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.
b axis
qd
q axis
q axis
d axis
qmq
qmq
qm
qa
g max
ror,min
qm
a axis
ror,max
g min
c axis
P
qm
2
P

qr  qm q  qm e 
2
2
qm e 
qd
qa
ris
a axis
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.
S dq0  KS abc
where
or
S abc  K 1S dq0
S dq 0
Sd 
Sa 
  S q , S abc   Sb 
 S 0 
 S c 
cosq m e  2 / 3
cosq m e  2 / 3 
 cosq m e 
2
K   sin q m e   sin q m e  2 / 3  sin q m e  2 / 3
3
 1 / 2

1/ 2
1/ 2
cosq m e 
 sin q m e 
1

K 1  cosq m e  2 / 3  sin q m e  2 / 3 1
cosq m e  2 / 3  sin q m e  2 / 3 1
(MIT’s notation)
 sin q r  sin q r  2 / 3 sin q r  2 / 3
2
K  cosq r  cosq r  2 / 3 cosq r  2 / 3
3
 1 / 2

1/ 2
1/ 2
sin q r 
cosq r 
1

K 1  sin q r  2 / 3 cosq r  2 / 3 1
sin q r  2 / 3 cosq r  2 / 3 1
(Purdue’s notation)
Voltage Equations (1)
Under motor reference convention for currents
(i.e. the positive reference direction for currents is into the machine):
For stator windings
K

d
v abc  R S i abc  λ abc
dt
1





K  K 1v dq0  KR S  K 1i dq0



d
K 1λ dq0
dt
d
 K  K 1λ dq0
dt
v dq0  R S K 1i dq0 
1 0 0
R S  Rs 0 1 0
0 0 1




d
d

v dq0  KR S K 1 i dq0  KK 1 λ dq0  K  K 1 λ dq0 
dt
 dt





v dq0  R S i dq0 
d
d

λ dq0  K  K 1 λ dq0
dt
 dt

Voltage Equations (2)
We derive the derivative of K-1:
cosq m e 
0
 sin q m e 
  cosq r 
d 1
K  m e sin q m e  2 / 3 cosq m e  2 / 3 0  r  cosq r  2 / 3
dt
sin q m e  2 / 3 cosq m e  2 / 3 0
 cosq r  2 / 3

me

dq me

dt

Then, we get
0
d

K K 1   r
 dt

 0
r 
 r
0
0
0
0
0
sin q r 
0
sin q r  2 / 3 0
sin q r  2 / 3 0
dq r
P
 me  m
dt
2
And for stator voltage, we
get
d


R
i





q r
 s d dt d
vd  

v    R i  d     
d r
 q   s q dt q

 v0  
d

R
i


s 0
0


dt
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.
v fkd kq  R r i fkd kq 
v fkd kq
d
λ fkd kq
dt
v f 
  0 
 0 
R f

Rr   0
0

λ fkd kq
0
Rkd
0
f 
 
 kd 
 
 kq 
0 

0 
Rkq 
Voltage Equations (4)
In summary:
d


R
i





q
r
d 
 s d
dt


d
 vd   Rs iq  d r 
q 
v  
dt

q
d
  

R
i


 v0   s 0 dt 0

 

d
v
f
  R f i f 
f

dt
0 

d
  

R
i


0
k
k
k


   d d

dt d


d
kq
 Rk q ik q 

dt


Dynamical Equations for Flux Linkage
 d  vd
   v
 q  q
d  0  


dt   f  
k  
 d 
k q  
Let
λ dqf
we have
 d 
 
 q
 0 



 f 
k 
 d
k q 
dλ dqf
dt
 Rs id  qr 
 Rs iq  d r 

v0  Rs i0

v f  Rf if


 Rk d ik d

 Rk q ik q

vd
v
 q

V




 Rs id  qr 
 Rs iq  d r 

v0  Rs i0

v f  Rf if


 Rk d ik d

 Rk q ik q

V
The derivations so far are valid for both linear and nonlinear models.
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
L ss
L abcf   T
L sr
 Laa Lab Lac 
L sr 
L L L 
L

ss
ba
bb
bc  L sr


L rr 
 Lca Lcb Lcc 

L  L  L  L cos 2q
ls
A
B
me
 aa

2 

L

L

L

L
cos
2
q



 bb
ls
A
B
me
3 



2 

Lcc  Lls  LA  LB cos 2q m e  
3 


 Laf Lak Lak 
d
q


  Lbf Lbkd Lbkq  L rr
L L

L
cf
ck
ck
d
q 

 Lf

  Lkd f
L
 kq f
L fkd L fkq 

Lkd Lkd kq 
Lkq kd Lkq 

L  L  L  L cos 2q
ls
A
B
r
 aa

2 

or:
L

L

L

L
cos
2
q



 bb
ls
A
B
r
3 

 
qr  qm e  
2 L  L  L  L cos 2q  2 
cc
ls
A
B
r
3 


Flux Linkage vs. Current (2)

