School of Electrical and Electronic Engineering

AC Vector Controlled Drives Induction Motor DrivesGreg AsherProfessor of Electrical Drives and Control Greg.asher@nottingham.ac.uk

Part II Concepts in Vector Control2.1 2.2 2.3 2.4 2.5 Introducing the High Performance Drive Concept of vectors Vectors in a DC machine Vector in an Induction machine Stationary and Rotating Frames

Introduction the high performance drive Some definitions A motor drive in which the motor torque obeys a torque demand within a few ms is called a torque controlled drive An IM driven by a V-f PWM converter is not a torque controlled drive. Why?Industry name Speed/position Servo drive Torque servo Servo drive Servo drive Control Engineers name Servo drive Servo drive Tracking drive Regulator drive

All the following are torque controlled drives Drive designed to go from one speed to another (or from one position to another) as quickly as possible Drive designed to go from one torque to another as quickly as possible Drive designed to follow a speed or position trajectory Drive designed to follow a fixed speed

Servo drives are often called high performance drives

Examples of high performance drives Examples of speed and position servo drives are speed and position servos in many applications machine tools (lathes, milling, laser cutting, 3-axis positioning etc) Torque servos are: traction drives, dynamometers drives for railways, ships, electric cars etc dynamometers (programmable torque loads for testing) Lifts, hoists, cranes paper feeds, paper mills, coil winding, conveyors object position tracking (radar, telescopes, robots)

Examples of torque servo drives are

Examples of tracking and regulator drives are

High performance induction drives - 1 Traditionally, all torque controlled drives have been DC machinesDC machine torque can be controlled directly and effectively AC machines traditionally steady-state AC machines driven from mains

AC vector controlled IM drives replacing DC drives for all high performance applicationsinduction motor simpler, less expensive induction motor low maintenance (no brushes) capable of faster torque response capable of faster speed response due to lower rotor inertia designs

Cost of AC-AC PWM power converters more than AC-DC thyristor converters but silicon products reducing in pricecosts recovered through cheaper machine, maintenance, higher performance

High performance induction drives - 2 Vector drives need much higher P power, but now not a problemNow so cheap, that vector controlled drives even replacing V-f PWM drives Can run machine more efficiently (flux and torque control means lower losses), vector drives common for large MW drives, even if not servo drives At lower powers, Manufacturers offer universal drives, for V-f, vector IM, vector PM etc Only extra cost is that of one extra current transducer (but even this can be overcome)

What is main competition to the vector controlled IM?Permanent magnet machine has better torque density (torque:volume) PM drives in automotive and aerospace PM cost disadvantage (magnets!) reduces at lower power, so common at O > R Expression

T = k ( i s ) assumes no leakage

Recall IM fields Total field and field due to rotor and stator currents

C:\acdrives\elc0022.swf

Recall IM fields Total field and field due to rotor and stator currents

C:\acdrives\elc0023.swf

Recall IM fields Total field and field due to rotor and stator currentsPosition of maximum stator current

Field due to stator currents alone

Position of maximum mmf due to Is - The is vector

Total Flux = S = RStator current IS Position of maximum rotor current

Total

Field due to rotor currents alone

Position of maximum mmf due to IR The iR vector

Torque and rotor and stator current vectors

T = k ( i s ) = k is sin

T = kisq d

All vectors, is , iR , rotate at e relative to the stator Remember iR rotates at sl relative to rotor; rotor rotates at R relative to statorq axis Position of isd current Position of isq current Fix d (direct) on total flux as shown The dq axis exists as a concept in the control P

isd isq d axis

isq known as the torque current isd known as the field producingcurrent

Total = d

Position of iRq current

iR vector = irq iRd = 0

Induction machine flux definitions Have leakage must choose a flux to fix dq axis on For reasons to be shown, we select the ROTOR flux as the direction of the direct axis This is called Rotor Flux Orientation (RFO) Can choose Stator Flux Orientation (SFO) isq plays the same role as Ia in a DC machine isd plays the same role as If in a DC machine SN

O

S O R(b)

d axis (RFO)

Rq axis (RFO) (a)

