DC Motor Transient Running Operation This demo simulates the running of a DC motor with constant field or pm excitation 5 hp 240V 1200 rpm rated torque Tr = 30 N.m BLOCK DIAGRAM MODEL OF
Trang 1DC Motor Control: Theory and Implementation
By Chung Tan Lam
CIMEC Lab.
I Introduction
A DC motor speed drive
The mathematical model of dc motor (permanent magnet type) can be expressed by these equations
where
a
V : Armature voltage [V]
a
i : Armature current [A]
e
T : Electromagnetic torque [N.m]
L
T : Load torque [N.m]
a
L : Armature inductance [H]
a
R : Armature resistance [Ω ]
K : Coupling coefficient [N.m/A]
J : Momen of inertia [Kg.m 2 ]
m
B : Damping coefficient
The block diagram of a cascade closed-loop speed control of the dc motor is shown below.
DC MOTOR MODEL
Trang 2CASCADE SPEED CONTROL OF A DC MOTOR DRIVE
Typical dynamic responses are also shown The motor is initially at standstill and
at no load when a step command in speed is applied; when steady-state conditions are reached, a reversal of speed is commanded followed by a step load application
The system is highly nonlinear due to the introduction of saturation needed to limit both the current delivered and the voltage applied to the motor The system is in the saturation mode when the errors are large; as a consequence, the controller functions as a constant current source, that is torque, resulting in the ramping of the speed since the load
in this example is a pure inertia The inclusion of saturation limits on the PI integrator is therefore necessary to provide antiwindup action The presence of the signum function in the torque expression is required in order to insure that the load is passive whether the speed is positive or negative (as is the case here)
Trang 31 DC Motor Transient Running Operation
This demo simulates the running of a DC motor with constant field or pm excitation (5 hp 240V 1200 rpm) rated torque Tr = 30 N.m
BLOCK DIAGRAM MODEL OF A DC MOTOR
Trang 4MOTOR TRANSIENT RUNNING OPERATION
Electrical system equation:
va = Ra.ia + La.dia/dt + ea
where ea = K.wm
Mechanical system equation:
Te = J.dwm/dt + Tl.signum(wm) + f(wm)
where Te = K.ia
Simulation Results
MOTOR SPEED [rad/s]
Trang 5MOTOR CURRENT [A]
2 DC Motor with bipolar PWM excitation
This demo simulates the running of a DC motor with PWM (bipolar) excitation (5
hp 240V 1200 rpm) rated torque Tr = 30 N.m
Trang 6DC Motor With Bipolar PWM Excitation
DC Motor Model
Simulation Results
Trang 7Voltage Average Applied To Motor
MOTOR SPEED (rad/s)
Motor Current (A)
3 DC Motor with unipolar PWM excitation
Trang 8This demo simulates the running of a DC motor with PWM (unipolar) excitation (5
hp 240V 1200 rpm) rated torque Tr = 30 N.m
DC Motor Model
Simulation Results
Trang 9Voltage Average Applied To Motor
Motor Speed (Rad/S)
Motor Current (A)
4 Automatic Starter of a DC Motor
Trang 10This demo simulates the starting of a DC motor with automatic starter
