Machine Design

How to Apply Direct-Drive Linear Servomotors

A novel direct-drive linear servomotor with only one moving part needs special design considerations.

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 = 4.5s/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,

This LM 1247 motor from Faulhaber is a typical example of small directdrive linear servomotors. A 3Φ delta-wound coil drives the output shaft that has embedded permanent magnets. Halleffect sensors monitor shaft position.

The linear motor must react against all forces attempting either to move the mass, m, or to prevent its motion.

The triangle speed profile has only an acceleration and deceleration leg.

The trapezoid speed profile incorporates a period of constant velocity between acceleration and deceleration. To simplify calculations, the segments are equally divided across the time period, t.

The total time of 100 msec for the example calculation is divided equally into three segments of the trapezoid speed profile. A fourth time period, t4, represents an idling period of no motion.

This table lists all of the motion parameters for each section of the speed profile.

A force table holds all force values for each segment of the motion profile. The polarities of several forces change during each segment. For example, the system must overcome friction. Yet, when stopped, friction helps hold the load in place.

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