emmc steps daac lesson3(rev3)

8/10/2019 Emmc Steps Daac Lesson3(Rev3)

1/42

ARBITRARY REFERENCE FRAME

DYNAMIC ANALYSIS AND CONTROL

OF AC MACHINES

ERASMUS MUNDUS MASTER COURSE on

SUSTAINABLE TRANSPORTATION AND

ELECTRIC POWER SYSTEMS


2/42

Dynamic Analysis and Control of AC Machines - Lesson 3 Pagina 2

2

ARBITRARY REFERENCE FRAME

A+

B-

B+

C-

C+

a

b

c

S

A-

N

d

q

0

+=t

dt0

)0()()(

==

=

c

b

a

abcd

q

qd

f

f

f

f

f

f

KfKf

0

0

( )

( )

=hhh

k 3

4

sin3

2

sinsin

3

4cos

3

2coscos

K


3/42


3

CHOICE OF THE ARBITRARY CONSTANTS

( ) ( )

=

13

4sin

3

4cos

13

2

sin3

2

cos

1sincos

1

K

( )

( )

=

2

1

2

1

2

13

4

sin3

2

sinsin

3

4cos

3

2coscos

3

2

K

CONSTANT AMPLITUDE TRANSFORMATION

== 2

1;3

2hk

This transformation is conservative with respect to

the amplitude.

The 0component always represents the

homopolar component


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5

CHOICE OF THE ARBITRARY CONSTANTS

T

KK =1

( )

( )

=

2

1

2

1

2

13

4

sin3

2

sinsin

3

4cos

3

2coscos

3

2

K

CONSTANT POWER TRANSFORMATION

== 2

1;3

2hk

This transformation is orthogonal, but is NOT

conservative with respect to the amplitude.

The 0component always represents the

homopolar component


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6

CHOICE OF THE ARBITRARY CONSTANTSCONSTANT POWER TRANSFORMATION

This transformation is intrinsically conservative

with respect to the power.

ccbbaaabc ivivivtp ++=)(

)()( 0 tptp qdabc =

000 )( ivivivtp ddqqqd ++=

Instantaneous power in the

abcreference frame

Instantaneous power in the

qd0reference frame

== 2

1;3

2hk


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7

TRANSFORMATION OF A RESISTIVE CIRCUIT

The transformation does not modify the structure

of the equation

=

=

R

R

R

abc

abcabcabc

00

00

00

R

iRv abcabcabcqd iRKvKv ==0

01

qdabc iKRK =

00 qdqd iR =

==

R

R

R

abcqd

00

00

00

0 RR

000 qdqdqd iRv =

abc qd0


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( ) 000000

001

010

qdqdqddt

dv

+=

8

TRANSFORMATION OF AN INDUCTIVE CIRCUIT

But:

( )

abcabcabc

abcabcdtd

iL

v

=

= ( )abcabcqd dtd KvKv ==0

abc qd0

( )01 qddt

dKK =

( ) ( ) 0101 qdqddt

d

dt

dKKKK +=

( ) ( )dt

d

d

d

dt

d

= 11 KKKK

( )

=

000001

0101

KK dt

d


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9

TRANSFORMATION OF AN INDUCTIVE CIRCUIT

( )

abcabcabc

abcabcdtd

iL

v

=

=

abc qd0

=LMMMLM

MML

abcL

abcabcabcqd iLKK ==0

01

qdabc iKLK =

00 qdqd iL =

+

=

=

ML

ML

ML

qd

abcqd

200

00

00

0

1

0

L

KLKL

=

0

q

d

dq

( ) dqqdqddt

dv += 00

000 qdqdqd iL =


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10

TRANSFORMATION BETWEEN REFERENCE FRAMES

A+

B-

B+

C-

C+

a

b

c

S

A-

N

x

dx

qx x

0

dy

q

y

y

y

xqd

yx

yqd 00 fKf =

abcxx

qd fKf =0

abcyy

qd fKf =0

abcx

y

x

y

qd

fKKf =0

xyxy KKK =


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12/42


12

ROTATING VECTOR (OR SPACE VECTOR)

)(3

23

4

3

2 j

c

j

ba efeff ++=f

The rotating vector is defined as:

The rotating vector describes a complete 3-phase system

(is not a phasor). A symmetrical 3-phase system withpulsation

e, generates a space vector of constant

amplitude which rotates at angular speede.

