Lecture 2: Scalar-Controlled Induction Motor Drive
ELEC-E8402 Control of Electric Drives and Power Converters
Marko Hinkkanen
Spring 2019
1 / 14
Learning Outcomes
After this lecture and exercises you will be able to:
I
Express the dynamic inverse-Γ model in synchronous coordinates
I
Calculate steady-state operating points and
draw the corresponding vector diagrams
I
Explain the operating principle of the scalar control
2 / 14
Outline
Motor Model
Scalar Control
3 / 14
Model in Stator Coordinates
I
Voltage equations
u ss
=
Rs i ss
+
dψ ss
u sR = RR i sR +
I
dt
dψ sR
dt
i ss
− jωm ψ sR = 0
Flux linkages
u ss
Rs
dψ ss
dt
RR
Lσ
i sM
LM
dψ sR
dt
i sR
jωm ψ sR
ψ ss = Lσ i ss + ψ sR
ψ sR = LM (i ss + i sR )
I
Steady state: d/dt = jωs
4 / 14
Model in Synchronous Coordinates
I
Synchronous (dq) coordinates rotate at
the angular speed ωs
I
Coordinate transformation i ss = i s ejϑs ,
where no superscript is used in
synchronous coordinates
I
Voltage equations become
u s = Rs i s +
dψ s
+ jωs ψ s
dt
dψ R
u R = RR i R +
+ jωr ψ R = 0
dt
I
q
β
d ωs
ϑs
ωm
α
Angular speed of the coordinate system
I
I
ωs with respect to the stator
ωr = ωs − ωm with respect to the rotor
5 / 14
Model in Synchronous Coordinates
I
Voltage equations
dψ s
+ jωs ψ s
dt
dψ R
u R = RR i R +
+ jωr ψ R = 0
dt
u s = Rs i s +
I
Flux linkages
is
Lσ
iR
iM
dψ s
dt
LM
dψ R
dt
ψ s = Lσ i s + ψ R
ψ R = LM (i s + i R )
I
Steady state: d/dt = 0
6 / 14
Power Balance
3
3
3
dWf
ωm
Re {u s i ∗s + u R i ∗R } = Rs |i s |2 + RR |i R |2 +
+ TM
2
2
2
dt
p
I
Electromagnetic torque
TM =
I
n
o
3p
Im i s ψ ∗s
2
Rate of change of the magnetic field energy
dψ
dψ
1
dWf
3
3 d 1
= Re i ∗s s + i ∗R R =
Lσ |i s |2 + LM |i M |2
dt
2
dt
dt
2 dt 2
2
is zero in the steady state
7 / 14
Vector Diagram: Currents and Flux Linkages
is
I
Airgap and leakage flux
paths are sketched
I
All vectors are constant in
synchronous coordinates
in the steady state
(but the rotor slips at −ωr )
ψs
ψσ
ψR
iR
8 / 14
Steady-State Torque
I
TM
Torque in the steady state
Tb
2Tb
TM =
ωr /ωrb + ωrb /ωr
I
Breakdown torque
Tb =
I
3p LM
ψs2
2 LM + Lσ 2Lσ
0
ωrb
ωr
Breakdown slip
ωrb =
RR
σLM
where σ =
Lσ
LM + Lσ
See the derivation in Exercise 1.
9 / 14
Outline
Motor Model
Scalar Control
10 / 14
Stator Voltage vs. Stator Frequency
I
I
Steady-state stator voltage
u s = Rs i s + jωs ψ s
us /umax
ψs /ψsN
Approximate magnitude
1
ψs ≈ ψsN
us = umax
ψs ∝ 1/|ωs |
us = |ωs |ψs
0.5
us ≈ |ωs |ψsN
where us = |u s | and ψs = |ψ s |
I
Maximum voltage is limited
us < umax
0
0
0.5
1
1.5
2 ωs /ωsN
11 / 14
Scalar Control or Constant-Volts-per-Hertz Control
I
Based on the steady-state equations
I
Supply frequency ωs,ref corresponds
to the desired rotor speed
I
Some speed error due to the slip
(can be partly compensated for)
I
Slow or oscillating dynamics
I
Torque cannot be controlled
I
Current cannot be limited
I
For simple applications
Udc
ψs,ref
ωs,ref
Scalar
control
u ss,ref
PWM
M
us,ref = ωs,ref ψs,ref (+Rs is compensation)
R
ϑs = ωs,ref dt
u ss,ref = us,ref ejϑs
12 / 14
TM /TN
Breakdown
torque Tb
Example
load torque TL
2
Steady-state
operating point
−2
−1
0
1
2
ωm /ωmN
−1
−2
ωs = ωsN
ψs = ψsN
ωs = 0
ψs = ψsN
13 / 14
TM /TN
Breakdown
torque Tb
Continuous
weakened-field
operating area
2
ωs = 2ωsN
ψs = ψsN /2
−2
Continuous
full-field
operating area
0
1
2
ωm /ωmN
−2
ωs = ωsN
ψs = ψsN
ωs = 0
ψs = ψsN
14 / 14