slides electrical drives e

8/2/2019 Slides Electrical Drives E

1/32

1

Electrical drives

Prof. A.GENON

Universit de Lige (Belgique)

2010

[email protected]

Power electronic systems

Analogic command

Numerical commnd

LAPLACE TRANSFORM

s is a complex variable

Definition

[ ]

==

0

)()()( dtetftfLsFst

)(tf )(sF

Linarit )()( 2211 tfktfk + )()( 2211 sFksFk +

Translation retarde )0()( 00 tttf )(0 sFest

Drivation)(tf

dt

d

)(ssF

Intgration

t

dttf0

)( s

sF )(

Valeur initiale )(lim0

tft

)(lim ssFs

Valeur finale )(lim tft

)(lim0

ssFs

Computational rules

= dsesFj

tfst

)(2

1)(

[ ] == )(

s)(Res)(

2

1)(

i

sFdeples

ststesFdsesF

jtf

[ ]( )

[ ]iss

stk

ik

k

esFssds

d

ksF

=

= )()(

!1

1)(Res

1

1

si

Inverse transformation

NoteNote : is a counterclockwise contour containing the origin ot thecomplex plane and all the poles of the function F(s)(Bromwich

contour .

Electrical Drives 2010 - A.GENON 1


2/32

2

TABLES

Inverse transformation

partial fraction expansion

=i

i

ps

rsF )(

=

)()(lim sFpsr i

psi

i

,

= tpi iertf )( .

)()(

)(sH

sE

sS=

Transfer functionContinuous system stability

I m

R e

1

Routh-Hurwitz criterion

1

71

6

13

1

51

4

12

1

31

2

11

=

=

=n

nn

nn

n

nn

nn

n

nn

nn

a

aa

aadtm

Aa

aa

aadtm

Aa

aa

aadtm

A

11

1411

71

23

11

1311

51

22

11

1211

31

21A

AA

aadtm

AA

AA

aadtm

AA

AA

aadtm

A

nnnnnn

=

=

=

21

2421

1411

33

21

2321

1311

32

21

2221

1211

31A

AA

AAdtm

AA

AA

AAdtm

AA

AA

AAdtm

A

=

=

=

01

1

1 ... asasasaDsoitn

n

n

n +++=

...

......

......

......

...

...

...

...

321

232221

131211

531

42

nnn

nnn

nnn

AAA

AAA

AAA

aaa

aaa

The necessary and sufficientcondition to ensure the stability of asystem is that

the coefficients of the denominatorofthe transfer function

and

the elements of the first column of theRouth table

are positive or have the same sign.

Rout-Hurwitz criterion



3/32

3

Z TRANSFORMATIONNotion of sampled signal

[ ]

=

=0

)()(k

EkTtkftf

Definition

[ ] [ ]

[ ]

[ ] )00(

)(

)()()(

0

00


4/32

4

TABLES

Inverse Z transform

partial fraction expansion

because

[ ]

=

=

=

1

)()(limavec

)(

k

ii

izz

i

i

i

zrkf

zFzzr

zz

rzF

i

ii

k

i

i

k

i

zzzz

zzzF

zz

zzF

=

=

=

1)(

)(

11

Inverse Z transform

Polynomial division

IfF(z) can be written as the ratio of two

polynomials and if the degree of the

numerator is lower or equal of the degree of

he denominator the polynomial division

gives :

[ ] kfkf

zfzffzDzNzF

=

+++== K

22

110

)()()(

Stability of sampled-data systems in

open loopSampling does not modify the stability of a system inSampling does not modify the stability of a system in

open loop.open loop.

A continuous system is stable in open loop when the real part of

all the poles is negative. As :

( ) ( )EiEiTTjTs

iTjTeeez EiEiiEi

sincos ===

1


5/32

5

Closed loop stability :

Routh-Hurwitz criterionRH allows to determine if the roots of a polynomial have

negative real parts.

Appying to the characteristic polynomials the following

transformation : z=(1+w)/(1-w) | w=(z-1)/(z+1) one

transforms the inside part of the unit circle onto the negative

real part of the w plane.

Indeed, if :

( )

( )( ) ( )

0sin1cos

sin21

1sincos

1sincos

1

1

10

22

2


6/32

1

Chapter 3

Mixed systems

Entranements lectriques

Electrical drivesProf. A. GENON, ULg

Is it possible to consider discrete

systems as if they were

continuous ?

If yes, under what conditions?

Pad approximation2/1

2/12/

2/

E

E

sT

sTsT

sT

sT

e

eez

E

E

E

+

==

We observe that the gray areas overlap satisfactorily. We

deduce that the approximation is valid for:

4/4/5,00 EE TetT

Example of a discrete regulator:

the numerical PID

Transfer function of a numerical PID

[ ] [ ] [ ] [ ] [ ]( )

)1(

/)/2()/(

)(

)()(

)(1

)(1

)()(

1

2

0

++++

==

+

+=

++= =

zz

TKzTKKzTKTKK

zE

zYzG

zEz

z

T

KzE

z

zTKzEKzY

kekeT

KieTKkeKky

EDEDPEDEIPRR

E

DEIPR

E

D

k

i

EIPR

1

)(

)(

)()(

+

==z

KzTKK

zE

zYzG PEIPRRPI :

Continuous treatment of a discrete system :

example of the numerical PID

( ) ( )

+++++

+=

+

=

++++

=

4/2/2/1

1)(

2/1

2/1

)1(

/)/2()/()(

2

2

EIEPDI

EIP

E

R

E

E

EDEDPEDEIPR

TKTKKss

KTKK

sTsG

sT

sTz

zz

TKzTKKzTKTKKzG

Hypothesis : Pad conditions are met

If we compare with an analog PID,

we obtain*

***

)(

)()( D

IP

RR Ks

s

KK

sE

sYsG ++==

4/2/ 2*

*

*

EIEPDD

II

EIPP

TKTKKK

KK

TKKK

++=

=

+=



7/32

2

Q: What must be done if a continuous

system is controlled by a discrete regulator?

