dq theory
DESCRIPTION
DQ theory for electrical machinesTRANSCRIPT
dq Theory for Synchronous Machine
with Damper Winding
Relevance to Synchronous Machine
dq means direct and quadrature. Direct axis is aligned with
the rotor’s pole. Quadrature axis refers to the axis whose
electrical angle is orthogonal to the electric angle of direct
axis.
a axis
d axisq axis
b axis
c axis
qm
qa
qd
qmq
22
2
qqq
memqr
mme
P
Pming
max,orr
min,orrmaxg
isr
qd
a axis
qm
qa
q axis
qmq
Park’s 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
qqq
qqq
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
qqq
qqq
rr
rr
rr
rrr
rrr
K
K
where
or
(MIT’s notation)
(Purdue’s notation)
c
b
a
abcq
d
dq
S
S
S
S
S
S
SS ,
0
0
Voltage Equations (1)
abcabcSabcdt
dλiRv
0
1
0
1
0
1
dqdqSdqdt
dλKiKRvK
0
1
0
1
0
1
dqdqSdqdt
dλKKiKKRvKK
0
1
0
1
0
1
0 dqdqdqSdqdt
d
dt
dλKKλKKiKKRv
0
1
000 dqdqdqSdqdt
d
dt
dλKKλiRv
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):
Voltage Equations (2)
We derive the derivative of K-1:
Then, we get
000
00
001
r
r
dt
d
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
q
And for stator voltage, we
get
2
P
dt
dmme
rr
q
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 kfkkfkrkfkdt
dλiRv
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
λ
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:
Dynamical Equations for Flux Linkage
dd
q
d
kk
kk
fff
s
rdqsq
rqdsd
k
k
f
q
d
iR
iR
iRv
iRv
iRv
iRv
dt
d 000
The derivations so far are valid for both linear and nonlinear models.
Let
we have Vλ
dt
d dqf
q
d
k
k
f
q
d
dqf
0λ
dd
kk
kk
fff
s
rdqsq
rqdsd
iR
iR
iRv
iRv
iRv
iRv
00
V
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
3
22cos
2cos
q
q
q
meBAlscc
meBAlsbb
meBAlsaa
LLLL
LLLL
LLLL
or:
3
22cos
3
22cos
2cos
q
q
q
rBAlscc
rBAlsbb
rBAlsaa
LLLL
LLLL
LLLL
2
qq 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
Flux Linkage vs. Current (2)
3
22cos
2
1
2cos2
1
3
22cos
2
1
q
q
q
meBAcaac
meBAcbbc
meBAbaab
LLLL
LLLL
LLLL
3
2cos
3
2cos
cos
q
q
q
mesffccf
mesffbbf
mesffaaf
LLL
LLL
LLL
or:
3
22cos
2
1
2cos2
1
3
22cos
2
1
q
q
q
rBAcaac
rBAcbbc
rBAbaab
LLLL
LLLL
LLLL
3
2sin
3
2sin
sin
q
q
q
rsffccf
rsffbbf
rsffaaf
LLL
LLL
LLL
2
qq mer
Flux Linkage vs. Current (3)
3
2cos
3
2cos
cos
q
q
q
meskckck
meskbkbk
meskakak
ddd
ddd
ddd
LLL
LLL
LLL
or:
3
2sin
3
2sin
sin
q
q
q
rskckck
rskbkbk
rskakak
ddd
ddd
ddd
LLL
LLL
LLL
2
qq mer
3
2sin
3
2sin
sin
q
q
q
meskckck
meskbkbk
meskakak
qqq
qqq
qqq
LLL
LLL
LLL
3
2cos
3
2cos
cos
q
q
q
rskckck
rskbkbk
rskakak
qqq
qqq
qqq
LLL
LLL
LLL
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, 6th Edition, pages 660-661.