(5 hp 240V 1200 rpm) rated torque Tr = 30 N.m The starter is simulated by a speed-dependent effective armature resistance using a look-up table Mathematical model of the dc machine
=
=
+ +
=
+ +
=
a e
m a
L m
m e
a a a
a a a
Ki T
K e
T B dt
d J T
e i R dt
di L V
ω ω
where
a
V : Armature voltage [V]
a
i : Armature current [A]
e
T : Electromagnetic torque [N.m]
L
T : Load torque [N.m]
a
L : Armature inductance [H]
a
R : Armature resistance [Ω ]
K : Coupling coefficient [N.m/A]
J : Momen of inertia [Kg.m 2 ]
m
B : Damping coefficient
Simulation Results
Trang 11MOTOR SPEED
MOTOR CURRENT
Trang 12H-BRIDGE DRIVER MODULE
T itle
F u ll B rid g e D riv e r
A 4
W e d n e s d a y , M a y 1 1 , 2 0 0 5
R 5
1 K
A N 1
1
C T
E 1 7
A N 2
3
C T
4
E 2 5
C 2 6
U 7 P C 8 1 7
A N 1
1
C T
E 1 7
A N 2
3
C T
4
E 2 5
C 2 6
U 6 P C 8 1 7
1 2 3 4 5 6 7 8 9
1 0
1 1
1 2
1 3
1 4
J 5
C O N 1 4
O u t2
O u t1
R 1 3
4 x 4 7 0
D 1
R C 0 /T 1 C K I
C 6
1 0 3
V c c
1
IN
2
C O M
3
L O
4
V b 8
H O 7
V s 6
N C 5
U 1
IR 2 1 0 5
D 5 _ P W M 1
2 4 V D C
1 5 V D C
R 6
5 0 K
O u t1
D 7 _ P W M 2
2 4 V D C
1 5 V D C
O u t2
H O 1
H O 1
R C 2 /C A P 1
3
2 1
-+
U 3 A
L M 3 5 8
5
6 7
-+
U 3 B
L M 3 5 8
V s 1
2 3 4 5
R 2 4
4 x 4 7 0
D 1 3
D 0
D 1
D 1 4
D 2
D 3
D 1 5 D 1 6
L O 1
+ C 1 2
3 3 u F -6 4 V
D 5
C 1
1 0 4
H O 2
R E 2
V s 2
2 3 4 5
R 1 4
4 x 2 2 K
L O 1
L O 2
5 V D C
In 4
In 3
D 6
V s 1
L O 2
5 V D C
In 2
D 7
In 1
+ C 1 3
3 3 u F -1 6 V
5 V D C
R 1 5 4 7 K
S e n s in g
D 8
In 1
R E 1
5 V D C
D 2
D 1
C 9
1 0 4
D 4
D 3
D 6
D 5 _ P W M 1
R 1 6 4 7 K
R C 4 /S D I
D 7 _ P W M 2
R B 0 /IN T 0
R B 4
R B 1 /IN T 1
R C 5 /S D O
In 2
A N 0
R E 2
R E 1
R B 5
2 4 V D C
G N D
R C 1
R C 2 /C A P 1
1
2
3
4
5
6
7
8
9
1 0
1 1
1 2
1 3
1 4
J 1
C O N 1
1 2 3 4 5 6 7 8 9
1 0
1 1
1 2
1 3
1 4
J 2
C O N 1
P O R T V C C
R C 0 /T 1 C K I
V C C _ IO
E X T V C C
R E 0 /R D
R E S E T #
R C 3 /S C K
D 0
S e n s in g
In 4
R 1 8 4 7 K
A N 0
D 2
V c c
1
IN
2
C O M
3
L O
4
V b 8
H O 7
V s 6
N C 5
U 2
IR 2 1 0 5
C 2
1 0 4
H O 2
IN
1
O U T 3
U 4 K I7 8 1 5
V s 2
1
2
3
4
5
J 3
E n c o d e r
R B 0 /IN T 0
2 3 4 5
R 1 9
4 x 4 7 0
2 4 V D C
R 2 0
4 x 1 5 0
R B 5
2 3 4 5
R 2 1
4 x 2 2 K
1 5 V D C
2 3 4 5
R 2 2
4 x 4 7 0
A N 1
1
C T
E 1 7
A N 2
3
C T
4
E 2 5
C 2 6
U 5 P C 8 1 7
R B 4
D 4
D 3
O u t2
C 7
1 0 4
O u t1
5 V D C
R 4
1 K
R B 1 /IN T 1
R C 1
5 V D C
Trang 13(H-Bridge Driver + 2 output +4 input + 4 Indicator LED’s)
T it le
H -B rid g e P o w e r M O S F E T A
W e d n e s d a y , M a y 1 1 , 2 0 0 5
G N D
O u t p u t 2
O u t p u t 1
I n p u t 3
G N D 3
I n p u t 4
G N D 4
1
2
3
4
5
6
7
8
9
1 0
1 1
1 2
1 3
1 4
J 3
T o H -B rid g e D riv e r
O u t 2
O u t 1
V s 1
H O 1
V s 2
I n p u t 1
I n p u t 2
I n p u t 3
C u r rA m p
1
T 1
I R F 5 4 0
1
Q D
I R F 5 4 0
O u t p u t 1
O u t 1
1
Q C
I R F 5 4 0
M 1
C 3
1 0 4
V S 1
V S 2
H V
H V
1
Q B
I R F 5 4 0
+
C 4
1 0 0 0 u F -6 4 V
1
T 2
I R F 5 4 0
O u t p u t 2
O u t 2
1
Q A
I R F 5 4 0
R 1
0 0 1 -1 W + C 53 3 u F -6 4 V
F 1
F U S E
1 2 3 4 5 6 7 8
J 1
O u t p u t 1
1 2 3 4 5 6 7 8
J 2
O u t p u t 2
M 2
H V
G N D
H V
C 6
1 0 4
R 2
0 0 1 -1 W
C u rrA m p