1
2 

L

L


L

L
cos
2
q

 r

ba
A
B
 ab
2
3




1
L

L


LA  LB cos2q r 
 bc
cb
2


1
2 

L

L


L

L
cos
2
q



ca
A
B
r
 ac
2
3 



1
2 

L

L


L

L
cos
2
q

 me

ba
A
B
 ab
2
3




1
L

L


LA  LB cos2q m e 
 bc
cb
2


1
2 

L

L


L

L
cos
2
q



ca
A
B
me
 ac
2
3 



L  L  L cosq 
fa
sf
me
 af

2 

L

L

L
cos
q



 bf
fb
sf
me
3 



2 


Lcf  L fc  Lsf cosq m e 
3 


or:
qr  qm e 

2

L  L  L sin q 
fa
sf
r
 af

2 

L

L

L
sin
q

 r

 bf
fb
sf
3




2 

L

L

L
sin
q

 r

 cf
fc
sf
3



Flux Linkage vs. Current (3)

L  L  L sin q 
kd a
sk d
r
 akd

2 

L

L

L
sin
q



 bkd
kd b
sk d
r
3 



2 


Lckd  Lkd c  Lskd sin q r 
3




 L  L  L cosq 
kd a
sk d
me
 akd

2 


 Lbkd  Lkd b  Lskd cosq m e 
3




2 


 Lckd  Lkd c  Lskd cosq m e 
3



or:

 Lak  Lk a   Lsk sin q m e 
q
q
 q

2 


 Lbkq  Lkqb   Lskq sin q m e 
3




2 


 Lckq  Lkq c   Lskq sin q m e 
3



qr  qm e 

2

 Lak  Lk a  Lsk cosq r 
q
q
 q

2 

L

L

L
cos
q



 bkq
kqb
sk q
r
3




2 


 Lckq  Lkq c  Lskq cosq r 
3



Flux Linkage vs. Current (4)
L f  Llf  Lm f

Lkd  Llkd  Lm kd

Lkq  Llkq  Lm kq
L fk  Lk f
d
 d
L fkq  Lkq f  0

Lkd kq  Lkq kd  0
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 52
and 195.
Flux Linkage vs. Current (5)
This matrix can be transformed into dq form and used to
find flux linkage.
 i abc
 λ abc 
L ss L sr  i
From λ abcf  Labcf i abcf with λ abcf  λ
 L abcf  LT L  abcf  i fk k

r 
 sr
 fk k 

d
d q
λ abc  L ss i abc  L sr i fkd kq
K

1


λ dq0  Lss K 1i dq0  Lsr i fkd kq


λ dq0  KLss K 1 i dq0  KLsr i fkd kq
λ fkd kq  LTsr i abc  Lrr i fkd kq
λ fkd kq  LTsr K 1i dq0  Lrr i fkd kq
λ dqf  Ldqf i dqf
where λ dqf
 λ dq0 


λ
 fkd kq 
L dqf
KL ss K 1 KL sr 
  T 1

L rr 
 L sr K
i dqf
 i dq0 


i
 fkd k q 


q 

Inductance Matrix in dq Frame
Through derivations, we have
L dqf
where
and
From
0
 Ld
 0
Lq

 0
0
3
0
  Lsf
2

3 L
0
 2 skd

3
0
Lskq

2

Ld  Lls  Lm d
Lq  Lls  Lm q
3
Lm d  ( LA  LB )
2
3
Lm q  ( LA  LB )
2
0
0
L0
Lsf
0
0
Lskd
0
0
0
Lf
L fkd
0
Lkd f
Lkd
0
0
0
L0  Lls
λ dqf  Ldqf i dqf
0 
Lskq 
0 

0 

0 


Lkq 

L dqf
KL ss K 1 KL sr 
  T 1

L
K
L
rr 
 sr
d  Ld id  Lsf i f  Lskd ikd
  L i  L i
q q
sk q k q
 q
0  L0i0

  L i  3 L i  L i
f f
sf d
fkd k d
 f
2

3
kd  Lkd ikd  Lskd id  L fkd i f
2

  L i  3 L i
kq kq
sk q
 kq
2 q
Dynamical Equation in terms of Current
For linear model
from
di dqf
dt
where
dλ dqf
dt
V
1
 Ldqf
V
vd  Rs id  qr 
v  R i    
s q
d r 
 q
v0  Rs i0

