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DC Drives

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DC MACHINES

DYNAMIC ANALYSIS AND CONTROL OF AC MACHINES

ERASMUS MUNDUS MASTER COURSE on SUSTAINABLE TRANSPORTATION AND

ELECTRIC POWER SYSTEMS

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

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

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

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=

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

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

TYPES OF DRUM WINDINGS

Lap winding Wave winding

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

LAP WINDING

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

WAVE WINDING

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

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 ×⋅=

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

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

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)

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

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

DYNAMIC ANALYSIS OF DC MACHINES

DYNAMIC ANALYSIS AND CONTROL OF AC MACHINES

ERASMUS MUNDUS MASTER COURSE on SUSTAINABLE TRANSPORTATION AND

ELECTRIC POWER SYSTEMS

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+=−

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+

+

-

-

-

-

+ -

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=

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)

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 =

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

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 +−

∆≅ −

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

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

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

DC MACHINE TRANSIENTS VOLTAGE STEP

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

DC MACHINE TRANSIENTS LOAD STEP

DC DRIVES

DYNAMIC ANALYSIS AND CONTROL OF AC MACHINES

ERASMUS MUNDUS MASTER COURSE on SUSTAINABLE TRANSPORTATION AND

ELECTRIC POWER SYSTEMS

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)

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

\

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

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

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

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 =

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.

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

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.

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.

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

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:

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)

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

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 =≤

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

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 =

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

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

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=

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

CURRENT LOOP SYNTHESIS EXAMPLE

Current response at zero speed

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

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

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 ^

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

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

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 ^

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

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

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

+-

-

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 =

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.

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.

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.

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

DC DRIVE WITH FIELD WEAKENING COMPLETE SCHEME

Field weakening control scheme

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

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