Induction Motor Vector Controlmotor.feld.cvut.cz/sites/default/...11oz.jz-ind-motor-vect-control.pdf · 4 Disadvantages of scalar (U/f) control: Very low dynamics Changes in the asynchronous

Digital Control of Electric Drives

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Induction Motor Vector Control

Czech Technical University in Prague – Faculty of Electrical Engineering

O. Zoubek, J. Zdenek

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● Scalar control● Vector control

Induction Motor Control Methods

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Three-phase Inverter

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● Disadvantages of scalar (U/f) control:

● Very low dynamics

Changes in the asynchronous machine take place at the speed given by the rotor time constant (ie, up to seconds on large machines)

● The input is the frequency - it is not possible to directly control of IM torque

● Not suitable for traction

● The motor torque is dependent on the slip frequency - it is still the same (the percentage with the lower output frequency of the drive increases)

● Advantages of scalar (U / f:) control:

● Simplicity of management (especially development)

Scalar Control Properties (U/f Method)

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Scalar Control - U/f Method)

● Scalar control:● Only one variable, mostly frequency, is controlled. Voltage is tied

to frequency, so the name "U / f"

Speed [1/min]

Voltage

Rated speed

Ratedvoltage

IMI1

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+

CR

Prescaler(Předdělička)

Up-Down Counter(Obousměrný čítač)

Comparator(Komparátor )

Compare Register(Porovnávaná hodnota )

Program

DeadTime

&OutputLogic

Clock(Takt)

Int

Carrier Wave Code(Kód nosné vlny)

Modulation Wave Value(Kód modulačního průběhu )

Žádost o obsluhu

Generování mrtvých doba výstupní logika

PWM Generation hw Support


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PWM Generation (One possible option)

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+

Carrier Wave(Nosná vlna)

Modulation Wave(Modulační průběh)

Interrupt Request(Žádost o obsluhu)

PWM Output(Výstup PWM)

Time

+1

-1

0

PWM Output(Výstup PWM)


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Vector Control

● Advantages of vector control versus scalar:● Unmatched higher dynamics● Working from zero revolutions (for some types of vector

control including standing rotor)● The drive input is a torque requirement (gas lever)● Fit for traction● The induction motor current is fully controlled● Possibility of short-term overload and work with a higher

torque than the maximum torque point● To switch from scalar to vector control, in most cases, only

software change is sufficient

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Control Dymamics

● Torque is related to currents● Torque of any electric machine ≈ current● Dynamics of electric machine ≈ fast current changes● Rapid changes in currents require a voltage (the winding of

the machine is inductive)● In order to increase the dynamics, it is necessary to directly

control the currents (for induction motor with the short-circuit armature only stator currents are possible to control), which can not be done by the U / f control

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Vector Control

● Vector control:● Two quantities of the induction machine are controlled separately● Mostly, this is a stator current divided into two components:

– Current component affecting the magnetic flux– Current component affecting the torque

IMI1

IM

IM

IΨ→ =

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DC machineIM machine

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Space Vector Idea

● The space vector can express the effect of all currents flowing through all phases of the electric machine stator

● Symmetrical winding is assumed for a three-phase, two-pole machine (the angle of rotation is 120 °)

● The expression can be modified provided the center point is not connected (i.e. i

a+i

b+i

c= 0) and after modifiying ej120° a ej240°

I 1

I=K i aib⋅ej⋅120 °i c⋅e

j⋅240 °

i =ℜ I =32⋅K⋅i a

I=K 32

i a j⋅32⋅i a2ib

i =ℑI =32⋅K⋅i b�i c=

32⋅K⋅i a2 i b

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Space Vector Idea

● The space vector can express the effect of all currents flowing through all phases of the electric machine stator

● Symmetrical winding is assumed for a three-phase, two-pole machine (the angle of rotation is 120 °)

