dynamic modeling, simulation and control of electric machines for
TRANSCRIPT
INTERNATIONAL JOURNAL OF CONTROL, AUTOMATION AND SYSTEMS VOL.1 NO.2 APRIL 2013
ISSN 2165-8277 (Print) ISSN 2165-8285 (Online) http://www.researchpub.org/journal/jac/jac.html
30
Farhan A. Salem*
Abstract- The mathematical models, corresponding simulink
models, analysis and control solutions of basic open loop
electric machines most used in mechatronics applications are
introduced; the introduced models are intended for research
purposes, as well as, for the application in educational process.
Index Terms—Mechatronics, Electric machine,
Modeling/Simulation.
I. INTRODUCTION
echatronics is defined as the synergistic integration of
sensors, actuators, signal conditioning, power electronics,
decision, control algorithms, computer hardware and software
to manage complexity, uncertainty, and communication in
engineered systems. The key element in mechatronics design
is the concurrent synergetic integration, instead of sequential,
analysis and optimization of these areas and balance between
modeling/analysis and experimentation / hardware
implementation, through the design process resulting in
products with more synergy [1].
Modeling, simulation, dynamics analysis and control of
electric machines most used for mechatronics motion control
applications are of concern, since the accurate control of
motion is a fundamental concern in mechatronics
applications, where placing an object in the exact desired
location with the exact possible amount of force and torque at
the correct exact time is essential for efficient system
operation, the accurate control of motion depends on many
factors including; the accuracy of applied control strategy
design, the accuracy of derived mathematical model, the
accuracy of interpreting simulation and analysis results. This
paper propose derivation of mathematical models, building
corresponding simulink models, dynamic analysis and
introduce control solutions of main DC machines used in
mechatronics applications. DC machines are electrical
machines that consume DC electrical power and produce
mechanical torque [2]. Due to precise, wide, simple, and
continuous control characteristics, small and large electric
machines are used in mechatronics applications, large electric
machines are used in machine tools, printing presses,
conveyor fans, pumps, hoists, cranes, paper mills, textile
mills, Small DC motors (in fractional power rating) are used
in control devices such as tacho-generators for speed sensing
and servomotors for positioning and tracking [3,4].
DC Machines can be classified according to the electrical
connections of the armature winding and the field windings,
the different ways in which these windings are connected
lead to machines operating with different characteristics.
The field winding can be either self-excited or separately-
excited, that is, the terminals of the winding can be
connected across the input voltage terminals or fed from a
separate voltage source. Further, in self-excited motors, the
field winding can be connected either in series or in parallel
with the armature winding. These different types of
connections give rise to very different types of machines.
Each electric machine is designed by a manufacturer to
operate in a certain range of voltages and currents, the
parameter quoted by the manufacturer is known as rating of
the machine.
The electric machines, considered in this paper are PMDC
motor, separately excited DC motor, armature controlled
DC motor, shunt DC motor and Series DC motor, for each
machine mathematical models are to be derived,
corresponding simulink models to be built and finally
control solutions are proposed.
II. BASIC EQUATIONS FOR MODELING ELECTRIC
MACHINES
Because of the ease with which they can be controlled,
systems of electric machines have been frequently used in
many applications requiring a wide range of motor speeds
and a precise output motor control [5,6]. The selection of
motor for a specific application is dependent on many
factors, such as the intention of the application, allowable
variation in speed and torque and ease of control, etc.
The dynamic equations of electric machines can be
derived, mainly, based on the Newton’s law combined
with the Kirchoff’s law. The fundamental system of
electromagnetic equations for any electric motor is given
by [7,8] Eq.(1)
( )
ks
s s s s
kR
s R R b m R
s s s R
R R R S
du R i j
dt
du R i j P
dt
L i L i
L i L i
(1)
Where : k the angular speed of rotating coordinate
system (reference frame), Depending on motor
construction (AC or DC), the method of the supply and
the coordinate system (stationary or rotating with the
rotor or stator flux) the above mentioned model becomes
transformed to the desirable form[9], the complement of
Eqs. (1), is equations describing mechanical part of
eclectic motor, without any load attached (that is total
Dynamic Modeling, Simulation and Control of Electric Machines for Mechatronics Applications
M
Farhan Atallah Salem AbuMahfouz , with Taif University, 888, Taif,
Saudi Arabia .He is now with the Department of Mechanical engineering
, Faculty of Engineering, Mechatronics Sec. and with Alpha Center for Engineering Studies and Technology researches (e-mail:
INTERNATIONAL JOURNAL OF CONTROL, AUTOMATION AND SYSTEMS VOL.1 NO.2 APRIL 2013
ISSN 2165-8277 (Print) ISSN 2165-8285 (Online) http://www.researchpub.org/journal/jac/jac.html
31
inertia of motor and load reduced to motor shaft), and given
by:
e Load
dJ T T
dt
(2)
In the aim of the synthesis of the electric machine control
system, control engineers most frequently use mathematical
models in form of transfer-function, the errors of the
parameters identification, nonlinearities and the temperature
influence (resistances of the windings) are usually omitted
in these models (motor and power converter). Thus, the
additional tests of the control system robustness should be
realized [9]. Electrical equivalent scheme of a DC motor is
shown in Fig. 1(a) , a nonlinear block diagram of a DC
motor is shown in Fig. 1(b).
Fig. 1(a) Electrical equivalent scheme of a DC motor
Fig. 1(b) A nonlinear block diagram of a DC motor
III. MODELING AND SIMULATION OF PMDC
MOTOR
In [1] Modeling, simulation and dynamics analysis issues of
electric motor, using different approaches and verification
by MATLAB/Simulink, are introduced. In [10] controller
selection and design for electric motor using different
control strategies and verification using
MATLAB/Simulink are introduced. Based on Eq.(1) and
last two references, the PMDC motor open loop transfer
function relating the input voltage, Vin(s), to the motor shaft
output angular velocity, ωm(s), given by Eq.(3), by
assuming that the armature inductance, La is low and can
be ignored (La =0) compared to the armature resistance, Ra.
