AC Vector Drives 4 Implementation

7/29/2019 AC Vector Drives 4 Implementation

1/41

School of Electrical and Electronic Engineering

AC Vector Controlled Drives

I nduction Motor Drives

Greg AsherProfessor of Electrical Drives and Control

[email protected]


2/41

Part IV

Implementation of Vector Control

Direct rotor flux control

Indirect Rotor Flux control

Controller design

Discussion


3/41

Fundamental structure of vector control

All vector controllers first transform measured currents to dq domain

- Measured 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

vs*

vs*

isd

isq3/2

is

is

isa

isb

isc

je

vsd*

vsq*

je 2/3

v*sabc

PWM IM

Vector

Controller

MicroprocessorVector controller needs to calculate


4/41

Finding the Rotor Flux angle

r

d

q

)()( tdt

dt

e

=

= dttt e )()(

dq axis frame rotates at instantaneousspeed e:

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


5/41

Vector Control using

Direct Rotor Flux Orientation (DRFO)


6/41

Vector Control using DRFO

Finding the rotor flux angle

Consider the dynamic equations of the two stator coils in the frame:

d

dRiv ssss

+dt

dRiv

ssss

+( ) dtiRv ssss ( ) dtiRv ssss

We want rotor fluxes

Remember the relation between rotor and stator fluxes:

rosss iLiL + rror iLiL + rrsor iLiL + rosss iLiL +

( ){ } ssssso

rr iLdtiRv

L

L=

( ){ } ssssso

rr iLdtiRv

LL =


7/41


Finding the rotor flux angle

( ){ } ssssso

rr iLdtiRv

LL =

( ){ } ssssso

rr iLdtiRv

L

L= q

r

is

r

r

d

Note: the above can be drawn as two

separate diagrams one for

and one for

r- -

++

is

vs

o

r

L

L

sL

s

R

1

r

rtan

R

is Flux &flux angle

calculatoroL

1Rvs imrd


8/41


The current controllers

isq

isd

vs*

3/2

is isabcje

vsq*

vsd*

2/3

v*sabc

PWM IM

vsabc3/2

vs

Flux &

flux angle

calculator

je

isq*

isd*PI

PI

isq and isd are compared with demand values

Ifisq* > real isq,then want more q-axis voltage to increase isq

isq*means a torque demand,

isd*means a flux current demand

The PI controllers acting on the current errors are called the current controllers

Real voltages are difficult to measure obtain from DC-link voltage & switching states


9/41


The speed controller

isq

isd

vs*

3/2

is isabcje

vsq*

vsd*

2/3

v*sabc

PWM IM

vsabc3/2

vs

Flux &

flux angle

calculator

je

isq*

isd*

PI

PI

PIr*

r

+

-

r is measured and compared with demand value

Ifr * > real r ,then want more isq. But must limit isq* to e.g. 2x rated The PI controller acting on the speed error is called the speed controller


10/41


The flux controller

isq

isd

vs*

3/2

is isabcj

e

vsq*

vsd*

2/3

v*sabc

PWM IM

vsabc3/2

vs

Flux &

flux angle

calculator

je

isq*

isd*

PI

PI

PIr*

r

+

-PI

oL

1r

-+

imd* imd

r

imrd is derived and compared with demand value

Ifimrd

* > real imrd,

then want more isd

to increase the flux

imrd* is derived from a field weakening block The PI controller acting on the imrd error is called the field or flux controller


11/41


Problem integrator drift

DRFO is NOT used for high performance vector control

r- -

++

is

vs

o

r

L

L

sL

sR

1

r

rtan

R

Ideal vs (similar waveform forvs)

Let e.g. vscontain (small) dc offset. This will be integrated to give increasing output Called integrator drift - situation is shown in diagrams below

Int i/p Int o/p Int i/pInt o/p

With sin input


12/41


Problem integrator drift

Blue line is frequency response of ideal integrator see that it has infinite dc

gain this is what causes integrator drift

1Hz 10Hz 100Hz

0

-90

-45

Must limit output. Do this by

replacing 1/s by 1/(sc+1)

Cut-off frequency fc (=1/2c) isabout 1Hz

Phase errors introduced below 1Hz

This means that will be wrong

But OK if e above 2Hz


13/41


Problem Stator resistance

r- -

++

is

vs

o

r

L

L

sL

sR

1

r

rtan

R

Remember: Vs proportional to frequency

At high frequencies vs >> isRs

Below 2Hz, vs or even < than isRs Since Rs varies thermally, it is not an accurate parameter

Therefore signals and can be inaccurate below 2Hz

Integrator drift and poor knowledge ofRs means that low speedperformance is poor. In fact, a DRFO drive often doesnt even start!