V

v

R
i
f f
 f

 Rk ik

d
d


 Rkq ik q

and
λ dqf  Ldqf i dqf
dynamical equation
in terms of current
d  Ld id  Lsf i f  Lsk ik
d
q  Lq iq  Lsk ik
q
q
d
Power

Electrical instantaneous Input Power on Stator can also be
expressed through dq0 theory.
pin  vaia  vbib  vcic  vTabci abc  vTdq0 (K 1 )T K 1i dq0
1 0 0
3
1 T
1
(K ) K  0 1 0
2
0 0 2
3
pin  vd id  vqiq  2v0i0 
2
Torque
From
d


R
i





q r
 s d dt d
vd  

v    R i  d     
d r
 q   s q dt q

 v0  
d

R
i


s 0
0


dt
3
pin  vd id  vqiq  2v0i0 
2
we have
dq
d0  3 P
3
3  dd
2
2
2
 
pin  Rs id i q 2i0   id
 iq
 2i0
m (d iq  q id )
2
2  dt
dt
dt  2 2


Copper Loss
Magnetic Power in
Windings
Therefore, electromagnetic torque on rotor
Te 
pm ech
m

3P
(d iq  qid )
22
Mechanical Power
pm ech
Equivalent Circuit on d Axis (1)
d axis of stator, field winding and d axis damper of rotor can form an equivalent
circuit.
Let
Ld  Lls  Lm d
L f  Llf  Lm f
Lkd  Llkd  Lm kd
From L  3 C N
ˆ2
md
d
a
Lm f
2
 Cd Nˆ 2f
Lsf  Cd Nˆ a Nˆ f
Lm kd  Cd Nˆ k2d
Cd 
8 0

g
Dl
(
1

)
2
g av P
2
Nˆ a , N̂ f and N̂ k are effective number of turns of
d
armature, field and d axis
damper windings, respectively.
Lsk d  Cd Nˆ a Nˆ kd
L fkd  Cd Nˆ f Nˆ kd
(Details @ Inductance for SM.ppt)
Equivalent Circuit on d Axis (2)
Define
Lsf
2
i 
if 
Lm d
3
'
f
Nˆ f
if
ˆ
N
and i
'
kd
a

Lskd
Lm d
ikd
2 Nˆ kd

ikd
ˆ
3 Na
dikd
di f
did
vd  Rs id  qr  Ld
 Lsf
 Lskd
dt
dt
dt
d (id  i 'f  ik' d )
did
 Rs id  qr  Lls
 Lm d
dt
dt
Define
Nˆ a
'
v f  Nv f and N  ˆ
Nf
dikd
di f 3
did
v f  Rf i f  Lf
 Lsf
 L fkd
dt 2
dt
dt
Nv f  NR f i f  NLlf
di f
dt
 NLm f
dikd
did
3
 NLsf
 NL fkd
dt 2
dt
dt
di f
'
di f
did  Nˆ a 3 Nˆ a
 3 2  '  3 2  di f
Nv f   N R f i f   N Llf 
 Lsf
 Lm d

L fkd
ˆ
ˆ

dt
dt  N f 2 N kd
2

2
 dt
 dik'
 d
 dt

Equivalent Circuit on d Axis (3)
v 'f  R 'f i 'f  L'lf
where
di 'f
dt
3  Nˆ a
'
Rf 
2  Nˆ f
dt
2

 Rf


2
3  Nˆ a
L 
2  Nˆ f

 Llf


0  Rkd ikd  Lkd
dikd
'
lf
From
 Lm d
d (id  i 'f  ik' d )
 Rkd ikd  Llkd
L fkd
Lm d
2 Nˆ f Nˆ kd

3 Nˆ a2
di f
did
3
 Lskd
 L fkd
dt
2
dt
dt
dikd
dikd 3
di f
did
 Lm kd
 Lskd
 L fkd
dt
dt
2
dt
dt
 2 Lm d

 3 Lsk
d

  Nˆ a

  Nˆ
  kd

  above


next page
Equivalent Circuit on d Axis (4)
3  Nˆ a
0
2  Nˆ kd
2

 ˆ
 R i'  3  Na
 kd kd 2  Nˆ k

 d
0R i L
dik' d
3  Nˆ a
'
Rkd 
2  Nˆ kd

 R
 kd

3  Nˆ a

2  Nˆ kd

 L
 lkd

' '
kd kd
'
lkd
2
2
'

