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DC MACHINES DYNAMIC ANALYSIS AND CONTROL OF AC MACHINES ERASMUS MUNDUS MASTER COURSE on SUSTAINABLE TRANSPORTATION AND ELECTRIC POWER SYSTEMS

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Page 1: Emmc Steps Daac Lesson2 (Rev0)

DC MACHINES

DYNAMIC ANALYSIS AND CONTROL OF AC MACHINES

ERASMUS MUNDUS MASTER COURSE on SUSTAINABLE TRANSPORTATION AND

ELECTRIC POWER SYSTEMS

Page 2: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 2

STRUCTURE OF A DC MACHINE

• EXCITATION (FIELD) WINDING in the stator • ARMATURE WINDING in the rotor • The armature is a closed winding and portions of it

are accessible through a collector and brushes

Page 3: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 3

EXCITATION WINDING

• A DC current, fed to the excitation winding, allows to establish a flux density field of constant amplitude which crosses radially the airgap

• The excitation field has fixed direction (polar axis)

B

Page 4: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 4

EXCITATION FIELD

• Field distribution induces a voltage in rotor moving conductors

• The induced voltage in each conductor is proportional to B and ω

• All the conductors under one pole are subject to voltages of the same sign

Page 5: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 5

BACK-EMF

• The back-emf at armature terminals is

R

L

τ

v

Bi

∑=

=2/N

1iiLvBE

∑=

=2/N

1iriLBEπτω

πτωravg L

2NBE =

( ) ravg LB2NE ωτπ

=

reEKE ωΦ=

Page 6: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 6

ARMATURE WINDING

• Originally, it was a ring-type winding (Pacinotti ring) • Now, it is a drum-type winding, with two layers

Page 7: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 7

TYPES OF DRUM WINDINGS

Lap winding Wave winding

Page 8: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 8

LAP WINDING

Page 9: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 9

WAVE WINDING

Page 10: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 10

ARMATURE CURRENT

• Current is the same for all the conductors under the same pole • As a conductor crosses the interpolar axis, the current in it goes

to zero and reverses sign (commutation) • However, the commutator and brushes system keeps the

current distribution constant under a pole

POLAR AXIS

INTERPOLAR AXIS

Page 11: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 11

TORQUE

• Torque derives from the combination of all the forces acting on the conductors

• Macroscopically, torque can be seen as the interaction (cross product) of the excitation and armature fields

• The commutator and brushes keep the two fields fixed in space and in quadrature

INTERPOLAR AXIS

POLAR AXIS Φe

Ia

BldIFd ×⋅=

Page 12: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 12

STEADY STATE EQUATIONS

Va

Ia Ra

E Ve

IeRe

EIRV aaa +=

eee IRV =

reEKE ωΦ=

aeT IKT Φ=)I(f ee =Φ

Page 13: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 13

MECHANICAL CHARACTERISTICS

Va

Ia Ra

E Ve

IeRe

( )a

aa R

EVI −=

a

aeT R

EVKT −Φ=

a

reEaeT R

KVKT ωΦ−Φ=

a

aeTr

a

2eET

RVK

RKKT Φ

−= ω

T

ωr

Tn

Tstart

ω0

a

aeTstart R

VKT Φ=

eE

a0 K

Page 14: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 14

VOLTAGE REGULATION

a

aeTr

a

2eET

RVK

RKKT Φ

−= ω

a

aeTstart R

VKT Φ=

eE

a0 K

T

ωr

Tn

Va increasing

• When the armature voltage is increased from 0 to nominal voltage, the mechanical characteristic moves parallel to itself

• The usable region is delimited by the nominal torque (constant torque)

Page 15: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 15

FIELD REGULATION

a

aeTr

a

2eET

RVK

RKKT Φ

−= ω

a

aeTstart R

VKT Φ=

eE

a0 K

• When the excitation field is decreased from nominal value to 0, the mechanical characteristic inclines itself.

• All characteristics are tangent to an hyperbola (constant power curve)

T

ωr

Φe decreasing

Page 16: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 16

COMBINED VOLTAGE AND FIELD REGULATION

T, Φe

ωr

ωn

Va, PE

Ia = const.

