dynamic modeling, simulation and control of electric machines for

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:

[email protected]).



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



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



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



Armature

inductance

mutual

inductance


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]


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



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.



35

fi led current motor torquemotor angular

speed

Motor l inear

speed

armature

Current,i Motor

Torque

Armature

Field

current



Armature

inductance

mutual

inductance


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]


-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:



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:



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.



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



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



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


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


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



41

Wm

W*

Ia

Current sensor

Step

Vin_armmature

Vin_f ield

Tload


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

REFERENCES

[1] Ahmad A. Mahfouz, Mohammed M. K., Farhan A. Salem,

Modeling, Simulation and Dynamics Analysis Issues of Electric

Motor, for Mechatronics Applications, Using Different Approaches and Verification by MATLAB/Simulink", IJISA, vol.5, no.5,

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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

dynamic modeling, simulation and control of electric machines for

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