLL
LLLL
LLLL
Note: Higher order harmonics are neglected.
or:
3
22cos
322cos
2cos
rBAlscc
rBAlsbb
rBAlsaa
LLLL
LLLL
LLLL
2
mer
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Flux Linkage vs. Current (2)
3
22cos
2
1
2cos21
3
22cos
2
1
meBAcaac
meBAcbbc
meBAbaab
LLLL
LLLL
LLLL
Note: Higher order harmonics are neglected.
or:
3
22cos
2
1
2cos21
3
22cos
2
1
rBAcaac
rBAcbbc
rBAbaab
LLLL
LLLL
LLLL
2
mer
Ref: 1. A. E. Fitzgerald, C. Kingsley, Jr., and S. D. Umans, Electric Machinery,6th Edition, pages 660-661.
2. P. C. Krause, O. Wasynczuk, and S. D. Sudhoff,Analysis of Electric Machinery and Drive Systems, 2nd 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 to
find flux linkage.
PMabcabcabcabc iL
PMabcdqabcfdq iKLK 0101
PMabcdqabcdq KiKKLKK 0101
0000 PMdqdqdqdq iL
PMabcdqabcdq KiKKL
0
1
0
where
)3/2cos(
)3/2cos(
)cos(
me
me
me
PMPMabc
or:
2
mer
)3/2sin(
)3/2sin(
)sin(
r
r
r
PMPMabc
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Inductance Matrix in dq0 Frame
Therefore, we get the following inductance matrix in dq0
frame:
0
1
0
0000
00
LL
L
q
d
abcdq KKLL
where
)(2
3
)(2
3
BAmq
BAmd
LLL
LLL
ls
mqlsq
mdlsd
LL
LLL
LLL
0
and
From
00 iL
iL
iL
ls
qqq
PMddd
0000 PMdqdqdqdq iL
000
PM
PMabcPMdq
K
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Dynamical Equation in terms of Current
V
dt
d dq0For linear model from
VLi
1
0
0 dq
dq
dt
d dynamical equation
in terms of current
00 iRv
iRv
iRv
s
rdqsq
rqdsd
V
and
where
qqq
PMddd
iL
iL
0000 PMdqdqdqdq iL
0
0
00
00
00
L
L
L
q
d
dqL
0000 /)(/)(
/)(
LiRvLiLiRv
LiLiRv
ii
i
dt
d
s
qPMrddrqsq
dqqrdsd
q
d
For Y connected winding, since , only need to considerthe first two equations for idand iq.
0)(
3
10 cba iiii
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Power
Electrical instantaneous Input Power on Stator can also be
expressed through dq0 theory.
0
11
0 )( dqTT
dqabc
T
abcccbbaain ivivivp iKKviv
00223 ivivivp qqddin
200010
001
2
3
)(
11KK
T
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Dynamical Equations of Motion
mm
dampLem
dt
d
TTTdt
dJ
where
qTqdqdqPMe iKiiLLiP
T )(22
3
For round rotor machine, qd LL qPMe iPT 43
mmdamp DT Dm is combined damping coefficient of rotorand load.
dqdPMq
eT iLL
P
i
TK )(
4
3 torque constant
PMTPK 43
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Parks Transformation
Stator quantities (Sabc) of current, voltage, or flux can be
converted to quantities (Sdq0) referenced to the rotor.
This conversion comes through the K matrix.
0
1
0
dqabc
abcdq
SKS
KSS
13/2sin3/2cos
13/2sin3/2cos
1sincos2/12/12/1
3/2sin3/2sinsin
3/2cos3/2coscos
3
2
1
meme
meme
meme
mememe
mememe
K
K
13/2cos3/2sin
13/2cos3/2sin
1cossin
2/12/12/1
3/2cos3/2coscos3/2sin3/2sinsin
3
2
1
rr
rr
rr
rrr
rrr
K
K
where
or
(MITs notation)
(Purdues notation)
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Voltage Equations (2)
We derive the derivative of K-1:
Then, we get
000
00
00
1r
r
dtd
KK
00
0
dt
diR
dt
diR
dt
diR
v
vv
s
rdqqs
rqdds
q
d
03/2sin3/2cos
03/2sin3/2cos
0sincos
03/2cos3/2sin
03/2cos3/2sin
0cossin1
rr
rr
rr
r
meme
meme
meme
medt
dK
dt
d meme
And for stator voltage, we
get
2
P
dt
dmme
rr
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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.