S

e

e

(a) No leakage: magnitudes equal S = O = R ; all point in same direction and all rotate together (b) Leakage: all rotate together, but magnitudes and directions slightly different; situation shown is that of motoring

Summary of Rotor Flux Orientation Remember: All vectors, is , iR , r rotate at e = d/dt relative to the stator dq axis now fixed on rotor flux vector - will rotate at e = d/dt Angle is the instantaneous angle between peak of rotor flux and the phase AA. It is called the rotor flux angle

isq is the torque current, and isd is the field producing currentq axis d axis

isq

isdTotal r

Stationary and rotating frames2-axis frame fixed to stator

Red vector is voltage (or current or flux) due to phase A Blue/yellow vector for phase B and C respectively

C:\acdrives\3phabc.avi

Stationary and rotating frames2-axis frame fixed to stator

Add blue and red together Add in yellow vector Note resultant is 1.5 times peak of phase vector

C:\acdrives\3phabc+.avi

Stationary and rotating frames2-axis frame fixed to stator B At a given instant in time, let

ia = 3A, ib = 1A, ic = 3A Adding these together gives the current vector i

ib = 1

i = 1

Resultant can be written i= i+ji where i = 1; i = -1.73 ia = 3Aej 2 3

= cos

ic = 3

2 2 + j sin = 0 .5 + j 0 .866 3 3

i = -1.73

i = i + ji = ia e j 0 + ib eCIn example above:or in real and imag parts:

j

2 3

+ ic e

j

4 3

i = ia + ib cos120o + ic cos 240o

i = ib sin 120o + ic sin 240o

i = i + ji = i a (1 + j 0 ) + ib ( 0 . 5 + j 0 . 866 ) + i c ( 0 . 5 j 0 . 866 )i = i + ji = (i a 0 . 5 ib 0 . 5 ic ) + j (0 . 866 ib 0 . 866 ic )i = i + ji = 1 . 5 i a + j (0 . 866 ib 0 . 866 ic ) since i a + ib + ic = 0

Stationary and rotating frames2-axis frame fixed to stator In steady state motor operation, is , is are sinusoidal and 90 apart For a given set of 3 phase currents, one can always find the equivalent 2phase currents (in a fictitious 2-phase winding) to give the same mmf and flux conditions as the 3 phase currents The flux linking the winding will be , . Similarly for voltages etc

is= 4.2

is

is

iis= 4.3 r

iis= 4.3

t= t1r

t= t2

is= 4.2 is=2.0

is= -1.6 is=2.0 is= -1.2

t= t1 t= t2

Stationary and rotating framesRotating dq axis frame

Look at the projection of the current vector onto two axis (at 90 degrees!) which are rotating at the same speed as all the vectors Call these axis d and q. The components of the current vector on these two axes will have constant values in steady state. As shown below But the dq axis is placed at an arbitrary angle to the rotor flux Therefore the dq components of is dont mean anything is is q isq= 3.6 i i

d r

t= t1

d

q isq= 3.6 r

t= t2isd= 3.8

isd= 3.8

d

isd = 3.8A isq = 3.6A

isd = 3.8A isq = 3.6At= t1 t= t2

Stationary and rotating framesRotating dq axis frame

Look at the projection of the current vector onto two axis (at 90 degrees!) which are rotating at the same speed as all the vectors Call these axis d and q. The components of the current vector on these two axes will have constant values in steady state. As shown below But the dq axis is placed at an arbitrary angle to the rotor flux Therefore the dq components of is dont mean anything is is q r isd= 4.0 isd= 3.8 isq= 2.6 d isd= 4.0 r d

t= t1

d

t= t2

q

isq= 2.6

i i

isd = 4.0A isq = 2.6A

isd = 4.0A isq = 2.6At= t1 t= t2

Stationary and rotating frames2-axis frame fixed to stator Rotating current and flux in fixed reference frame Vector components are sinusoidal

Stationary and rotating framesDq rotating frame Dq rotating frame Vector components now dc values, but not field orientated values

Stationary and rotating framesDq rotating frame Dq rotating frame Vector components now dc values, and field orientated values