● The expression can be modified provided the center point is not connected (i.e. i

a+i

b+i

c= 0) and after modifiying ej120° a ej240°

I 1

I=K i aib⋅ej⋅120 °i c⋅e

j⋅240 °

i =ℜ I =32⋅K⋅i a

I=K 32

i a j⋅32⋅i a2ib

i =ℑI =32⋅K⋅i b�i c=

32⋅K⋅i a2 i b

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Při vhodné volbě Kiα = ia

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Clark Transformation

Clark's transformation converts the three phases (a, b, c) intotwo phases (α, β)For K = 2/3 Clark's transformation is expressed as:

ia

ibi

c

iα

iβ

Three phases of the machine →

→

Two axis

iα=i a iβ=√33⋅(i a+2 i b)

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Rotation of Coordinates

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Space vector I in the coordinate system tied to the stator

Space vector I in coordinate system k

Where θk is the angle of rotation of the coordinate system k

I 1

I k

I k=

I 1⋅e

� j k

ik=iαcosθ

k+iβsinθ

ki l=�iαsinθk+iβcosθk

If the coordinate system (k, l) is rotated by the angle θk against (α, β):

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Park Transfomation

α

→

ω1

id=icosi sin

iq=�isini cos

θ

β

d

q

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Used Coordinate Systems

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Coordinates Designation Rotationspeed

Coordinates tied to the stator α,β 0

Coordinates tied to the rotor k, l ω

Coordinates tied to the magnetic flux of the rotor d, q ω1

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Induction Motor Torque

P=32∣U 1∣⋅∣I 1∣⋅cos=

32⋅ℜ U 1⋅

I 1∗Pro K=

23

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P=32∣U 1∣⋅∣I 1∣⋅cos=

32⋅ℜ U 1⋅

I 1∗For K=

23

U 1=R1⋅I 1

d 1

dt=R1⋅

I 1 j1 1

Voltage equation of induction motor stator windingwritten by space vectors:

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P=32∣U 1∣⋅∣I 1∣⋅cos=

32⋅ℜ U 1⋅

I 1∗Pro K=

23

P=32⋅[R1i1

2 i12 11 i11 i1]

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P=32∣U 1∣⋅∣I 1∣⋅cos=

32⋅ℜ U 1⋅

I 1∗Pro K=

23

P=32⋅[R1i1

2 i12 11 i11 i1]

Power dissipation in resistance:

P=R∣I∣2=Ri a2ib

2

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P=32∣U 1∣⋅∣I 1∣⋅cos=

32⋅ℜ U 1⋅

I 1∗Pro K=

23

P=32⋅[R1i1

2 i12 11 i11 i1]

Losses in the stator

P=R∣I∣2=Ri a2ib

2

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P=32∣U 1∣⋅∣I 1∣⋅cos=

32⋅ℜ U 1⋅

I 1∗Pro K=

23

P=32⋅[R1i1

2 i12 11 i11 i1]