Constant torque

Constant power

Φe = ΦenVa increasing

Φe decreasingVa = Van

Page 17: Emmc Steps Daac Lesson2 (Rev0)

DYNAMIC ANALYSIS OF DC MACHINES

DYNAMIC ANALYSIS AND CONTROL OF AC MACHINES

ERASMUS MUNDUS MASTER COURSE on SUSTAINABLE TRANSPORTATION AND

ELECTRIC POWER SYSTEMS

Page 18: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 18

DYNAMIC EQUATIONS

edtdiLiRv a

aaaa ++=dtdiLiRv e

eeee +=

reEKe ωϕ=

aeT iKT ϕ=)i(f ee =ϕ

Va

Ia Ra

E Ve

IeReLeLa

rr

L Bdt

dJTT ωω+=−

Page 19: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 19

BLOCK DIAGRAM

∫1/Le

Re

∫1/La

Ra

× 1/J ∫

B

×

ve ie ϕe

iava KT

KE

e

T

TL

ωr

dωr/dtdia/dt

die/dt+

+

-

-

-

-

+ -

Page 20: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 20

BLOCK DIAGRAM (FIXED EXCITATON)

∫1/La

Ra

1/J ∫

B

iava Kt

Ke

e

T

TL

ωr

dωr/dtdia/dt

+

-

-

-

+ -

• The model can be linearized if we consider φe = constant

eEe KK Φ= eTt KK Φ=

Page 21: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 21

LAPLACE DOMAIN BLOCK DIAGRAM

• This model can be used to represent a DC machine with fixed excitation under linearity hypothesis.

1/(Ra+sLa)iava Kt

Ke

e

T

TL

ωr+ -

-

+ 1/(B+sJ)

Page 22: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 22

SPEED TRANSIENT

• Speed transient can be analyzed using effect superposition principle

)s(T)s(G)s(V)s(G)s( L2a1r ⋅+⋅=Ω

0)s(LTa

r1 )s(V

)s()s(G=

Ω=

0)s(aVL

r2 )s(T

)s()s(G=

Ω=

• Simplifying hypotheses:

0B = mJJ =

Page 23: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 23

SPEED TRANSIENT

• Speed transient caused by voltage variation

sJ1K

sLR1K1

sJ1K

sLR1

)s(G

mt

aae

mt

aa1

⋅⋅

+

⋅+

⋅⋅

+

=

( )

++

=++

=1

KKJRs

KKJLsK

1KKsLRsJ

K)s(G

et

ma

et

ma2e

etaam

t1

( )1ssK1)s(G

mam2

e1 ++

=τττ

et

mam KK

JR=τ

a

aa R

L=τ

Page 24: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 24

SPEED TRANSIENT

• Usually, τm >> τa

( )( )ame1 s1s1K

1)s(Gττ ++

( )amm ss τττ +≅

( )me1 s1K

1)s(Gτ+

• Applying a voltage step of amplitude ∆Va

( ) sV

s1K1)s( a

mer

∆⋅

+≅Ω

τ

• In time domain:

( ) )0(e1KV)t( r

m/t

e

ar ωω τ +−

∆≅ −

Page 25: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 25

SPEED TRANSIENT

• Similarly

( )( )( )am

a

et

a2 s1s1

s1KK

R)s(Gττ

τ++

+−≅

• Applying a load step of amplitude ∆TL

• In time domain:

( ) )0(e1KKRT)t( r

m/t

et

aLr ωω τ +−

∆−≅ −

( )( )

( )

++

+⋅−=

+++

−=1

KKJRs

KKJLs

s1KK

RKKsLRsJ

sLR)s(G

et

ma

et

ma2

a

et

a

etaam

aa2

τ

( ) sT

s11

KKR)s( L

met

ar

∆⋅

+−≅Ω

τ

Page 26: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 26

CURRENT TRANSIENT

• Applying a voltage step:

( ) ( ))s(K)s(Vs11

R1)s(I rea

aaa Ω−⋅

+=

τ

( )

Ω−∆

⋅+

= )s(KsV

s11

R1)s(I re

a

aaa τ

• In time domain (assuming no load torque):

( ) ( )