Stationary and rotating frames2-axis frame fixed to statoro

i = ia + ib cos120 + ic cos 240i = ib sin 120o + ic sin 240oAt t=0 , let phase A be max

o

i = ia i =

3 3 ib ic 2 2

1 3 ib + ic = ia 2 2

(

)

ia =ib = ic =

~ 2 I ph cos 100 t =~ 2 I ph cos 2 ~ 2 I ph cos 2 2 = 3 4 = 3

~ 2 I ph

i =

3 ~ 2 I ph 2

At t= t1 = /2 later; phase A is zero

~ 2 I ph cos 6 5 ~ 2 I ph cos 6

is =

3 2

~ 2 I ph 3 2 ~ 2 I ph

is =

3 3 3 3 3 ~ ~ ~ = 2 I ph 2 I ph 2 I ph i = 2 2 2 2 2

t= t1

i

i

Hence

is =

ist= 0

ia

3 ~ 2 I ph 2

t= 0

t= t1

Stationary and rotating frames

xs =

3 ~ 2 X ph 2

The magnitude of the current vector is 3/2 x peak of the phase stator current Called the 3/2 times peak convention The voltage vector will also be 3/2 x peak of the phase stator voltageFor any vector:

3 ~ 2 X ph 2 2 P L0 2 The torque is: T = isqisd 3 2 Lr xs =

The scaling of the transformation is arbitrary: FOUR conventions are in use in the world today.

For alternative rms convention see Worked Example 2 To calculate the rated values of is , isd, isq etc see Worked Example 2

Transforming from 3 phase to 2 phase Transforming from 2-phase to dqFirst transform measured 3-phase currents into 2-phase currents Numbers are for 3/2 times peak convention. For rms x all by 2/33 i sa (t ) 2 3 3 i s (t ) = i sb (t ) i sc (t ) 2 2 i s (t ) =

is3/24 3

isa isb isc

i = i + ji = ia e

j0

+ ib e

j

2 3

+ ic e

j

is

Then transform 2-phase currents into dq currentsi sd (t ) = i s (t ) sin + i s (t ) cos

isd isq

e j

is is

isd (t ) = is (t ) cos + is (t ) sin

Inverse transformationsCan transform from dq currents into 2-phase currentsi s (t ) = i sd (t ) cos i sq (t ) sin i s (t ) = i sd (t ) sin + i sq (t ) cos

isd isq

e j

is is

Can transform 2-phase currents into 3-phase currentsi sa (t ) = 2 i s (t ) 3

is2/3

isa isb isc

1 i sb (t ) = i s (t ) + 1 i s (t ) 3 31 i sc (t ) = i s (t ) 1 i s (t ) 3 3

is

Numbers are for 3/2 times peak convention. For rms x all by 3/2

Inverse transformations

ALL transformations can be applied to voltages and fluxes etc e.g. from dq voltages into 2-phase voltages:

vs ( t ) = vsd ( t ) cos vsq ( t ) sin

vsd vsq

e j

vs vs

vs ( t ) = vsd ( t ) sin + vsq ( t ) cos

And from 2-phase voltages into 3-phase voltages:v sa ( t ) = 2 v s ( t ) 3

vs2/3

vsa vsb vsc

1 v sb ( t ) = v s ( t ) + 1 v s ( t ) 3 31 v sc ( t ) = v s ( t ) 1 v s ( t ) 3 3

vs

Fundamental structure of vector control

All vector controllers first transform measured currents to dq domainMeasured voltages can be transformed to dq if necessary (not shown below)

Vector controller controls the currents in the dq domain and outputs dq voltage demands Voltage demands are inversed transformed into 3-phase demand voltages for PWM The transformations need the angle at every point in time

vsd* vsq*

vs*

v*sabc 2/3PWM IM

e j

vs*

Vector Controller

isd isq

is

isa 3/2 isb isc

e

j

is

Microprocessor Vector controller needs to calculate

Finding the Rotor Flux angle

q r

d

dq axis frame rotates at instantaneous speed e: d e (t ) = (t ) dt

(t ) = e (t )dt

-

DIRECT VECTOR CONTROL in which the rotor flux angle is derived from measured stator voltages and currents INDIRECT VECTOR CONTROL in which is derived from the vector controlled constraint equation