Eq.(3) can be simplified to have form given by Eq.(4)
3
2
( )( )
( )
( )( ) ( )
t
speed
in a a m m t b
t
speed
a m a m m a a m t b
KsG s
V s L s R J s b K K
KG s
L J s R J b L s R b K K
( )( )
( )
( )1
1
t
speed
in a m a m t b
b t
a t b B
speed
a
a t b
KsG s
V s R J s R b K K
K K
R b K K KG s
sR Js
R b K K
4
The PMDC motor open loop transfer function without
any load attached relating the input voltage, Vin(s), to the
motor shaft output angle, θm(s), is given by Eq.(5), this
equation can be simplified to have form given by Eq.(6)
3 2
( )( )
( ) ( ) ( )
t
angle
in a m a m m a a m t b
KsG s
V s L J s R J b L s R b K K s
5
2
( )( )
( ) ( ) ( )
/( )
( ) 1
t
angle
in a m a m t b
t a a
in t b
m
m a
KsG s
V s R J s R b K K s
K R Js
V s K Ks s b
J R
6
A. Simulation of PMDC motor open loop system using
Simulink
The main, simulink models of PMDC are introduced in
[1] including models based on simplified models and for
Speed/time, Torque/time, Position/time and Current/time
curves are shown in Fig. 2
current ,id/dt
d2/dt2(theta) d/dt(theta)
Output angleOutput speed
sum
anlge
12
Vin
Torque.
0
Torque load
Motor4.mat
To File3
Motor3.mat
To File2
Motor2.mat
To File1
Motor1.mat
To File
Sum
Step,
Vin=12
Ra
Rresistance, Ra
-K-KtKb
Kb
1
s
Integrator1
1
s
Integrator,
1
s
Integrator
1/Jm
Inertia , 1/Jm
1/La
Inductance, 1/La
bm
Damping, b
Current
Angular speed
(a) Simulink model based on state space representation.
armature
Current,ia
Motor
Torque
Armature
Va
-K-
rad2mps
V=W*r2
Kt
motor
constant1
linear speed1
1/n
gear ratio
n=3.2
Vin
motor.mat
To File..1
Kb
TloadLoad
torque1
Angular speed1
1
La.s+Ra
1
Jequiv.s+bequiv
,1
(b) a suggested full block diagram model of PMDC motor open loop system with introduced saturation and coulomb friction.
TorqueCurrent
Torque
Motor3.mat
To File7
Motor2.mat
To File5
Motor1.mat
To File4
TL
Tload
Sum.3
Sum.2
Step,
Vin=12
Kt
Kt.
Kt
Kt
1
s
Integrator..2
1
s
Integrator
1/Jm
Inertia ,
1/Jm1
1/Ra
Inductance,
1/La
bm
Damping,bm
Current Angular speed
Angular position
Motor4.mat
. .
(c) Simulink model based on simplified mathematical model
INTERNATIONAL JOURNAL OF CONTROL, AUTOMATION AND SYSTEMS VOL.1 NO.2 APRIL 2013
ISSN 2165-8277 (Print) ISSN 2165-8285 (Online) http://www.researchpub.org/journal/jac/jac.html
32
TorqueCurrent
speed.
angle..
Torque
Motor4.mat
To File3
Motor3.mat
To File2
Motor2.mat
To File1
Motor1.mat
To File
0
TloadSum.1
Sum.
Step,
Vin=12.
Ra
Rresistance, Ra.
Kt
Kt..
Kt
Kt. 1
s
Integrator..1
1
s
Integrator..
1
s
Integrator.1/Jm
Inertia ,
1/Jm
1/La
Inductance,
1/La1
bm
Damping, b.
Current.
(d) Simulink model based on simplified mathematical model
Fig. 2 PMDC simulink models
VI. MODELING AND SIMULATION OF BRUSHED DC
MOTOR
There are four classical types of self excited brushed DC
machines with field windings; series, shunt, separately
excited windings and compound DC machine
A. Modeling and simulation of separately excited DC motor
In separately excited DC motor the field magnet has a
power supply that is separate from the armature
electromagnet; this means motor field strength is
completely independent from the armature field strength.
An Armature Controlled DC Motor is a separately excited
DC motor where the field current is usually constant and
the armature current controls the motor torque, the speed of
a separately excited dc motor could be varied from zero to
rated speed mainly by varying armature voltage in the
constant torque region. A Field Controlled DC Motor is a
separately excited DC motor where the field current
controls the motor torque. Separately excited DC motor
allows having independent control of both the magnetic
flux and the supply voltage, which allows the required
torque at any required angular speed to be set with great
flexibility; the biggest drawback is they are noisy. The
separately excited motor allows one to have independent
control of both the magnetic flux and the supply voltage,
which allows the required torque at any required angular
speed to be set with great flexibility [11].
A simplified equivalent representation of the separately
excited DC motor's two components are shown in Fig. 3, it
consists of independent two circuits, armature circuit and
field circuit, in which loads are connected to the armature
circuit The voltage is applied to both to field and armature
terminals, as shown , there are two currents, filed current,
if(t) and armature current, ia(t) in order to have linear
system, one of these two currents most held constant.
Fig. 3 Schematic of a simplified equivalent representation of the field
controlled DC motor's electromechanical components.
Performing the energy balance on the DC motor
system (Fig. 3) the sum of the torques must equal zero,
we have: 2 2T J * J *d / dt
– – 0e EMFT T T T
Setting, Te t a fK i i , substituting values , considering
shaft output position gives Eq.(7) and considering shaft
output position gives Eq.(8): 2
20t a f Load m m
d dK i i T J b
dtdt
(7)
0t a f Load m m
dK i i T J b
dt
(8)
Taking Laplace transform and rearranging yields Eqs.(9),
for each output speed and angle.