2. P. C. Krause, O. Wasynczuk, and S. D. Sudhoff, Analysis of Electric Machinery and Drive Systems, 2nd Edition, pages 52and 195.
qqq
ddd
mklkk
mklkk
mflff
LLL
LLL
LLL
0
0
dqqd
dd
kkkk
fkfk
fkfk
LL
LL
LL
Flux Linkage vs. Current (5)
This matrix can be transformed into dq form and used to
find flux linkage.
abcfabcfabcf iLλ
qdkfksrdqssdq iLiKLλK
0
1
0
1
dqfdqfdqf iLλ
qdkfksrdqssdq iKLiKKLλ
0
1
0
r
T
sr
srss
abcfLL
LLL
qd kfk
abc
abcf λ
λλ
qd kfk
abc
abcf i
ii
qd kfksrabcssabc iLiLλ
qdqd kfkrrabc
T
srkfk iLiLλ qdqd kfkrrdq
T
srkfk iLiKLλ
0
1
rr
T
sr
srss
dqfLKL
KLKKLL
1
1
qd kfk
dq
dqf λ
λλ
0
qd kfk
dq
dqf i
ii
0
From with
where
Inductance Matrix in dq Frame
ddd
d
q
d
ksk
kfksk
fkfsf
skq
sksfd
dqf
LL
LLL
LLL
L
LL
LLL
0002
30
0002
3
0002
3
00000
0000
000
0
L
where
)(2
3
)(2
3
BAmq
BAmd
LLL
LLL
mqlsq
mdlsd
LLL
LLL
and
dqfdqfdqf iLλ From
qskkkk
ffkdskkkk
kfkdsffff
kskqqq
kskfsfddd
iLiL
iLiLiL
iLiLiL
iL
iLiL
iLiLiL
qqqq
ddddd
dd
dd
2
32
32
3000
Through derivations, we have
rr
T
sr
srss
dqfLKL
KLKKLL
1
1
lsLL 0
Dynamical Equation in terms of Current
Vλ
dt
d dqf
For linear model
from
VLi
1 dqf
dqf
dt
d dynamical equationin terms of current
dqfdqfdqf iLλ
dd
kk
kk
fff
s
rdqsq
rqdsd
iR
iR
iRv
iRv
iRv
iRv
00
V
and
where
dd
kskqqq
kskfsfddd
iLiL
iLiLiL
Power
Electrical instantaneous Input Power on Stator can also be
expressed through dq0 theory.
0
11
0 )( dq
TT
dqabc
T
abcccbbaain ivivivp iKKviv
0022
3ivivivp qqddin
200
010
001
2
3)( 11 KK T
Torque
0022
3ivivivp qqddin
00
0
dt
diR
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 Loss Mechanical PowerMagnetic Power in Windings
Therefore, electromagnetic torque on rotor
)(22
3dqqd
m
meche ii
PpT
m echp
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
ˆˆ
ˆˆ
ˆ
ˆˆ
ˆ
ˆ2
3
2
2
2
)2
1(8
2
0 g
av
dPg
DlC
From
(Details @ Inductance for SM.ppt)
,ˆaN
fN̂ and are effective number of turns of armature, field and d axis damper windings, respectively.
dkN̂
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
ˆ
ˆ
3
2' Define
d
d
d
d
d k
a
k
k
md
sk
k iN
Ni
L
Li
ˆ
ˆ
3
2' and
dt
diL
dt
diL
dt
diLiRv d
d
k
fkd
sf
f
ffff 2
3
dt
diNL
dt
diNL
dt
diNL
dt
diNLiNRNv d
d
k
fkd
sf
f
mf
f
lffff 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