Losses in the statorPower transmitted by

air gap from the stator to the rotor

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P=32∣U 1∣⋅∣I 1∣⋅cos=

32⋅ℜ U 1⋅

I 1∗Pro K=

23

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P=32∣U 1∣⋅∣I 1∣⋅cos=

32⋅ℜ U 1⋅

I 1∗Pro K=

23

1�s

=1

pp

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P=32∣U 1∣⋅∣I 1∣⋅cos=

32⋅ℜ U 1⋅

I 1∗Pro K=

23

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ABc

I1 I1 I1 I1 I2 I2 I2 Ψ1I2 Ψ1 Ψµ Ψ2 Ψ1 Ψµ Ψ2 Ψ2

Lh Lh / L2 Lh / L1 Lh / σL1L2

M i=32

pp⋅c⋅∣A∣⋅∣B∣⋅sin

sin γ is the angle between the vectors A a B

M i=32⋅pp⋅ 1 i1 1 i1

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ABc

I1 I1 I1 I1 I2 I2 I2 Ψ1I2 Ψ1 Ψµ Ψ2 Ψ1 Ψµ Ψ2 Ψ2

Lh Lh / L2 Lh / L1 Lh / σL1L2

M i=32

pp⋅c⋅∣A∣⋅∣B∣⋅sin

sin γ is the angle between the vectors A a B

M i=32⋅pp⋅ 1 i1 1 i1

M i=32⋅pp⋅ 2d i1q 2q i1d ⋅

Lh

L2

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Stator Current Components

M i=32⋅pp⋅ 2d i1q 2q i1d ⋅

Lh

L2

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M i=32⋅pp⋅ 2d i1q 2q i1d ⋅

Lh

L2Ψ

2q is equal to zero because the (d, q) axis direction is given just

by the direction of Ψ2

M i=32⋅pp⋅ 2 i1q⋅

Lh

L2

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M i=32⋅pp⋅ 2d i1q 2q i1d ⋅

Lh

L2Ψ



M i=32⋅pp⋅ 2 i1q⋅

Lh

L2

M i≈i1q 2 i1q

stator current component

which affect the torque

Ψ2 is excitation

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M i=32⋅pp⋅ 2d i1q 2q i1d ⋅

Lh

L2Ψ



M i=32⋅pp⋅ 2 i1q⋅

Lh

L2

M i≈i1q 2i1q


which affects the torque

Ψ2 is excitation

d 2d

dt=

R2

L2Lh i1d�2d => 2 Lh i1d

i1d


which affects magnetic flux

32

IM versus DC machine

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AM

i1q

i1d

ia

ib

i1q

≈ ia

i1d

≈ ib

The stator current of the three-phase induction short-circuit machinehas two stages of freedom (the third one is deleted because centerof windings is drawn outside) and can be divided into two componentsthat correspond to the armature current and the excitation of the DC motor.It is therefore possible to control independetly the torque and the excitationof induction machine.

DC machineIM machine

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Voltage Vector Control of an Induction Machine

IMInverterPI

PI

IRCDecodern

θIRC

α,β

a,b,c

α,β

d,qIMModel

θr

α,β

d,q

θ

i1d

*

i1q

*

α,β

a,b,c

i1d

i1q

ia,i

b,(i

c)

iα

iβ

θ

u1d

u1q

u1α

u1β

u1a,b,c

∫ω

2

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IMInverterPI

PI

IRCDecodern

θIRC

α,β

a,b,c

α,β

d,qIMModel

θr

α,β

d,q

θ

i1d

*

i1q

*

α,β

a,b,c

i1d

i1q

ia,i

b,(i

c)

iα

iβ

θ

u1d

u1q

u1α

u1β

u1a,b,c

∫ω

2

Clark Transformation

i=i a i =33⋅i a2 i b

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AMstřídačPI

PI

IRCDecodern

θIRC

α,β

a,b,c

α,β

d,qIMModel

θr

α,β

d,q

θ

i1d

*

i1q

*

α,β

a,b,c

i1d

i1q

ia,i

b,(i

c)

iα

iβ

θ

u1d

u1q

u1α

u1β

u1a,b,c

∫ω

2

Park Transformation

id=i cosi sin

iq=�isini cos

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AMstřídačPI

PI

IRCDecodern

θIRC

α,β

a,b,c

α,β

d,qIMModel

θr

α,β

d,q

θ

i1d

*

i1q

*

α,β

a,b,c

i1d

i1q

ia,i

b,(i

c)

iα

iβ

θ

u1d

u1q

u1α

u1β

u1a,b,c

∫ω

2

Induction machine I1-n model

dΨ2

dt=

R2

L2(Lhi1d�Ψ2) ω2=

Lh R2

L2

i1qΨ2

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AMstřídačPI

PI

IRCDecodern

θIRC

α,β

a,b,c

α,β

d,qIMModel

θr

α,β

d,q

θ

i1d

*

i1q

*

α,β

a,b,c

i1d

i1q

ia,i

b,(i

c)

iα

iβ

θ

u1d

u1q

u1α

u1β

u1a,b,c

∫ω

2

Induction machine I1-n model

d 2

dt=

R2

L2Lhi1d� 2 2=

Lh R2

L2

i1q

2

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AMstřídačPI

PI

IRCDecodern

θIRC

α,β

a,b,c

α,β

d,qIMModel

θr

α,β

d,q

θ

i1d

*

i1q

*

α,β

a,b,c

i1d

i1q

ia,i

b,(i

c)