14/41



Finds the flux angle from a Flux estimator with measured voltages and

currents gives rotor flux estimation in coordinates

Angle obtained using inverse tan function

Flux and angle estimator doesnt use speed measurement

Flux and angle estimator uses integrator and is dependent on RS

Poor performance below 2Hz due to integrator drift and RS errors

becoming important at low speed

Main use is in speed sensorless vector drives due to lack of speed

sensor for flux estimator; other methods used to overcome integrator

drift and RS problem


15/41


Indirect Rotor Flux Orientation(IRFO)

V t C t l i I di t R t Fl O i t ti


16/41

Vector Control using Indirect Rotor Flux Orientation

Getting the rotor flux angle

Consider the vector control equations

sdmrdmrdr

r iiidt

d

R

L = D-axis vector control equation :

sdmrdmrdr iisi = ( ) sdmrdr iis =1 ( ) sdrmrd isi 11

+ imrd can be obtained from isdas follows:

sqmrdr

sqmrdr

rsl ii

iiL

R

1 = q-axis vector control (constraint) equation :

dti

idtdt

mrdr

sqrslre )()(

+=+==

We will therefore derive as :

with imrd obtained above

Vector Control using IRFO


17/41


Implementing the rotor flux angle calculator

( ) sdrmrd isi 11

+dtii

dtdtmrdr

sqrslre )()(

+=+==

isq

isdisabcje

vsq*

vsd*2/3 PWM IM

je

isq*

isd*

PI

PI

PIr*

+

r

-

3/2

sl

imrd

1

1

+r

s

P/2

re

+

+mrdr

sq

i

i



18/41


Adding field weakening

isq

isdisabcje

vsq*

vsd*2/3 PWM IM

je

isq*

isd*

PI

PI

PIr*

r

+-

mrdr

sq

i*i

3/2

sl

P/2

re

+

+

irmd

1

1

+rs

imd*

PI

r-+

Demand isq* can be used for slip gain calculator (cleaner than isq) Field weakening added as shown

If no field weakening action, isd* can be used in slip gain calculator



19/41


Why IRFO works

Proof for base speed when imrd= isd Let the real flux angle in the machine be

Let the real flux angle in the controllerbe

The current vector is the same since currents are measured

The rotation speed e is the same But the angle error in the d-axis will produce different dq current components

e

q

is

r

d

sqi

sdi

eis

dq

r

sdisqi

d

er

What the controller thinksSituation in the real machine


20/41



21/41


Why IRFO works

sd

sq

rsq

mrdr

rsl i

iiiLR

1 = q-axis vector control (constraint) equation :

But in the controller we also have:

sd

sq

r

sq

mrdr

rsl

i

i

i

iL

R

1 =

slsl = We know that So the following is true:

sd

sq

rsd

sq

r i

i

i

i

1

1 =

But:2222

sqsdsqsds iiiii +

If rr = then =is

d

q

sdi

sqi

d

sqi

sdi

qHence = But if rr then and

r Errors in cause orientation lossrR Unfortunately, can change by 50% to 100% due to heating


22/41


Indirect Rotor Flux Orientation (IRFO)

Finds the flux angle by forcing the vector control constraint equation on the

system. This is only true if q-axis flux is zero (ie. orientation)

Needs to have speed measurement

Performance is excellent and is an industry standard

Only problem is variation of rotor time constant due to changes of Rr

Rotor time constant also changes due to LR in field weakening

For best performance, requires a r tuning mechanism

Under field weakening common to have ra function ofimrd (look up table)


23/41

Controller Design

(Either DRFO or IRFO)

C t ll d i i i


24/41

Controller design - revision

)(

)(

pssaskk)s(G ii ++=

)(aswrittenbecan

ps

k

+

222 2

)(

)(

)(

1 nn

ii

iii

iicl

ss

askk

akkskkps

askk

G(s)

)s(G

)s(*x

)s(x)s(G +

+=++=+

sask ii )( +

1 +pps kx*(s) x(s)

skk intprop +

222 nnss +

System is digital with a samp. Normally samp is about 10-20 x n [for s-plane design]

n can be related to step response rise time 3/n ; orsettling time 4/n Given n and , know ; therefore can find ki and ai

samp=2fsampP

n is natural frequency in rad/s often called bandwidth chosen by designer is the damping - chosen by designer

controller plant


25/41

Controller speeds and sampling processes


26/41

Controller speeds and sampling processes

Note update is fast motor position can be sampled at 200 s Motor speed doesnt change very fast - cannot be sampled at 200 s

Seen by considering [r(k)- r(k-1)]/Tsamp - position will not have changed in 200s!

r derived e.g 2ms (every 10th sample). Hence processing isq* imd* sl also at 2ms

Slow sampling shown in black

isq

isdisabcje

vsq*

vsd* 2/3PWM IM

je

isq*

isd*

PI

PI

PI

r*

r

+-

mrdr

sq

i

*i

3/2

sl

P/2

re

+

+

irmd

1

1

+r

s

imd*

PI

r-+

d/dtr

Control block diagram for the Current Controllers


27/41

Control block diagram for the Current Controllers

The isd and isq P+I controllers will be identical one design for both Current loops are fast. Choose n as 2 x 100-200Hz e.g. 1000rad/s