ˆ
ˆ
2
3
ˆ
ˆ
2
3
2
3
Define
ff Nvv 'ˆ
ˆa
f
NN
Nand
Equivalent Circuit on d Axis (3)
dt
iiidL
dt
diLiRv dkfd
md
f
lffff
)( '''
''''
lf
f
alf
f
f
af
LN
NL
RN
NR
2
'
2
'
ˆ
ˆ
2
3
ˆ
ˆ
2
3
where
2ˆ
ˆˆ
3
2
a
kf
md
fk
N
NN
L
Ldd
dt
diL
dt
diL
dt
diL
dt
diLiR
dt
diL
dt
diL
dt
diLiR
f
fkd
sk
k
mk
k
lkkk
f
fkd
sk
k
kkk
dd
d
d
d
ddd
dd
d
ddd
2
3
2
30From
aboveˆ
ˆ
3
2
dd k
a
sk
md
N
N
L
Lnext page
Equivalent Circuit on d Axis (4)
dt
diL
NN
N
dt
diL
dt
diL
N
N
dt
diL
N
NiR
N
N f
fk
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
ˆˆ
ˆ
2
3
ˆ
ˆ
2
3
ˆ
ˆ
2
3
ˆ
ˆ
2
30
dt
iiidL
dt
diLiR dd
ddd
kfd
md
k
lkkk
)(0
'''
'''
where
d
d
d
d
d
d
lk
k
alk
k
k
ak
LN
NL
RN
NR
2
'
2
'
ˆ
ˆ
2
3
ˆ
ˆ
2
3
dd
dd
kf
a
fk
md
k
a
mk
md
NN
N
L
L
N
N
L
L
ˆˆ
ˆ
2
3
ˆ
ˆ
2
3
2
2
Equivalent Circuit on d Axis (5)
From
we get
mdmddls
kskfsfddd
iLiL
iLiLiLdd
''
dkfdmd iiii
dt
iiidL
dt
diLiRv dkfd
mdd
lsrqdsd
)( ''
dt
iiidL
dt
diLiRv dkfd
md
f
lffff
)( '''
''''
dt
iiidL
dt
diLiR dd
ddd
kfd
md
k
lkkk
)(0
'''
'''
'
dki
Equivalent Circuit on q Axis (1)
q axis equivalent circuit and q axis damper equivalent circuit can be combined:
Let mqlsq LLL
qqq mklkk LLL
kaqsk
kqmk
aqmq
NNCL
NCL
NCL
ˆˆ
ˆ
ˆ2
3
2
2
)2
1(8
2
0
Pg
DlC
av
q
From
(Details @ Inductance for SM.ppt)
sN̂ and are effective number of turns of stator and q axis damper windings, respectively.
dkN̂
Equivalent Circuit on q Axis (2)
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 iN
Ni
L
Li
ˆ
ˆ
3
2'
where
dt
diL
dt
diL
dt
diLiR
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
aboveˆ
ˆ
3
2
qq k
a
sk
mq
N
N
L
Lnext page
Equivalent Circuit on q Axis (3)
dt
diL
dt
diL
N
N
dt
diL
N
NiR
N
N q
mq
k
mk
k
ak
lk
k
akk
k
a q
q
q
q
q
q
q
'2
'2
'
2
ˆ
ˆ
2
3
ˆ
ˆ
2
3
ˆ
ˆ
2
30
dt
iidL
dt
diLiR
qqq
kq
mq
k
lkkk
)(0
''
'''
where
q
q
q
q
q
q
lk
k
alk
k
k
ak
LN
NL
RN
NR
2
'
2
'
ˆ
ˆ
2
3
ˆ
ˆ
2
3
kf
a
fk
mq
k
a
mk
mq
NN
N
L
L
N
N
L
L
ˆˆ
ˆ
2
3
ˆ
ˆ
2
3
2
2
Equivalent Circuit on q Axis (4)
From
we get
mqmqqls
kskqqq
iLiL
iLiLqq
'
qkqmq iii
dt
iidL
dt
diLiRv
qkq
mq
q
lqrdqsq
)( '
dt
iidL
dt
diLiR
qqq
kq
mq
k
lkkk
)(0
''
'''
'
qki
Equivalent Circuit on 0 Axis
0 axisdt
diLiRv s
0000
This circuit is not necessary for Y connected windings since i0=0.
