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DC DrivesTRANSCRIPT
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
+Φ
−= ω
T
ωr
Tn
Tstart
ω0
a
aeTstart R
VKT Φ=
eE
a0 K
VΦ
=ω
Dynamic Analysis and Control of AC Machines - Lesson 2 Pagina 14
VOLTAGE REGULATION
a
aeTr
a
2eET
RVK
RKKT Φ
+Φ
−= ω
a
aeTstart R
VKT Φ=
eE
a0 K
VΦ
=ω
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
+Φ
−= ω
a
aeTstart R
VKT Φ=
eE
a0 K
VΦ
=ω
• 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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