2– – 0
t a f load m m
t a f load m m
t a f load m m
K I s I s T J s s b s s
K I s I s T J s b s s
K I s I s T J s b s
(9)
Further rearranging to separate angular speed gives
Eq.(10)
( ) 1
* ( ) * ( ) ( )
* ( ) * ( ) ( ) ( )
t a f Load m m
t a f L
m m
s
K I s I s T s J s b
K I s I s T ss
J s b
(10)
Applying Kirchoff’s law around the field electrical loop
by summing voltages throughout the R-L circuit gives:
( )( ) f
f f f f
di tV R i t L
dt
Taking Laplace transforms, rearranging to separate the
field current, if gives:
( ) ( ) f
f f f f f f
f f
VV R I t L sI I s
R L s
Applying Kirchoff’s law around the armature electrical
loop by summing voltages throughout the R-L circuit,
taking Laplace transform, gives:
– – 0in R LV V V V EMF
Setting, EMF ( ) /b fK i d t dt , gives:
( ) ( )( )
a
in a a a b f
in a a b f
di t d tV R i t L K i
dt dt
V s R I s L s I s K i s s
Rearranging to separate the armature current, ia and field
current, if, gives:
_
1( ) ( ) * * ( )
( ) ( )
a a b f
a a
in f
f
f f
I s V s K i sL s R
V sI s
L s R
Substituting armature current, ia in Eq.(9) gives:
21
( ) ( ) ( ) ( )t f in b f load m m
a a
K i V s K i s T J s s b s sL s R
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33
Rearranging, the transfer function relating input armature
voltage to output motor angular speed given by:
2
2
2
/( )
( )1
t f a f
armature b fielda a
a a a f
K I R R bs
V s K VL J L Js s
R b R b R R b
The following state space equation and matrix form can be
written, as:
_
_
1*
1
*
a f
a a f in a
a a a
f
f f in f
f f
loadf
a f
R Ldi i i V
dt L L L
Rdi i V
dt L L
TLd bi i
dt J J J
10 0 0 0
0
10 0 0 0
10 0 0 0
f
f ff f f
a f
a a f a
a a a
L
f
f a
R
L Li i V
R Ldi i i V
dt L L LT
b Li i
J JJ
Using derived equation of separately excited DC motor,, the
simulink model shown in Fig. 4 (a), can be built.
fi led current motor torquemotor angular
speed
Motor l inear
speed
armature
Current,ia Motor
Torque
Armature
if, Field
current
the armature current IS maintained constant ia(t) = ia= constant
SEPARETLY EXCITED DC MOTOR
Armature
inductance
mutual
inductance
Table: Parameters of the DC Motor.
Vf=240[V]
La=0.012[mH]
Va=240[V]
Lmutual=1.8[mH]
Rf=240[W]
J=1[Kg.m2]
Ra=0.6[W]
Cr=29.2[N.m]
Lf=120[mH]
Fc=0.0005[N.m.Sec/Rad]
Vf
Va
armature
Current,ia
Motor
Torque
Armature
Va
-K-
rad2mps
V=W*r2
-K-
rad2mps
V=W*r1
Kt
motor
constant1
Kt
motor
constant
linear speed1
linear speed
1/n
gear ratio
n=3.2
1/n
gear ratio
n=3.1
Fc
friction
coefficient
1
Lf.s+Rf
filed
Transfer Fcn
1
Lf.s+Rf
field
angular
speed
Vin.
fi led
Vin
armature1
Vin
armature
V armature
V Field
1
J.s+B
Transfer Fcn
motor1.mat
To File..1
motor.mat
To File..
Step
12 V
Kb
Kb
Scope
Product1
ProductTloadLoad
torque1
TloadLoad
torque
1
s
Integrator.,
1
s
Integrator.
1
s
Integrator,
1
s
Integrator
1/J
Inertia
Km
Gain
1/Lf
Field
inductance.
Cr
Couple
resisting
Angular speed1
Angular speed
1
La.s+Ra
-K-
.1
.,
1
La.s+Ra
1
Jequiv.s+bequiv
,1
,.
1
Jequiv.s+bequiv
,
-K-
mutual
Inductance
La
armature
inductance1/Lf
Field
inductance
Rf
Field
resistance
Ra
Armature
resistance
1/La
Fig. 4 (a) separately excited DC motor model
B. Modeling and simulation of the field current controlled
DC motor,
In the field current controlled DC motor, the armature
current must maintained constant ia(t) = ia= constant , and
the field current, if varies with time ,t, to cause the motor to
rotate, this yields; the air-gap flux, Φ is proportional to the
field current and given by:
*f fK i (11)
The back EMF voltage is given by:
EMF K - Im in a aV R
The torque developed by the motor is related linearly to air-
gap flux, Φ and the armature current ia(t), and given by
Eq.(12):
1 ( )m aMotor Torque T K i t (12)
Substituting (12) in (11), we yields:
1 ( ) ( )m f a fT K K i t i t
The armature current must maintained constant ia(t) = ia=
constant, rearranging, yields:
1 ( ) ( ) ( )m f a f m fT K K i i t K i t
Where Km : the motor constant. Applying Kirchoff’s law,
Ohm's law, and Laplace transform to the stator field
yields mathematical model describing the electrical
characteristics of field controlled DC motor and given
by:
_ _ _– – 0in field R field L fieldV V V V
Applying Ohm's law, substituting and rearranging, we
have differential equation that describes the electrical
characteristics, given by:
_
( )f
in field f f f
di tV R i L
dt
Where: Lf, stator inductance, Rf, stator resistance. Taking
Laplace transform and rearranging, gives:
in _ field
in _ field f f f f
f f
V sV s L s R I s I s
L s R
(13)
The Mechanical characteristics of filed controlled DC
motor is obtained by performing the energy balance on
the motor system, where the sum of the torques must
equal zero, we have: 2 2T J * J *d / dt – – 0mT T T
The motor torque Tm, is related to the load torque, by:
2 2
2
m m
– * /
* ( ) – b *s s J *s s
m
m f
T T J d dt
K i t
Substituting If given by (13) and rearranging, gives:
in _ field 2
m m
f f
V s * – b s s J s s
L s RmK
(14)
Rearranging Eq.(14),where the electrical and mechanical
field current controlled DC motor components are
coupled to each other through an algebraic the motor
constant , Km , we obtain the transfer function relating
input filed voltage Vin_field(s), and motor output angle
θm(s), and given by:
_
( )( )
( )
m
angle
in filed f f
KsG s
V s s L s R Js b
The simulink model of the filed current controlled DC
motor is shown in Fig. 4 (b), here note that the armature
controlled DC motor is in nature closed loop system,
while filed current controlled DC motor is open loop.