Need to find plant transfer function. Look at all delays in system (modelled by 1st

order time constant)

s

ask ii )( +1

1

+Is

1

1

+fs

)s(Gp

plantinvertercontroller

Anti-aliasing filter

isq* isqvsq* usq*

I= 200s

f= 200s

Inverter delay: delay before demand voltage actually gets on to lines. Normally

assume 1/fsamp. In our example, this is 200s

Filter filters away PWM current ripple, cut-off < fsamp/2 e.g. 1kHz f 200 s What is Gp(s)?

Finding the plant for the current controller design


28/41

Finding the plant for the current controller design

Consider the dynamic equations for the stator coils in dq frame

rdr

osqsesdssdssd dt

d

L

LiLi

dt

dLiRu +

rdr

oesdsesqssqssq LLiLi

dtdLiRu +

It would be good if we had only the first two terms e.g. for the d-axis equation:

sdssdssd iddLiRu sdssdssd siLiRu sdsssd iRLsu )( +

sd

ss

sd uRLs

i)(

1

+ 11 +s

s

s

R/usd isd

sqssqssq id

dLiRu

11 +s ss R/

isqusqsq

sssq uRLsi

)(1+

And the 1st two terms of the q-axis equation:

Giving:

Idea of voltage compensation terms


29/41

(also called feedforward or decoupling terms)

rdr

osqsesdssdssd dt

d

L

LiLi

dt

dLiRu +

rdr

o

esdsesqssqssq L

L

iLidt

d

LiRu +

Of course we had:

isabc

2/3 PWM IM

3/2je

je

isqisd

vsq*

vsd*isd*

PI

PI

isq*

+

The terms in the red box are the demand

voltages to overcome changing flux andthe back-emf in the machine.

We can add these as feed-forward terms

into the controller

imdro L/sL 2

se L-

+

se L

roe L/L2

++

+

Current Controllers


30/41

Current Controllers

The feedforward terms means that the PI control only supplies the voltages toincrease or decrease the current

The plant transfer function Gp(s) is now 1st order and given below

The value of s is about 10-25ms and is much slower than the delays of filter and

inverter

s

ask ii )( +

1

1

+Is

1

1

+fs

plantinverter

Anti-aliasing filter

isq

1

1

+ss

s

R/

f= 200s

f= 200s f= 20ms

isq*

Current Controllers


31/41

Current Controllers

The feedforward terms means that the PI control only supplies the voltages toincrease or decrease the current

The plant transfer function Gp(s) is now 1st order and given below

The value of s is about 10-25ms and is much slower than the delays of filter and

inverter

plant

s

ask ii )( +isq* isq

1

1

+ss

s

R/

5050+s Control design is now easy. Let s = 0.02 and Rs = 1. Hence plant

Assume wish for n = 1000rad/s with =0.75

Then we have:

From which PI parameters are derived

622 10150050)5050( +ssaksks iii

Control block diagram for the Speed Controller


32/41

g p

The machine will develop torque Te and this will cause a speed response

The response will be defined by:

2

rarvrLoade BBd

d

JTT+ J = moment of inertia B

v

= viscous friction coefficient

Ba = aerodynamic friction coefficient

Tload is associated with the application it is unknown and is treated as a disturbance

Bv and Ba may be included as part ofTload if they are not known

J is that of the motor and load

Ba normally neglected to make design linear (but can be significant at high speeds)

rvrLoade BJ sTT + )(1 Loadev

r TTJ/Bs J/ + Writing using Laplace operator and ignoring aerodynamic friction:

J/Bs

J/

v

1Te

TLoad

r Often Bv very small and put to zero

Control block diagram for the Speed Controller


33/41

g p

ktcalled the torque constant derived from

We can now put a torque demand isq* on to machine. The machine will develop

torque according to isq this will be within 2-10ms of the torque demand fast.

sask ii )( +

plant

J/BsJ/v

1

TLoadisq* isqf= 5ms

sqmrdr

o iiL

LPT

= 2

23 (replace 3 by 2/3 for

3/2 x peak convention)

Te rtk

r*

Speed Controller Design


34/41

p g

ktcalled the torque constant derived from

We can now put a torque demand isq* on to machine. The machine will develop

torque according to isq this will be within 2-10ms of the torque demand fast.

sqmrdr

o iiL

LPT

= 2

23 (replace 3 by 2/3 for

3/2 x peak convention)

plant

sask ii )( + isq*= isq

J/BsJ/kv

t+ rr*

Current loop effectively neglected since it speed will not change much in a few ms!

