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UN
IVERS
ITYOFRIJE
KA
FACULTYOF EN
GINE
ERING
Student: Stoi Dino
Index number: 0069051394
Lab Exercise Control of Electric Machines
powered by
Rijeka, January 2016
Topic:
Synchronous Machines
Ass. Prof. Dipl. Ing. Dr. Wolfgang Gruber
Johannes Kepler University Linz
Institute for Electrical Drives and Power Electronics
Altenbergerstr. 69, A-4040 Linz, Austria
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3.7 Lab Exercise Control
Use the 3-phase motor model in all the following simulations to create the proper control
scheme.
My parameter percentage changes are:
Student
numberStudent name
PM Flux
(Lmd*If)
Inductivity
(Lsd/Lsq)
Resistance
(Rs)
Moment of
Inertia (Jall)
Battery
Voltage
(UDC)
56 Stoi Dino 10 % 20% 0 % 0 % -20%
Table 2: Percentage changes
of 3-phase motor model parameters.
3.7.1 Feedback voltage control
Use the phase angle of the stator flux space vector as phase angle feedback. Add (for a
constantpmax) different offsets to the phase angle feedback. Check the influence of this offset
angle in the phase currents. For which offset do you get optimal field orientated control with
maximum efficiency in steady state? Try to identify this offset angle also with the voltage
space vector diagram of the PMSM.
u_ind
60/2/pi
rad/s -> rpm
0.6
p_max
n
currents
abc
thetadq0
abc_to_dq0
Transformation1
abc
thetadq0
abc_to_dq0
Transformation
Voltages
p_max [0 .. 1]
offs et [rad]
theta_el [rad]
[p1 p2 p3]
Voltage Generation
(3.6+88.5).*pi/180 %5.45*pi/180
Us_offset
Torque
-88.5*pi/180
Rotor angle correction
p(1,2,3) u(1,2,3)
Power Electronics
M_an [mNm]
theta_el [rad]
omega_mech [rad/s]
Mechanic Model
u(1,2,3) [V]
theta_el [rad]
omega_mech[rad/s]
i(1,2,3) [A]
M [mNm]
uind(1,2,3) [V]
ECI2442 - 3 phase model
Add1
Figure 1: Feedback voltage control scheme (Simulink scheme:
feedback_voltage_control.mdl)
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pmax 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Optimum offset angle
offset()
(isd= 0 A)
2.4 2.7 3.1 3.3 3.4 3.6 3.8 4 4.3 4.6
Stator voltage angle
(offset+ 88.5) ()90.9 91.2 91.6 91.8 91.9 92.1 92.3 92.5 92.8 93.1
n(rpm) 241.5 507.5 776.5 1048 1320 1591.5 1862 2128 2397 2663.5
Tav(mNm) 18 24 28 31 33 36 39 43 47.5 52.5
Umax(V) 0.96 1.92 2.88 3.84 4.8 5.76 6.72 7.68 8.64 9.6
isq(A) 0.365 0.48 0.56 0.62 0.66 0.72 0.79 0.87 0.96 1.06
Cos 1 1 1 1 1 1 1 1 1 1
0.8661 0.9227 0.9411 0.9527 0.9599 0.9645 0.9550 0.9561 0.9583 0.9593
Table 1: Simulation results.
Motor efficiency formula:
cos45
cos2
90cos
2
360
2
cos3)statesteady(sqmax
av
maxmax
av
maxmax
avav
iU
nT
IU
nT
IU
nT
UI
T
A constant value of 0.6 is chosen for pmaxand several different offsets are added to the phase
angle of feedback signal theta_el. Offset value that has been added to the phase angle
feedback signal theta_el was 3.6 and overall value of phase angle feedback was 3.6 +
88.5 (fixed value during the experiments) = 92.1 which is approximately equal to 1.6074
radians. Direct component of stator current isdmust be equal to zero to achieve optimum field
oriented control, but this condition is not enough to achieve maximum efficiency of a motor
in steady state and that can be seen in Table 3.
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Offset value of 3.6 and pmax = 0.6 ensure optimal field oriented control with maximum
efficiency (= 0.9645) in steady state. Under these conditions, direct component of stator
current isdresponse is shown on the following figure and it is represented by yellow curve:
Figure 2: Current responses isd(yellow curve), isq(purple curve) and i0(cyan) for pmax= 0.6
and offset= 3.6.
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3.7.2 Field oriented dq-current control
Design a simple dq-current control. For this reason use PI-controllers. Normally only two
motor phases are measured, so assume that only two current signals (i1and i2) are available.
Adjust the current controller in such a way, that an iq-jump from 0A to 8A is reached in 0,01s
without overshoot and with an maximum deviation of 5% can be achieved.