R: We must determine the discrete transfer

function of the continuous system ....

Discrete transfer function of the continuous

system preceded by a holding element

( )

( )( )( )

( )( )

==

=

=

=

=

==

=

=

==

i i

i

i

i

i

i

i

i

i

Tp

i

i

tp

i

i

i ii

i

i i

iS

Ts

S

Ts

Ts

EEE

zz

z

p

rzGzzG

zzz

zz

p

rzG

ez

etup

rtg

pssp

r

ps

r

ss

sGsG

sGesGs

esG

s

esGTtututg

Ei

i

E

E

E

1)()1()(

1

1)(

1)()(

111)()(

)()1()(1

)(

1)()()()(

1

1

1

1

1

1

ExampleExample : command

of a DC motor by a

C

Discrete transfer function of the continuous

system preceded by a holding element :

EXAMPLE : 2nd order filter

( )( )

( )( )

21

2211

2211

122211

/

2

/

1

21

21212121

)1()1(

)1()1(

)1()1(

)(

/1

11

/1

11

11

1)(

2

1

TT

zTzTK

zTzT

zzTzzTn

ez

ez

zzzz

nzKzG

TsTTTsTTsTsTsG

TT

TT

S

E

E

=

=

=

=

=

++

+=

++=

Analog to digital conversion

Proper sampling requires that the sampling frequency

is much higher than the highest frequency the signal tobe sampled (3 to 4 times).

If necessary, we must first filter the signal to be

sampled.

Shannon theorem


Quantization error[ ]

( )

( )

NSBBS

VBS

VV

qeVAR

qdVVpVqdVVpVeeVAR

dVVpVqdVVpVeeE

Vq

V

dVdVVp

dB

N

q

q

N

V

q

q

N

V

N

eff

eff

02,626,4log10/

28

3/

22

12)(

12)()(12)()()(

0)()(12)()()(

12

)dVVV,signal(Prob)(

10

2

2

max

22/3

2/

2

0

2

2/3

2/0

max

max

max,

max,

max

max

+==

=

=

=

==

===

===

=

+==

S/B= signal to noise ratio


Signal to noise ratio

NSBBSdB 02,626,4log10/ 10 +==

Nombre de bits Rapport signal/bruit

4 21 dB

8 45 dB

16 93 dB



8/32

3

Parallel analog to digital converter

10 bits 1024 comparators !!!

very fast

Successive approximations ADC

necessits a sample and hold element

works by successive comparisons and accuracy increase with each iteration

Ve > Vref/2, bit1=1

Ve< Vref/2+Vref/4, bit2=0

Ve> Vref/2+Vref/8, bit3=1

etc

slower conversion (10us for 12 bits)

Ramp compare ADC

1. Integration of Ve during a fixed time t1-t02. Negative integration with a known voltage Vrefand measure of elapsed time

t2-t13. Relative slow, but can be very precise

01

12

tt

ttVVrefe

=

Digital to analog converter

Much easier to achieve than the reverse

+++=16842

0123 BBBBVVrefS



9/32

1

Chapter 4

COMMAND CIRCUITS



Conventional circuits

Transistor amplifier : common emitter


Transistor amplifier : common collector


Push-pull


Use of a triac

Transistor chopper command

optocoupler

comM UU #

PWM command



10/32

2

Transistor chopper command

The LM555 IC

PWM command1

)2(

44.1

CRRf

BA +=

astable monostable

Transistor inverter command

2sin

221,

UU

UU

M

M

=

=

THYRISTOR command THYRISTOR chopper command

THYRISTOR rectifier command

One compares the control signal Ucom to a

sinusoidUrwhose value is maximum at the

instant of natural switching

(ex : -V2 for Th1).

Then

com

r

C

r

comC

Cc

UU

U

U

UU

UU

max

0

max

0

0

arccoscos

cos

=

=

==



11/32

1

Chapter 5

Analogic controllers



Introduction

C o m p a r a t e u r

R g u l a t e u r

C i r c u i t s d e

c o m m a n d e

C o n v e r t i s s e u r

l e c t r o n i q u e

d e p u i s s a n c e

O r g a n e s

d e m e s u r e

S y s t m e

r g l e r

+

-

U

c

The comparator and the analog controller are often combined

in one element.

The purpose of analogic controllers is to compare a reference

and a measurement and to develop a corrective signal applied

to the command circuits.

The analog controllers are mostly designed with operational

amplifiers.

The operational amplifierExample : The uA709

+

R

R

R

1

2

3

Q

Q

Q

Q

Q

Q

Q

Q

Q Q

Q

Q

Q

Q

1

2

3

4

5

6

7

8

9

1 0

1 1

1 2

1 3

1 4

C

C

+

V

E -

V

A -

V

S

V

A +

V

E +

The operational amplifier

+

U

s

U

e

-

In its field of linearity, the ideal OA has the following

characteristics:

infinite gain, therefore no differential input voltage;

infinite input impedance, therefore no input current.