ˆ2
dik' d 3  Nˆ a 
dik' d
di
di
N
3
a
 L
 L
 
 Lm d d 
L fkd f
lkd
m kd

dt
2  Nˆ kd 
dt
dt 2 Nˆ f Nˆ kd
dt

dt
 Lm d
d (id  i 'f  ik' d )
dt
where
L'lkd
2
Lm d
Lm kd
2
Lm d
L fkd
3  Nˆ a

2  Nˆ kd
3 Nˆ a2

2 Nˆ Nˆ
f




kd
2
Equivalent Circuit on d Axis (5)
From
d (id  i 'f  ik' d )
did
vd  Rs id  qr  Lls
 Lm d
dt
dt
v R i L
'
f
' '
f f
'
lf
0  Rk' d ik' d  L'lkd
di 'f
dt
'
kd
di
dt
 Lm d
 Lm d
d (id  i 'f  ik' d )
dt
d (id  i 'f  ik' d )
dt
we get
ik' d
im d  id  i 'f  ik' d
d  Ld id  Lsf i f  Lsk ik
d
 Llsid  Lm dim d
d
Equivalent Circuit on q Axis (1)
q axis equivalent circuit and q axis damper equivalent circuit
can be combined:
Let
Lq  Lls  Lm q
Lkq  Llkq  Lm kq
From
3
Cq Nˆ a2
2
 Cq Nˆ k2q
Lm q 
Lm kq
Lsk q  Cq Nˆ a Nˆ k
Cq 
q
8 0

Dl

(
1

)
2
g av P
2
N̂ s and N̂ k are effective number of turns of
d
stator and q axis damper
windings, respectively.
(Details @ Inductance for SM.ppt)
Equivalent Circuit on q Axis (2)
vq  Rs iq  d r  Lq
 Rs iq  d r  Llq
 Rs iq  d r  Llq
From
diq
dt
diq
dt
diq
dt
 Lsk q
 Lm q
 Lm q
0  Rkq ikq  Lkq
 Rkq ikq  Llkq
dikq
dt
diq
 Lsk q
dikq
dt
dt
d (iq  ik' q )
dt
where
ik' d 
Lsk q
Lm d
ik q
2

3
Nˆ k q
ik q
ˆ
N
a
dikq
diq
3
 Lskq
dt
2
dt
dikq
dikq 3
diq
 Lm kq
 Lskq
dt
dt
2
dt
 2 Lm q   Nˆ

 a
 3 Lsk   Nˆ k
q 

 q

  above


next page
Equivalent Circuit on q Axis (3)
3  Nˆ a
0
2  Nˆ kq

2

 ˆ
 R i'  3  Na
 kq kq 2  Nˆ k

 q
0  Rk' q ik' q  L'lkq
dik' q
dt
 Lm q
2

dik' q 3  Nˆ a
 L
 
lk q

dt
2  Nˆ kq


2

dik' q
diq
 L

L
mq
 m kq dt
dt

d (iq  ik' q )
dt
where
2
3  Nˆ a
'
Rk q 
2  Nˆ k q


 R
 kq

3  Nˆ a

2  Nˆ k q


 L
 lkq

'
lk q
L
2
Lm q
Lm kq
Lm q
L fkq
3  Nˆ a

2  Nˆ kq

3 Nˆ a2

2 Nˆ Nˆ
f




kq
2
Equivalent Circuit on q Axis (4)
From vq  Rs iq  d r  Llq
0  Rk' q ik' q  L'lkq
diq
dt
dik' q
dt
 Lm q
 Lm q
d (iq  ik' q )
dt
d (iq  ik' q )
dt
we get
ik' q
im q  iq  ik' q
q  Lq iq  Lsk ik
q
q
 Llsiq  Lm qim q
Equivalent Circuit on 0 Axis
0 axis
v0  Rs i0  L0
di0
dt
This circuit is not necessary
for Y connected windings
since i0=0.
Dynamical Equations from
Equivalent Circuits (1)
vd  Rsd id  r q  Llsd
vq  Rsq iq  r d  Llsq
0  Rk' d ik' d  L'lkd
0  Rk' q ik' q  L'lkq
v 'f  R 'f i 'f  L'lf
v0  Rs 0i0  Lls 0
where
dik' d
dt
dik' q
dt
di 'f
dt
di0
dt
im d  id  i 'f  ik' d
im q  iq  ik' q
did
di
 Lm d m d
dt
dt
diq
di
 Lm q m q
dt
dt
 Lm d
 Lm q
 Lm d
dim d
dt
dim q
dt
dim d
dt
Dynamical Equations from
Equivalent Circuits (2)
The equations can be written in matrix form as:
L
where
 Llsd  Lm d

0


0
L
 Lm d
 Lm d

0

 id 
i 
 q
 i0 
I ' 
if 
ik' 
 'd 
ik q 
dI
V
dt
0
Llsq  Lm q
0
0
0
Lm q
or
0
0
Lls 0
0
0
0
vd
v
 q