Which rotational speeds can be reached for a isq-current of 0.4A, 0.8A, 1.2A and 2A (forisd= 0A in all cases)? What is the motor efficiency in these points of operation?
voltages1 voltages
u_ind
60/2/pi
rad/s -> rpm
0.5
p_max
n
8
iq_set [A]
0
id_set [A]
dq0
thetaabc
dq0_to_abc
Transformation currents
abc
thetadq0
abc_to_dq0
Transformation1
abc
thetadq0
abc_to_dq0
Transformation
0
U0 Torque
Terminator
-88.5*pi/180
Rotor angle correction
p(1,2,3) u(1,2,3)
Power Electronics0.5
PWM_offset
M_an [mNm]
theta_el [rad]
omega_mech [ rad/s]
Mechanic Model
u(1,2,3) [V]
theta_el [rad]
omega_mech[rad/s]
i(1,2,3) [A]
M [mNm]
uind(1,2,3) [V]
ECI2442 - 3 phase modelDemux
id_soll
iq_soll
id_ist
iq_ist
Ud_lim
Uq_lim
Current Control
Add1
Add
Figure 3: Field oriented dq-current control system (Simulink
scheme:current_control_isq_jump.mdl)
Chosen parameters of current controller (Current control) for currents isdand isqare:
Kp,isq= 2.5
Ki,isq= 10
Kp,isd= 2.5Ki, isd= 10
1
y
Zero-Order
Hold
2.5
P-Anteil
1
s
10
I-Anteil
2
disable
1
e
Figure 4: PI controllers for isdand isq-current.
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Figure 5: isq-jump (purple) from 0 to 8 A in less than 10 miliseconds without overshoot.
Figure 6: isqcurrent (purple) deviation after jumping from 0 to 8 A.
Deviation: (8 7.9) / 8 = 0.0125 = 1.25 %
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a) n= 315 rpm at isq= 0.4 A b) n= 1920 rpm at isq= 0.8 A
(Simulink schemes: current_control_isq04.mdl and current_control_isq08.mdl)
c) n= 2665.5 rpm at isq= 1.2 A d) n= 2665.5 rpm at isq= 2 A
(Simulink schemes: current_control_isq12.mdl and current_control_isq2.mdl)
Figure 7: Achieved rotational speeds a), b), c) and d) at different values of isqunder
condition isd= 0 for all cases.
isq 0.4 0.8 1.2 2
n (rpm) 315 1920 2665.5 2665.5
Tav(mNm) 20 40 52.5 52.5
Umax(V) 9.6 9.6 9.6 9.6
cos 1 1 1 1
0.1145 0.6981 0.8481 0.5088
Table 2: Motor efficiency for various points of operation (isd= 0 A).
Formula used for motor efficiency calculation is:
cos2
90cos
2
90cos
2
360
2
cos3sqmax
av
maxmax
av
maxmax
avav
iU
nT
IU
nT
IU
nT
UI
T
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3.7.3 Field oriented speed control with maximum efficiency
Extend the model of the dq-current control with an outer speed controller (simple PI-
controller. Limit the output of the current controller to 8A.
Simulate rotational speed jumps from 1000 rpm up to 3000 rpm.
Try to get a very fast settling time of the real rotational speed by tuning the control parameters
accordingly. What rise time is achievable? What is the efficiency of the drive at 3000 rpmnow?
u_lim -> p
voltagesu_lin
u_ind
60/2/pi
rad/s-> rpm
0.5
p_max
3000
n_set2 [rpm]
n
0 id_set [A]
dq0
thetaabc
dq0_to_abc
Transformation0
disable
currents
abc
thetadq0
abc_to_dq0
Transformation1
abc
thetadq0
abc_to_dq0
Transformation
0
U0
Torque
Terminator
e
disable
y
Speed controller
-88.5*pi/180
Rotor angle correction
p(1,2,3) u(1,2,3)
Power Electronics0.5
PWM_offset
M_an [mNm]
theta_el [rad]
omega_mech [rad/s]
Mechanic Model
Id
Iq
id_lim
iq_lim
Limitation
u(1,2,3) [V]
theta_el [rad]
omega_mech[rad/s]
i(1,2,3) [A]
M[mNm]
uind(1,2,3) [V]
ECI2442 - 3 phase model
Demux
id_soll
iq_soll
id_ist
iq_ist
Ud_lim
Uq_lim
Current Control
Add2
Add1
Add
Figure 8: Extended model of the dq-current control with an outer speed controller
(Simulink scheme: speed_control.mdl).
Speed controller parameters are:
1
y
Zero-Order
Hold
0.04
P-Anteil
1
s
0.003
I-Anteil
2
disable
1e
Figure 9: Speed controller parameters.
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Achieved rise time trfor speed reference jump from 0 to 3000 rpm:
Figure 10: Speed response after speed controller parameters tuning
(Kp= 0.04 andKi= 0.003).
Achievable speed response rise time tr=0.44 seconds.
Fast settling time tsafter achieving a speed of 3000 rpm:
Figure 11: Settling time tsfor speed response (Kp= 0.04 andKi= 0.003).
