For precise automated positioning, think stepper-motor-based linear actuators.

Linear actuators essentially generate force and motion through a straight line. In a typical mechanical system, a device’s output shaft would provide linear motion using a rotary motor through gears, a belt and pulley, or other mechanical components. Trouble is, these components must be coupled and aligned. Worse yet, they add wear elements such as friction and backlash to a system. For finer positioning needs, a more-effective and straightforward alternative comes from stepper-motor-based linear actuators.

These devices simplify the design of a machine or mechanism that requires precise linear positioning because they provide rotary-to-linear conversion directly inside the motor. The actuators move a given degree of rotary motion for every electrical-input pulse. This so-called “stepping” feature and the use of a precise leadscrew provides precise and repeatable positioning.

Stepper-motor basics
To see how the actuators work, it is helpful to understand the basics of stepper motors. Different kinds of stepper motors include variable reluctance (VR), permanent magnet (PM), and hybrid. This discussion centers on the hybrid stepper, which provides high torque and fine positioning resolution (1.8 or 0.9° step). In linear actuator systems, hybrids are found in devices such as X-Y tables, blood analyzers, HVAC equipment, small gantry robots, valve-control mechanisms, and automated stage-lighting systems.

Under the hood of a hybrid stepper sits a permanent-magnet rotor and a steel stator wrapped with a coil winding. Energizing the coil creates an electromagnetic field with north and south poles. The stator conducts the magnetic field, causing the rotor to align itself with the field. Because sequentially energizing and deenergizing the coil windings alters the magnetic field, every input pulse or step causes the rotor to incrementally move 0.9 or 1.8 rotational degrees, depending on the hybrid model. In a stepper-motor linear actuator, a threaded precision nut embedded in the rotor engages with the leadscrew (which replaces a conventional shaft).

The leadscrew provides a linear force using the simple mechanical principle of the inclined plane. Imagine a steel shaft with a ramp or inclined plane wrapped around it. The mechanical advantage or force amplification is determined by the angle of the ramp which is a function of screw diameter, lead (axial distance a screw thread advances in a single revolution), and pitch (axial distance measured between adjacent thread forms).

The leadscrew threads translate a small rotational force into a large load capability, depending on the ramp steepness (thread lead). A small lead provides a higher force but lower linear speeds. A large lead gives a lower force but a higher linear speed from the same source of rotary power. In some designs, the power nut embedded in the rotor is made of a bearing-grade bronze that lends itself to the machining of internal threads. But bronze is an engineering compromise between lubricity and physical stability. A better material is a lubricated thermoplastic with a much lower coefficient of friction at the nut-screw thread interface.

Stepping sequences
Schemes for driving a stepper motor include “one phase on” stepping and “two phase on” stepping.

In a “one phase on” sequence for a simpli ed two-phase motor, Step 1 shows Phase A of the energized stator. This magnetically locks the rotor since unlike poles attract. Turning Phase A o and B on makes the rotor move 90° clockwise (step 2). In Step 3, Phase B is o and Phase A on, but with the polarity reversed from Step 1. The causes the rotor to rotate another 90°. In Step 4, Phase A is turned o and Phase B is turned on, with polarity reversed from Step 2. The repetition of this sequence causes the rotor to move clockwise in 90° steps.

In the “two phase on” sequence, both motor phases are always energized, and only the polarity of one phase switches. This causes the rotor to align itself between the “average” north and the “average” south magnetic poles. Because both phases are always on, this method provides 41.4% more torque than “one phase on” stepping.

Unfortunately, although plastic works well for the threads, it is not stable enough for the bearing journals in the hybrid stepper design. That’s because under a continuous full-load condition, plastic journals can expand four times as much as brass journals. This amount is unacceptable because the motor design requires the stator-to-rotor air gap be only a few thousandths of an inch. A way around this problem is to injection mold plastic threads inside a brass sleeve that will be inserted into the permanent-magnet rotor. This approach increases motor life and provides low friction while maintaining bearing-journal stability.

Of the different types of Haydon actuators, “captive” devices have a built-in antirotation mechanism. This configuration provides a maximum stroke of up to 2.5 in. and suits applications such as precision fluid dispensing, throttle control, and valve movement. Other kinds of Haydon linear actuators are the “noncaptive” and “external linear” which suit applications needing a longer stroke such as the transfer of blood tubes by small gantry robots, X-Y motion systems, and imaging systems.

Sizing an actuator
An application example best shows how to size an actuator. Consider the following parameters:

Linear force required to move the load = 15 lb (67 N)
Linear distance, m, the load needs to be moved = 3 in. (0.0762 m)
Time, t, needed to move the load in seconds = 6 sec
Target number of cycles = 1,000,000

There are four steps to sizing a stepper-motor linear actuator: 1) Determine the actuator’s initial force rating needed to meet the required life; 2) Determine velocity in millimeters/second; 3) Choose the proper actuator frame size; and 4) Determine proper screw resolution based on force requirements.

The best way to predict life is through application testing, which is highly recommended. A technique using the Percent Load Versus Number of Cycles curve serves as a good first approximation. Stepper motors have no brushes to wear out, and they use precision, long-life ball bearings, so the main wear component is the power nut. Therefore, the number of cycles a device lasts while still meeting design specifications is a function of load.

Refer to the Percent Load Versus Number of Cycles chart to determine the correct sizing factor for the actuator to withstand the 1,000,000 cycles. This turns out to be 50% — a factor of 0.5. The initial rated force, N, required to meet the load after 1,000,000 cycles is therefore 15 lb/0.5 = 30 lb or 133 N.

Now determine the required linear mechanical power in watts:

P_linear = (N × m)/t

In our example, this becomes (133 × 0.0762)/6 = 1.7 W

With this data, use the Actuator frame size table to select the correct frame size. All stepper-motor linear actuators require a drive to send pulses to the motor. Note that the table lists power for both an L/R drive (constant voltage) and a chopper drive (constant current). Unless the application is battery powered (as in a handheld portable device), manufacturers highly recommend a chopper drive for maximum performance. In this example, a review of the chopper drive power specs in the table reveals the Haydon 43000 Series (Size 17 Hybrid) most closely meets the 1.7-W requirement. This selection meets the load requirements without over designing the system.

Next, calculate linear velocity (ips). This is given by m/t and comes to 3 in./6 sec = 0.5 ips. With optimized frame size (Size 17 Hybrid) and linear velocity (0.5 ips) in hand, use the appropriate Force Versus Linear Velocity curve to determine the proper resolution of the actuator leadscrew. In this case, the leadscrew resolution needed is 0.00048 in.

Recall that the leadscrew advances based on the number of input steps to the motor. Performance curves are expressed in both “ips” and “steps/sec.” To verify your selection, check the force at the required step rate by referring to the Force Versus Pulse Rate curve, where: Resolution chosen = 0.00048 in./step Required linear velocity = 0.5 ips Required step rate = (0.5 ips)/ (0.00048 in./step) = 1,041 steps.

Plotting 1,041 as the X-axis value (pulse rate) and drawing a perpendicular line from this point to the curve shows the Y-axis value (force) is 30. Therefore, the selection is correct.