Filed
current
Motor
torque
motor angular
speed
Motor l inear
speed
armature
Current,ia Motor
Torque
Armature
if, Field
current
the armature current IS maintained constant ia(t) = ia= constant
SEPARETLY EXCITED DC MOTOR
Armature
inductance
mutual
inductance
Table: Parameters of the DC Motor.
Vf=240[V]
La=0.012[mH]
Va=240[V]
Lmutual=1.8[mH]
Rf=240[W]
J=1[Kg.m2]
Ra=0.6[W]
Cr=29.2[N.m]
Lf=120[mH]
Fc=0.0005[N.m.Sec/Rad]
Vf
Va
armature
Current,ia
Motor
Torque
Armature
Va
-K-
rad2mps
V=W*r2
-K-
rad2mps
V=W*r1
Kt
motor
constant1
Kt
motor
constant
linear speed1
linear speed
1/n
gear ratio
n=3.2
1/n
gear ratio
n=3.1
Fc
friction
coefficient
1
Lf.s+Rf
filed
Transfer Fcn
1
Lf.s+Rf
field
angular
speed
Vin.
fi led
Vin
armature1
Vin
armature
V armature
V Field
1
J.s+B
Transfer Fcn
motor1.mat
To File..1
motor.mat
To File..
Step
input V
Kb
Kb
Scope
Product1
ProductTloadLoad
torque1
TloadLoad
torque
1
s
Integrator.,
1
s
Integrator.
1
s
Integrator,
1
s
Integrator
1/J
Inertia
Km
Gain
1/Lf
Field
inductance.
Cr
Couple
resisting
Angular speed1
Angular speed
1
La.s+Ra
-K-
.1
.,
1
La.s+Ra
1
Jequiv.s+bequiv
,1
,.
1
Jequiv.s+bequiv
,
-K-
mutual
Inductance
La
armature
inductance1/Lf
Field
inductance
Rf
Field
resistance
Ra
Armature
resistance
1/La
Fig. 4 (b) Simulink model of the filed current controlled DC motor
C. Modeling and simulation of the armature controlled
DC Motor (1)
In armature-current controlled DC motor, the field
current if is held constant, and the armature current ia is
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34
controlled through the armature voltage Vin, different
approaches to derive mathematical model
Approach (1) : The motor equations can be written to have
the following form:
_ _
m m
m m
*
0
b + J s
b + J s
m t a
EMF R a L a
a b a a a
m Load
m Load
T K I
V V V V
V K I R L s
T T
T T
Based on these equations, the transfer function given by
Eq.(15) is derived and the simulink model shown in Fig. 4
(c), is built. Equation (15) can be simplified to be first order
transfer function given by Eq.(16), by assuming La=0
(electrical time constant is much smaller than the time
constant of the load dynamics), this yields:
/( )
( ) / / / /
t a
a a a b t a
K L Js
V s s R L s b J K K L J
(15)
/( )
( ) /
t a
a a b t a
K LR Js
V s s bR K K R J
(16)
armature
Current,ia
Motor
Torque
Va
EMF constant
-K-
rad2mps
V=W*r2linear speed1
1/n
gear ratio
n=3.2
Vin
armature1
Kt
Torque constant
motor.mat
To File..1
Kb
TloadLoad
torque1
Angular speed1
La
s+Ra/La
1/Jequiv
s+b/Jequiv
,1
Fig. 4 (c) the armature controlled DC Motor
Approach (2) : The back EMF is given by :
EMF bV K and motor torque is given by:
m b aT K I . The motor equations can be written to have
the following form
_ _
2
m m
0
b s s + J s s
EMF R a L a
m Load
V V V V
T T
Assuming absence (negligible) of friction in rotor of motor,
yields:
2
m m J s s J s sm Load LoadT T T
Since equation for back EMF is given by : EMF bV K
and motor torque is given by: m b aT K I , substituting
back EMF and motor torque in motor equation, rearranging
and solving for armature current and angular speed, gives:
_ _ 0EMF R a L a b a a aV V V V K I R L sI
1 /
a
a in b
a a
RI V K
L s R
m
J s
m LoadT T
Based on these equation simulink model shown in Fig. 4
(d), is built, the transfer function relating input voltage
and output angular speed is given by:
2 2
1 /( )
( ) /1
1 /
b
in b a
a a
Ks
V s K R
Js L R
armature
Current,
ia
Motor
Torque
Va
field flux
linear speed11/n
gear ratio
n=
Vin
armature
phai
TloadLoad
torque1
Angular speed
Ra
s+La/Ra
1
Jequiv.s
,1
phai
field flux
-K-
V=W*r2
Fig. 4 (d) separately excited DC motor model considering flux
Approach (3) : The motor equations can be written to
have the following form :
*Load t a
a
in b a a a
dT K i J b
dt
diV K L R i
dt
Further solving these equations will result in PMDC
transfer function, since the field current is kept constant,
result in permanent magnetic filed, the simulink model is
identical to model given in Fig. 2(b)
2
( )( ) ,
( )
( )
( ) ( ) ( )
t
speed
in a a m m t b
t
in a m a m m a a m t b
KsG s
V s L s R J s b K K
Ks
V s L J s R J b L s R b K K
D. Accurate modeling and simulation of separately
excited DC motor.
Accurate characteristic equations of separately excited
DC motor can be represented as follows:
_
_*
*
in aa a
a
a a
in amutualf f
f a
f f f
mutual r
a f
VR idi
dt L L
VLR idi i
dt L L L
L Cd bi i
dt J J J
Based on these equations, simulink model shown in Fig. 4
(e) is built [12], in this model the couple resisting Cr,
mutual inductance Lmutual , are introduced.
