5.1: Mechatronic Actuator Background

Learning Objectives

After completing this section, you will be able to:

1. Describe the fundamental working principles of a number of mechatronic actuators and their applications.
2. Distinguish the rotational and linear actuators and describe their differences and similarities.

In general, a mechatronic actuator converts one kind of energy (e.g., electrical energy) to another kind of energy (e.g., mechanical energy) based upon the command signal provided by the information domain. For example, a DC motor (see Chapter 2) converts the electrical energy (armature current) into rotational mechanical energy (motor shaft torque) based upon the desired torque (or armature current) controlled by the PWM (pulse-width-modulated) with its duty-cycle regulated by the command output from the information domain. A mechatronic actuator is defined below.

Definition 5.1

A mechatronic actuator is a device that generates force, motion, heat, or other forms of physical effects based upon a control command produced in the information domain, utilizing the supplied electric, hydraulic, or other forms of energies.

Remark 5.1

In a mechatronic system, actuators are essential for the mechanical components to track the desired trajectories such as displacement, speed profile, and temperature.

The majority of mechanical actuators can be divided into two main categories: rotational and linear. Rotational actuators are the most common and simple ones. Linear actuators are more expensive and complicated. Rotational actuators typically consist of a rotary shaft connected to an armature. A typical example is the brushed or brushless DC motor or step motor. The DC motor was briefly discussed in Chapter 2 regarding its modeling technique; it will be addressed in more detail in Sections 5.2 and 5.3. The permanent magnetic synchronous motor will be discussed in detail in Section 5.6. Next, however, we will discuss the step motor.

A step motor [13], also called stepper motor, is an electric motor that rotates in a series of small angular steps, activated by control pulses, instead of rotating continuously. Step motors are often operated in an open-loop mode without position feedback. They can be set to any desired position constrained by the motor step resolution. Note that the step-motor position can be rapidly increased or decreased, creating continuous rotation. The motor can also be controlled to hold its position at a given step.

Figure 5.1 shows a schematic diagram of a bipolar step motor and two associated winding pairs. These are the upper-lower (green winding #1) and left-right (red winding #2), where the upper-lower windings are energized. The generated magnetic polarities are shown just below and above the winding number of 1. Note that other forms of the step motors can be found in reference [14]. The step-motor drive circuit for a bipolar step motor is a little bit more complicated than the single one. This is because the H-bridge drive is needed so that the directions of applied voltages to windings #1 and #2 can be reversed in order to move the step motor in the desired direction. Section 5.3 will discuss the H-bridge drive in detail.

Figure 5.2 presents the drive control signals for a bipolar step motor with a single step. The blue-dashed lines represent the voltages applied to winding terminals 1a and 2a. The green lines are associated with the voltages applied to winding terminals 1b and 2b. Note that the horizontal axis represents the time increment from left to right.

The diagram in Figure 5.1 corresponds to the #1 location in Figure 5.3. Here, winding #1 is energized with positive voltage at 1a and negative at 1b, resulting in the N (north) and S (south) magnetic fields indicated in Figure 5.1. Meanwhile, winding #2 is not energized. Both terminals (2a and 2b) of winding #2 are connected to the battery negative terminal, leading to no magnetic field formed.

Figure 5.3 shows a clockwise rotation of 90 degrees for the bipolar step motor when the pulse sequence #1 to #4, shown in Figure 5.2, is applied to windings #1 and #2. We can see that for the #1 plot in Figure 5.3, the upper-lower windings are energized, with positive connected to 1a and negative to 1b. Left-right windings are not energized, leading to a locked position shown with the red-dot on top. Next, in the #2 plot, the left-right windings are energized, with positive connected to 2a and negative to 2b. Upper-lower windings are not energized, rotating the rotor clockwise by 30 degrees, and resulting in a locked position shown with the red-dot turned clockwise by 30 degrees.

Then, in the #3 plot, the upper-lower windings are energized, with positive connected to 2b and negative to 2a. Left-right windings are not energized, rotating the rotor clockwise by another 30 degrees, and resulting in a locked position shown with the red-dot turned clockwise by another 30 degrees. Last, in the #4 plot, the left-right windings are energized with positive connected to 2b and negative to 2a. Upper-lower windings are not energized, rotating the rotor clockwise by 30 degrees, resulting in a locked position shown with the red-dot turned clockwise by yet another 30 degrees.