P controller

KR

R

UU

UsG

RRR

emc

sR

e

==

=

+=

1

10

)(

121

PI controller

+=

+=

+=

=

=

ni

n

i

n

emc

sR

e

sTT

T

Ts

Ts

CRs

CRs

UU

UsG

RR

11

11)(

21

1

11

0

Tn=R1C1Ti=ReC1

frequency response step response



12/32

2

PI controller PI

+=

+=

+=

=

=

ni

n

i

n

emc

sR

e

sTT

T

Ts

Ts

CRs

CRs

UU

UsG

RR

11

11)(

21

1

11

0

If the values ofTn=R1C1 and Ti=ReC1are known :

selectRe between 10 and 100 kohms

C1=Ti/ReR1=Tn/C1

PI controller (adjustable coefficients)

( )ss U

sCRRR

sCRR

UU

sCRRSi

++

+

=

+


13/32

3

PI controller (adaptation of 1 time constant)

111

1

11

11

11)(

0/1

CRUKTCRT

Ts

Ts

CRUKs

CRssG

sCR

UUK

R

UU

UUKU

evmin

i

n

evm

R

svm

e

cm

svmi

==

+=

+=

=+

+

=

PI controller (adaptation of 2 time constants)

CRUKT

CRUKT

Ts

Ts

CRUKs

CRUKssG

UU

U

evimi

vnmn

i

n

evim

vnmR

mc

s

2

111

2

111 11)(

=

=

+=

+==

Adaptive controller with variable structure

i

n

e

RsT

sT

RR

RsCR

RRRRsC

sG+

=

+

++

=1

1

)(

12

12

122

ie

RsTCsRR

RsG

1)( 3 ==

FET1 ON

FET2 ON

Adaptive controller with variable structure

Note:The sources of the JFETs are connected to the input of the OA (virtual ground)

The diode between drain and ground prevents the drain potential to become too

high

When the JFET leads, the diode does not because UDS is very low

First order passive filter

sTsRCU

U

e

s

+=

+=

1

1

1

1

Active filter of second order

( )( ) ( )( )2121 111

14/1

1

sTsTsRCsRCU

U

e

s

++=

++=



14/32

4

PID + second order active filter

( )( )( )( ) ( )[ ]

( )( )( )( )

i

vn

eeemc

s

sT

sTsT

sTsT

CsR

CRRsCsRCsRCsRCsR

CsRCsRUU

U

++

++=

++++++

++=

11

11

1

111.

14/1

1

21

1

12123122211

23

PID + reference filter

PID with clipping circuit

If one LED leads, the voltage Us is clamped as shown on

the diagram

The diodes do not

conduct if:

( )( ) 01

01

222

111

>+=


15/32

1

Chapter 6

Digital controllers



Main configurations

Classic method

Consist to use a digital PI or PID

Suitable for systems with 1 or 2 dominant

time constants and who have a well damped

oscillating behaviour

Main configurations.

State control The orders issued by the regulator depends

- from the values of the state variables of the system thatare either measured or estimated (state estimator)

- sometimes from the measurable or estimabledisturbances

The control signal (CS) is a linear combination of

quantities of orders and disturbances The SC is usually an integral component to cancel

the error in steady state

Suitable for control of all systems

Main configurations.

State control

EXAMPLE : the BOOST converter

State variables : Ue et Us

The reference is Usc ; Uem and Usm are measured

The duty cycle to impose is The integral term Ki is added, so

sc

em

U

U=1

=

1

es

UU

)(

1

1 smscii

i

sc

em

UUkKK

KU

U

+=

+=

Main configurations

Sliding mode control

Suitable for systems with a

pulse device (chopper or

inverter)

The output of the controller is

a logical variable depending

from the reference and the

state variables

Main configurations

Cascade control



16/32

2

Main configurations

Trajectory pursuit control

For fine tuning (robots)

Example: if acceleration and speed are good but position

is too late, R3 acts to increase the speed

RST controlClassic control:

RST control:

More possibilities, numeric, etc

+Um

UsUcN(z)

D(z)

R(z)

S(z)

+Um

UsUcN(z)

D(z)

1T(z)

R(z)

S(z)

)()()()(

)()(

)(

)(

zSzDzRzN

zTzN

zU

zU

c

m

+=

)()()()(

)()(

)(

)(

zSzDzRzN

zRzN

zU

zU

c

m

+=

RST control : effect of disturbances

+ ++++

Um

P2P1

UsUcN(z)

D(z)

1T(z)

R(z)

S(z)

For a null static error in case of disturbance, 1/ S (z) must

contain at least one integrator, and :

)(')1()( zSzzS =

RST control : effect of disturbances

+ ++++

Um

P2P1

UsUcN(z)

D(z)

1T(z)

R(z)

S(z)

Against disturbances, the RST behaviour is the same as

its of a conventional controller

)()()()(

)()()(

)()()()(

)()()(

)()()()(

)()()()( 21

zDzSzNzR

zDzSzP

zDzSzNzR

zNzSzP

zDzSzNzR

zNzTzUzU cm

++

++

+=

RST control : causality+ +

+++Um

P2P1

UsUcN(z)

D(z)

1T(z)

R(z)

S(z)

tztztztzT

rzrzrzrzR

szszszzS

zUzRzUzTzUZS mcs

++++=

++++=

++++=

=

......)(

......)(

......)(

)()()()()()(

2

2

1

10

2

2

1

10

2

2

1

1

( ) ( )( ) )(......