R 3
1 0
R 4
1 0
R 5
1 0
R 6
1 0
H O 1
L O 1
H O 2
L O 2
I n p u t 1
G N D 1
I n p u t 2
G N D 2
Trang 14Power MOSFET Module
T itle
C A N _ U S B In te rf a c e
A 4
W e d n e s d a y , J u ly 2 8 , 2 0 0 4
D B 2
D B 1
P O R T V C C
E X T 5 V
D B 4
D B 3
1 2
J P 2
U S B 5 V
D B 6
D B 5
1 2
J P 3
E x t 5 V
D B 7
R C 4 /S D I
R B 4
R C 1
R B 0 /IN T 0
R C 5 /S D O
V C C
R B 1 /IN T 1
(11,12,13,14): Only for USB Controller
1 2 3
C N 1 3
A D 2
V C C
A N 2
C 7
1 0 4
M C L R
R E 2 /C S
R e s e t_ S W
A N 0
1 2 3
C N 1 1
A D 0
R B 5
R E 1 /W R
R C 0
S W 2
S ta rt
1 2 3
C N 1 4
A D 3
V C C
A N 3
C 8
1 0 4
1
2
3
4
C N 5
A u x _ P o w e r
M C L R
(10): V+=12 or 24VDC depend on Extended board
R 6
1 0 K
R C 3 /S C K
1 2 3 4 5 6
C N 1 0
IC D 2
V C C
1 2 3
C N 1 5
A D 4
V C C
A N 4
C 9
1 0 4
R 7
1 5 0
1
2
3
C N 7
E x t_ 5 V
R B 3 /C A N R X
Power Management
R C 3 /S C K
A N 3
R B 2 /C A N T X
R A 4
1 2
9
1 0
R L Y 1
R e s e t
R A 4
V +
A N 4
R E 0 /R D
R 1
1 0 K
T X D
1
G N D
2
V C C
3
R X D
4
R s 8
C A N H 7
C A N L 6
V re f 5
U 2
M C P 2 5 5 1
V C C
V C C
S W 1
R e s e t
1 2 3 4 5
C N 8
S P I
C 2
E X T 5 V
V C C
V C C
R C 7 /R X
R C 2 /P W M 1
SDO
E X T 5 V
SDI
V C C
SCK
R E 1 /W R
R C 3 /S C K
R E 2 /C S
R B 4
CA NL
CA NH
C 4
1 0 4
CA NL
CA NH
R B 0 /IN T 0
R e s e t_ S W
R A 0 /A N 0
2
R A 1 /A N 1
3
R A 2 /A N 2
4
R A 3 /A N 3
5
R A 4
6
R A 5 /A N 4
7
R B 0 /IN T 0 3 3
R B 1 /IN T 1 3 4
R B 2 /C A N T X 3 5
R B 3 /C A N R XR B 4 3 6
3 7
R B 5 /P G M 3 8
R B 6 /P G C 3 9
R B 7 /P G D 4 0
R C 0
1 5
R C 1
1 6
R C 2 /C C P 1
1 7
R C 3 /S C K
1 8
R C 4 /S D I 2 3
R C 5 /S D O 2 4
R C 6 /T X 2 5
R C 7 /R X 2 6
G N DV D D 3 1
3 2
M C L R
1
O S C 1
1 3
O S C 2
1 4
V D D
1 1
G N D
1 2
R E 0 /R D /A N 5
8
R E 1 /W R /A N 6
9
R E 2 /C S /A N 7
1 0
R D 0 /P S P 0
1 9
R D 1 /P S P 1
2 0
R D 2 /P S P 2 2 1
R D 3 /P S P 3 2 2
R D 4 /P S P 4 2 7
R D 5 /P S P 5 2 8
R D 6 /P S P 6 2 9
R D 7 /P S P 7 3 0
U 1
P IC 1 8 F 4 5 8
V C C
8
O U T 5
G N D 4
N C
1
Y 2
O S C 4 0
V C C
D 1
L E D
R 2
1 2 0
R B 1 /IN T 1
D B 2
D B 3
D B 4
C 3
1 0 4
D B 5
D B 6
1 2
C N 1
R e m o te C P U R e s e t
D B 7
R 3
3 3 0
R C 7 /R X
R B 2 /C A N T X
1 2 3 4
C N 2
R S 2 3 2
R C 4 /S D I
R 4
1 5 0
SCK
D B 0
J P 1
J U M P _ 1
R B 3 /C A N R X
D B 1
R 5
1 5 0
SDI
1 2 3 4 5
C N 6
S P I
V C C RX
SDO
TX
A N 0
A N 1
V +
C 1
1 0 4
A N 2
V C C
G N D
R C 4 /S D I
V C C
1 2 3 4
C N 4
C A N _ O U T
R B 5
V C C
V +
1 2 3 4
C N 3
C A N _ IN
G N D
R B 7
A N 0
R C 6 /T X
R C 5 /S D O
E X T 5 V
1 2 3 4
C N 9
A u x _ P o w e r
G N D
V +
R e s e t_ S W
R C 6 /T X
E X T V C C
R C 5 /S D O
R B 6
R C 2 /P W M 1
C 5
1 0 4
R C 1
1 2 3 4 5 6 7 8 9
1 0
1 1
1 2
1 3
1 4
J 1
C O N 1 4
1 2 3 4 5 6 7 8 9
1 0
1 1
1 2
1 3
1 4
J 2
C O N 1 4
R B 7
R C 4 /S D I
R C 0
R E S E T #
G N D
R C 5 /S D O
E X T V C C
P O R T V C C
R E S E T #
V C C _ IO
R B 6
1 2 3
C N 1 2
A D 1
R E 0 /R D
V C C
A N 1
C 6
1 0 4
V C C _ IO
D B 0
R C 3 /S C K
Motor Controller (PIC18F458 + 5AD’s + CAN Comm.)
Trang 15MEASURING MOTOR PARAMETERS