V




dI
 L1V
dt
Lm d
0
0
L'lf  Lm d
Lm d
0
Lm d
0
0
Lm d
L'lkd  Lm d
0
 Rsd id  r q 
 Rsq iq  r d 

v0  Rs 0i0

v 'f  R 'f i 'f

'
'

 Rk d ik d

 Rk' q ik' q

0

Lm q 

0

0


0

L'lkq  Lm q 
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.
b axis
qd
q axis
q axis
d axis
qmq
qmq
qm
qa
g max
ror,min
qm
a axis
ror,max
g min
c axis
P
qm
2
P

qr  qm q  qm e 
2
2
qm e 
qd
qa
ris
a axis
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.
S dq0  KS abc
where
or
S abc  K 1S dq0
S dq 0
Sd 
Sa 
  S q , S abc   Sb 
 S 0 
 S c 
cosq m e  2 / 3
cosq m e  2 / 3 
 cosq m e 
2
K   sin q m e   sin q m e  2 / 3  sin q m e  2 / 3
3
 1 / 2

1/ 2
1/ 2
cosq m e 
 sin q m e 
1

K 1  cosq m e  2 / 3  sin q m e  2 / 3 1
cosq m e  2 / 3  sin q m e  2 / 3 1
(MIT’s notation)
 sin q r  sin q r  2 / 3 sin q r  2 / 3
2
K  cosq r  cosq r  2 / 3 cosq r  2 / 3
3
 1 / 2

1/ 2
1/ 2
sin q r 
cosq r 
1

K 1  sin q r  2 / 3 cosq r  2 / 3 1
sin q r  2 / 3 cosq r  2 / 3 1
(Purdue’s notation)
Voltage Equations (1)
Under motor reference convention for currents
(i.e. the positive reference direction for currents is into the machine):
For stator windings
K

d
v abc  R S i abc  λ abc
dt
1





K  K 1v dq0  KR S  K 1i dq0



d
K 1λ dq0
dt
d
 K  K 1λ dq0
dt
v dq0  R S K 1i dq0 
1 0 0
R S  Rs 0 1 0
0 0 1




d
d

v dq0  KR S K 1 i dq0  KK 1 λ dq0  K  K 1 λ dq0 
dt
 dt





v dq0  R S i dq0 
For field winding:
d
d

λ dq0  K  K 1 λ dq0
dt
 dt

v f  Rf i f 
d
λf
dt
Voltage Equations (2)
We derive the derivative of K-1:
cosq m e 
0
 sin q m e 
  cosq r 
d 1
K  m e sin q m e  2 / 3 cosq m e  2 / 3 0  r  cosq r  2 / 3
dt
sin q m e  2 / 3 cosq m e  2 / 3 0
 cosq r  2 / 3

me

dq me

dt

Then, we get
0
d

K K 1   r
 dt

 0
r 
 r
0
0
0
0
0
sin q r 
0
sin q r  2 / 3 0
sin q r  2 / 3 0
dq r
P
 me  m
dt
2
And for voltage, we get
d


R
i





q r
 s d dt d
vd  

d
 v   Rs iq  q  d r 
dt
 q 

d
 v0  

R
i


s 0
0
  

dt
v
 f  

d
 Rf if  f

dt


Dynamical Equations for Flux Linkage
d  vd
  
d  q   vq

dt  0  
  
 f  
Let
λ dqf
we have
d 
 
q
 
 0 
 
 f 
 Rs id  qr 
 Rs iq  d r 

v0  Rs i0

v f  Rf i f

vd
v
q
V



dλ dqf
dt
 Rs id  qr 
 Rs iq  d r 

v0  Rs i0

v f  Rf i f

V
The derivations so far are valid for both linear and nonlinear models.
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
L ss L sf 
L abcf   T

L
L
sf
f


 Laa Lab Lac 
L ss   Lba Lbb Lbc 
 Lca Lcb Lcc 

L  L  L  L cos 2q
ls
A
B
me
 aa

2 

L

L

L

L
cos
2
q



 bb
ls
A
B
me
3 



2 

Lcc  Lls  LA  LB cos 2q m e  
3 


 Laf 
 
L sf   Lbf 
 Lcf 
 

L  L  L  L cos 2q
ls
A
B
r
 aa

2 

or:
L

L

L

L
cos
2
q



 bb
ls
A
B
r
3 

 
qr  qm e  
2 L  L  L  L cos 2q  2 
cc
ls
A
B
r
3 


Flux Linkage vs. Current (2)


1
1
2 
2 


L
L


L
L




L
L


L
L
cos
cos
2
2
q
q






ab
ba
ba
A
A
B
B
me
r
 ab

2
2
3 
3 






1
1
Lbc  Lcb   LA  LB cos2q m e 
Lbc  Lcb   LA  LB cos2q r 
2
2




1
1
2 
2 


L
L


L
L




L
L


L
L
cos
cos
2
2
q
q






ca
ca
A
A
B
B
me
r
 ac
 ac
2
2
3
3







or:

L  L  L sin q 
L  L  L cosq 

fa
sf
r
fa
sf
me
 af
qr  qm e 
 af
2

2 


2 

L

L

L
sin
q



 bf
fb
sf
r

Lbf  L fb  Lsf cosq m e 
3 

3 




2 


2 

L

L

L
sin
q




cf
fc
sf
r

Lcf  L fc  Lsf cosq m e 
3 


3 


Note: Higher order harmonics are neglected.
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.
i abc 
L ss L sf 
λ abc 
From λ abcf  Labcf i abcf with λ abcf    L abcf  LT L  i abcf   i 
f 
 f 
 sf
 f 

λ abc  Lss i abc  Lsf i f
K

1


λ dq0  L ss K 1i dq0  L sf i f


λ dq0  KL ss K 1 i dq0  KL sf i f
λ f  LTsf i abc  L f i f
λ f  LTsf K 1i dq0  L f i f
λ dqf  Ldqf i dqf
where λ dqf
λ dq0 



 f 
L dqf
KL ss K 1 KL sf 
  T 1

L
K
L
 sf
f 

i dqf
i dq0 
 
 if 
Inductance Matrix in dq0 Frame
Through derivations, we have
L dqf
where
and
From
 Ld
 0

 0
3
 Lsf
2
0
Lq
0
0
0
L0
0
0
Lsf 
0 
0

Lf 

L dqf
KL ss K 1 KL sf 
  T 1

L
K
L
 sf
f 

Ld  Lls  Lm d
Lq  Lls  Lm q
L0  Lls
3
Lm d  ( LA  LB )
2
3
Lm q  ( LA  LB )
2
λ dqf  Ldqf i dqf
d

q
0


 f
 Ld id  Lsf i f
 Lq iq
 Lls i0
3
 L f i f  Lsf id
2
Dynamical Equation in terms of Current
For linear model
from
di dqf
dt
dλ dqf
dt
V
1
 Ldqf
V
and
λ dqf  Ldqf i dqf
dynamical equation
in terms of current
where
vd
v
q
V



 Rs id  qr 
 Rs iq  d r 

v0  Rs i0

v f  Rf i f

d  Ld id  Lsf i f
q  Lq iq
Power

Electrical instantaneous Input Power on Stator can also be
expressed through dq0 theory.
pin  vaia  vbib  vcic  vTabci abc  vTdq0 (K 1 )T K 1i dq0
1 0 0
3
1 T
1
(K ) K  0 1 0
2
0 0 2
3
pin  vd id  vqiq  2v0i0 
2
Torque
From
d


R
i





q r
 s d dt d
vd  

v    R i  d     
d r
 q   s q dt q

 v0  
d

R
i


s 0
0


dt
3
pin  vd id  vqiq  2v0i0 
2
we have
dq
d0  3 P
3
3  dd
2
2
2
 
pin  Rs id i q 2i0   id
 iq
 2i0
m (d iq  q id )
2
2  dt
dt
dt  2 2


Copper Loss
Magnetic Power in
Windings
Therefore, electromagnetic torque on rotor
Te 
pm ech
m

3P
(d iq  qid )
22
Mechanical Power
pm ech
Equivalent Circuits (1)
d


R
i





q r
d  Ld id  Lsf i f
 s d dt d

vd  

d
q  Lq iq
 v   Rs iq  q  d r 
dt
 q 
0  Lls i0

d
 v0  


R
i


s 0
0
  L i  3 L i
  

dt
f f
sf d
v f  
 f

2
d
 Rf if  f

dt


di f
did
d axis
vd  Rs id  qr  Ld
 Lsf
dt
dt
Equivalent Circuits (2)
q axis
0 axis
vq  Rs iq  d r  Lq
v0  Rs i0  L0
diq
dt
di0
dt
This circuit is not necessary
for Y connected windings
since i0=0.
Equivalent Circuits (3)
Field winding
v f  Rf i f  Lf
di f
dt

di
3
Lsf d
2
dt
Combined Equivalent Circuit on d Axis (1)
d axis equivalent circuit and field winding equivalent circuit can be combined:
Ld  Lls  Lm d
L f  Llf  Lm f
From
3
Cd Nˆ a2
2
 Cd Nˆ 2f
Lm d 
Lm f
Cd 
8 0

g
Dl
(1 
)
2
g av P
2
N̂ a and N̂ f are effective number of
turns of armature and
(Details @ InductanceSM.ppt)
field windings.
ˆ
L
N
Let N  a  2 Lm d  sf
3 Lsf
Lm f
Nˆ f
di f
did
vd  Rs id  qr  Ld
 Lsf
dt
dt
Lsf
2i f
'
'
d
(
i

i
)
di
if 
if 
d
f
 Rs id  qr  Lls d  Lm d
Lm d
3N
dt
dt
Lsf  Cd Nˆ a Nˆ f
Combined Equivalent Circuit on d Axis (2)
v f  Rf i f  Lf
di f

dt
Nv f  NR f i f  NLlf
di
3
Lsf d
2
dt
di f
dt
 NLm f
if 
3 '
Ni f
2
di f
Lsf
2 Lm d
N