)(......)(......1

2

2

1

10

2

2

1

10

2

2

1

1

zUzrzrzrzr

zUztztztztzUzszszs

m

cs

++++

++++=++++

[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]

++++++

+++++++++++

+=+

1......211

1......211

1.....11

210

210

21

kUrkUrkUrkUr

kUtkUtkUtkUt

kUskUskUskU

mmmm

cccc

ssss

often ==

RST control : causality+ +

+++Um

P2P1

UsUcN(z)

D(z)

1T(z)

R(z)

S(z)

( )

tztztztzT

rzrzrzrzR

szszszzS

zUzRzUzTzUZS

zUzRzUzTzS

zU

mcs

mcs

++++=

++++=

++++=

=

=

......)(

......)(

......)(

)()()()()()(

)()()()()(

1)(

2

2

1

10

2

2

1

10

2

2

1

1

[ ] [ ] [ ] [ ][ ] [ ] [ ] [ ][ ] [ ] [ ] [ ]

++++++

+++++++++++

+=+

1......211

1......211

1.....11

210

210

21

kUrkUrkUrkUr

kUtkUtkUtkUt

kUskUskUskU

mmmm

cccc

ssss

1

1

+

+

If at position k+1, Uc and/or Um are not available :

often 1==



17/32

3

RST control : Closed loop transfer fuction

The closed loop transfer function may be selected with some

liberties but :

- avoid to delete zeros of N (z) too close to the unit circle

- meet the following condition :

+Um

UsUcN(z)

D(z)

1T(z)

R(z)

S(z)

)(

)(

)()()()(

)()(

)(

)(

zD

zN

zSzDzRzN

zTzN

zU

zU

m

m

c

m =+

=

[ ] [ ] [ ] [ ])()()()( zNzDzNzD mm

RST control : Closed loop transfer fuction+

Um

UsUc

N(z)

D(z)

1T(z)

R(z)

S(z)

1)1(

)1(=

m

m

D

N

)1/()( = zzzUc

1)1(

)1(

)(

)(

1)1(lim)(

1

==

=

m

m

m

m

zm

D

N

zD

zN

z

zzU

If a null static error against the reference is suited,

the following condition must be met :

Indeed, if UC (z) isa unit step :

RST control : Summary+

UmUsUc

N(z)

D(z)

1T(z)

R(z)

S(z)

)(

)(

)()()()(

)()(

)(

)(

zD

zN

zSzDzRzN

zTzN

zU

zU

m

m

c

m =+

=

== 1== or

)(')1()( zSzzS = if a null static error against disturbances is wished

[ ] [ ] [ ] [ ])()()()( zNzDzNzD mm

1)1(

)1(=

m

m

D

N

ATTENTION : avoid to delete badly damped zeros

if a null static error against reference is wished

RST control : synthesis example

Consider a small motor with a transfer

function between the reference and the

speed is :

( )( )( )45.1264.18

29.1445.20)(

++

+=

ss

ssG

s

It is proposed to control this engine with a digital controller having a sampling period of

100ms.

It is proposed to use a RST controller with a zero static error against the disturbances and

command.

In closed loop, a time constant of 50ms is wished

REMARKS

The transfer function has the following characteristics:

A first time constant of 54ms (time constant electric)

A second time constant of 690ms (mechanical time constant)

A zero corresponds to a delay of 775ms

We note that the sampling period is not negligible against the time constants of the system.


( )( )( ) ( ) ( )45.1

195.0

264.18

640.20

45.1264.18

29.1445.20)(

+

+=

++

+=

ssss

ssG

s

865.045.1195.0

161.0264.18640.20

1

1

222

111

====

====

E

E

Tp

Tp

ezpr

ezpr

( )( )( ) )(

)(

161.0865.0

878.0930.0

865.0

018.0

161.0

948.01)(

zD

zN

zz

z

zzzz

z

p

rzG

i i

i

i

i =

=

=

=

Transfer function of the engine preceded by a holding element

RST control : synthesis example+

Um

UsUc

N(z)

D(z)

1

T(z)

R(z)

S(z)

A first-order behaviour is wanted (possible because the difference of

degrees between the numerator and denominator in OL is equal to 1).

Moreover, the zero of the numerator OL (0.878) is sufficiently distant

from the unit circle so it can be removed safely:

am

m

c

m

zz

a

zD

zN

U

U

==

)(

)(

( )( )( )161.0865.0878.0930.0

)()(

=zz

zzDzN



18/32

4


135.045.0/1.0/ ===

eezTBFT

aE

135.0)(

)(

==z

a

zD

zN

U

U

m

m

c

m

A time constant of 450 ms is expected

135.01

)1(

)1(

==z

a

D

N

m

m 865.0=a135.0

865.0

)(

)(

==zzD

zN

U

U

m

m

c

m

A null static error against the reference is expected

.

+Um

UsUcN(z)

D(z)

1T(z)

R(z)

S(z)


+Um

UsUc

N(z)

D(z)

1T(z)

R(z)

S(z)

)()()()(

)()(

)(

)(

zDzSzNzR

zNzT

zD

zN

U

U

m

m

c

m

+==

( )

( )( )161.0865.0

878.0930.0

)(

)(

=

zz

z

zD

zN

( )( ) ( )( ) 135.0

865.0

161.0865.0)(878.0)(930.0

878.0)(930.0

=

+

=

zzzzSzzR

zzT

U

U

c

m

To eliminate the numerators zero and to have a zero static error against

disturbances, S(z) must have the following form :

( )( )

878.0

878.1

1

878.01)(

2

1

0

=

=

=

=

s

s

s

zzzS

21

2

0

21

2

0

)(

)(

rzrzrzR

tztztzT

++=

++=

( )( ) ( )( ) 135.0

865.0

161.0865.0)1(930.0

930.0

21

2

0

21

2

0

=

+++

++=

zzzzrzrzr

tztzt

U

U

c

m

( )( ) ( )( ) ( ) 2

2

21

2

0

21

2

0

135.0

865.0

161.0865.0)1(930.0

930.0

zz

z

zzzrzrzr

tztzt

U

U

c

m

=

+++

++=


+Um

UsUcN(z)