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35
fi led current motor torquemotor angular
speed
Motor l inear
speed
armature
Current,i Motor
Torque
Armature
Field
current
the armature current IS maintained constant ia(t) = ia= constant
SEPARETLY EXCITED DC MOTOR
Armature
inductance
mutual
inductance
Table: Parameters of the DC Motor.
Vf=240[V]
La=0.012[mH]
Va=240[V]
Lmutual=1.8[mH]
Rf=240[W]
J=1[Kg.m2]
Ra=0.6[W]
Cr=29.2[N.m]
Lf=120[mH]
Fc=0.0005[N.m.Sec/Rad]
-K-
rad2mps
V=W*r1
Km
motor
constant
linear speed
1/n
gear ratio
n=3.1
Fc
friction
coefficient
1
Lf.s+Rf
filed
Transfer Fcn
1
Lf.s+Rf
field
angular
speed
Vin.
fi led
Vin
armature
V armature
V Field
1
J.s+B
Transfer Fcn
motor.mat
To File..
Step
12 V
Kb
Scope
Product1
Product
TloadLoad
torque
1
s
Integrator.,
1
s
Integrator.
1
s
Integrator,
1
s
Integrator
1/J
Inertia
Km
Gain
1/Lf
Field
inductance.
Cr
Couple
resisting
Angular speed
-K-
.1
.,
1
La.s+Ra
,.
1
Jequiv.s+bequiv
,
-K-
mutual
Inductance
La
armature
inductance1/Lf
Field
inductance
Rf
Field
resistance
Ra
Armature
resistance
1/La
Fig. 4 (e) Accurate modeling of separately excited
E. Simplified Modeling and simulation of separately
excited DC motor
Ignoring armature reactions effects, to minimize the
effects of armature (compensating winding), this, a linear
model of a simplified separately excited DC motor consists
of a mechanical equation and electrical equation as
determined in the following equations:
_
m n a load
a
a in a a a b
dJ K i b T
dt
diL V i R K
dt
Where Kn: motor constant, Based on these equation, the
simplified simulink model shown in Fig. 4 (f), is built.
Ia
W
If
Kn
armature current
angular speed
Vin
Vin Armature
TL
TL
Motor torque
Field current
K
,
Kb
''
La
'
J
.
1
s
,
B
'
Ra
1
s
Fig. 4 (f) Simplified separately excited DC motor model
For parameters specified, running simulation of separately
excited DC motor open loop model will result in
Torque/time, Speed/time , Position/time Current/time,
angular acceleration/time curves for 12 V step input shown
in Fig. 4 (g)
0 2 4 60
5
10
15
Time (sec)
Am
p
Current Vs Time
0 2 4 60
0.1
0.2
0.3
0.4
Tims (sec)
N/m
Torque Vs Time
0 2 4 60
20
40
60
Tims (sec)
Rad
angle Vs Time
0 2 4 60
5
10
Tims (sec)
Rad/m
Angular speed VS time
0 5 100
2
4
6
8
Tims (sec)
Rad/m
2
Angular acceleration VS time
0 5 100
5
10
Tims (sec)
Rad/m
2
Angular acceleration VS time
0 5 10 15 2011.7
11.8
11.9
12
12.1
Tims (sec)
Am
p
Current VS time
Fig. 4 (g) Torque/time, Speed/time , Position/time Current/time, angular
acceleration/time curves for 12 V step input,
V. MODELING AND SIMULATION OF SHUNT DC
MOTOR
A shunt wound DC motor has the armature and field
(stator) coils connected in parallel (or shunt) across the
power source, in result the same voltage is applied to both
coils, the transient in the armature circuit is simultaneous
with the transient in the field circuit , [13] a shunt excited
machine is essentially the same as a separately excited
machine, with the constraint that the field winding supply
voltage Vin_f is equal to the armature winding supply
voltage, Vin_a. this is shown in Fig. 5. Shunt wound DC
motor is designed for applications where constant speed
characteristics under varying load conditions are
important such as pumping fluids and fans, shunt motor
speed varies only slightly with changes in load. A shunt
wound DC motor is difficult to control, as reducing the
supply voltage also results in a weakened magnetic field,
thus reducing the back EMF, and tending to increase the
speed [ 14].
Fig. 5 Two circuit representations of shunt wound DC motor [ 14]
The stator and rotor circuits have the same voltage supply
and therefore the same voltage drop, and the current
drawn by the motor, im is the sum of the field current, if
and armature current ia, this all can be expressed as:
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36
in a fV V V , m f ai i i
The DC shunt motor has the same dynamics equations for
torque as for the separately excited motor, with constraint
that Vin_f = Vin_a and given (including matrix form) by the
following equations:
_
_
_
_
1 *
,
1
a
in a a a a b f
a f
a a f in a
a a a
f
in f f f f
f
f f in f
f f
m b a f Load
diV R i L K i
dt
R Ldi i i V
dt L L L
diV R i L
dt
Rdi i V
dt L L
dT K i i T b J
dt
d
dt
* loadf
a f
TLbi i
J J J
_
_0
in aa b
aa aa a
f f in af
f f
VR K
LL Li i
i i VR
L L
At steady state, currents are given by:
* *, in t f in
a f
a f
V K i Vi i
R R
Substituting in torque equations, we have:
2
_ 1t t
m t a f m in a
a f f
K KT K i i T V
R R R
Further substituting, and rearranging the load torque is
given by:
2
m m
2
m m
2
_ m m
T = b *s s J *s s
= T - b *s s J *s s
1 - b J
m Load
Load m
t t
Load in a
a f f
T
T
K KT V s
R R R
Based on these, the simulink models of shunt DC motor are
built and shown in Fig. 6(a)(b).
armature
Current,ia Motor
Torque
Armature
if, Field
current
Vf
Va
-K-
rad2mps
V=W*r1
Kt
motor
constant
linear speed
1/n
gear ratio
n=3.1
1
Lf.s+Rf
field
Vin
armature
motor.mat
To File..
