Automatic Control Systems

Lecture- DC Motor Modelling and Control

1

Emam FathyDepartment of Electrical and Control Engineering

email: emfmz@yahoo.comhttp://www.aast.edu/cv.php?disp_unit=346&ser=68525

2

D.C Drives

• Speed control can be achieved usingDC drives in a number of ways.

• Variable Voltage can be applied to thearmature terminals of the DC motor .

• Another method is to vary the flux perpole of the motor.

• The first method involve adjusting themotor’s armature while the lattermethod involves adjusting the motorfield. These methods are referred to as“armature control” and “field control.”

Armature Controlled D.C Motor

u

iaT

Ra La

J

B

eb

System constants

ea = armature voltageeb = back emfRa = armature winding resistanceLa = armature winding inductanceia = armature winding currentKb = back emf constantKt = motor torque constantJm = moment of inertia of the motorBm=viscous-friction coefficients of the motorJL = moment of inertia of the loadBL = viscous friction coefficient of the loadN1/N2 = gear ratio

oltageback-emf v e,edt

diLiRu bb

a

aaa

Mechanical Subsystem

BωωJTmotor

Input: voltage uOutput: Angular velocity

Electrical Subsystem

Armature Controlled D.C Motor

uia

T

Ra La

J

B

eb

Torque-Current:

Voltage-Speed:

atmotor iKT

The previous equations result in the following mathematical model:

Power Transformation:

ωKe bb

0at

baa

a

a

iKBωJ

uωKiRdt

diL

Kt: torque constant,

Kb: velocity constant For an ideal motor bt KK

Armature Controlled D.C Motor

uia

T

Ra La

J

B

eb

Taking Laplace transform of the system’s differentialequations with zero initial conditions gives:

Eliminating Ia yields the input-output transfer function

btaaaa

t

KKBRsBLJRJsL

K

U(s)

Ω(s)

2

0(s)IΩ(s)-KBJs

U(s)Ω(s)K(s)IRsL

at

baaa

Armature Controlled D.C Motor

Reduced Order Model

Assuming small inductance, La 0

abt

at

RKKBJs

RK

U(s)

Ω(s)

Armature Controlled D.C Motor

If output of the D.C motor is angular position θ then:

abt

at

RKKBJss

RK

U(s)

(s)

Which yields the following transfer function

Armature Controlled D.C Motor

)()( sssordt

d

uia

T

Ra La

J

θ

B

eb

Example

• Assume the following values for the physical parameters.

– moment of inertia of the rotor (J) = 0.01 kg.m^2/s^2

– damping ratio of the mechanical system (B) = 0.1 Nms

– electromotive force constant (K=Kb=Kt) = 0.01 Nm/Amp

– electric resistance (Ra) = 1 ohm

– electric inductance (La) = 0.5 H

– input (U): Source Voltage

– output (theta): position of shaft

– The rotor and shaft are assumed to be rigid 10

Design requirements

• The uncompensated motor can rotate at 0.1rad/sec with an input voltage of 1.

• For the unit step input , then the motor speedoutput should have:

– Settling time less than 2 seconds

–Overshoot less than 5%

– Steady-state error less than 1% 11

Uncompensated System

12

Uncompensated System

13

From the plot, when 1 volt is applied to the system; the

motor can achieve:

maximum speed of 0.1 rad/sec, ten times smaller

than our desired speed.

it takes the motor 3

seconds to reach its

steady-state speed; this

does not satisfy the 2

seconds settling time

criterion.

An armature controlled D.C motor runs at 5000 rpm when 15v applied at thearmature circuit. Armature resistance of the motor is 0.2 Ω, armatureinductance is negligible, back emf constant is 5.5x10-2 v sec/rad, motor torqueconstant is 6x10-5, moment of inertia of motor 10-5, viscous friction coefficientis negligible, moment of inertia of load is 4.4x10-3, viscous friction coefficientof load is 4x10-2.

1. Drive the overall transfer function of the system i.e. ΩL(s)/ Ea(s)

2. Determine the gear ratio such that the rotational speed of the load isreduced to half and torque is doubled.

Example

15 via

T

RaLa

Jm

Bm

eb

JL

N1

N2

BL

L

ea

System constants

ea = armature voltage

eb = back emf

Ra = armature winding resistance = 0.2 Ω

La = armature winding inductance = negligible

ia = armature winding current

Kb = back emf constant = 5.5x10-2 volt-sec/rad

Kt = motor torque constant = 6x10-5 N-m/ampere

Jm = moment of inertia of the motor = 1x10-5 kg-m2

Bm=viscous-friction coefficients of the motor = negligible

JL = moment of inertia of the load = 4.4x10-3 kgm2

BL = viscous friction coefficient of the load = 4x10-2 N-m/rad/sec

gear ratio = N1/N2

Since armature inductance is negligible therefore reduced order transferfunction of the motor is used.

Example

15 via

T

RaLa

Jm

Bm

eb

JL

N1

N2

BL

L

ea

btaeqaeqaeq

tL

KKRBsLBRJ

K

U(s)

(s)Ω

Lmeq JN

NJJ

2

2

1

Lmeq B

N

NBB

2

2

1

A field controlled D.C motor runs at 10000 rpm when 15v applied at the fieldcircuit. Filed resistance of the motor is 0.25 Ω, Filed inductance is 0.1 H,motor torque constant is 1x10-4, moment of inertia of motor 10-5, viscousfriction coefficient is 0.003, moment of inertia of load is 4.4x10-3, viscousfriction coefficient of load is 4x10-2.

1. Drive the overall transfer function of the system i.e. ΩL(s)/ Ef(s)

2. Determine the gear ratio such that the rotational speed of the load isreduced to 500 rpm.

Example 2

ifTm

Rf

Lf

JmωmBm

Ra La

eaef

JL

N1

N2

BL

L

+

kp

-

JL

_

ia

eb

RaLa

+

Tr c

ea

_

+

e

_

+

N1

N2

BL

θ

if = Constant

JM

BM

Example 2

Numerical Values for System constants

r = angular displacement of the reference input shaftc = angular displacement of the output shaftθ = angular displacement of the motor shaftK1 = gain of the potentiometer shaft = 24/πKp = amplifier gain = 10ea = armature voltageeb = back emfRa = armature winding resistance = 0.2 ΩLa = armature winding inductance = negligibleia = armature winding currentKb = back emf constant = 5.5x10-2 volt-sec/radK = motor torque constant = 6x10-5 N-m/ampereJm = moment of inertia of the motor = 1x10-5 kg-m2

Bm=viscous-friction coefficients of the motor = negligibleJL = moment of inertia of the load = 4.4x10-3 kgm2

BL = viscous friction coefficient of the load = 4x10-2 N-m/rad/secn= gear ratio = N1/N2 = 1/10

e(t)=K1[ r(t) - c(t) ]or

E(S)=K1 [ R(S) - C(S) ]

Ea(s)=Kp E(S)

Transfer function of the armature controlled D.C motor Is given by

(1)

(2)

θ(S)

Ea(S)=

Km

S(TmS+1)

System Equations

System Equations (contd…..)

Where

And

Also

Km =K

RaBeq+KKb

Tm =RaJeq

RaBeq+KKb

Jeq=Jm+(N1/N2)2JL

Beq=Bm+(N1/N2)2BL