emmc steps daac lesson5
TRANSCRIPT
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VECTOR CONTROL OF INDUCTION MACHINES
DYNAMIC ANALYSIS AND CONTROLOF AC MACHINES
ERASMUS MUNDUS MASTER COURSE on
SUSTAINABLE TRANSPORTATION AND
ELECTRIC POWER SYSTEMS
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Outline
Dynamic Analysis and Control of AC Machines - Lesson 5 Pagina 2
Principles of Vector Control for induction machines(steady state analysis)
Vector control problem and solutions:
Indirect VC and Direct VC Dynamic behaviour
Flux Observers
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Classical IM equivalent circuit (steady-state)
Dynamic Analysis and Control of AC Machines - Lesson 5
3
The circuit shows the two leakage inductances
The circuit gives explicit account for the air-gap flux
r r e
r r
e
I E P
I sr P
T
33 2
Pagina 3
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Generalized IM equivalent circuit (steady-state)
Dynamic Analysis and Control of AC Machines - Lesson 5
4
a is a generalized constant
a=N s /Nr yields the classical equivalent circuit
r E a~
Pagina 4
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Rotor flux IM equivalent circuit (steady-state)
Dynamic Analysis and Control of AC Machines - Lesson 5
5
r E a~
r
m
L L
a
Only one leakage inductance
The circuit gives explicit account for the rotor flux
Pagina 5
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Torque control (steady-state)
Dynamic Analysis and Control of AC Machines - Lesson 5
6
Stator current can be decomposed into two components.
The two components are in quadrature.
One component is proportional to (establishes) rotor flux. The other component, given the rotor flux, is proportional
to the torque.
s sT s I j I I ~~~
Pagina 6
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Rotor flux component of stator current
Dynamic Analysis and Control of AC Machines - Lesson 5
7
me
r
m
r
mr
m
r r
m
s L j E
jX E
X L
L
j
E L L
I
~~~
~
r er j E ~~
smr I L ~~
Pagina 7
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Torque component of stator current
Dynamic Analysis and Control of AC Machines - Lesson 5
8
sT r
mr I L
L I
~~
)()(33 sT r
m sme
er r
e
I L L
I L P
I E P
T
s sT r
m I I L
L P T 2
3
Pagina 8
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Slip frequency equation (steady state)
Dynamic Analysis and Control of AC Machines - Lesson 5
9
r r m
r
r
r
m
r r m
sT E r s
L L
sr
L
L
E L L
I ~
~~
2
2
s
sT
r
r e I
I Lr
s
smer I L j E ~~
ser
r sT I sr
L j I
~~ Fixing the current components
fixes also the slip frequency.
In fact, fixing the synchronousfrequency and the rotor fluxdefines the mechanicalcharacteristic. Therefore, fixingthe torque defines also the slip.
Pagina 9
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Phasor diagram (steady state)
Dynamic Analysis and Control of AC Machines - Lesson 5
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Stator current can be viewed as the phasor sum of twocomponents in quadrature.
Is is proportional to rotor flux.
IsT is proportional to the torque (and is the stator image ofrotor current).
Pagina 10
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Phasor diagram (steady state)
Dynamic Analysis and Control of AC Machines - Lesson 5
11
+
Pagina 11
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Vector control for IM
Dynamic Analysis and Control of AC Machines - Lesson 5
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To achieve vector control for IM means to determine thetwo components of the stator current I s and I sT which areproportional to the rotor flux and the torque.
As a consequence, also the amplitude Is and angle (andfrequency , of course) of the stator currents are determined.
Chosing the currents means also that the equivalent circuitcan be completely solved: it is possible to determine whichis the voltage that allow to achieve the required current.
Finally, also the slip frequency (and therefore themechanical speed) is determined by the currents.
Pagina 12
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From phasors to synch. reference frame
Dynamic Analysis and Control of AC Machines - Lesson 5
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The synchronous reference frame is obtained from thearbitrary reference frame, when: )()( t t e
and the d-axis is aligned with the rotor flux: 0qr r dr
The phase relations that hold for phasors are still valid whenwe consider rotating vectors.