The projections of the space vector fon the q- and d- axes

gives the same results as applying the coordinatetransformation matrix K.


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13

ROTATING VECTOR (OR SPACE VECTOR)

dqqdo jff =f

The rotating vector in qd0 coordinates can be expressed

as a complex number:

Alternatively, it can be expressed as the projection of the

vector on the q- and d- axes. jqdo ej

== fff ))sin()(cos(

which yields:

))sin()(cos(32 3

4

3

2

jefeffj

c

j

baqdo

++=f


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14

TRANSFORMATION EXAMPLE

0 0.02 0.04 0.06 0.08 0.1-400

-300

-200

-100

0

100

200

300

400

Va

Vd

Vq

0 0.01 0.02 0.03 0.04 0.05-400

-300

-200

-100

0

100

200

300

400

Iq

Ia

Ib

Ic


15/42

PM MACHINES

DYNAMIC ANALYSIS AND CONTROL

OF AC MACHINES

ERASMUS MUNDUS MASTER COURSE on

SUSTAINABLE TRANSPORTATION AND

ELECTRIC POWER SYSTEMS


16/42


16

OUTLINE

Classification

PM Machines modelling

Vector control of PM Machines

Flux weakening


17/42


17

CLASSIFICATION (BY OPERATION)

PM motors

Inverter fedLine start

Cage rotor Cageless rotor

Rectangular fedSinusoidal fed

SensorlessWith position

sensors

AC BRUSHLESS

DC BRUSHLESS


18/42


18

CLASSIFICATION (BY MACHINE TYPE)

Surface PM (SPM)

Interior PM (IPM)

Inset PM


19/42


19

Surface PM machines (SPM)

Magnets are mounted on rotor surface Mechanical stress at high speeds

The machine is magnetically

ISOTROPIC (magnetair)

No inductance variation with rotor position

Relatively low inductance (high airgap)

Low demagnetization risk

Trapezoidal back-emf (DC BRUSHLESS)

Can have Sinusoidal back-emf (AC BRUSHLESS) with

proper stator winding arrangement


20/42


20

Interior PM machines (IPM)

Magnets are mounted inside rotor Mechanically resistent at high speeds


ANISOTROPIC (magnetiron)

Reluctance and cogging torque

Inductance variation with rotor position

Relatively high inductance (low airgap)

High demagnetization risk

Sinusoidal back-emf (AC BRUSHLESS)


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21

Inset PM machines

Magnets are placed inside hollows on rotor surface Mechanically resistent at high speeds


ANISOTROPIC (magnetiron)

Reluctance and cogging torque

Inductance variation with rotor position

Sinusoidal back-emf (AC BRUSHLESS)

In between SPM and IPM


22/42


22

AC brushless modelling

Modelling hypotheses:

Sinusoidal flux distribution along the airgap;

Saturation phenomena neglected;

Hysteresis and eddy current losses neglected;

Temperature effects neglected;

No rotor cage.