Kb
TloadLoad
torque
Angular speed
.,
1
La.s+Ra
,.
1
Jequiv.s+bequiv
,
(a)
Ia
If
Te Ia*If
W
If
W
Separately Excited D.C. Motor fi led
current
motor
torque
motor angular
speed
Motor angular
position
Vin field
1
Lf.s+Rf
filed
Transfer Fcn
armature current
angular speed
Vin_a
Vin Armature
1
J.s+B
Transfer Fcn
TL
TL
Step
12 V
Scope1
Motor torque
1
s
Integrator
Km
Gain
Field current
.,
.
Kb
,
1
1
s
,
1/J
'1
B
'
Kb/La
.
Rf/Lf
1/Lf
1/La
Ra/La
1
s
1
s
(b)
Fig. 6 Shunt DC motor models
VI. MODELING AND SIMULATION OF SERIES DC
MACHINES
In a series wound DC motor the field and armature
circuits are connected in series, in result the same current
flows is applied to both coils, this is shown in Fig. 7
Fig. 7 Two circuit representations of Series wound DC motor
A series wound DC motor is easy to use, will generate a
larger torque increase (provide startup torque) compared
with a shunt wound DC motor for given increase in
current. Series motors cannot be used where a relatively
constant speed is required under conditions of varying
load." this means series wound DC might not climb hills
with varying slope briskly and smoothly. The voltage
supply is divided between stator and rotor circuits and a
common current flow through the field and armature coils
current ia,[ 14] this all can be expressed as:
in a fV V V , ,m f ai i i
Applying Kirchoff’s law around the electrical loop,
yields:
( )
( )
a f
in a f a f a a
a f
a f a f b aa
di diV s L L I R I R EMF
dt dt
di diL L I R R K i
dt dt
These equations can be rewritten, to have the following
form:
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37
2
2
( ) ( ) * *
*
in a a f a f a mutual n
m t a Load
m mutual a Load
d dV s i L L i R R L i
dt dt
dT K i T b J
dt
dT L i T b J
dt
Where :Lmutual :is the mutual inductance between the
armature winding and the field winding .Under steady state
condition, induction (L=0), gives:
( )in f a a aV s R I R I EMF
( )in a f aV s I R R EMF
The torque developed in the rotor is:
* * *m f fT K i K i
2 *m tT K i
The back EMF, also, can be expressed as:
* * ( * )b n b f a nEMF K I K K
Substituting, we have the armature current given by:
( )in
a
a f b m
V sI
R R K
And the developed torque given by:
2
2
*in t
a f t m
V KT
R R K
Based on this equation, if the input voltage Vin is kept
constant, the output angular speed is almost inversely
proportional to the square root of the torque, therefore a
high torque is obtained at low speed and a low torque is
obtained at high speed. Finally, the dynamic equations, for
simulation, can be written as follows:
* *( ) f a b nin
a a
a f a f a f
Loadm
R R K iV si i
L L L L L L
TTd b
dt J J J
Based on these, the simulink models shown in Fig. 8, are
built.
ia
Motor
Torque
Va
Lmutual
-K-
rad2mps
V=W*r2linear speed1
1/n
gear ratio
n=3.2
Vin
armature1
motor.mat
To File..1Kb
TloadLoad
torque1
Angular speed1
1
La.s+Ra.
1
Jequiv.s+bequiv
,1
'
-K-
Fig. 8 (a) Considering the mutual inductance
ia
Motor
Torque
Va
-K-
rad2mps
V=W*r2
Kt
motor
constant1
linear speed1
1/n
gear ratio
n=3.2
Vin
armature1
motor.mat
To File..1Kb
TloadLoad
torque1
Angular speed1
1
La.s+Ra.
1
Jequiv.s+bequiv
,1
'
Fig. 8 (b) Considering the torque constant
Ia
W
fi led
current
motor
torque
motor angular
speed
Motor angular
position
Tm
W
1
Lf.s+Rf
filed
Transfer Fcn
armature current
angular speed
Vin_a
Vin Armature
1
J.s+B
Transfer Fcn
TL
TL
Step
12 V
Scope1
Motor torque
1
s
Integrator
Km
Gain
.
1
1
s
,
1/J
'1
B
'
Kb/(La+Lf)
.
Kt
1/(La+Lf)
(Ra+Rf)/(La+Lf)
1
s
Fig. 8(c) Series DC motor models
VII. COMPOUND DC MACHINES
It is a combination of shunt wound and series wound
configurations, so it can run as a shunt motor, a series
motor, or a hybrid of the two, as shown in Fig. 9. This
allows the compound motor to be used in applications
where high starting torque and controlled operating speed
are both required. The total motor voltage drop is the sum
of the series field and armature voltage drops, so the
series field coil is usually made out of a few turns of
heavy wire to keep the series field voltage drop to a
minimum. A shunt field coil is usually wound with many
turns of thin wire to minimize the shunt field current. In
most compound wound DC motors the field windings
have separate connections so they can be switched in or
out as desired [15]. The speed of a DC compound motor
can be easily controlled. It is enough if we change just the
voltage supplied to it
Fig. 9 Two circuit representation of compound DC motor
VIII. BRUSHLESS DC MOTOR (BLDC) MACHINES
The rotor (armature) is composed of one or more
permanent magnets and coils for the stator (field). The
rotor, being a permanent magnet, simply follows the
stator magnetic field around. The speed of the motor is
controlled by adjusting the frequency of the stator power.
In the BLDC motor, the electromagnets do not move;
instead, the permanent magnets rotate and the armature
remains static. The BLDC motor is actually an AC motor.
The wires from the windings are electrically connected to
each other either in delta configuration or WYE ("Y"-
shaped) configuration. The main disadvantages of
brushed DC motor is that they need a commutator and
brushes which are subject to wear and require
maintenance, therefore have low life-span.
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38
The kinetics of the motor can be described as: 2
EMF 2e – T – T T 0 0e Load m m
d dT T J b
dtdt
The generated electromagnetic torque, Te is given by:
* * *P a a b b c ce
e
n m
EMF i EMF i EMF iT
Where : Pe electromagnetic power of the motor, ea, eb, ec :
the back EMF in each phase . ia, ib, ic stator phase currents.