The associated control signals can be found in Figure 5.2 for the first four control pulses. By repeating these pulses, we can realize continuous rotation. The anticlockwise rotation can be realized by reversing the pulse sequence.

For this bipolar step motor with three pairs of permanent magnets, the resolution for each step is 30 degrees. To increase the step-motor step resolution, either the number of winding polarities or the number of permanent magnet pairs (or both) needs to be increased. On the other hand, a speed-down gear can also be used to improve the step-motor step resolution.

From this discussion, we can see that step motors are used mainly to control or regulate the rotational position. They should not be used for controlling the rotational speed in open-loop because it is difficult for the step motor to generate a torque to achieve a desired speed with unknown load, while a DC motor can regulate motor output torque using closed-loop speed control by regulating its armature current. One advantage of using a step motor to control rotational position is that there is no need for position feedback; the resulting rotational position can be found by counting the applied number of pulses. However, to reach the desired position with applied pulse sequence, the generated torque during the rotational transition needs to be greater than the load torque.

Example 5.1

Consider the bipolar step motor shown in Figure 5.1. Find the formula for calculating the step angle and then calculate the step motor’s step angle. Note that the step angle of a given step motor can be calculated using the following formula.

\[SA=\dfrac{360{}^\circ }{2\times n_p\times n_m} \label{5.1}\]

where \(n_p\) is the number of polars and \(n_m\) is the number of permanent magnet pairs. For the step motor shown in Figure 5.1, the step angle \(SA\) can be calculated as follows, with \(n_p=2\) (bipolar) and \(n_m=3\) (three pairs of magnets).

\[SA=\dfrac{360{}^\circ }{2\times 2\times 3}=SA=30{}^\circ \label{5.2}\]

As a result, for the bipolar step motor shown in Figure 5.1, each stepping pulse leads to a thirty-degree rotation.

There are two main types of linear actuators. One uses a rotational actuator (DC motor or step motor) to drive a screw with a nut sliding on a rail (see Figure 5.4). The other employs the linear motor described in detail in this section.

The advantages of the linear actuator shown in Figure 5.4 are its simple concept and the utilization of the rotational actuator that can be easily controlled for regulating speed and displacement. However, the disadvantages are twofold. First, the response speed could be slow due to the conversion of the rotational speed to a linear one. Second, the gap between the screw and nut could lead to unwanted regulation error. Example 5.2 clarifies the relationship between motor rotational angle and the linear motion of the nuts.

Example 5.2

Consider the linear actuator shown in Figure 5.4. Assuming that the screw has one start with a screw pitch \(p_s\) of one-twentieth inch (or 1.27 mm), find the gain that converts the motor rotational angle to the linear displacement. Also find the linear displacement if the motor rotates \(180{}^\circ\). The gain can be calculated using the following formula.

\[G_{r2l}=\dfrac{p_s}{360{}^\circ }=\dfrac{1.27}{360{}^\circ }=0.003528\ \ mm/deg \label{5.3}\]

If the motor rotates \(180{}^\circ\) (\(θ_r\)), the linear displacement \(d_l\) can be calculated as follows.

\[d_l=G_{r2l}\times θ_r=0.003528\times 180=0.635\ mm \label{5.4}\]

Therefore, the resulting linear displacement is 0.635 (mm) if the motor rotates 180 degrees.

Let’s now discuss linear motors, which we will refer to as linear actuators. A linear actuator is an electric motor that has its state and rotor “unrolled” to a straight line. As a result, instead of producing a rotational torque, it produces a linear force (displacement) along its length. To illustrate the working principle of a linear actuator, we will use a linear step motor as an example. A continuous linear actuator can be constructed in a similar way. However, to produce a smooth linear force, the control of actuation solenoids is more complicated in order to minimize the ripple of the actuation force. Linear position sensors are often required to have precise control. Details of linear actuators can be found in references [15] and [16].

Figure 5.5 shows an eight-pole linear step motor. It can be viewed as an unrolled rotational step motor in Figure 5.1 with double the number of poles. The rotational step motor shown in Figure 5.1 has four poles. In Figure 5.5, the magnet in the middle moves horizontally. The solenoid for each pole can be driven by an H-bridge circuit (see Section 5.3); therefore, each pole can be set to either North or South. Here, both solenoids #1 and #3 are energized (marked in red). The pole associated with solenoid #1 is set to North and the pole associated with solenoid #3 is set to South. The remaining solenoids are not energized. In this figure, the magnet in the middle (actuator moving piece) is locked in place.