23/42


23

AC brushless equations in phase coordinates

abcabcabcdt

diRv +=

abcabcPMabc iL +=

=

)34sin(

)32sin(

)sin(

e

e

e

PMPM

=

)()()(

)()()(

)()()(

)(

rccrcbrca

rbcrbbrba

racrabraa

rabc

LLL

LLL

LLL

L

=

R

R

R

00

00

00

R

T

abc

e

abcPT i

=

dtdJTT rL =

re P =

re P =

A+

B-

B+

C-

C+

a

b

c

S

A-

N

R

r

PM


24/42


24

Inductance matrix in phase coordinates (IPM)

abc

[ ]

[ ]

+++++

++

++

++

+

++

+++

=

)3

2(2cos)(2cos

2

1)

3(2cos

2

1

)(2cos2

1)

3

2(2cos)

3(2cos

2

1

)3

(2cos2

1)

3(2cos

2

1)2cos(

)(

rBAlrBArBA

rBArBAlrBA

rBArBArBAl

rabc

LLLLLLL

LLLLLLL

LLLLLLL

L

A+

B-

B+

C-

C+

a

b

c

S

A-

N

R

r

PM


25/42


25

Torque equation in phase coordinates

abcabcPMabc iL +=

abc

T

abcabc

T

abcabc

T

abc dt

diiRivi

+=

( ) ( ) ( )dt

d

tdt

d eabc

e

T

abcabc

T

abcabc

T

abc

+

= iii

( )dt

d

PTT eabc

e

T

abce

r

== i

( )abc

e

T

abcPT i

=


26/42


27/42


27

Synchronous reference frame

A+

B-

B+

C-

C+

a

b

c

S

A-

N

R

e

PM

d

qe

0

The synchronous referenceframe is obtained from the

arbitrary reference frame,

when:

)()( tt e =

and the d-axis is aligned

with the PM flux rotating

vector:

0,

,

=

=

qPM

PMdPM


28/42


28

AC brushless equations in synchr. reference frame

dqeqdqdqddt

diRv ++= 000

000,0 qdqdqdPMqd iL +=

=

0

0

0, PMqdPM

=

0

0

00

00

00

L

L

L

d

q

qdL

=

R

R

R

00

00

00

R

)(2

3dqqd iiPT =

dtdJTT rL =

re P =

re P =

=

0

q

d

dq

A+

B-

B+

C-

C+

a

b

c

S

A-

N

R

e

PM

d

qe

0


29/42


29

Matrix inductance in synchr. reference frame

qd0

=

=

0

0

10

0000

00

LL

L

d

q

qd

abcqd

L

KLKL

l

BAld

BAlq

LL

LLLL

LLLL

=

+=

++=

0

23

2

3

[ ]

[ ]

+++++

++

++

++

+

++

+++

=

)

3

2(2cos)(2cos

2

1)

3

(2cos

2

1

)(2cos2

1)

3

2(2cos)

3(2cos

2

1

)3

(2cos2

1)

3(2cos

2

1)2cos(

)(

rBAlrBArBA

rBArBAlrBA

rBArBArBAl

rabc

LLLLLLL

LLLLLLL

LLLLLLL

L


30/42


30

Torque equation in synchr. reference frame

000,0 qdqdqdPMqd iL +=

( ) dqe

T

qdqd

T

qdqd

T

qdqd

T

qd dt

diiiRivi ++=

0000000 2

3

2

3

2

3

2

3

( ) 000 =

dt

d eqd

e

T

qd

i

dqe

T

qde

rP

TT i ==

02

3

dq

T

qdPT i = 02

3


31/42


31

Torque equation

))((2

3qdqdqPM iiLLiPT +=

)(2

3

dqqd

iiPT = ddPMd iL +=

qqq iL =

qPM iPT = 2

3

The torque equation shows a first term which is

proportional to PM flux (synchronous torque) and a

second term which is proportional to the anisotropy

(reluctance torque).

For an IPM machine, since Ld


32/42


32

Maximum Torque per Ampere (MTPA)

))((2

3qdqdqPM iiLLiPT +=

MTPA is a control strategy that seeks to maximize thetorque for a given amount of current.