D(z)

1T(z)

R(z)

S(z)

( )( ) ( )( ) 135.0

865.0

161.0865.0)1(930.0

930.0

21

2

0

21

2

0

=

+++

++=

zzzzrzrzr

tztzt

U

U

c

m

By identifying the coefficients :

( )( ) ( )( ) ( ) 22120

221

20

135.0161.0865.0)1(930.0

865.0930.0

zzzzzrzrzr

ztztzt

=+++

=++

049.10

253.10

033.2930.0

22

11

00

==

==

==

rt

rt

rt


[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]111111 21021021 +++++=+ kUrkUrkUrkUtkUtkUtkUskUskU mmmcccsss

+Um

UsUcN(z)

D(z)

1T(z)

R(z)

S(z)

[ ] [ ] [ ] [ ] [ ] [ ] [ ]1049.1253.11033.21215.01878.0878.11 ++++=+ kUkUkUkUkUkUkU mmmcsss

049.10

253.10

033.2930.0

22

11

00

==

==

==

rt

rt

rt

878.0

878.1

1

2

1

0

=

=

=

s

s

s

Numerical command circuits

Microprocessor :

fas t (3GHz)

32 to 128 bits logic

requires external devices

expensive

Microcontroller

contains CPU + itsminimum environment

autonomy

relatively slow (10MHz, 8 bit)

inexpensive

Numerical command circuits

DSP (Digital Signal Processor)

Evolved microcontroller, rapidity

Contains special functions (FFT, ...)

FPGA (Field Programmable Gate Array)

Circuit with logic functions which may be interconnected byprogramming (VHDL = Very High Speed Description Language)

ASIC (Application Specific Integrated Circuit)

For large series



19/32

1

Chapter 7

Electrotechnical applications

of regulators



Cascade control

Rectifiers 4 quadrants AC/DC controller with

current circulation

4 quadrants AC/DC controller with

current circulation

4 quadrant AC/DC controller without current

circulation

LOGICLOGIC : the 2 bridges may never

conduct at the same time dead time



20/32

2

1 quadrant chopper with 2 position controller1 quadrant chopper with PWM controller

4 quadrant chopper with PWM controller

7

1

6

!

6

"

6

*)

6

8

-

1

J

J

7

J

J

J

J

8

)

J

8

*

J

6

@ 6

,

!

6

"

,

"

6

!

,

!

6

"

,

6

,

6

,

6

VVVU )12()1( ==

(other command modes exist)

Inverter with 2 position controller

Inverter with PWM controller



21/32

1

Chapter 8

Command of switched power supplies



Boost converter command

v

o

i

o

i

D

i

L

v

i

C

R

i

T

L

D

If the switching frequency is sufficiently high, the system can be studied as a

continuous process although command is discreet

[ ]

+=

+=

))(1)(()(1

))(1)(()(1

tgR

vitg

R

v

Cdt

dv

tgvvtgvLdt

di

oi

oo

oiiL g=1 : switch ON

g=0 : switch OFF

To study this kind of system, we proceed by averaging over one sample period :

=t

TtdttX

TtX )(

1)(


v

o

i

o

i

D

i

L

v

i

C

R

i

T

L

D

[ ]

+=

+=

))(1)(()(1

))(1)(()(1

tgR

vitg

R

v

Cdt

dv

tgvvtgvLdt

di

oi

oo

oiiL

=t

TtdttX

TtX )(

1)(+

[ ]

+

+=

=

C

i

RC

v

C

tgi

RC

v

dt

vdL

v

L

vtgvv

Ldt

id

LoLoo

oioi

L

)1())(1(

)1())(1(

1

v

i

( 1 - @ )

v

o

i

o

i

L

( 1 - @ )

i

L

v

i

C

R

L

Boost converter command[ ]

+

+=

=

C

i

RC

v

C

tgi

RC

v

dt

vdL

v

L

vtgvv

Ldt

id

LoLoo

oioi

L

)1())(1(

)1())(1(

1

v

i

( 1 - @ )

v

o

i

o

i

L

( 1 - @ )

i

L

v

i

C

R

L

This model is not linear.

Thus, we proceed to the linearization in the vicinity of a static equilibrium point:

+=

+=

+=

+=

~

~

~

~

(

(

(

(

LLL

ooo

iii

iii

vvv

vvv

R

vi

vv

oL

io

(

((

(

(

(

=

=

)1(

1

+=

+

=

C

i

C

i

RC

v

dt

vdL

v

L

v

L

v

dt

id

LLoo

ooiL

~)1(

~~~

~)1(~~

~

((

(

(

By eliminating the static current between these relationships and assuming

constant vi, we obtain the transfer function:

LCs

RCs

LC

v

C

is

voL

o

22 )1(1

)1(

~

~

(

(

(

(

++

=

+


This transfer function has 1 positive zero and 2 complex poles generally weakly

damped.

This system is very difficult to regulate.

Therefore generally one prefer to use a current control rather than a duty cycle

control.

LCs

RCs

LC

v

C

is

voL

o

22 )1(1

)1(

~

~

(

(

(

(

++

=

v

o

i

o

i

D

i

L

v

i

C

R

i

T

L

D

Current control

principle

Let T be the period of hash and the conduction time. The switch is closed at the

beginning of the period and the current increases. The switch is opened when the

current reaches the value ip-mc .

The slope mc is chosen to ensure stable operation.

v

o

i

o

i

D

i

L

v

i

C

R

i

T

L

D

iL

T=

ip -mc

m1 -m2

0

To ensure stable operation, it is necessary that the

current is amortized if any disturbance happens

=

>

1

12

22112

mmmmc

mc is not necessary if 0,5



22/32

2

Current control

Transfer function

v

o

i

o

i

D

i

L

v

i

C

R

i

T

L

DThe transfer function between the outputvoltage and the current iL can be found after

local averaging and linearization of the

equations:iL

T=

ip -mc

m1-m2

0

( )( )csbsas

Ki

v

p

o

++

=~

~

This transfer function has a positive zero and two real poles (one at low frequency

and one at higher frequency).

Under these conditions, the system is much easier to

control

Current control

The IC LM5021

Example : forward converter

with current control



23/32

1

Chapter 9

COMMAND OF INDUCTION MOTORS



Speed variation history.

1. Ward-Lonard

2. DC motor drives

3. AC motor drives

Dynamics of induction motor :

3 phase model

For each coil :

iiiidt

dIRU +=

+=6

ijjijiii IMIL

Dynamics of induction motor :

3 phase model

;

;

.

;

;.

=

xyz

XYZ

RtSR

SRS

xyz

XYZ

I

I

ZZ

ZZ

U

U

3,3

33;;;;

=

=

=

=

z

y

x

xyz

Z

Y

X

XYZ

z

y

x

xyz

Z

Y

X

XYZ

I

I

I

I

I

I

I

I

U

U

U

U

U

U

U

U

+

+

+

=

SSSS

SSSS

SSSS

S

sLRsMsM

sMsLRsM

sMsMsLR

Z

3333

3333

3333

3

++

+

=RRRR

RRRR

RRRR

R

sLRsMsMsMsLRsM

sMsMsLR

Z

3333

3333

3333

3

+

+

+

=

)cos()3/2cos()3/2cos(

)3/2cos()cos()3/2cos(

)3/2cos()3/2cos()cos(

333

333

333

3

RSRRSRRSR

RSRRSRRSR

RSRRSRRSR

SR

sMsMsM

sMsMsM

sMsMsM

Z

Dynamics of induction motor : 2 phase model

.

xyzt

XYZtAB

ICI

ICI

*

*

=

=

=

2/3

2/3

0

2/1

2/1

1

3

2*C

Dynamics of induction motor : 2 phase model

.

xyzt

XYZtAB

ICI

ICI

*

*

=

=

=

2/3

2/3

0

2/1

2/1

1

32*C

( )( )( )

( )

( )

( )( )( )

( )

( )tgII

tgII

tgII

tgII

tgII

tII

tII

tII

tII

tII

MaxR

MaxR

MaxRz

MaxRy

MaxRx

MaxSB

MaxSA

MaxSZ

MaxSY

MaxSX

cos2

3

sin2

3

240sin

120sin

sin

cos2

3

sin2

3

240sin

120sin

sin

,

,

,

,

,

,

,

,

,

,

=

=

+=

+=

=

=

=

+=

+=

=



24/32

2

Induction motor : axes related to magnetic field

=

=

B

A

tS

B

A

aa

aa

qs

ds

I

IP

I

I

I

I,

)cos()sin(

)sin()cos(

=

=

I

IP

I

I

I

ItR

RaRa

RaRa

qr

dr

,)cos()sin(

)sin()cos(


=

=

B

A

tS

B

A

aa

aa

qs

ds

I

IP

I

I

I

I,

)cos()sin(

)sin()cos(

=

=

I

IP

I

I

I

ItR

RaRa

RaRa

qr

dr

,)cos()sin(

)sin()cos(

MaxRqr

MaxRdr

MaxSqs

MaxSds

II

II

II

II

,

,

,

,

2

3

2

3

2

3

2

3

=

=

=

=

( ) tgt

t

Ra

a

==

=

0


=

qr

dr

qs

ds

r

s

RtSR

SRS

t

t

tr

ts

qr

dr

qs

ds

I

I

I

I

P

P

C

C

ZZ

ZZ

C

C

P

P

U

U

U

U

0

0

0

0

0

0

0

0

3,3

33

,

,


=

qr

dr

qs

ds

r

s

RtSR

SRS

t

t

tr

tsqs

ds

I

I

I

I

P

P

C

C

ZZ

ZZ

C

C

P

PU

U

0

0

0

0

0

0

0

0

0

03,3

33

,

,

qrdrR

qsdsS

qsdsS

jIII

jUUU

jIII

+=

+=

+=En posant:

On obtient

Le couple vaut dsqrqsdr IMIIMIC =

dt

dgjIR

dt

djIRU

RRRR

S

SSSS

++=

++=

0

RRSR

RSSS