∆⋅−+−

∆+= −− m/t

ae

ae

a/t

a

aaa e1

RKVKe1

RV)0(i)t(i ττ

Page 27: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 27

DC MACHINE TRANSIENTS VOLTAGE STEP

Page 28: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 28

DC MACHINE TRANSIENTS LOAD STEP

Page 29: Emmc Steps Daac Lesson2 (Rev0)

DC DRIVES

DYNAMIC ANALYSIS AND CONTROL OF AC MACHINES

ERASMUS MUNDUS MASTER COURSE on SUSTAINABLE TRANSPORTATION AND

ELECTRIC POWER SYSTEMS

Page 30: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 30

FORMING A DC DRIVE

va

ia Ra

e ve

ieReLeLa=

=

Vdc

PWMSpeed regulator+ -

T

Speed signal conditioning

ωr* va

* d

ωr

TACHODC-DC

CONVERTER

• This drive structure conceptually works, since there is a direct relation between voltage and speed.

• However, in this structure the current is not controlled: – during transients current can exceed limits – torque is not controlled

ARMATURE VOLTAGE REGULATION (FIXED EXCITATION)

Page 31: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 31

FORMING A DC DRIVE

• This drive structure controls also the current • It is composed by two nested loops • The inner loop must have higher bandwidth, so that

its closed loop transfer function can be considered as “instantaneous” from the outer loop

• A third, position loop can be added outside

ARMATURE VOLTAGE REGULATION (FIXED EXCITATION)

va

ia Ra

e ve

ieReLeLa=

=

Vdc

PWMSpeed regulator+ -

T

Current signal conditioning

ωr* va

* d

ωr

TACHODC-DC

CONVERTER

+ -

ia*Current

regulator1/KtT*

ia*

Speed signal conditioning

\

Page 32: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 32

MODELLING OF A DC DRIVE

• So far, the following models have already been derived: – DC machine

ARMATURE VOLTAGE REGULATION (FIXED EXCITATION)

1/(Ra+sLa)iava Kt

Ke

e

T

TL

ωr+ -

-

+ 1/(B+sJ)

– Transducers

• The regulators are generally formed with PID controllers

1/(1+sτ)x xmeas

KP

∫KI

KD d/dt

ε u+

+

+KP+KI/s+sKD

ε u

Page 33: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 33

MODELLING OF A DC DRIVE

• A model for the converter is needed:

ARMATURE VOLTAGE REGULATION (FIXED EXCITATION)

va

=

=

Vdc

PWMva

* d

DC-DC CONVERTER

???va

* va

Page 34: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 34

MODELLING OF DC-DC CONVERTER

ton

Ts

ton = d Ts

1

va

Vdc

-Vdc

va,avg

va,pu*

vamax,pu*

Triangular carrier

• Different time-scales can be considered: – Instantaneous – Switching period average

• Also, non-ideal switches can be considered: – Turn on, turn off times – Dead time – On-state voltage drop

Page 35: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 35

MODELLING OF DC-DC CONVERTER

• Instantaneous model

<−

≥=

carrier*

pua,dc

carrier*

pua,dca v vif ,V

v vif ,V)t(v

dcavga,dc

*aavg,a

VvV

,v)t(v

≤≤−

=

• Switching period average model

carrier*

pua,carrier

dccarrier

*pua,

avg,a

VvV

,VVv

)t(v

≤≤−

= carrierdc

*a*

pua, VVv vif =

Page 36: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 36

MODELLING OF DC-DC CONVERTER

• In general, the switching period average model is used to determine the controller parameters of the drive.

• This method neglects the harmonics at (multiples of) the switching frequency. Therefore, the armature current calculated with this model will neglect harmonics which actually exist.

• This implies a limitation in the bandwidth of the current loop.

• Can this limitation be overcome? No, because we can vary the voltage reference va

* only once (maximum twice) per switching period.

Page 37: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 37

MODELLING OF DC-DC CONVERTER SWITCHING PERIOD AVERAGE MODEL

va*

Vdc

-Vdc

va

LINEARIZED SWITCHING PERIOD AVERAGE MODEL

va* va1

Page 38: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 38

MODEL OF A DC DRIVE

Vdc

-Vdc

1/(Ra+sLa)iava Kt

Ke

e

T

TL

ωr+ -

-

+1/(B+sJ)Speed

regulator+ -

ωr* va

*

ωr

+ -

ia*Current

regulator1/KtT*

ia

1/(1+sτ1)

1/(1+sτ2)

MOTOR MODELCONVERTER

MODEL

TRANSDUCERS MODELS

• Having modeled all components, it is now possible to analyze the regulators of the two control loops.

