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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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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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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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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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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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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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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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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000 iRv

iRv

iRv

dt

d

s

rdqsq

rqdsd

q

d


0

0

q

d

dq

00 iRv

iRviRv

s

rdqsq

rqdsd

V

Let

we haveV

dt

d dq0

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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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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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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



0

11

0 )( dqTT

dqabc

T



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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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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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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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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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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qq

dd

q

d

kk

kk

fff

s

rdqsq

rqdsd

k

k

f

q

d

iR

iRiRv

iRv

iRv

iRv

dt

d 000


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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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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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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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



0

11

0 )( dqTT

dqabc

T



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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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


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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


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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


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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


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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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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


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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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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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15 dq Theory