Combine kt with plant, and control design follows in same way as before

n difficult to chose generally a few Hz for machines 5-50kW In practice, J and mechanics of plant not known, PI is normally a PID and is set bytrial and error by the commissioning engineer

The field or flux controller


35/41

Field controller takes error ofimrd (or flux) and outputs signal isd* Plant will take isd* and outputs imrd. Must identify all delays between isd* and imrd

Know that isd* follows isd with delay 2-10ms (like q-component)

isq

isdisabcje

vsq*

vsd*2/3 PWM IM

je

isq*

isd*

PI

PI

PI

r*

r

+-

mrdr

sq

i*i

3/2

sl

P/2

re

+

+

irmd

1

1

+rs

imd*

PI

r-+

d/dtr

The field or flux controller


36/41

Field controller takes error ofimrd (or flux) and outputs signal ids* Plant will take isd* and outputs imrd. Must identify all delays between ids* and imrd

Know that isd* follows isd with delay 2-10ms (like q-component)

plant

s

ask ii )( +1

1

+r

s

isd* isdf= 5ms

imrdimrd*

To choose n: note imrd* comes from the field weakening function driven by speed.Thus imrd* cannot change by more than the CL speed bandwidth (a few Hz). ChooseCL flux n as e.g. 8-10 times that of the speed.

Eg let r = 100ms, n = 20Hz, damping factor = 0.8

Require: s2+200s+15800; have s2+10(1+k)s+10ka

k=19, a =83

BUT in field weakening r increases slower plant. Do not try for too much n

Understanding Closed Loop responses

1 Current Loop


37/41

1. Current Loop

If a step isq

* is applied to a machine in

Bv is zero or very low, then machine

will accelerate in an uncontrolled

manner (see right)

The acceleration depends on the

difference between ktisq* andTLoad

isabc

2/3 PWM IM

3/2je

je

isqisd

vsq*

vsd*isd*

PI

PI

isq*

+

Assume no speed or flux loop see below

isqisq*

2-10ms

A torque demand or flux producing current demand may

be applied. The actual isd and isq will track the demandswithin milliseconds.

isq

isq*

J

ik

d

d sqtr =

J

Tik

d

d Loadsqtr =

(i) idq applied with no load torque

(ii) idq applied with constant Tload but < Te

(i)

(ii)

Understanding Closed Loop responses

1 Action of Speed loop


38/41

1. Action of Speed loop

Speed settling - n_speed

Current settling - n_cur

isq

r*

r

t10 t2 t3 t4 t5

t=0: step speed demand (e.g. 1000rpm); isq* hits limit at 2 x rated value (2x rated Te) t1: isq reaches demand current (after a few milliseconds); motor accelerates constant rate

t2: speed approaches final value; isq* leaves limit

t3: speed settles at demand value; isq drops to low value (e.g. because low Bv)

t4: rated Tload applied; speed drops; isq* responds to error

t5: r reaches r*, output of speed PI is constant

imrd*

isq*

IRFOcurrentcontrol

IMPI+ -

r

r*

Tload= 0, then rated at t4


39/41

IRFO Vector Control

Nearly all vector controlled commercial drives are IRFO

All will have field weakening and voltage compensation terms

All will have rotor time constant tracking similar to the scheme used

in the reference section of Worked Example 3

No products use DRFO for sensored drives

Sensorless drives (no speed sensor on shaft) may use DRFO or IRFO

structures but speed/torque/flux control performance is poor for f


40/41

As Masters students should know the physics of IM operation

Understand V-f drive and why their dynamic performance is not good

Be able to derive machine parameters from manufacturers data Understand the concept of high-performance drive and servo performance

for speed, torque and flux

Understand the concept of space vector applied to AC machines

Understand how a 3-phase machine can be transformed into a 2-phase

one, and the transformations used to do this

Understand how a rotating measurement frame can measure the rotating

vector as DC quantities and the concept of field orientation and torque andfield current components

Be familiar with two scaling conventions (3/2 time peak and rms) and be

able to derive rated values of isd and isq from machine data

What you need to know


41/41

y

Understand in principle how the dq-axis field orientated equations are

obtained (but you will not be asked to derive these)

Understand DRFO and IRFO as the two ways of tracking the angle of rotor

flux in real time

Know the structure of DRFO and IRFO controllers and be able to reproduce

them; understand their advantages and disadvantages

Understand the IRFO slip and torque equations and be able manipulate themto find speeds, torques, currents etc when machine is driving known loads

Understand orientation errors in IRFO and why the dq current components in

the machine and controller may be different. For the 20-credit course, you

should also understand how the rotor time constant is tracked in real time Understand how to design current, field and speed controllers given the

machine and load parameters

Understand how the machine speed and currents behave when step changes

in speed and load torque are applied

Documents

AC Vector Drives 4 Implementation