Page 39: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 39

BANDWIDTHS

log(f)

fs BW1BWfilter,1BWiBWωBWθ

x 10

• At least one decade is needed between frequencies, in order to allow enough attenuation.

• The switching frequency imposes all others, since it is usually also the sampling frequency and corresponds to the rate of change for the innermost control variable va

*. • The current ripple at switching frequency must be

filtered to avoid aliasing.

Page 40: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 40

BANDWIDTHS

• Example: – Switching frequency: 10 kHz – BW of current transducer: 100 kHz – Cutoff frequency of filter: 1 kHz – BW of current regulator: 100 Hz – BW of speed regulator: 10 Hz – BW of position regulator: 1 Hz

Page 41: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 41

CURRENT LOOP

Vdc

-Vdc

1/(Ra+sLa)iava

Ke

e ωr

+ -

va*

+ -

ia*

ia

1/(1+sτ1)

KP

∫KI

KD d/dt

+

+

+

1/(1+sτfilt1)

iaia*

Wi(s)

• At the end, the current loop will be represented by:

Page 42: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 42

CONSIDERATIONS ABOUT CURRENT LOOP

• Inner loop has the highest bandwidth • Current loop is also a torque loop

• Current HAS to be the inner loop due to the cause and effect relationships – Voltage is the cause that makes current change – Torque is the cause that makes speed change

ia KtTia*

1/KtT*

Wi(s)

Page 43: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 43

SPEED LOOP

ia KtT

TL

ωr-

+1/(B+sJ)Speed

regulator+ -

ωr*

ωr

+ia*

1/KtT*

1/(1+sτfilt2)

Wi(s)

1/(1+sτ2)

• Methods for the synthesis of the regulator parameters of the speed loop have already been investigated within the “Control of electromechanical systems” module

Page 44: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 44

INCREASING CURRENT LOOP BANDWIDTH

• It is possibile to eliminate the filter by using a particular PWM technique

• The sampling must be done twice per switching period and sampling instants should be at the peaks of the triangular carrier.

• The resulting PWM is the so-called asimmetric PWM, updated twice per switching period

• It is possible to demonstrate that in this way only the average value of the current is sampled.

• Therefore the filter can be eliminated and the bandwith of the current regulator can be increased.

3.10f

7.20f

BW switchingsamplingmax,i =≤

Page 45: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 45

CURRENT LOOP SYNTHESIS

1/(Ra+sLa)iava

e

+ -

va*

+ -

ia*

ia

KP

∫KI +

+

1

1

1/(Ra+sLa)iava

e

+ -

va*

+ -

ia*

ia

1Kp+Ki/s

Page 46: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 46

CURRENT LOOP SYNTHESIS

• Requirements: – Zero steady state error – Specified closed loop bandwith BWi

• Feasible desired closed loop transfer function:

s11)s(W *

i

*i τ+

with

i

*i BW

2πτ =

• Ideal closed loop transfer function:

1)s(W *i =

Page 47: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 47

CURRENT LOOP SYNTHESIS

• Hypothesis: ideal zero-pole cancellation

• Open loop transfer function

s

sKK1K

)s(G I

PI

+

=a

a

I

PRL

KK

=

s1

RK)s(F

a

I ⋅=

1/Ra(1+sτa)

iava

e=0

+ -

va*

+ -

ia*

ia

1Ki(1+Kp/Kis)s

• Set the back-emf equal to zero

Page 48: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 48

CURRENT LOOP SYNTHESIS

• Closed loop transfer function

sKR1

1)s(F1

)s(F)s(W

I

ai

+=

+=

• The transfer function has zero steady state error for step input and can achieve the desired bandwidth if:

*i

I

aKR τ= *

iaI BW2RK ⋅⋅= π

• EXERCISE: Determine what happens if the zero-pole cancellation is not exact

Page 49: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 49

CURRENT LOOP SYNTHESIS

• Design the current regulator with bandwidth equal to 300 Hz for a DC machine having the following parameters

EXAMPLE

Ω= m100Ra

mH3La =

Hz300BW *i =

6549.5KP =

5.188KI =V10Vdc =

)(@V5E nω=

Page 50: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 50

CURRENT LOOP SYNTHESIS EXAMPLE

Current response at zero speed

Page 51: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 51

CURRENT LOOP SYNTHESIS

• The back-emf will act as a slowly changing (mechanical time constant) disturbance to the current regulator.