Dynamical Equations from
Equivalent Circuits (1)
dt
diLiRv
dt
diL
dt
diLiRv
dt
diL
dt
diLiR
dt
diL
dt
diLiR
dt
diL
dt
diLiRv
dt
diL
dt
diLiRv
lss
mdmd
f
lffff
mq
mq
k
lkkk
mdmd
k
lkkk
mq
mq
q
lsqdrqsqq
mdmd
dlsdqrdsdd
q
qqq
d
ddd
00000
'
''''
'
'''
'
'''
0
0
where
'
''
q
d
kqmq
kfdmd
iii
iiii
Dynamical Equations from
Equivalent Circuits (2)The equations can be written in matrix form as:
VI
L dt
d
where
mqlkmq
mdlkmdmd
mdmdlfmd
ls
mqmqlsq
mdmdmdlsd
LLL
LLLL
LLLL
L
LLL
LLLL
q
d
'
'
'
0
0000
000
000
00000
0000
000
L
'
'
'
0
q
d
k
k
f
q
d
i
i
i
i
i
i
I
''
''
'''
000
dd
kk
kk
fff
s
drqsqq
qrdsdd
iR
iR
iRv
iRv
iRv
iRv
V
VLI 1
dt
dor
dq Theory for Synchronous Machine
without Damper Winding
Relevance to Synchronous Machine
dq means direct and quadrature. Direct axis is aligned with
the rotor’s pole. Quadrature axis refers to the axis whose
electrical angle is orthogonal to the electric angle of direct
axis.
a axis
d axisq axis
b axis
c axis
qm
qa
qd
qmq
22
2
qqq
memqr
mme
P
Pming
max,orr
min,orrmaxg
isr
qd
a axis
qm
qa
q axis
qmq
Park’s 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
qqq
qqq
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
qqq
qqq
rr
rr
rr
rrr
rrr
K
K
where
or
(MIT’s notation)
(Purdue’s notation)
c
b
a
abcq
d
dq
S
S
S
S
S
S
SS ,
0
0
Voltage Equations (1)
abcabcSabcdt
dλiRv
0
1
0
1
0
1
dqdqSdqdt
dλKiKRvK
0
1
0
1
0
1
dqdqSdqdt
dλKKiKKRvKK
0
1
0
1
0
1
0 dqdqdqSdqdt
d
dt
dλKKλKKiKKRv
0
1
000 dqdqdqSdqdt
d
dt
dλKKλiRv
100
010
001
sS RR
For stator windings
For field winding:
ffff λdt
diRv
Under motor reference convention for currents
(i.e. the positive reference direction for currents is into the machine):
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
dt
diR
dt
diR
dt
diR
dt
diR
v
v
v
v
000
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
q
And for voltage, we get
2
P
dt
dmme
rr
q
Dynamical Equations for Flux Linkage
fff
s
rdqsq
rqdsd
f
q
d
iRv
iRv
iRv
iRv
dt
d
000
The derivations so far are valid for both linear and nonlinear models.
f
q
d
dqf
0
λ
fff
s
rdqsq
rqdsd
iRv
iRv
iRv
iRv
00
V
Let
we haveV
λ
dt
d dqf
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
3
22cos
2cos
q
q
q
meBAlscc
meBAlsbb
meBAlsaa
LLLL
LLLL
LLLL
or:
3
22cos
3
22cos
2cos
q
q
q
rBAlscc
rBAlsbb
rBAlsaa
LLLL
LLLL
LLLL
2
qq mer
f
T
sf
sfss
abcf LL
LLL
cccbca
bcbbba
acabaa
ss
LLL
LLL
LLL
L
Flux Linkage vs. Current (2)
3
22cos
2
1
2cos2
1
3
22cos
2
1
q
q
q
meBAcaac
meBAcbbc
meBAbaab
LLLL
LLLL
LLLL
Note: Higher order harmonics are neglected.
3
2cos
3
2cos
cos
q
q
q
mesffccf
mesffbbf
mesffaaf
LLL
LLL
LLL
or:
3
22cos
2
1
2cos2
1
3
22cos
2
1
q
q
q
rBAcaac
rBAcbbc
rBAbaab
LLLL
LLLL
LLLL
3
2sin
3
2sin
sin
q
q
q
rsffccf
rsffbbf
rsffaaf
LLL
LLL
LLL
2
qq 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.