Pagina 13
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From phasors to synch. reference frame
Dynamic Analysis and Control of AC Machines - Lesson 5
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It is possible to define the equivalent circuit in thesynchronous reference frame, valid in steady state:
Pagina 14
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Vector control problem for IM
Dynamic Analysis and Control of AC Machines - Lesson 5
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Knowing the desired rotor flux and torque, it is possible todefine the desired stator current rotating vector.
By using the synchronous reference frame, aligned with therotor flux, the q- and d- axes components of the desiredcurrent are identifed.
PROBLEM
How do we know where the rotor flux is?In PM machines measuring rotor position was enough;
for IM slip must be considered!Pagina 15
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Indirect Vector Control (VSI inverter)
Dynamic Analysis and Control of AC Machines - Lesson 5
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*
**
ds
qs
r
r e I
I
L
r s
Estimatedvalues!!!
Rotor flux position
Pagina 17
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Indirect Vector Control (CSI inverter)
Dynamic Analysis and Control of AC Machines - Lesson 5
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*
*
*
ds
qs
r
r e I
I
Lr s
Estimatedvalues!!!
CR network =derivative operator
The derivation isneeded to add aterm whichcompensates forphase anglevariations
Pagina 18
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Indirect Vector Control (CSI inverter)
Dynamic Analysis and Control of AC Machines - Lesson 5
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be detected. Therefore, the new reference will be I s * instead ofIqds2 *, with a phase error . If not compensated, this errorwill still go to zero eventually, but the dynamic perfomanceswill be deteriorated.
When there is a change inthe reference from I qs1 * toIqs2 *, the resolver blockwill identify the new
amplitude, while the slipcalculator block will helpdetermine the frequency.
The phase change will not
Pagina 19
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Direct Vector Control (VSI inverter)
Dynamic Analysis and Control of AC Machines - Lesson 5
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Rotor flux position
Rotor flux position isidentified through the blockCFO (flux observer).
The same block estimatesalso the rotor fluxamplitude, which is usefulto estimate also the torque.
s sT r
m
I I L L
P T
2
3
qsr r
m I L L
P T 23
Pagina 20
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Direct Vector Control (CSI inverter)
Dynamic Analysis and Control of AC Machines - Lesson 5
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Pagina 21
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Transient analysis of Vector Controlled IM
Dynamic Analysis and Control of AC Machines - Lesson 5
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Machine equations in the synchronous reference frame
)(23
dsqr qsdr r
m ii L L
P T
dseqsqs sqs dt d
ir v
qsedsds sds dt d
ir v
dr r eqr qr r dt d
ir )(0
qr r edr dr r dt
d ir )(0
qr mqs sqs i Li L
dr mds sds i Li L
qr r qsmqr i Li L
dr r dsmdr i Li L
Pagina 22
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Transient analysis of Vector Controlled IM
Dynamic Analysis and Control of AC Machines - Lesson 5
qr r edr dr r dt
d ir )(0
dr r eqr qr r dt d
ir )(0
23
Machine equations in the synchronous reference frame withthe d-axis aligned with rotor flux ( r= dr )
= 0
)(23
dsqr qsdr r
m ii L L
P T
dseqsqs sqs dt d
ir v
qsedsds sds dt d
ir v
qr mqs sqs i Li L
dr mds sds i Li L
qr r qsmqr i Li L
dr r dsmdr i Li L
Pagina 23
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Transient analysis of Vector Controlled IM
Dynamic Analysis and Control of AC Machines - Lesson 5
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Equations affected by the condition of rotor flux alignment:
qsdr r
m i L L
P T 23
dr r eqr r ir )(0
dr dr r
dt
d ir 0
0 qr r qsm i Li L
(1)
(2)
(3)
(4)
From (3) we have:
qs
r
mqr i
L
Li
extension of:
r m
r sT I L
L I
~~
and consistent with (2) insteady state:
0dr I Pagina 24
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Transient analysis of Vector Controlled IM
Dynamic Analysis and Control of AC Machines - Lesson 5
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Equations affected by the condition of rotor flux alignment:
qsdr r
m i L L
P T 23
dr r eqr r ir )(0
dr dr r
dt
d ir 0
0 qr r qsm i Li L
(1)
(2)
(3)
(4)
From (1) we have:
dr
qsm
r
r e
i L
Lr
s
extension of:
dr
qr r er e
ir s
or also:
s
sT
r
r e I
I Lr
s
Pagina 25
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Rotor current on d-axis
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Combining (2) and (5) :
From (2) : 0dr I
during transients.
dr r dsmdr i Li L r
dsmdr dr L
i Li
(5)
dr dr r
dt
d ir 0 in steady state.