Under normal operation, only two phases are in conduction,
therefore the voltage balance equation, cross the two
windings under conduction, is given by:
( )( ) w
w w w w w
di tV R i t L EMF
dt
IX. CONTROL SYSTEM SELECTION AND DESIGN
There are many well known motor control system
design strategies that may be more or less appropriate to a
specific type of application, each has its advantages and
disadvantages; the designer must select the best one for
specific application [16]. Different resources introduce,
different models, designs and verifications of different
control strategies for DC motors, In [] DC motor control
applying Proportional-Integral PI, Proportional-Integral-
Derivative PID or bipositional are introduced. [16]
Discussed modeling and controller design for electric
motor, using different control strategies and verification
using MATLAB/Simulink. In [17] different closed loop
control strategies and compensator designs were compared
to eliminate the steady state error and enhance the DC
motor system transient response in terms of output speed,
similar approach will be applied in this paper in terms of
output position. In [18,19] a good description of the optimal
control design, including linear state regulator control, the
output regulator control and linear quadratic tracker. [20]
covered how it is possible to improve the system
performance, along with various examples of the technique
for applying cascade and feedback compensators, using the
methods root locus and frequency response. It also covered
some methods of optimal linear system design and
presentation of eigenvalues assignments for MIMO system
by state feedback. A negative feedback control system with
forward controller shown in Fig. 10 (a)(b) is most used for
controlling DC machines used.
A. Current controller in a DC drive system
There is a need to control current in motor armature,
this is because of the fact that mechanical time constant is
very large, compared with electric time constant, and initial
speed of motor, when started from zero, is zero, this will
result in maximum error, and hence given maximum
voltage, resulting in very large current flow at starting time,
correspondingly, because back EMF is zero when motor
started from zero, this current may exceeds the motor
maximum current limit and can damage the motor
windings. By applying current controller, the applied
voltage Vin will now not dependent on the speed error only,
but also on the current error, this all will result in two
loop motor control, speed control and current control.
B. Two loops control
As shown in Fig. 11(a), two loops are used to control,
the motor, inner and outer loops. the motor torque is
controlled by the armature current Ia, which is regulated
by inner current control loop. The motor speed is
controlled by an external loop, which provides the current
reference Ia_R , for the current control loop. A current
sensor with gain Ks is used to measure the armature
current and a speed sensor (tachometer) with gain Ktach is
used to measure the angular speed.
A chopper is a high speed “ON" or “OFF” semiconductor
switch, it connects source to load and load and disconnect
the load from source at a fast speed (PWM), chopper
takes a fixed DC input voltage and gives variable DC
output voltage, chopper works on the principle Pulse
Width Modulation technique, there is no time delay in its
operation, therefore it can be represented by a simple
constant gain Kc.Most suitable controller for both inner
and outer loops are; PI controllers, considering that
mechanical time constant is much larger than electric
time constant.
Applying this method to control the motion of cuboide
platform using PMDC motor simulated as shown in Fig.
11(b) to be prime mover, ( such application example
include mobile robot and small electric vehicle) and
considering load disturbance torque, which is the total
resultant torque generated by the acting resistive forces
including rolling resistance, aerodynamic drag, lift, hill
climbing and coulomb friction, which are modeled in
simulink as shown in Fig. 11(b). The overall system
simulink model is shown in Fig. 11(d).
Running the model for desired output speed of 24 rad/s,
with Ktach =1, and step input of 24, will result in a suitable
output speed response curve shown in Fig. 11(e), with
zero steady state error , the response shows that the
system reaches desired output in 7 seconds without
overshoot and with less time.
Improving this model as shown in Fig. 11(f), to include
speed sensor, current sensor and chopper, assigning
tachometer constant Ttach=1, running the model for
desired output speed of 12 rad/s well result in response
curves shown in Fig. 11(g).these response curves show
that the system reaches the desired output speed in 1.5
seconds, with generated motor torque equal to 9 Nm.
B.1 Proposed control method
To minimize the negative current characteristics, and
maintain generated motor torque, the simulink model
shown in Fig. 11(f), can be modified to be as shown in
Fig. 11(h), where we can relate the load torque
corresponding desired current to overcome this torque,
by dividing load torque value over torque constant, and
multiplying result by current sensor constant Ks, will
result in approximate current required to overcome load
torques, running the model for desired output speed of 12
rad/s well result in response curves shown in Fig. 11(i),
comparing these resulted response curves with response
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39
curves shown in Fig. 11(g) show, that applying this new
theoretical technique , will result in more smooth response ,
less current (61 Amp), and with the same motor torque.
Controller
(angle, speed)Control voltage,
Vc
Angle or Speed measure e,.g
Potentiometer, Tachometer
Sensor
+-
Error, VoltAngle or Speed
reference (desired)
Volt
Motor shaft
ω or θ
Fig. 10 (a) Block diagram representation of PMDC motor control
Error Angle, speed
-
PID-controller
-K-
sensor
In1 Out1
electric Motor
Subsystem
aaaaa.mat
To File
Step input
Output
P(s)
Controller to be
selected
Fig. 10 (b) Preliminary simulink model for negative feedback with forward
compensation
ChopperDC Vin
Speed
Sensor
Speed ω ,
angle θ
Current
controller
-
+
Speed
controller
+
-
Current
Sensor
Desired speed
Armature
Current, Ia
ωmIa
(a)
0.5*ru*A*Cd*V^2*r
Coloum friction
1 Load torque
r
wheel radius,
V=W*r1
-K-
r^2m/2. -K-
r*m*g/2
-K-
aerodaynamic torque,
sin(u)
cos(u)
SinCos.
-K-M*g,
60
Inclination
angle
du/dt
Derivative,
-K-
.1
r/2.
.