ILIM

IMIL

+=

+=


+

+

=

=

=

qs

ds

a

a

a

a

a

a

qs

ds

S

B

A

Z

Y

X

I

I

I

IPC

I

IC

I

I

I

)3/2sin(

)3/2sin(

)sin(

)3/2cos(

)3/2cos(

)cos(**

+

+

=

=

=

qs

ds

a

a

a

a

a

a

qs

ds

S

B

A

Z

Y

X

U

U

U

UPC

U

UC

U

U

U

)3/2sin(

)3/2sin(

)sin(

)3/2cos(

)3/2cos(

)cos(**

Induction motor

In steady state conditions, we find the classical

equations of the equivalent circuit

)( MLX SS =

)( MLXRR =

MX =



25/32

3

Command of the induction motor

GENERAL STRATEGY :

We try to maintain constant and maximum the flux in the

machine because the torque is proportional to flux.

To act on the couple, we act on the stator current because :

),sin(.. IIkC =

Command of the induction motor

Four techniques are considered:

1. Scalar control on voltage

2. Vector control on voltage

3. Scalar control on current

4. Vector control on current

Scalar control on voltage

( )

( ) ( )),(

1

11

2

22222

2

22

22

SS

sRs

R

RR

SS

SSS

RR

SS

SSSSSS

gfMLLgLR

RMgpC

LgjR

MgLjR

RU

LgjR

MgLjR

R

j

UIRU

j

R

=+

=

+++

=

+++

==

Scalar control on voltage : simplified version

c

a

a

a

ctt

Z

Y

X

UUCP

U

U

U

+

==

)3/2cos(

)3/2cos(

)cos(

Scalar control on voltage :

action on couple

( ) ( )),(

)3/2cos(

)3/2cos(

)cos(

2

22222

2

SS

sRs

R

c

a

a

a

ctt

Z

Y

X

gfMLLgLR

RMgpC

UUCP

U

U

U

R

=+

=

+

==


sophisticated version

( ) ( )),(

)3/2cos(

)3/2cos(

)cos(

2

22222

2

SS

sRs

R

c

a

a

a

ctt

Z

Y

X

gfMLLgLR

RMgpC

UUCP

U

U

U

R

=+

=

+

==



26/32

4


speed control

( ) ( )),(

)3/2cos(

)3/2cos(

)cos(

2

22222

2

SS

sRs

R

c

a

a

a

ctt

Z

Y

X

gfMLLgLR

RMgpC

UUCP

U

U

U

R

=+

=

+

==

Vector control on voltage

( )

( ) ( )),(

1

11

2

22222

2

1

22

22

SS

sRs

R

RR

SS

SSqsdsS

RR

SS

SSSSSS

gfMLLgLR

RMgpC

LgjR

MgLjR

RjjUUU

LgjR

MgLjR

R

j

UIRU

j

R

=+

=

+++

=+=

+++

==

Vector control on voltage

+

+

==

qs

ds

a

a

a

a

a

a

ctt

Z

Y

X

U

UUCP

U

U

U

)3/2sin(

)3/2sin(

)sin(

)3/2cos(

)3/2cos(

)cos(

Scalar control on current

( ) ( )),(

2

22222

2

12

2

SS

sRs

R

S

RR

SS

S

RR

SRSSS

gfMLLgLR

RMgpC

LgjR

MgjLI

ILgjRMgjLIMIL

R

=+

=

+=

+=+=

Scalar control on current Scalar control on current



27/32

5

Scalar control on current

t ( s e c )

C o u p l e

21 , 510 , 5

2

1 , 5

1

0 , 5

0

Vector control on current

dt

dgjIR

dt

djIRU

R

RRR

S

SSSS

++=

++=

0

RRSR

RSSS

ILIM

IMIL

+=

+=

dsqrqsdr IMIIMIC =

In an axes system related to the rotor flux (d in the direction of the rotor flux):

( )dtg

gMR

LIC

gMR

LI

R

Ls

MI

a

R

RRqsR

R

RRqs

R

RRds

+=

==

=

+=

2

)1(


( )dtg

I

I

I

I

I

a

qs

ds

a

a

a

a

a

a

Z

Y

X

+=

+

+

=

)3/2sin(

)3/2sin(

)sin(

)3/2cos(

)3/2cos(

)cos(


( )dtg

I

I

I

I

I

a

qs

ds

a

a

a

a

a

a

Z

Y

X

+=

+

+

=

)3/2sin(

)3/2sin(

)sin(

)3/2cos(

)3/2cos(

)cos(

To reach overspeed

conditions, one reduces the

flux



28/32

1

Chapter 10

Preparing the lab.

Controlling a DC machine.



Transfer function of a DC motor with

independent excitation

( )

IkC

CIksJN

NkIsLRU

r

=

=++=

k

1

s J

1 N

Cr

CIU

R(1 + s Ta)k

+-

ou :

( ) ( )

( )r

a

CCsJ

N

IkC

NkUsTR

NkULsR

I

=

=

+

=+

=

1

)1(

11

Transfer function of a DC motor with

independent excitation

k

1

s J

1 N

Cr

CIU

R(1 + s Ta)k

+-

( ) ( ) ra

aC

k

sTR

k

U

k

RsTsJN

22

)1(1)1(

+

=

+

+

( )

=

2k

RJTm

( )( ) r

aam C

k

sTR

k

UsTsTN

2

)1()1(1

+

=++

We have also :

Principle of cascade control

Niref

+-

+-

Nrefmoteurrgulateur

de vitessergulateurde courant

convertisseur

I

General scheme

Niref

+-

+-

Nrefmoteurrgulateur

de vitessergulateurde courant

convertisseur

I

Technics

To determine the parameters of the regulators,

there exists 2 methods:

analytical method based on model knowledge

experimental method



29/32

2

ANALYTICAL METHOD FOR

DETERMINING THE

REGULATORS PARAMETERS

Current regulator (1)

K

m i

1 + s T

m i

K

c o n v

1 + s T

c o n v

1

IU

K

i

1 + s T

i

s T

i

i

r e f

R ( 1 + s T