Flux Linkage vs. Current (3)
This matrix can be transformed into dq0 form and used to
find flux linkage.
abcfabcfabcf iLλ
fsfdqssdq iLiKLλK
0
1
0
1
dqfdqfdqf iLλ
fsfdqssdq iKLiKKLλ
0
1
0
f
T
sf
sfss
abcf LL
LLL
f
abc
abcf
λλ
f
abc
abcf i
ii
fsfabcssabc iLiLλ
ffabc
T
sff iLλ iL ffdq
T
sff iLλ
0
1iKL
f
T
sf
sfss
dqfL1
1
KL
KLKKLL
f
dq
dqf
0λλ
f
dq
dqf i
0ii
From with
where
Inductance Matrix in dq0 Frame
fsf
q
sfd
dqf
LL
L
L
LL
002
3
000
000
00
0L
where
)(2
3
)(2
3
BAmq
BAmd
LLL
LLL
ls
mqlsq
mdlsd
LL
LLL
LLL
0
and
dqfdqfdqf iLλ From
dsffff
ls
qqq
fsfddd
iLiL
iL
iL
iLiL
2
3
00
Through derivations, we have
f
T
sf
sfss
dqfL1
1
KL
KLKKLL
Dynamical Equation in terms of Current
Vλ
dt
d dqf
For linear model
from
VLi
1 dqf
dqf
dt
d dynamical equationin terms of current
dqfdqfdqf iLλ
fff
s
rdqsq
rqdsd
iRv
iRv
iRv
iRv
00
V
and
where
qqq
fsfddd
iL
iLiL
Power
Electrical instantaneous Input Power on Stator can also be
expressed through dq0 theory.
0
11
0 )( dq
TT
dqabc
T
abcccbbaain ivivivp iKKviv
0022
3ivivivp qqddin
200
010
001
2
3)( 11 KK T
Torque
0022
3ivivivp qqddin
00
0
dt
diR
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 Loss Mechanical PowerMagnetic Power in Windings
Therefore, electromagnetic torque on rotor
)(22
3dqqd
m
meche ii
PpT
m echp
Equivalent Circuits (1)
fff
s
rdqqs
rqdds
f
q
d
dt
diR
dt
diR
dt
diR
dt
diR
v
v
v
v
000
dsffff
ls
qqq
fsfddd
iLiL
iL
iL
iLiL
2
3
00
d axisdt
diL
dt
diLiRv
f
sfd
drqdsd
Equivalent Circuits (2)
q axisdt
diLiRv
q
qrdqsq
0 axisdt
diLiRv s
0000
This circuit is not necessary for Y connected windings since i0=0.
Equivalent Circuits (3)
Field winding
dt
diL
dt
diLiRv d
sf
f
ffff2
3
Combined Equivalent Circuit on d Axis (1)
dt
iidL
dt
diLiR
dt
diL
dt
diLiRv
fd
mdd
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
ˆ
ˆ
N
ii
L
Li
f
f
md
sf
f3
2'
fadsf
fdm f
admd
NNCL
NCL
NCL
ˆˆ
ˆ
ˆ2
3
2
2
)2
1(8
2
0 g
av
dPg
DlC
From
(Details @ InductanceSM.ppt)
Let
aN̂ fN̂and are effective number of turns of armature and field windings.
Combined Equivalent Circuit on d Axis (2)
dt
diL
dt
diLiRv d
sf
f
ffff2
3 '
2
3ff Nii
dt
diNL
dt
diNL
dt
diNLiNRNv d
sf
f
mf
f
lffff2
3
mf
sf
sf
md
L
L
L
LN
3
2
dt
diL
dt
diL
dt
diLNiRNNv d
md
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
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
dq Theory for Permanent Magnet
Synchronous Machine (PMSM)
Relevance to PM Machine
dq means direct and quadrature. Direct axis is aligned with
the rotor’s pole. Quadrature axis refers to the axis whose
electrical angle is orthogonal to the electric angle of direct
axis.