Kt
bm
0.8
2 angular speed1 Cuurent, ia
(b)
Fig. 11Load torque sub-system of a mobile platform
C. Combined armature and field currents control
Combined armature and field currents control using
PI controllers is shown in Fig. 12, running this model for
defined values of field, running this model for armature
volt input of 200 and field volt input of 100 will result in
response curves shown in Fig. 12 (b) the response curves
show that the output angular speed of 75 rad/s, is
achieved, with motor torque of 100 Nm. Applied PI
controllers can be tuned for better performance
CONCLUSIONS
The mathematical models, corresponding simulink
models, analysis and control solutions of basic open loop
electric DC machines most used in mechatronics
applications are introduced. Two loops current and speed
control of eclectic machines are introduced and tested. A
proposed, yet theoretical, control method relating the load
torque with desired armature current and torque constant
to minimize current drawn while attaining generated
desired motor torque, is proposed and theoretically tested,
resulting in reducing controlling current in acceptable
ranges. Proposed models are intended for research
purposes, as well as, for the application in educational
process.
As future work, a practical implementation of the
proposed current and torque control is to be held tested,
and compared with theoretical result and proved
physically.
Tmcurrent angular
speedError
4To SPEED
controller3 To CURRENT controller 2 armature current, Ia 1 angular speed, W
-K-
speed feedbacK
speed
1/n
gear ratio.
Kb
EMF constant
du/dtDerivative1
du/dtDerivative
Angular acceleration
1
La.s+Ra
1/(Ls+R)3
1
den(s)
1/(Js + b)3
Ks
current sensor, Ks
Kt
1/r
4Input
3
from
SPEED
controller1
2TL1
from
CURRENT
controller
Fig. 11(c) PMDC motor sub-system
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40
Iaspeed
two_loops.mat
To File5
f rom CURRENT controller
TL
f rom SPEED controller1
Input
angular speed, W
armature current, Ia
To CURRENT controller
To SPEED controller
Subsystem
Step1
s
PI speed controller
1
s
PI current controller1
Cuurent, ia
angular speed
Load torque
Load sub-system
KcChopper gain Kc
Armature current -K-
Ki
-K-
Kp
Load torque
Fig. 11(d) Overall motion control system including load torque.
I
speed sensor
Current sensor
Ks
current sensor
StepVin_armmature
Tload
armature current, Ia
motor torque, Tm
Angular speed, W
angular Position,
PM DC Motor Subsystem1
1
s
PI speed controller
1
s
PI current controller.Cuurent, ia
angular speed
Load torque
Load sub-system1Kc
Kc
Field current
Armature current
Angular speed
Angular position
1/Kt
1/Kt
exited3.mat
'.
exited2.mat
'''
exited1.mat
''
exited.mat
'
-K-
1
-K-
1
-K-
-K-
Ks
Ktach
Fig. 11(f) Overall motion control system including load torque.
I
speed sensor
Current sensor
Ks
current sensor
StepVin_armmature
Tload
armature current, Ia
motor torque, Tm
Angular speed, W
angular Position,
PM DC Motor Subsystem1
1
s
PI speed controller
1
s
PI current controller.Cuurent, ia
angular speed
Load torque
Load sub-system1Kc
Kc
Field current
Armature current
Angular speed
Angular position
1/Kt
1/Kt
exited3.mat
'.
exited2.mat
'''
exited1.mat
''
exited.mat
'
-K-
1
-K-
1
-K-
-K-
Ks
Ktach
Fig. 11(h) Proposed model
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41
Wm
W*
Ia
Current sensor
Step
Vin_armmature
Vin_f ield
Tload
armature current, Ia
Field current, If
Motor torque, Tm
angular speed, W
Separately excited DC Motor Subsystem1
1
s
PI speed controller1
1
s
PI field controller.
1
s
PI armature controller
Motor torqueCuurent, ia
angular speed
Load torque
Load sub-system1Kc
Kc
Field current
12
Desired angular speed1
V
Constant
Armature current
Angular speed
exited3.mat
'.
exited2.mat
'''
exited1.mat
''
exited.mat
'
Kc
chopper gain
-K- 3
Kp
2Ki
1
-K-
1
-K-
1
-K-
-K-
Ks
Ks
Fig. 12 (a) Block diagram of combined armature and field currents control.
0 5 100
10
20
30
Time(s)
O
mega
Two loops control;Speed and current
0 5 10-10
0
10
20
30
Time(s)
M
gnitude
Two loops control ;Error signal
Fig. 11(e) speed step response curve applying two loops control, speed
and current
0 1 2 30
5
10
15
.
, Angular speed Vs time
0 1 2 30
200
400
600
Time(s)
A
mpere
Armature current
0 2 4 60
5
10
.
N
/m
Motor torque Vs time
0 1 2 30
10
20
30
40
Time(s)
Angular position Vs time
Fig. 11(g) angular speed/time, motor torque/time, armature current/time, angular position/time response curves , running model given in Fig.
11(f)
0 20 40 600
5
10
15
.
, Angular speed Vs time
0 20 40 600
20
40
60
80
Time(s)
A
mpe
re
Armature current
0 2 4 60
5
10
.
N
/m
Motor torque Vs time
0 20 40 600
200
400
600
Time(s)
Angular position Vs time
Fig. 11(i) Speed/time, torque/time, armature current/time, position/time
response curves.
0 20 40 600
20
40
60
80
.
O
mega
, Combined control;Armature and field
0 20 40 600
50
100
150
200
Time(s)
A
mpere
Armature current
0 20 40 600
50
100
Time(s)
A
mpere
Field current
0 20 40 600
50
100
.
N
/m
Torque
Fig. 12 (b) applying combined armature and field currents control
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Farhan Atallah Salem AbuMahfouz
B.Sc., M.Sc and Ph.D., in
Mechatronics of production systems,
Moscow, 2000. Now he is ass.
professor in Taif University,
Mechatronics program, Dept. of
Mechanical Engineering and gen-
director of alpha center for engineering studies and
technology researches. Research Interests; Design,
modeling and analysis of primary Mechatronics
Machines, Control selection, design and analysis for
Mechatronics systems. Rotor Dynamics and Design for
Mechatronics applications