)()(
)( si sLr Lr
s dsr r
mr dr
From :
Deriving the rotor flux equation and substituting into (2) :
)()()( si s L si sLr dsmdr r r
Following a step change in i ds , the rotor fluxwill reach its steady state value L mids with afirst order transient defined by the rotor timeconstant r = L r /r r .
A transient current on rotor d-axis may existonly if there is a change in the stator d-axiscurrent. After a first order transient defined by
r = L r /r r , idr will go to zero. Pagina 26
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Indirect control implementation (i ds *,iqs *)
Dynamic Analysis and Control of AC Machines - Lesson 5
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)()1(
)( si s
L s dsr
mdr
dr
qsm
r
r e
i L
Lr
s
*
**
11
1
dsr
qs
r
e
i s
i s
Pagina 28
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Indirect control implementation (i ds *,iqs *)
Dynamic Analysis and Control of AC Machines - Lesson 5
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Assuming infinte BW of the current controller, a step change in i ds * will cause firstorder transient in the rotor flux dr , with a final value proportional to the current.
)()1(
)( si s
L s ds
r
mdr
*
**
11
1
dsr
qs
r e
i s
i s
Steady state Dynamic
Pagina 29
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Indirect control implementation ( dr *,iqs *)
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Using dr * as the controlledvariable, instead of i ds *, means thatnow the controller has to calculateids * using (1) and therefore willincorporate a compensation forthe rotor transient behaviour.
)()
1(1
)( ** s s L
si dr r m
ds
*
**
11
dr m
qs
r e
L
i s
(1)
Pagina 30
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Flux Observers for Direct Vector Control
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Inputs to Flux Observers aremeasurements at machineterminals (typically voltages andcurrents). Using the equations ofthe dynamic model for the IM it is
possible to estimate rotor flux inqd stationary reference frame.
Finally, from these components, itis possible to determine rotor fluxangle and magnitude.
Flux Observers can be open loopor closed loop
Pagina 31
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Flux Observers Voltage model
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Three parameters are needed: r s , Ls , L r /Lm
It is difficult to calculate the integral (1) at low frequencies (open loop observer) If the rotor has closed slots, then L s , even if it is a leakage inductance, will
depend on i r
In general, this observer gives good results at medium and high speeds.
dt r S qds sS qdsS qds iv
S
qdsr
m s
S qds
m
r S qdr
L
L L
L
Li
2
(1)
vqdsS
iqdsS
(r s)^
(Ls-Lm2/Lr )
^
(Lr /Lm )^
( qdsS )^
( qdsS )^
( qdr S )^
Pagina 33
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Flux Observers Current model
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Variables are expressed in the rotor qd reference frame (superscript R )
Rqdr
Rqdr r dt
d r i 0
Rqdr
r
r Rqdsm
Rqdr dt
d r
L L i
R
qdr r
R
qdsm
R
qdr L L ii
Rqdr
r
Rqdr dt
d r
i 1
Rqds
r
m Rqdr s
L i )
1(
Pagina 34
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Flux Observers Current model
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Two parameters are needed: r , Lm
Mechanical position is required in order to identify the rotor reference frame This observer is sensitive to saturation
This observer is suitable also for low speeds
Rqds
r
m Rqdr
s
Li
)
1(
q r
iqdsS
KR (Lm)^/(1+ ^
s)
( qdr R)^ ( qdr
S )^iqdsR
KR-1
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Flux Observers Closed loop
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A closed loop form for the Flux Observer can be obtained by combining thevoltage model and the current model The gains of the PI regulator will determine a closed loop BW Below this BW, the output if the observer will follow the current model. Above the BW, the regulator will not have effect and therefore the output will
follow the voltage model
q r
iqdsS
KR (Lm)^/(1+ ^
s)
( qdr R)^
( qdr,IS)^
iqdsR
KR-1
vqdsS
iqdsS
(r s)^
(Ls -Lm2/Lr )
^
(Lr /Lm )^
( qdsS)^
( qdsS)^
( qdr,VS)^
PI
CURRENT MODEL VOLTAGE MODEL
Pagina 36