Achieved settling time is: 0.465 tr= 0.465 0.44 = 0.025 s = 25 ms.
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Drive efficiency at speed of 3000 rpm:
cos290cos
290cos
23
60
2
cos3sqmax
av
maxmax
av
maxmax
avav
iU
nT
IU
nT
IU
nT
UI
T
12.16.945
rpm3000Nm061.0
AV
1090.1 (Paradox!)
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3.7.4 Field oriented speed control with field weakening operation
What happens if field weakening is applied to the current and speed controlled system at 3000
rpm, the load torque stays constant and the drive has not reached its maximum power? Verify
your opinion with simulation results.
Is it possible for the speed to change in this case? What must be fulfilled for a speed change?
Try to speed up the motor to a speed of 4000 rpm by the help of field weakening.
Which id-current is necessary in this case? Now compute the efficiency of the drive chain inthis point of operation.
Load torque dependance on speed was neglected and friction was also neglected. Instead of
variable load torque inside of mechanical model a constant load torque was implemented
(Tload= 52 mNm).
2
omega_mech [rad/s]
1
theta_el [rad]
2
pz
1/1000
mNm
mod
Math
Function
1
s
Integrator1
1
s
Integrator
1/Jall
Gain
52
Constant_Load_Ml
[mNm]
2*pi
Constant
Add
1
M_an [mNm]
Omega_mech [rad/s]
Theta_el
Figure 12: Modified mechanical model.
Mechanical model shown on Figure 12 will be used in all cases.
Case #1: applied field weakening (isdreference is set to -3.9 A) at 0.5 seconds before reaching
speed of 3000 rpm
u_lim -> p
voltagesu_lin
u_ind
60/2/pi
rad/s -> rpm
0.5
p_max
3000
n_set2 [rpm]
n
i_sd
dq0
thetaabc
dq0_to_abc
Transformation0
disable
currents
abc
thetadq0
abc_to_dq0
Transformation1
abc
thetadq0
abc_to_dq0
Transformation
0
U0
Torque
Terminator
e
disable
y
Speed controller
-88.5*pi/180
Rotor angle correction
p(1,2,3) u(1,2,3)
Power Electronics0.5
PWM_offset
M_an [mNm]
theta_el [rad]
omega_mech [rad/s]
Mechanic Model
Id
Iq
id_lim
iq_lim
Limitation
u(1,2,3) [V]
theta_el [rad]
omega_mech[rad/s]
i(1,2,3) [A]
M[mNm]
uind(1,2,3) [V]
ECI2442 - 3 phase model
Demux
id_soll
iq_soll
id_ist
iq_ist
Ud_lim
Uq_lim
Current Control
Add2
Add1
Add
Figure 13: Speed and current controlled system in case #1
(field_weakening_speed_and_current_controlled_system_1.mdl).
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Figure 14. Speed response in case #1
(nref= 3000 rpm, isd,ref= -3.9 A at t= 0.5 sec, constant load torque of 52 mNm,Kp= 0.04,Ki=
0.003, current controller output limits 8A).
At time instant t= 0.5 seconds, field weakening command (isdreference jumps from 0 to -3.9
A) was applied and after that motor achieves speed of 3000 rpm. In this case speed changes,
i.e. increases with negative values of isdand if isdis big enough, motor can achieve speed of
3000 rpm. In other words, for speed change isdreference must be different than zero.
Case #2: motor speed up to 4000 rpm by the help of field weakening
u_lim -> p
voltagesu_lin
u_ind
60/2/pi
rad/s-> rpm
0.5
p_max
4000
n_set2 [rpm]
n
-11.5
i_sd_set [A]
dq0
thetaabc
dq0_to_abc
Transformation1
disable
currents
abc
thetadq0
abc_to_dq0
Transformation1
abc
thetadq0
abc_to_dq0
Transformation
0
U0
Torque
Terminator
e
disable
y
Speed controller
-88.5*pi/180
Rotor angle correction
p(1,2,3) u(1,2,3)
Power Electronics0.5
PWM_offset
M_an [mNm]
theta_el [rad]
omega_mech [rad/s]
Mechanic Model
Id
Iq
id_lim
iq_lim
Limitation
u(1,2,3) [V]
theta_el [rad]
omega_mech[rad/s]
i(1,2,3) [A]
M [mNm]
uind(1,2,3) [V]
ECI2442 - 3 phase model
Demux
id_soll
iq_soll
id_ist
iq_ist
Ud_lim
Uq_lim
Current Control
Add2
Add1
Add
Figure 15: Speed and current controlled system in case #1(field_weakening_speed_and_current_controlled_system_2.mdl).
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Figure 15: Speed response in case #2
(nref= 4000 rpm, isd,ref= -11.5 A at t= 0.5 sec, constant load torque of 52 mNm, Kp= 0.04,Ki
= 0, current controller output limits 15A).
In this case, a greater current isdis needed to speed up motor form 0 to 4000 rpm.