Robert Repas
Associate editor
There’s a growing demand
for smaller linear
motors. Manufacturers
have developed new direct-
drive linear servos in
response. Standard rotary
servomotors spin leadscrews
or pull belt drives
to produce linear motion.
Direct-drive linear motors
are more likely to use
self-supporting 3Φ coils
to position a linear shaft
filled with permanent
magnets. Typical of motors
taking this approach
is the Quickshaft Series
from Faulhaber Corp.,
Germany.
The servomotor design lets the
controller drive the shaft to new positions
with almost no residual static
force. This makes these types of motors
good candidates for micropositioning.
Performance life is mostly
influenced by the effectiveness of
the sleeve bearings determined by
the operating speed and the applied
load.
Built-in Hall-effect sensors monitor
shaft position. For the Quickshaft
Series, the lack of an external
encoder lets the overall size of the
motor housing be about 0.5-in. wide,
0.75-in. tall, and 2-in. long. The short
2-in. housing can produce linear
shaft motions from ±10 to ±40 mm
at peak forces greater than 9 N.
Needless to say, such a novel
motor also requires great attention
to detail when planning applications.
For example, say the
motor positions a load on an
inclined surface. Obviously, to
move such a mass the motor must
overcome all forces opposing the
movement. Aside from external
forces pushing the load down
the incline, the motor must also
overcome the parallel force of the
mass sliding down the incline
from its own weight and the frictional
force of the mass moving
against the surface of the incline.
Among the first steps in selecting
a motor is the definition of a
speed profile representing various load movements. Equations
of uniform straight-line motion
(USM) and uniformly accelerated
motion (USAM) allow definitions
of various speed versus
time profiles.
Two of the more widely used
motion speed profiles are the
triangular and trapezoidal profiles.
The triangular speed profile
simply gives an acceleration
and deceleration time. For the factors
of displacement (s), speed (v),
acceleration (a), and time (t), the
formulas are:
For displacement:
s = 0.5vt = 0.25at^2 = v^2/a
For speed:
v = 2s/t = at/2 = SQR(as)
For acceleration:
a = 4s/t^2 = 2v/t = v^2/s
The trapezoidal speed profile divides the move into three parts: acceleration,
constant velocity or travel, and deceleration.
An optional fourth time period is used when
there’s no motion. For example, the lifting
of an object may force the motor to push
against gravity while the object remains stationary.
To simplify the calculations, the
three motion-related sections in the example
are all given equal times.
For displacement:
s = 2vt/3 = at^2/4.5 = 2v^2/a
For speed:
v = 1.5s/t = at/3 = SQR(as/2)
For acceleration:
a = 4.5s/t^2 = 3v/t = 2v^2/s
When defining the speed profile for motion,
the designer typically calculates the
maximum speed, how fast the mass should
accelerate, the distance of the movement,
and the length of the rest time.
The triangular or trapezoidal profile is
the usual choice if the movement parameters
are not clearly defined. For the purpose
of this example, assume a 500-g load must
move 20 mm in 100 msec along a slope at an
angle of 20°.
As the mass, distance, and time is known,
calculate the speed and acceleration:
Speed of the load:
vmax = 1.5s/t = 1.5(20x10^-3/100x10^-3) = 0.3
m/sec
Acceleration of the load:
a = 4.5
s/
t^2 = 4.5(20x10^-3/(100x10-3)^2) = 9
m/sec^2
Note for the t1 and t3 time periods, the
mass accelerates or decelerates respectively.
Therefore, the speed of the mass changes
from 0 to 0.3 m/sec or from 0.3 to 0 m/sec.
All values for the four time periods are recorded
into a table for easy tracking.
The parameters of speed, acceleration, distance, and time go
into calculating the amount of force needed to carry out each
action. Also needed is the coefficient of friction for the surface
the mass rests upon. For this example, assume the surface has a
0.2 coefficient of friction. The force due to friction is thus:
Ff = m x g x ∝ x cos(θ) = 0.5 x 10 x 0.2 x cos(20°) = 0.94 N
The parallel force, Fx, becomes:
Fx = m x g x sin(θ) = 0.5 x 10 x sin(20°) = 1.71 N
And the force due to acceleration is:
Fa = m x a = 0.5 x 9 = 4.5 N
Use the values calculated to complete a force table for the four
time periods as the mass moves up and down the incline. Watch
out for the polarity of the force signs, especially friction. When
the block is in motion, force must overcome friction so the sign
is positive.
However,
friction
helps hold
the block in
place when
it stops and
so the sign
becomes
negative.
Likewise,
the parallel force must be overcome when the block moves up the incline or stops, but
helps move the block down the ramp when descending.
With the forces of the four profile parts known, the peak and continuous
forces are next. The peak force is the highest absolute value of all of the total
forces. For this example, the highest force needed is 7.15 N.
Fp = max( |7.15|,|2.65|,|-1.85|,|0.77|,|3.73|,|-0.77|,|-5.27|,|-0.77|) = 7.15N
Continuous force is calculated by the equation:
Fe = SQR( SUM(
t x
Ft^2)/2 x SUM(
t) )
Plugging in all the values from the previous tables yields:
Fe = SQR( (0.033 x 7.15^2 + 0.033 x 2.65^2 + 0.033 x (-1.85)^2 + 0.1 x 0.77^2 +
0.033 x 3.73^2 + 0.033 x (-0.77)^2 + 0.033 x (-5.27)^2 + 0.1 x 0.77^2) /
(2 x (0.033 + 0.033 + 0.033 + 0.1) ) )
With the peak and continuous forces known, the selection of the motor is
simply a matter of looking up the two parameters in charts. The Faulhaber
motor specifications for this example show the LM 1247-020-01 motor has
a continuous force ability of 3.09 N, with a peak force of 9.26 N. The two
parameters from the example are well within those ratings.
There is one more step once the
motor has been chosen. The motor
will always push against the load
even when not moving, to prevent
it sliding down the incline. Designers
must assess the temperature of
the windings to assure the motor
will not suffer burnout.
Continuous motor current must
be calculated first to determine coil
temperatures. The force constant
(kf) of the motor is from its specification
sheet. The LM 1247 motor
for this example has a force constant
of 6.43 N/A. Divide the continuous
force value by the force constant to
determine continuous amps:
Ie = Fe/kf = 2.98/6.43 = 0.46 A
The final formula takes into
account the motor thermal resistance,
a reduced thermal resistance
when mounted in free air, ambient
temperature, and the resistance of
the coils.
Make Contact
MicroMo Electronics Inc., (800) 807-9166, micromo.com