a )

+

-


mi

mi

aconv

conv

i

ii

ref

mes

sT

K

sTRsT

K

sT

sTK

I

I

+++

+=

1)1(

1

1

1

amiconvcmi TTTT


30/32

3


22221

1

cmicmiref

mes

TssTI

I

++=

If we apply a step at system input, we can observe

an overshoot of4,3% a rise time of 4,7*Tcmi

K

m i

1 + s T

m i

K

c o n v

1 + s T

c o n v

1

IU

K

i

1 + s T

i

s T

i

i

r e f

R ( 1 + s T

a )

+

-

Current regulator (6b)

Time

0s 2ms 4ms 6ms 8ms 10ms 12ms 14ms 16ms 18ms 20ms

V(LAPLACE4:OUT) V(V13:+)

0V

0.2V

0.4V

0.6V

0.8V

1.0V

1.2V

Behaviour of the closed current loop

overshoot 4,3%

rise time : 4,7*Tcmi= 4,7ms


22221

1

cmicmiref

mes

TssTI

I

++=

Subsequently, we will approximate this transfer function with:

cmimiref sTKI

I

21

11

+=

K

m i

1 + s T

m i

K

c o n v

1 + s T

c o n v

1

IU

K

i

1 + s T

i

s T

i

i

r e f

R ( 1 + s T

a )

+

-

Speed regulator (1)

Speed regulator (2)

mncmic

c

mn

min

nn

mn

mn

cmimin

nn

ref

mes

TTTavec

sT

K

sJ

k

KsT

sTK

sT

K

sJ

k

sTKsT

sTK

N

N

+=

+

+

+

+

+=

2

1

11

121

111

min

min

In OL :

(by grouping small time constants)

Speed regulator (3)

To have a good performance in rejecting disturbances,

one choose ususaly :

min1

11

c

mn

min

nn

ref

mes

sT

K

sJ

k

KsT

sTK

N

N

+

+

mnn

mimn

cn

RKT

KTkK

TT

2

4 min

=

=



31/32

4

Speed regulator (4)

min1

11

c

mn

min

nn

ref

mes

sT

K

sJ

k

KsT

sTK

N

N

+

+

mnn

mimn

cn

RKT

KTkK

TT

2

4 min

=

=

3

min

32

min

2

min

min

8841

41

ccc

c

ref

mes

TsTssT

sT

N

N

+++

+=

In CL

As response to a step at the input, we observe:

an overshoot of 43%

a rise time of 3.2 * Tcmi(valid if the mechanical time constant is muchhigher thanother time

constants)

Speed regulator (4b)

Overshoot : 43%

Rise time : 3,2*Tcmin= 3,2ms

Time


V(R18:1) V(V13:+)

0V

0.5V

1.0V

1.5V

Closed Speed loop

Speed regulator (5)

3

min

32

min

2

min

min

8841

41

ccc

c

ref

mes

TsTssT

sT

N

N

+++

+=

In CL

As response to a step, we observe :

an overshoot of 8%

a rise time equal to 7,5*Tcmi

To reduce the overshoot (due to the zero in the

numerator), we can add a reference filter :

min41

1)(

csTsF

+=

3

min

32

min

2

min 8841

1

cccref

mes

TsTssTN

N

+++=

Speed regulator (5b)

Closed speed loop with reference filter

Overshoot : 8%

Rise time : 7,5*Tcmi = 7,5 ms

Time


V(LAPLACE5:OUT) V(V13:+)

0V

0.2V

0.4V

0.6V

0.8V

1.0V

1.2V

EXPERIMENTAL METHOD FOR

THE DETERMININATION OF

REGULATORS PARAMETERS

Reminder: transfer function of a

PID

++= v

n

p sTsT

KsH1

1)(



32/32

Critical oscillations method (1)

Initialy, the system is in CL with :

0

v

n

p

T

T

Klow

Kp is increased gradually until the limit of

oscillation. At this point, we note :

periodnoscillatio

pointat this

=

=

crit

pcrit

T

KK

pv

n

p KsTsT

KsH

++=

11)(

Critical oscillations method (2)

The controller parameters are given in the

following table (Ziegler-Nichols method) :

Kp Tn Tv

PcritK50.0

PIcritK45.0 critT85.0

PIDcrit

K59.0crit

T50.0 critT12.0

Response to a step (1)

The system is initialy in OL : a step K is applied

at the system input and the response is observed.

Tu,Tg and Ks are measured


Coefficients following Ziegler-Nichols :

Kp Tn Tv

P

us

g

TK

T

PI

us

g

TK

T90.0

uT33.3

PID

us

g

TK

T2.1

uT2

uT5.0


Coefficients following Chien, Hrones and Resewick :

Critre de qualit

Rgulation apriodique de trs

courte dure

Rgulation avec dpassement

de 20%

Rejet de

perturbation

Suivi de

consigne

Rejet de

perturbation

Suivi de

consigne

P

us

g

pTK

TK 3.0=

us

g

pTK

TK 3.0=

us

g

pTK

TK 7.0=

us

g

pTK

TK 7.0=

PI

un

us

g

p

TT

TK

TK

4

6.0

=

=

un

us

g

p

TT

TK

TK

2.1

35.0

=

=

un

us

g

p

TT

TK

TK

3.2

7.0

=

=

un

us

g

p

TT

TK

TK

=

= 6.0

PID

uv

un

us

g

p

TT

TT

TK

TK

42.0

4.2

95.0

=

=

=

uv

un

us

g

p

TT

TT

TK

TK

5.0

60.0

=

=

=

uv

un

us

g

p

TT

TT

TK

TK

42.0

2

2.1

=

=

=

uv

un

us

g

p

TT

TT

TK

TK

47.0

35.1

95.0

=

=

=


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