A+

B-

B+

C-

C+

a

b

c

S

A-

N

e

i

d

q

e

0

e

)cos(=Iiq

)sin(= Iid

[ ])sin()cos()()cos(2

3 2 = ILLIPT qdPM

MTPA looks for the angle

that satisfies:

0=

I

T


33/42


33

Vector control

To achieve vectorcontrol for PM machines means to

determine the amplitudeI and angle(and frequency,of course) of the current rotating vector, such that the

desired torque is obtained with the minimum current

amplitude (MTPA).

In the 3-phase stationary reference frame abc:

In the synchronous reference frame qd0:

+=

+=

+=

)34cos()(

)32cos()(

)cos()(

***

***

***

*

**

tIti

tIti

tItiI

T

ec

eb

ea

=

=

)sin(

)cos(

***

***

*

**

Ii

IiIT

d

q


34/42


35/42


35

Vector control for SPM machines (cont.)

For SPM machines, vectorcontrol means to control the

current such that its vector

is aligned with the q-axis.

In other words, it means tocontrol the current so that it

is in phase with the back-

emf.

(E=ePMis on the q-axis)

A+

B-

B+

C-

C+

a

b

c

S

A-

N

i*

e

PM

d

qe

0

E

NOTE: id=0 must be forced by the controller: idwill not stay

at zero naturally.


36/42


36

General control scheme for PM machines

PM MOTOR EPI, PI,i

Current

feedback

*

+

+

T* i*

i

Speed

feedback

v*

Speed Regulator Current Regulator =

Torque Regulator

INVERTER

Vdc

~


37/42


37

Current control in 3-ph. reference frame

PM MOTOR EReference

generator

PI,i+

T*

ia

va*

INVERTER

Vdc

~

PI,i+

vb*

PI,i+

vc*

ib

ic

MTPA

I*

*

P

W

M

ia*

ib*

ic*

Requires three current regulators


38/42


38

Current control in synchr. reference frame

Requires two current regulators

Requires two coordinate transformations Computationally more intense

Higher performances (current bandwidth not affected by

speed)

PM MOTOR EK-1

PI,i+

T*

ia

va*

INVERTER

Vdc

~

PI,i+

vb*

vc*

ib

ic

MTPA

P

W

M

iq*

id*

e

K

iq id


39/42


39

Flux weakening

As the speed increases, also the back-emf increases. At

some point (generally at nominal speed) the voltagerequired to balance the back-emf reaches the nominal

voltage of the machine or the voltage limit of the inverter.

)(ddPMeqqq

iLdt

d+++= iRv

qqq

iL =

qqeddd iLdt

d+= iRvddPMd iL +=

Above this speed, since the PM flux is constant, the only

possible thing is to weaken the linked flux on the d-axis

through an injecton of negative d-axis current (flux

weakening), so that the total induced voltage on the q-

axis remains constant.


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40

Flux weakening for SPM machines

PMbde E === 0const.

PMbPMedde EiL ==+ 0*

beup /.. =

PMbPMupbddupb iL =+

..

*

..

dup

upPM

dL

i

=

..

..*)1(

de

up

dL

Ei

=

)1( ..0*

Let bbe the base speed.

Below base speed MTPA requires id=0.

Above base speed the total induced voltage on the q-

axis is imposed to be constant:

If then

or


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41

Flux weakening for SPM machines (cont.)

de

up

dL

Ei

=

)1( ..0*

2*2max

*max, dq iIi =

)23/( **

PMq PTi =

>

=

>

=

b

de

up

b

d

bqq

bq

q

L

E

i

ii

ii

)1(

0

),min(

..0*

*

max,

*

*

*

Given id

*, respect of the maximum current amplitude Imax

must be enforced.

EQ 1

EQ 2


42/42


Flux weakening for SPM machines (cont.)

PM MOTOR EKPI,i

+

T*

ia

va*

INVERTER

Vdc

~

PI,i+

vb*

vc*

ib

ic

EQ. 2

PW

M

iq*

id*

e

K-1

iq id

PI,* +

EQ. 1id

*

Speedobserver

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