stosic dino pmsm report

7/24/2019 Stosic Dino PMSM Report

1/13

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

[email protected]


2/13

2

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)


3/13

3

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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4

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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5

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


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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7

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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8

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


p(1,2,3) u(1,2,3)


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]


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.


9/13

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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.


10/13

10

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


p(1,2,3) u(1,2,3)


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]


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


p(1,2,3) u(1,2,3)


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]


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.

stosic dino pmsm report

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