qdqdqd kfkkfkrkfk dt
diRv
q
d
k
k
f
r
R
R
R
00
00
00
R
0
0
f
kfk
v
qdv
q
dqd
k
k
f
kfk
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Voltage Equations (4)
qqq
ddd
kkk
kkk
fff
s
qrdqs
drqds
f
q
d
dt
diR
dt
diR
dt
diR
dt
diR
dt
diR
dt
diR
v
v
v
v
000
0
0
In summary:
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Dynamical Equations for Flux Linkage
qq
dd
q
d
kk
kk
fff
s
rdqsq
rqdsd
k
k
f
q
d
iR
iRiRv
iRv
iRv
iRv
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
dd
kk
kk
fff
s
rdqsq
rqdsd
iR
iR
iRv
iRv
iRv
iRv
00
V
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Flux Linkage vs. Current (1)
The next step is to relate current to flux linkage through
inductances. For salient pole rotor, the inductances can
be approximately expressed as
3
22cos
322cos
2cos
meBAlscc
meBAlsbb
meBAlsaa
LLLL
LLLL
LLLL
or:
3
22cos
322cos
2cos
rBAlscc
rBAlsbb
rBAlsaa
LLLL
LLLL
LLLL
2
mer
rr
T
sr
srss
abcfLL
LLL
cccbca
bcbbba
acabaa
ss
LLL
LLL
LLL
L
qd
qd
qd
ckckcf
bkbkbf
akakaf
sr
LLL
LLL
LLL
L
qdqq
qddd
qd
kkkfk
kkkfk
fkfkf
rr
LLL
LLL
LLL
L
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Fl Li k C t (3)
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Flux Linkage vs. Current (3)
3
2cos
3
2
cos
cos
meskckck
meskbkbk
meskakak
ddd
ddd
ddd
LLL
LLL
LLL
or:
3
2sin
3
2
sin
sin
rskckck
rskbkbk
rskakak
ddd
ddd
ddd
LLL
LLL
LLL
2
mer
3
2sin
3
2sin
sin
meskckck
meskbkbk
meskakak
qqq
qqq
qqq
LLL
LLL
LLL
3
2cos
3
2cos
cos
rskckck
rskbkbk
rskakak
qqq
qqq
qqq
LLL
LLL
LLL
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Fl Li k C t (5)
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Flux Linkage vs. Current (5)
This matrix can be transformed into dq form and used to
find flux linkage.
abcfabcfabcf iL
qdkfksrdqssdq iLiKLK 01
0
1
dqfdqfdqf iL
qdkfksrdqssdq
iKLiKKL 0
1
0
r
T
sr
srss
abcf
LL
LLL
qdkfk
abc
abcf
qdkfk
abc
abcfi
ii
qdkfksrabcssabc iLiL
qdqd kfkrrabc
T
srkfk iLiL qdqd kfkrrdqT
srkfk iLiKL
0
1
rr
T
sr
srss
dqfLKL
KLKKLL
1
1
qdkfk
dq
dqf
0
qdkfk
dq
dqfi
ii
0
From with
where
Inductance Matrix in dq Frame
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Inductance Matrix in dq Frame
qq
ddd
d
q
d
ksk
kfksk
fkfsf
skq
sksfd
dqf
LL
LLL
LLL
L
LL
LLL
000230
0002
3
0002
300000
0000
000
0
L
where
)(2
3
)(2
3
BAmq
BAmd
LLL
LLL
mqlsq
mdlsd
LLL
LLL
and
dqfdqfdqf iL From
qskkkk
ffkdskkkk
kfkdsffff
kskqqq
kskfsfddd
iLiL
iLiLiL
iLiLiL
iL
iLiL
iLiLiL
qqqq
ddddd
dd
qq
dd
2
32
32
3000
Through derivations, we have
rrTsr
srss
dqf LKL
KLKKLL
1
1
lsLL 0
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Power
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Power
Electrical instantaneous Input Power on Stator can also be
expressed through dq0 theory.
0
11
0 )( dqTT
dqabc
T
abcccbbaain ivivivp iKKviv
00223 ivivivp qqddin
200
010
001
2
3)( 11 KK T
Torque
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Torque
0022
3ivivivp qqddin
00
0
dtdiR
dt
diR
dt
diR
v
v
v
s
rdqqs
rqdds
q
d
From
we have
)(22
32
2
32
2
3 00
2
0
22
dqqdm
q
qd
dqdsin iiP
dt
di
dt
di
dt
diiiiRp
Copper LossMechanical PowerMagnetic Power inWindings
Therefore, electromagnetic torque on rotor
)(22
3
dqqdm
mech
e ii
Pp
T
mechp
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 mdlsd LLL
mflff LLL
ddd mklkk LLL
dd
dd
dd
kfdfk
kadsk
kdmk
fadsf
fdmf
admd
NNCL
NNCL
NCL
NNCL
NCL
NCL
2
3
2
2
2
)21(8
2
0 g
av
dPg
DlC
From
(Details @ Inductance for SM.ppt)
, aN fN and are effective number of turns of
armature, field and d axis
damper windings, respectively.