iα

iβ

θ

u1d

u1q

u1α

u1β

u1a,b,c

∫ω

2

Induction machine I1-n model after

suitable substitution is used i 2= 2

Lhd i 2

dt=

R2

L2i1d�i 2 2=

R2

L2⋅

i1q

i 2

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AMstřídačPI

PI

IRCDecodern

θIRC

α,β

a,b,c

α,β

d,qIMModel

θr

α,β

d,q

θ

i1d

*

i1q

*

α,β

a,b,c

i1d

i1q

ia,i

b,(i

c)

iα

iβ

θ

u1d

u1q

u1α

u1β

u1a,b,c

α,β

∫ω

2

Slip frequency integrator

r=∫2 dt

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40

IMstřídačPI

PI

IRCDecodern

θIRC

α,β

a,b,c

α,β

d,qIMModel

θr

α,β

d,q

θ

i1d

*

i1q

*

α,β

a,b,c

i1d

i1q

ia,i

b,(i

c)

iα

iβ

θ

u1d

u1q

u1α

u1β

u1a,b,c

α,β

∫ω

2

The IRC sensor decoder

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θ IRC=number of pulsespulses per rotation

⋅pp

2

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41

AMstřídačPI

PI

IRCDecodern

θIRC

α,β

a,b,c

α,β

d,qIMModel

θr

α,β

d,q

θ

i1d

*

i1q

*

α,β

a,b,c

i1d

i1q

ia,i

b,(i

c)

iα

iβ

θ

u1d

u1q

u1α

u1β

u1a,b,c

α,β

∫ω

2

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r≈∫2

≈∫1

IRC≈∫

Slip frequency

Feeding frequency of machine

El. angle speed of shaft

2=1⇒r IRC=

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IMInverterPI

PI α,β

d,q

θ

i1d

*

i1q

*

α,β

a,b,c

i1d

ia,i

b,(i

c)

iα

u1d

u1q

u1α

u1β

u1a,b,c

ut =k pet ∫k i et dt

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dekodérIRCn

θIRC

α,βa,b,c

α,β

d,qIMModel

θr

i1q

iβ

θ

∫ω

2

PI controllers regulate each component of the current separatelyBoth components are DC (does not change with ω, ω

1 or ω

2)

The input to the controllers are the error values of the currents.The outputs are a voltage requirements.

ut =k pet ∫k i et dt

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IMInverterPI

PI α,β

d,qi1d

*

i1q

*

α,β

a,b,c

u1d

u1q

u1α

u1β

u1a,b,c

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dekodérIRCn

θIRC

α,β

a,b,c

α,β

d,qmodelAM

θr

i1d

i1q

ia,i

b,(i

c)

iα

iβ

θ

α,β

∫ω

2

Inverse Park and Clark transformations

u=ud cos�uqsin

u=ud sinuqcos

Inverse: sins have inverse signs

ua=u

ub=�12

u32⋅u

uc=�12

u�32⋅u

44


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45

Three-phase Inverter

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Space Vector Modulation

Each of the three branches of the voltage converter always has only one of two transistor switched on. For the three branches it is 23 = 8 different states altogether.

a b c Ua-b Ub-c Ua-c výstup

0 0 0 0 0 0 nic

1 0 0 +UDC

0 -UDC

0°

1 1 0 0 +UDC

-UDC

60°

0 1 0 -UDC

+UDC

0 120°

0 1 1 -UDC

0 +UDC

180°

0 0 1 0 -UDC

+UDC

240°

1 0 1 +UDC

-UDC

0 300°

1 1 1 0 0 0 nic

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Six available active statesat the output of the inverter

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a

b

c

U0

U60U120

U180

U240 U300

50


To achieve a certain outputshould be combined over timeseveral output vectors,including zero vector.There are many ways,how to draw a vectorwhich differ in accuracy,computational difficultyand switching losses.However, the outputs alwayshave character PWM modulation.

The simplest method is:

PWMx=ux

2Umax

0.5

x∈{a ,b ,c}

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a

b

c

U0

U60

51


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Digital Control of Electric Drives

Induction Motor Vector Control

End

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Documents

Induction Motor Vector Controlmotor.feld.cvut.cz/sites/default/...11oz.jz-ind-motor-vect-control.pdf · 4 Disadvantages of scalar (U/f) control: Very low dynamics Changes in the asynchronous