Modeling of Induction Motor using dq0 Transformations First Semester 1431/1432

Modeling of Induction Motor using dq0 Transformations

First Semester 1431/1432

Steady state model developed in previous studies of induction motor neglects electrical transients due to load changes and stator frequency variations. Such variations arise in applications involving variable-speed drives.

Variable-speed drives are converter-fed from finite sources, which unlike the utility supply, are limited by switch ratings and filter sizes, i.e. they cannot supply large transient power.

Introduction

Thus, we need to evaluate dynamics of converter-fed variable-speed drives to assess the adequacy of the converter switches and the converters for a given motor and their interaction to determine the excursions of currents and torque in the converter and motor. Thus, the dynamic model considers the instantaneous effects of varying voltages/currents, stator frequency and torque disturbance.

Introduction (cont’d)

Circuit Model of a Three-Phase IM

1. Space mmf and flux waves are considered to be sinusoidally distributed, thereby neglecting the effect of teeth and slots.

2. The machine is regarded as group of linear coupled circuits, permitting superposition to be applied, while neglecting saturation, hysteresis, and eddy currents.

3. Ls : self inductance per phase of the stator windings.4. Ms: mutual inductance per phase of the stator windings.5. rs: resistance per phase of the stator windings.6. Lr : self inductance per phase of the rotor windings.7. Mr: mutual inductance per phase of the rotor windings8. rr: resistance per phase of the rotor windings.9. Msr: maximum value of mutual inductance between any

stator phase and any rotor phase.

Assumptions and Definitions:

ias

i bs

ics

iar

i br

icrvas+

+v

bs

+v cs

v ar+

+vbr

vcr +

axis of stator phase c

axis of stator phase a

axis of stator phase b

axis of rotor phase c

axis of rotor phase b

axis of rotor phase a

r

r

r

Circuit Model of a Three-Phase IM

r

Stator Voltage Equations:

asas as s

dv i r

dt

bsbs bs s

dv i r

dt

cscs cs s

dv i r

dt

Voltage Equations

ias

i bs

ics

iar

i br

icrvas+

+v

bs

+v cs

v ar+

+vbr

vcr +







r

r

r

r

Rotor Voltage Equations:

a ra r a r r

dv i r

dt

brbr br r

dv i r

dt

crcr cr r

dv i r

dt

Voltage Equations (cont’d)

ias

i bs

ics

iar

i br

icrvas+

+v

bs

+v cs

v ar+

+vbr

vcr +







r

r

r

r

cos( ) cos( 120 ) cos( 120 )sr sr sar r br r c

as s as s bs s c

r r r

s

M M

L i M i M i

i i M i

cos( 120 ) cos( ) cos( 120 )ar r

bs s as s bs s

br r csr s r rr s

cs

rM M

M i L i M i

Mi i i

Flux Linkage Equations

ias

i bs

ics

iar

i br

icrvas+

+v

bs

+v cs

v ar+

+vbr

vcr +







r

r

r

r

(

cos( ) cos( 120 ) cos 20 )

)

( 1sr sar r br r cr

as s as s bs cs

rr srM M

L i M

i i M

i i

i

cos( 120 ) cos( ) c

( )

os( 120 )

bs s as cs s b

sr sr sar r br r cr r

s

r

M i

M M Mi i

i L i

i


0as bs csi i i

In general, we can assume:ias

i bs

ics

iar

i br

icrvas+

+v

bs

+v cs

v ar+

+vbr

vcr +







r

r

r

r

ss s sL L M

Let:

cos( ) cos( 120 ) cos( 120 )

( )ar r r r

r ar r br c

as bs cs

r

sr sr sr

L i M

i i M

i i

M M i


0ar br cri i i

In general, we can assume:

ias

i bs

ics

iar

i br

icrvas+

+v

bs

+v cs

v ar+

+vbr

vcr +







r

r

r

r

cos( 120 ) cos( ) cos( 120 )

( )assr sr sbr r r r

r b

bs cs

r c

r

r ar r

M M M

L i M i

i i i

i

rr r rL L M

Let:


cos( 120 ) cos( 120 ) cos( )cs ar r bss c r r crs sr ri iM iL i

cos( ) cos( 120 ) cos( 120 )as ss sr ar r br r ra crsL i i i iM

cos( 120 ) cos( ) cos( 120 )bs ss bs ar r br r r rsr ci iL M ii

Stator:

Rotor:

cos( ) cos( 120 ) cos( 120 )as bs cssar r r r rrr ariM i ii L

cos( 120 ) cos( ) cos( 120 )as bs cbr r r rs rr brsr i i i L iM

cos( 120 ) cos( 120 ) cos( )ascr r r r rrbs s cr rcs i i i L iM

0 0

0

0 0

0

0

0

0

0

0 0

ar arrr

ar brrr

ar crr

sr

Ts

asssas

bsssb

r

s

ccs

r

sss

iL

iL

i

iL

L

L

L

iL

iL

cos( ) cos( 120 ) cos( 120 )

cos( 120 ) cos( ) cos( 120 )

cos( 120 ) cos( 120 ) cos( )

r r r

r rsr r

r

s

r r

rL M


asas as s

dv i r

dt

To build up our simulation equations, we could just differentiate each expression for , e.g.

But since Lsr depends on position ,

which will generally be a function of time, the trigonometric terms will lead to a mess!

