Download 10 lec10 EE418 DC motor control

Automatic Control Systems
Lecture- DC Motor Modelling and Control
Emam Fathy
Department of Electrical and Control Engineering
email: [email protected]
http://www.aast.edu/cv.php?disp_unit=346&ser=68525
1
D.C Drives
• Speed control can be achieved using
DC drives in a number of ways.
• Variable Voltage can be applied to the
armature terminals of the DC motor .
• Another method is to vary the flux per
pole of the motor.
• The first method involve adjusting the
motor’s armature while the latter
method involves adjusting the motor
field. These methods are referred to as
“armature control” and “field control.”
2
Armature Controlled D.C Motor
Ra
La
B
ia
u
eb
T
J

System constants
ea = armature voltage
eb = back emf
Ra = armature winding resistance
La = armature winding inductance
ia = armature winding current
Kb = back emf constant
Kt = motor torque constant
Jm = moment of inertia of the motor
Bm=viscous-friction coefficients of the motor
JL = moment of inertia of the load
BL = viscous friction coefficient of the load
N1/N2 = gear ratio
Armature Controlled D.C Motor
Ra
Input: voltage u
Output: Angular velocity 
La
B
u
ia
eb
T
J

Electrical Subsystem
dia
u  R a i a  La
 eb ,
dt
eb  back-emf voltage
Mechanical Subsystem
Tmotor  Jω  Bω
Armature Controlled D.C Motor
Ra
La
Power Transformation:
Torque-Current: Tm otor  K t ia
Voltage-Speed:
B
u
ia
eb
T
J

eb  K b ω
Kt: torque constant,
Kb: velocity constant For an ideal motor
Kt  Kb
The previous equations result in the following mathematical model:
di a

 Ra i a  K b ω  u
 La
dt

 Jω
   B  K t i a  0
Armature Controlled D.C Motor
Taking Laplace transform of the system’s differential
equations with zero initial conditions gives:
La s  Ra I a (s)  K b Ω(s)  U(s)

 Js  B Ω(s)-K t I a (s)  0
Eliminating Ia yields the input-output transfer function
Kt
Ω(s)

2
U(s) La Js  JRa  BLa s  BRa  K t K b
Armature Controlled D.C Motor
Reduced Order Model
Assuming small inductance, La 0

K t Ra 
Ω(s)

U(s) Js  B  K t K b Ra 
Armature Controlled D.C Motor
If output of the D.C motor is angular position θ then:
Ra
d

dt
or
La
B
( s )  s ( s )
u
ia
eb
J
T
θ
Which yields the following transfer function

K t Ra 

U(s) sJs  B  K t K b
(s)
Ra 
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 0.1
rad/sec with an input voltage of 1.
• For the unit step input , then the motor speed
output should have:
– Settling time less than 2 seconds
– Overshoot less than 5%
– Steady-state error less than 1%
11
Uncompensated System
12
Uncompensated System
 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.
13
Example
An armature controlled D.C motor runs at 5000 rpm when 15v applied at the
armature circuit. Armature resistance of the motor is 0.2 Ω, armature
inductance is negligible, back emf constant is 5.5x10-2 v sec/rad, motor torque
constant is 6x10-5, moment of inertia of motor 10-5, viscous friction coefficient
is negligible, moment of inertia of load is 4.4x10-3, viscous friction coefficient
of load is 4x10-2.
La
Ra
ea 15 v
ia
N1
Bm
eb
T
Jm
BL

JL
N2
L
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 is
reduced to half and torque is doubled.
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
Example
Since armature inductance is negligible therefore reduced order transfer
function of the motor is used.
Kt
ΩL(s)

U(s)
J eq Ra  Beq La s  Beq Ra  K t K b

La
Ra
ea 15 v

ia
N1
Bm
eb
T
Jm
BL

JL
N2
2
N 
J eq  J m   1  J L
 N2 
L
2
N 
Beq  Bm   1  BL
 N2 
Example 2
A field controlled D.C motor runs at 10000 rpm when 15v applied at the field
circuit. 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, viscous
friction coefficient is 0.003, moment of inertia of load is 4.4x10-3, viscous
friction coefficient of load is 4x10-2.
Ra
Rf
ef
if
Lf
La
ea
Tm
Bm ωm
N1
Jm
BL
JL
N2
L
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 is
reduced to 500 rpm.
Example 2
+
+
e
kp
r
_
La
Ra
+
ea
_
N1
+
ia
JM
BM
T
eb
_
BL
θ
JL
N2
-
if = Constant
c
Numerical Values for System constants
r = angular displacement of the reference input shaft
c = angular displacement of the output shaft
θ = angular displacement of the motor shaft
K1 = gain of the potentiometer shaft = 24/π
Kp = amplifier gain = 10
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
K = 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
n= gear ratio = N1/N2 = 1/10
System Equations
e(t)=K1[ r(t) - c(t) ]
or
E(S)=K1 [ R(S) - C(S) ]
(1)
Ea(s)=Kp E(S)
(2)
Transfer function of the armature controlled D.C motor Is given by
θ(S)
Ea(S)
=
Km
S(TmS+1)
System Equations (contd…..)
Where
K
Km =
RaBeq+KKb
And
Tm
RaJeq
=
RaBeq+KKb
Also
Jeq=Jm+(N1/N2)2JL
Beq=Bm+(N1/N2)2BL