3 Lsf
Lm f
di
3
 NLsf d
dt 2
dt
'
di f
di
 3 2  '  3 2  di f
Nv f   N R f i f   N Llf 
 Lsf
 Lm d d
dt
dt
2

2
 dt
v 'f  R 'f i 'f  L'lf
di 'f
dt
 Lm d
d (id  i 'f )
dt
v 'f  Nv f
3 2
N Rf
2
3
L'lf  N 2 Llf
2
R 'f 
Combined Equivalent Circuit on d Axis (3)
From
d (id  i 'f )
did
vd  Rs id  qr  Lls
 Lm d
dt
dt
v R i L
'
f
' '
f f
'
lf
di
'
f
dt
 Lm d
d (id  i )
'
f
dt
we get
im d  id  i 'f
d  Ld id  Lsf i f
 Lls id  Lm dim d
v 'f  Nv f
if 
3 '
Ni f
2
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.
b axis
qd
q axis
qmq
qm
d axis
qa
a axis
P
qm
2
P

qr  qm q  qm e 
2
2
qm e 
c axis
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.
S dq0  KS abc
where
or
S abc  K 1S dq0
S dq 0
Sd 
Sa 
  S q , S abc   Sb 
 S 0 
 S c 
cosq m e  2 / 3
cosq m e  2 / 3 
 cosq m e 
2
K   sin q m e   sin q m e  2 / 3  sin q m e  2 / 3
3
 1 / 2

1/ 2
1/ 2
cosq m e 
 sin q m e 
1

K 1  cosq m e  2 / 3  sin q m e  2 / 3 1
cosq m e  2 / 3  sin q m e  2 / 3 1
(MIT’s notation)
 sin q r  sin q r  2 / 3 sin q r  2 / 3
2
K  cosq r  cosq r  2 / 3 cosq r  2 / 3
3
 1 / 2

1/ 2
1/ 2
sin q r 
cosq r 
1

K 1  sin q r  2 / 3 cosq r  2 / 3 1
sin q r  2 / 3 cosq r  2 / 3 1
(Purdue’s notation)
Voltage Equations (1)
Under motor reference convention for currents
(i.e. the positive reference direction for currents is into the machine):
For stator winding
K

d
v abc  R S i abc  λ abc
dt
1





K  K 1v dq0  KR S  K 1i dq0



d
K 1λ dq0
dt
d
 K  K 1λ dq0
dt
v dq0  R S K 1i dq0 
1 0 0
R S  Rs 0 1 0
0 0 1




d
d

v dq0  KR S K 1 i dq0  KK 1 λ dq0  K  K 1 λ dq0 
dt
 dt





v dq0  R S i dq0 
d
d

λ dq0  K  K 1 λ dq0
dt
 dt

Voltage Equations (2)
We derive the derivative of K-1:
cosq m e 
0
 sin q m e 
  cosq r 
d 1
K  m e sin q m e  2 / 3 cosq m e  2 / 3 0  r  cosq r  2 / 3
dt
sin q m e  2 / 3 cosq m e  2 / 3 0
 cosq r  2 / 3

me

dq me

dt

Then, we get
0
d

K K 1   r
 dt

 0
r 
 r
0
0
0
0
0
sin q r 
0
sin q r  2 / 3 0
sin q r  2 / 3 0
dq r
P
 me  m
dt
2
And for voltage, we get
d


R
i





q r
 s d dt d
vd  

v    R i  d     
d r
 q   s q dt q

 v0  
d

R
i


s 0
0


dt
Dynamical Equations for Flux Linkage
d  vd  Rs id  qr 
d   



v

R
i



q
q
s q
d r


dt
 0  

v0  Rs i0
Let
vd  Rs id  qr 
V  vq  Rs iq  d r 


v0  Rs i0
d 
λ dq0  q 
0 
we have
dλ dq0
dt
V
The derivations so far are valid for both linear and nonlinear models.
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
 Laa Lab Lac 
L abc   Lba Lbb Lbc 
 Lca Lcb Lcc 