Table 5.1: Five-step right movement control sequence.

Moving Steps	Energized solenoids	Resulting Polarities
Step #0	Solenoid #1 & #3	N (#1) & S (#3)
Step #1	Solenoid #2 & #4	N (#2) & S (#4)
Step #2	Solenoid #3 & #5	N (#3) & S (#5)
Step #3	Solenoid #4 & #6	N (#4) & S (#6)
Step #4	Solenoid #5 & #7	N (#5) & S (#7)
Step #5	Solenoid #6 & #8	N (#6) & S (#8)

Let’s consider a control sequence that moves the actuator from the most left to the most right with a five-step movement. Because it is a linear step motor, we will move the actuator step-by-step. Table 5.1 presents the control sequence from Steps #1 to #5, where the solenoids are grouped as pairs. For example, for Step #0, solenoids #1 and #3 are paired together. Here, the positive is applied to solenoid #1, leading to a North pole for solenoid #1, while the negative is applied to solenoid #3, leading to a South pole for solenoid #3. In this case the rest of the solenoids are not energized.

Now let’s consider Step #2 control with solenoids #3 and #5 energized and the rest of the solenoids not energized. In this case, the positive is applied to solenoid #3, leading to a North pole for solenoid #3, while the negative is applied to solenoid #5, leading to a South pole for solenoid #5. Note that for the controls of Steps #0 and #2, the voltage applied to solenoid #3 has an opposite sign; solenoid #3 is negative for Step #0 and positive for Step #2. As a result, an H-bridge drive circuit is needed (see Section 5.3).

Figure 5.6 shows the process of moving the linear actuator from its most-left position in Step #0 to its most-right position in Step #5. Note that in Step #0, the linear step motor is in the initial locked state; solenoids #1 and #3 activated and the resulting polarities of the solenoids are shown on the top-left plot of Figure 5.6 (North and South for solenoids #1 and #3, respectively). In Step #1, solenoids #2 and #4 are activated with the rest of the solenoids not energized. This results in North and South polarities of solenoids #2 and #4, respectively, moving the center magnet to right with its locked position shown on the top-right plot of Figure 5.6.

Next, in Step #2, solenoids #3 and #5 are activated with the rest of the solenoids not energized. This results in North and South polarities of solenoids #3 and #5, respectively, moving the center magnet to right with its locked position shown on the middle-left plot of Figure 5.6. In turn, in Step #3, solenoids #4 and #6 are activated with the rest of the solenoids not energized. This results in North and South polarities of solenoids #4 and #6, respectively, moving the center magnet to right with its locked position shown on the middle-right plot of Figure 5.6.

Next, in Step #4, solenoids #5 and #7 are activated with the rest of the solenoids not energized. This results in North and South polarities of solenoids #5 and #7, respectively, moving the center magnet to right with its locked position shown on the bottom-left plot of Figure 5.6. Last, solenoids #6 and #8 are activated with the rest of the solenoids not energized. This results in North and South polarities of solenoids #6 and #8, respectively, moving the center magnet to right with its locked position shown on the bottom-right plot of Figure 5.6. This completes movement from left to right.

This description of the steps summarizes the operational principle of a linear step motor (actuator). The continuous linear motor (actuator) often has the same hardware as the linear step motor, but the control of these solenoids needs to be modified to produce a continuous linear force.

Now we are ready to discuss a class of discrete linear (solenoid) actuators that can be used alone as valve actuators or be used as the base solenoid actuator for an actuation system. For example, certain engine hydraulic intake/exhaust valve systems use two solenoids to control the hydraulic valve system [5], [17]. The discrete solenoid actuator normally has two states (displacement positions). One is the solenoid not energized state associated with the resting state of the actuator. The other is the solenoid energized state associated with the activated state of the actuator. A popular application of the discrete solenoid actuator is the internal combustion engine fuel injector, where the discrete solenoid actuator is used to activate the fuel injector when the solenoid is activated [18].

Figure 5.7 shows an example of a discrete linear solenoid actuator. Here, the left plot shows the resting state of the discrete solenoid actuator, and the right plot presents the activated state. Note that the rectangular blocks with cross represent the solenoid windings. On the left plot, the spring on the right of the solenoid winding is extended, leading to a retracted (closed) valve position, whereas on the right plot, the spring on the right of the solenoid winding is compressed due to the magnetic force produced by the activated solenoid, resulting in an extended (open) valve position.

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