a axis
d axisq axis
b axis
c axis
qa
qd
22
2
qqq
memqr
mme
P
P
qm
qmq
Park’s 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
qqq
qqq
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
qqq
qqq
rr
rr
rr
rrr
rrr
K
K
where
or
(MIT’s notation)
(Purdue’s notation)
c
b
a
abcq
d
dq
S
S
S
S
S
S
SS ,
0
0
Voltage Equations (1)
abcabcSabcdt
dλiRv
0
1
0
1
0
1
dqdqSdqdt
dλKiKRvK
0
1
0
1
0
1
dqdqSdqdt
dλKKiKKRvKK
0
1
0
1
0
1
0 dqdqdqSdqdt
d
dt
dλKKλKKiKKRv
0
1
000 dqdqdqSdqdt
d
dt
dλKKλiRv
100
010
001
sS RR
For stator winding
Under motor reference convention for currents
(i.e. the positive reference direction for currents is into the machine):
Voltage Equations (2)
We derive the derivative of K-1:
Then, we get
000
00
001
r
r
dt
d
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
q
And for voltage, we get
2
P
dt
dmme
rr
q
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
dqλ
00 iRv
iRv
iRv
s
rdqsq
rqdsd
V
Let
we haveV
λ
dt
d dq0
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
3
22cos
2cos
q
q
q
meBAlscc
meBAlsbb
meBAlsaa
LLLL
LLLL
LLLL
Note: Higher order harmonics are neglected.
or:
3
22cos
3
22cos
2cos
q
q
q
rBAlscc
rBAlsbb
rBAlsaa
LLLL
LLLL
LLLL
2
qq mer
Flux Linkage vs. Current (2)
3
22cos
2
1
2cos2
1
3
22cos
2
1
q
q
q
meBAcaac
meBAcbbc
meBAbaab
LLLL
LLLL
LLLL
Note: Higher order harmonics are neglected.
or:
3
22cos
2
1
2cos2
1
3
22cos
2
1
q
q
q
rBAcaac
rBAcbbc
rBAbaab
LLLL
LLLL
LLLL
2
qq 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.
Flux Linkage vs. Current (4)
This matrix can be transformed into dq0 form and used to
find flux linkage.
PMabcabcabcabc λiLλ
PMabcdqabcfdq λiKLλK
0
1
0
1
PMabcdqabcdq KλiKKLλKK
0
1
0
1
0000 PMdqdqdqdq λiLλ
PMabcdqabcdq KλiKKLλ
0
1
0
where
)3/2cos(
)3/2cos(
)cos(
q
q
q
me
me
me
PMPMabcλor:
2
qq mer
)3/2sin(
)3/2sin(
)sin(
q
q
q
r
r
r
PMPMabcλ
Inductance Matrix in dq0 Frame
Therefore, we get the following inductance matrix in dq0
frame:
0
1
0
00
00
00
L
L
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 λiLλ
0
00
PM
PMabcPMdq
Kλλ
Dynamical Equation in terms of Current
Vλ
dt
d dq0For linear model from
VLi
1
0
0 dq
dq
dt
d dynamical equationin terms of current
00 iRv
iRv
iRv
s
rdqsq
rqdsd
V
and
where
qqq
PMddd
iL
iL
0000 PMdqdqdqdq λiLλ
0
0
00
00
00
L
L
L
q
d
dqL
0000 /)(
/)(
/)(
LiRv
LiLiRv
LiLiRv
i
i
i
dt
d
s
qPMrddrqsq
dqqrdsd
q
d
For Y connected winding, since , only need to consider the first two equations for id and iq.
0)(3
10 cba iiii
Power
Electrical instantaneous Input Power on Stator can also be
expressed through dq0 theory.
0
11
0 )( dq
TT
dqabc
T
abcccbbaain ivivivp iKKviv
0022
3ivivivp qqddin
200
010
001
2
3)( 11 KK T
Torque
0022
3ivivivp qqddin
00
0
dt
diR
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 Loss Mechanical PowerMagnetic Power in Windings
Therefore, electromagnetic torque on rotor
)(22
3dqqd
m
meche ii
PpT
m echp
qqq
PMddd
iL
iL
qdqdqPMe iiLLiP
T )(22
3
Dynamical Equations of Motion
mm
dampLem
dt
d
TTTdt
dJ
q
where
qTqdqdqPMe iKiiLLiP
T )(22
3
For round rotor machine, qd LL qPMe iP
T 4
3
mmdamp DT Dm is combined damping coefficient of rotor and load.
dqdPM
q
eT iLL
P
i
TK )(
4
3 torque constant
PMT
PK
4
3