L  L  L  L cos 2q
ls
A
B
me
 aa

2 

L

L

L

L
cos
2
q



 bb
ls
A
B
me
3 



2 

Lcc  Lls  LA  LB cos 2q m e  
3 



L  L  L  L cos 2q
ls
A
B
r
 aa

2 

or:
L

L

L

L
cos
2
q



 bb
ls
A
B
r
3 

 
qr  qm e  
2 L  L  L  L cos 2q  2 
cc
ls
A
B
r
3 


Note: Higher order harmonics are neglected.
Flux Linkage vs. Current (2)

1
2 

L

L


L

L
cos
2
q

 me

ba
A
B
 ab
2
3




1
L

L


LA  LB cos2q m e 
 bc
cb
2


1
2 

L

L


L

L
cos
2
q



ca
A
B
me
 ac
2
3 



1
2 

L

L


L

L
cos
2
q

 r

ba
A
B
 ab
2
3




1
L

L


LA  LB cos2q r 
 bc
cb
2


1
2 

L

L


L

L
cos
2
q



ca
A
B
r
 ac
2
3 


or:
qr  qm e 

2
Note: Higher order harmonics are neglected.
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.
λ abc  L abci abc  λ PMabc

K
1



λ dq0  L abcf K 1i dq0  λ PMabc



K K 1λ dq0  KL abc K 1i dq0  Kλ PMabc


λ dq0  KL abcK 1 i dq0  Kλ PMabc
λ dq0  Ldq0i dq0  λ PMdq0
where
 cos(q m e)

 sin(q r )

or:
λ PMabc  PM cos(q m e  2 / 3)
λ PMabc  PM sin(q r  2 / 3)
cos(q m e  2 / 3) q r  q m e  
sin(q r  2 / 3)
2
Inductance Matrix in dq0 Frame

Therefore, we get the following inductance matrix in dq0
frame:
PM 
 Ld 0 0 
L dq0  KL abcK 1   0 Lq 0 
λ PMdq0  Kλ PMabc   0 
 0 0 L0 
 0 
where
Ld  Lls  Lm d
Lq  Lls  Lm q
L0  Lls
and
From
3
Lm d  ( LA  LB )
2
3
Lm q  ( LA  LB )
2
λ dq0  Ldq0i dq0  λ PMdq0
d  Ld id  PM

 q  Lqiq
  L i
0
ls 0

Dynamical Equation in terms of Current
For linear model from
di dq0
dt
where
dλ dq0
dt
V
1
 Ldq
0V
vd  Rs id  qr 
V  vq  Rs iq  d r 


v0  Rs i0
and
λ dq0  Ldq0i dq0  λ PMdq0
dynamical equation
in terms of current
d  Ld id  PM
q  Lqiq
L dq0
 Ld
  0
 0
0
Lq
0
0
0 
L0 
(vd  Rs id  r Lqiq ) / Ld
id  

d   

i

(
v

R
i


L
i



)
/
L
q
q
s q
r d d
r PM
q
dt   
i0  

(v0  Rs i0 ) / L0
1
i

(ia  ib  ic )  0 , only need to consider
For Y connected winding, since 0
3
the first two equations for id and iq.
Power

Electrical instantaneous Input Power on Stator can also be
expressed through dq0 theory.
pin  vaia  vbib  vcic  vTabci abc  vTdq0 (K 1 )T K 1i dq0
1 0 0
3
1 T
1
(K ) K  0 1 0
2
0 0 2
3
pin  vd id  vqiq  2v0i0 
2
Torque
From
3
pin  vd id  vqiq  2v0i0 
2
we have
d


R
i





q r
 s d dt d
vd  

v    R i  d     
d r
 q   s q dt q

 v0  
d

R
i


s 0
0


dt
dq
d0  3 P
3
3  dd
2
2
2
 
pin  Rs id i q 2i0   id
 iq
 2i0
m (d iq  q id )
2
2  dt
dt
dt  2 2


Copper Loss
Magnetic Power in
Windings
Mechanical Power
pm ech
Therefore, electromagnetic torque on rotor
Te 
pm ech
m
3P

(d iq  qid )
22
d  Ld id  PM
q  Lqiq

3P
Te 
PM iq  ( Ld  Lq )id iq
22

Dynamical Equations of Motion
dm
J
 Te  TL  Tdam p
dt
dq m
 m
dt
where


3P
Te 
PM iq  ( Ld  Lq )id iq  KT iq
22
KT 
For round rotor machine,
Tdamp  Dmm

Te 3P

PM  ( Ld  Lq )id
iq
4
Ld  Lq

torque constant
3P
Te 
PM iq
4
3P
KT 
PM
4
Dm is combined damping coefficient of rotor
and load.