First raw of the Matrixasas as s

d dv i r

dt dt

Model of Induction Motor

The Park’s transformation is a three-phase to two-phase transformation for synchronous machine analysis. It is used to transform the stator variables of a synchronous machine onto a dq reference frame that is fixed to the rotor.

The +ve d-axis is aligned with the magnetic axis of the field winding and the +ve q-axis is defined as leading the +ve d-axis by /2.

Park’s Transformation

Park’s Transformation (cont’d)

ias

i bs

ics

iar

i br

icrvas+

+v

bs

+v cs

v ar+

+vbr

vcr +







r

r

r

r

d-axis

q-axis

The result of this transformation is that all time-varying inductances in the voltage equations of an induction machine due to electric circuits in relative motion can be eliminated.

In induction machine, the d-axis is assumed to align on a-axis at t = 0 and rotate with synchronous speed ()

The Park’s transformation equation is of the form:

where f can be i, v, or .

0

0

f f

f T f

f f

d a

q dq b

c


0

2 2cos cos cos

3 3

2 2( ) sin sin sin

3 3

1 1 1

2 2 2

T

d d d

dq d d d dK


where K is a convenient constant. The current id and iq are proportional to the components of mmf in the direct and quadrature axes, respectively, produced by the resultant of all three armature currents, ia, ib, and ic. For balanced phase currents of a given maximum magnitude, the maximum value of id and iq can be of the same magnitude. Under balanced conditions, the maximum magnitude of any one of the phase currents is given by . To achieve this relationship, a value of 2/3 is assigned to the constant K.

2 2, , ,a peak b peak c peak d qi i i i i

The inverse transform is given by:

Of course, [T][T]-1=[I]

1

0

cos sin 1

2 2( ) cos sin 1

3 3

2 2cos sin 1

3 3

T

d d

dq d d d

d d


Thus,

and

0 0

0

d a

q dq b dq abc

c

v v

v T v T v

v v

0 0

0

d a

q dq b dq abc

c

i i

i T i T i

i i


ids

vds+

iqr +vqr

idr

vdr+

iqs +vqs

Induction Motor Model in dq0

d-axis

q-axis

Lets us define new “dq0” variables.

Our induction motor has two subsystems - the rotor and the stator - to transform to our orthogonal coordinates:

So, on the stator,

where [Ts]= [T()], ( = t)

and on the rotor,

where [Tr]= [T()], ( = - r = ( r) t )

0dq s s abcsT

0 [ ]dq r r abcrT

Induction Motor Model in dq0 (cont’d)


0

0 0

1

10

"abc": [ ]

"dq0": [ ]

[ ]

STATOR:

[ ]

[ ]

[ ]

abcs ss abcs

dq s s abcs ss s abcs s

dq s ss dq s s

sr

sr

sr

abcr

r r abcr

r dq r

L i

T L T i

L

L

L

T

L

i

T i

i TT

T

i

1

1

0

0 00

"abc": [ ]

"dq0": [ ]

ROTOR:

[ ]

[ ]

[ ][ ]

Tsr

T

abcs

s s abcs

s dq

abcr rr abcr

dq r r abcr r rr r abcr

dq s r rr dq rs

sr

Tsr

L iL

L

i

T T i

TL

T T L T i

T i L i

1 0 0

0 1 0

0 0 1ss ssL L

1 0 0

0 1 0

0 0 1rr rrL L

Now:

Just constants!!

Our double reference frame transformation eliminates the trigonometric terms found in our original equations.

1 1

30 0

23

0 02

0 0 0

sr

Tsr sr ss s rr rT

M

T L MT T T


Let us look at our new dq0 constitutive law and work out simulation equations.

0dq s s abcs s abcs s abcs

dv T v T R i T

dt

1 10 0s s dq s s s dq s

dT RT i T T

dt


10 0dq s s s dq s

dR i T T

dt

Using the differentiation product rule:

0 0 0

0 0

0 0

0 0 0

dq s dq s dq s

d

dtd d

R idt dt


10 0 0 0dq s dq s dq s s s dq s

d dv R i T T

dt dt

For the stator this matrix is:

For the rotor the terminal equation is

essentially identical but the matrix is:

0 0

0 0

0 0 0

0 ( ) 0

( ) 0 0

0 0 0

r

r


Simulation model; Stator Equations:

dsds ds s qs

dv i r

dt

qsqs qs s ds

dv i r

dt

00 0

ss s s

dv i r

dt


Simulation model; Rotor Equations:

( ) drdr dr r r qr

dv i r

dt

( ) qrqr qr r r dr

dv i r

dt

00 0

rr r r

dv i r

dt


Zero-sequence equations (v0s and v0r) may be ignored for balanced operation.

For a squirrel cage rotor machine,

vdr= vqr= 0.


We can also write down the flux linkages:

0

0

0

0

0

0

0 0 3 2 0 0

0 0 0 3 2 0

0 0 0 0 0

3 2 0 0 0 0

0 3 2 0 0 0

0 0 0 0 0

dr rr dr

qr rr qr

r r

ds ss ds

qs ss q

sr

sr

sr

sr

s

r r

s

s s s

L i

L

ML i

M

M

M i

i

L

L i

L

i


The torque of the motor in qd0 frame is given by:

where P= # of poles

F=ma, so:

where = load torque

3

2 2 qr dr d re r qi iP

( )re l

dJ

dt

l


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Modeling of Induction Motor using dq0 Transformations First Semester 1431/1432