dkN
Equivalent Circuit on d Axis (2)
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Equivalent Circuit on d Axis (2)
dt
iiidL
dt
diLiR
dt
diL
dt
diL
dt
diLiRv
d
d
d
kfd
mdd
lsrqds
k
sk
f
sfd
drqdsd
)( ''
f
a
f
f
md
sf
f iN
Ni
L
Li
3
2' Define
d
d
d
d
d k
a
k
k
md
sk
k iN
Ni
L
Li
3
2' and
dt
diL
dt
diL
dt
diLiRv d
d
k
fkd
sf
f
ffff 2
3
dtdiNL
dtdiNL
dtdiNL
dtdiNLiNRNv d
d
kfk
dsf
fmf
flffff 2
3
dt
diL
N
N
N
N
dt
diL
dt
diL
dt
diLNiRNNv d
d
d
k
fk
k
a
f
admd
f
sf
f
lffff
''
2'2
2
3
2
3
2
3
Define
ff Nvv '
a
f
NN
Nand
Equivalent Circuit on d Axis (3)
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Equivalent Circuit on d Axis (3)
dt
iiidL
dt
diLiRv d
kfd
md
f
lffff
)( '''''''
lf
f
alf
ff
a
f
LN
NL
RN
N
R
2
'
2
'
2
3
2
3
where
2
32
a
kf
md
fk
N
NNLL dd
dt
di
Ldt
di
Ldt
di
Ldt
di
LiR
dt
diL
dt
diL
dt
diLiR
f
fk
d
sk
k
mk
k
lkkk
f
fkd
sk
k
kkk
dd
d
d
d
ddd
dd
d
ddd
2
3
2
30From
above
3
2
dd 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
diL
NN
N
dt
diL
dt
diL
N
N
dt
diL
N
NiR
N
N ffk
kf
admd
k
mk
k
ak
lk
k
akk
k
a
d
d
d
d
d
d
d
d
dd
d
'2'2
'2
'
2
2
3
2
3
2
3
2
30
dt
iiid
Ldt
di
LiR dd
ddd
kfd
md
k
lkkk
)(
0
'''
'''
where
d
d
d
d
d
d
lk
k
alk
k
k
ak
LN
NL
R
N
NR
2
'
2
'
2
3
2
3
dd
dd
kf
a
fk
md
k
a
mk
md
NN
N
L
L
N
N
L
L
2
3
2
3
2
2
Equivalent Circuit on d Axis (5)
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Equivalent Circuit on d Axis (5)
From
we get
mdmddls
kskfsfddd
iLiL
iLiLiLdd
''
dkfdmd iiii
dt
iiidLdt
diLiRv dkfd
mdd
lsrqdsd)(
''
dt
iiidL
dt
diLiRv d
kfd
md
f
lffff
)( '''''''
dtiiidL
dtdiLiR dd
ddd
kfdmd
klkkk )(0
'''
'''
'
dki
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 LLL
qqq mklkk LLL
qq
qq
kaqsk
kqmk
aqmq
NNCL
NCL
NCL
23
2
2
)2
1(8
2
0
Pg
DlC
av
q
From
(Details @ Inductance for SM.ppt)
sN and are effective number of turns of
stator and q axis damperwindings, respectively.
dkN
Equivalent Circuit on q Axis (2)
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q q ( )
dt
iidL
dt
diLiR
dt
diL
dt
diL
dt
diLiR
dt
diL
dt
diLiRv
q
q
q
q
q
kq
mq
q
lqrdqs
k
sk
q
mq
q
lqrdqs
k
sk
q
qrdqsq
)(
'
q
q
q
q
d k
a
k
k
md
sk
k i
N
Ni
L
Li
3
2'
where
dt
di
Ldt
di
Ldt
di
LiR
dt
diL
dt
diLiR
q
sk
k
mk
k
lkkk
q
sk
k
kkk
q
q
q
q
qqq
q
q
qqq
2
3
2
30
From
above
3
2
qq k
a
sk
mq
N
N
L
L
next page
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Equivalent Circuit on q Axis (4)
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q q ( )
From
we get
mqmqqls
kskqqq
iLiL
iLiLqq
'
qkqmq iii
dt
iidLdt
diLiRv qkq
mq
q
lqrdqsq)(
'
dt
iidL
dt
diLiR
qq
qqq
kq
mq
k
lkkk
)(0
''
'''
'
qki
Equivalent Circuit on 0 Axis
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0 axisdt
diLiRv s
0000
This circuit is not necessaryfor Y connected windingssince i0=0.
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