These are the motor parameters that are need:
Note that the above values are stated for a single winding with dc motors, and are the phase values for a BLDC motor Brushless dc motors (BLDC) are usually 3 phase synchronous motors used in a configuration to be treated as dc drives Also note, it is assumed that dc motors being discussed have a permanent magnet field supply Wound field motors are not part of this discussion
MOTOR RESISTANCE
For the winding resistance use an ohmmeter For a DC motor measure the armature resistance between the 2 armature wires The ohmic value of the armature resistance will
be very small, thus a high sensitivity ohmmeter will be needed If it is a WYE connected BLDC motor, the armature resistance is the line-to-line resistance Thus divide the resistance (l-l) by 2 to get the phase resistance The ohmic value will also be very small
shaft at some speed [rpm] such as 1000rpm With a dc motor, use a DC voltmeter to
rad/sec Convert rpm to rad/sec as
] [
] [
or sec]
/ [
] [
sec sec
60
min radias
2 min
rpm speed
v Volts rad
speed
v Volts K
rad x
rev x
rev
= π
With a BLDC motor use an ac voltmeter to measure the voltage between any 2 wires of the 3 motor wires and then convert the line-to-line voltage to the phase voltage
Trang 1673 1
) (
) (phase K line to line
as-
=
rpm
volts K
amp
in lb
T
) (
00684 0
The derivation for the above equation is Equate Electrical Power to Mechanical Power converted to Watts
356 1
x 12
T
x
x 60
2
x
x
Where
N:rpm
Rearranging terms and simplifying:
I
T N
E
x 00684 0
= Where
=
=
=
A
in lb K
I
T
rpm
volts K
N
E
T
l l rms e
constant, Torque
constant, emf
Converting rpm to rad/sec
=
=
−
sec
554 9 2
x sec
60
x min
x 2 min
sec 60
x
rad
v rad
v rev
rev v
π π
Thus
Trang 17
=
−
=
=
−
sec
8 1396
00684 0
554 9
00684 0 554 9
) (
rad
v K amp
in lb K
K K
K K
l l e T
e T
T e
Motor rotor inertia can be measured by making an experiment The inertia can be calculated from the equation
=
sec
rad on accelerati
x sec -in -lb Inertia ]
[
on torque
Also (rearranging terms)
[ sec 2] on
accelerati
torque accel Inertia = lb−in−
DC MOTORS
To do this test it is necessary to measure the acceleration of the motor rotor and the acceleration torque of the motor rotor These two parameters are described as follows:
ACCELERATION – This parameter is determined by putting a step in current into the
motor winding to bring the motor up to rated speed The motor will accelerate exponentially The acceleration is a measure of the rate of change of velocity over a period of time To make this test, a dc tachometer should be connected to the motor shaft The output of the tachometer should be connected to a strip chart recorder When a step input in current is applied to the motor winding, the chart recorder will plot the rate of change of the motor shaft velocity as a function of the tachometer output voltage The tachometer calibration can be used to convert volts to rpm The acceleration is therefore the change in velocity for the linear part of the exponential curve (sometimes refered to