• It will not affect the zero steady-state error. • If there is a back-emf, the effective voltage acting

to change the current is ∆v=(va-e). • Therefore, the regulator (in particular the integral

term) will have to produce a bigger voltage reference va

*, in order to obtain the same ∆v. • This affects the transient response.

INCLUDING THE EFFECT OF THE BACK-EMF

Page 52: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 52

CURRENT LOOP SYNTHESIS INCLUDING THE EFFECT OF THE BACK-EMF

Current response at rated speed

Page 53: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 53

CURRENT LOOP SYNTHESIS

• In order to improve the transient response, we can add a feed-forward term, based on an estimate of the back-emf

INCLUDING THE EFFECT OF THE BACK-EMF

• Even a rough estimate can significantly improve the transient

1/Ra(1+sτa)

iava

e

+ -

va*

+ -

ia*

ia

1Ki(1+Kp/Kis)s

ωrKee^+

ωr ^

Ke ^

Page 54: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 54

CURRENT LOOP SYNTHESIS INCLUDING THE EFFECT OF THE BACK-EMF

Current response with estimated back-emf FF compensation

Page 55: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 55

CURRENT LOOP SYNTHESIS

• At high speeds, expecially if fast transients are required, the reference voltage va

* can exceed the maximum feasible voltage va

* = Vdc.

DC-DC CONVERTER NON LINEAR MODEL

Page 56: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 56

CURRENT LOOP SYNTHESIS DC-DC CONVERTER NON LINEAR MODEL

• In this case, we need to consider the non-linear model of the DC-DC converter.

1/Ra(1+sτa)

iava

e

+ -

va*

+ -

ia*

ia

Ki(1+Kp/Kis)s

ωrKee^+

ωr ^

Ke ^

Page 57: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 57

CURRENT LOOP SYNTHESIS

• When saturation is considered, a significant oscillation in the current appears.

INTEGRATOR TERM WINDUP

Page 58: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 58

CURRENT LOOP SYNTHESIS

• The oscillation is originated by the “integrator windup”.

• Initially, the output is saturated (current rises linearly, not with desired BW).

• When the error is zero, the output is still saturated.

• It takes time to exit saturation, which causes the oscillation.

INTEGRATOR TERM WINDUP

Page 59: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 59

CURRENT LOOP SYNTHESIS

• This phenomenon can be avoided by using a PI regulator scheme with “anti-windup”

ANTI-WINDUP

1/Ra(1+sτa)

iava

e

+ -

va*

+ -

ia*

ia

Ki(1+Kp/Kis)s

ωrKee^+

ωr ^

Ke ^

va,presat*

Kaw

+-

-

Page 60: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 60

CURRENT LOOP SYNTHESIS

• The response is still dominated by the saturation, but no oscillation occurs.

ANTI-WINDUP

PAW K

1K =

Page 61: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 61

CURRENT LOOP SYNTHESIS COMPLETE SCHEME FOR CURRENT REGULATOR

va*

+ -

ia*

ia

Ki(1+Kp/Kis)s

e^+

va,presat*

Kaw

+-

-

• The scheme, made of PI regulator, FF disturbance compensation and anti-windup, is used also for all other control loops.

Page 62: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 62

CURRENT LOOP SYNTHESIS – DIGITAL DOMAIN

• The scheme, made of PI regulator, FF disturbance compensation and anti-windup, is used also for all other control loops.

Page 63: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 63

MOTION CONTROLLER REFERENCE MODEL FEEDFORWARD

The reference model is used to enhance transient response.

Page 64: Emmc Steps Daac Lesson2 (Rev0)

Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 64

DC DRIVE WITH FIELD WEAKENING COMPLETE SCHEME

Field weakening control scheme

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Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 65

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