as the 66% response of the exponential change in velocity) divided by the time elapsed for the detected rate of change in velocity The resulting calculation of the acceleration must have the dimensions changed to be in units of rad/sec2
TORQUE- An ammeter must be inserted in series with the motor input winding When the step in input current is applied to the motor input, the maximum value of current should be noted This current must be converted to torque The torque is equal to the maximum value of current observed multiplied by the motor torque constant Torque [lbin]= amps[a] x KT [lb-in/a] The inertia is then the acceleration torque divided by the acceleration - Inertia= Accel torque[lb-in]/acceleration[rad/sec2]=lb-in-sec2
Trang 18BLDC MOTORS- The tests described thus far must be modified for a bldc motor A
bldc motor must be tested with its servo amplifier The step input will be a step in dc voltage to the servo amplifier input The voltage step should be large enough to cause the motor to reach rated speed With bldc motors it is not possible to measure the high frequency phase current in a WYE connected motor Thus some other procedure must be used to measure the current or torque and velocity Commercial bldc servo amplifiers have two dc output test points One output is a dc voltage proportional to velocity with a given calibration The voltage can be directly connected to a strip chart as described previously to measure the motor acceleration, A second test output provides a dc voltage proportional to torque or a percentage of rated torque This calibrated voltage can also be recorded with a stripchart recorder to observe its maximum value for a step voltage input
to the servo amplifier The inertia can thus be calculated as done previously
MOTOR INDUCTANCE L
This procedure assumes a model as a series equivalent R L circuit To measure the motor inductance uses a very low voltage ac source to the motor winding The resistance and inductance will be very small values For a magnetic flux field dc motor, apply the ac voltage to the armature winding For a BLDC motor apply the ac voltage to one pair of the three wires In both cases measure the voltage and the current
Remember that the BLDC motor is usually connected in WYE Thus the readings will be line-to-line You want the phase values for the voltage, so divide the voltage by 2 The impedance of the dc motor or the BLDC motor phase winding is
then-2 2
2 2
tan impedance
s]
var reactance[
] [ tan
reacctance impedance
The
] [ [amps]
current line
[volts]
voltage phase
or voltage Armature
Impedance
ce resis
ohms ce
resis
ohms
−
=
= +
=
=
=
Solve for the reactance from this equation Note that the phase resistance was measured previously and will be small in value The inductance can than be calculated from
inductance
x frequency
x 2 s]
var Reactance[ = π The frequency is probably going to be from the ac source at 60 Hz Thus
Hz]
frequency[
x 2
s]
var reactance[
[henries]
Inductance
π
=