5.2: Mechatronic Actuator Background

Learning Objectives

After completing this section, you will be able to:

1. Introduce the DC-motor operational principle.
2. Describe how the DC-motor torque is generated based upon the Lorentz principle.
3. Define the stall (maximum) torque and no-load (maximum) speed of a DC motor.

Note that there are mainly two types of motors: AC (alternating current) and DC (direct current) motors. In general, AC motors have high-power level and density; they are good for variable speed applications. DC motors have low-power, high torque, and variable speed; they are good for mechatronic applications.

In this section, we will address the DC-motor operational principle and its characteristics, especially the relationship between motor output torque and the resulting speed for a given armature voltage. Note that an electric motor converts electric energy to mechanical energy. Electric motors operate through electromagnetism, e.g. the interaction between the magnetic field and electric current. During the operational process, heat is generated because energy conversion efficiency is not 100%; see Figure 5.8 for an illustration. One of the heat sources is the armature winding resistance \(R_a\). The armature current \(i_a\) flow into the armature winding generates the heat of \(e_{heat}=R_ai^2_a\). In this section, we will focus on the permanent magnetic DC motors.

In Section 2.4, mechatronic system modeling, we described a DC motor by four differential equations associated with equations \ref{2.101}, \ref{2.103}, \ref{2.104}, and \ref{2.106}. These are listed in equation \ref{5.5} with the same variables defined.

\[\left\{ \begin{array}{c} \begin{array}{cc} e_a\left(t\right)= & R_ai_a\left(t\right)+L_a\dfrac{di_a(t)}{dt}+e_m\left(t\right) \ τ\left(t\right)= & K_τi_a\left(t\right) \end{array} \\ \begin{array}{cc} e_m\left(t\right)= & K_m\dot{θ}\left(t\right)=K_mω\left(t\right) \ \ J\ddot{θ}\left(t\right)= & τ\left(t\right)-b_m\dot{θ}\left(t\right)-τ_l\left(t\right) \end{array} \end{array} \right. \label{5.5}\]

Recall that both DC-motor torque and back EMF coefficients, \(K_τ\) and \(K_m\), were given without providing their derivations.

Theorem 5.1 [19], [20]

In a magnetic field with flux density \(B\), the Lorentz magnetic force law states that the magnetic force \(F\) applied to a cable with length \(L\) and supply current \(i_a\) can be calculated by integrating over the coil length, resulting in the following equation.

\[F=i_a\cdot L\cdot B \label{5.6}\]

The direction of generated force can be determined by Fleming’s left-hand rule, shown in Figure 5.9. The associated variables in equation \ref{5.6} are marked in Figure 5.9 and Figure 5.10.

Figure 5.9 shows two wires moving in a magnetic field \(B\) formed by two magnets, indicated by the red (north) and green (south) bars. The dot inside the circle indicates the current \(i_a\) flowing out of the surface; the circle represents the wire. The cross inside the circle indicates the current \(i_a\) flowing into the surface, where the direction of the generated force can be determined by the left-hand rule shown in Figure 5.9.

Now, let’s consider a single motor winding in a magnetic field \(B\) as shown in the left plot of Figure 5.10. Here, the winding radius is \(R\), winding depth is \(L\), the current flow into and out of the winging is \(i_a\), and the generated force is represented by \(F\). Using the left-hand rule shown in Figure 5.9, we can apply equation \ref{5.6} to calculate the resulting force due to the current flow \(i_a\) inside the wire with length \(L\), given magnetic field \(B\) with the force direction indicated on the left plot of Figure 5.10, and the associated torque generated by the single winding motor rotor.

\[τ=2\cdot i_a\cdot L\cdot B\cdot R \label{5.7}\]

For a motor rotor inside uniform magnetic field \(B\) with the number of winding equal to \(N\), we can calculate the generated torque by the following equation.

\[τ=2\cdot N\cdot i_a\cdot L\cdot B\cdot R=K_τi_a \label{5.8}\]

where \(K_τ=2NLBR\). However, in practice, for a given motor design, the magnetic field may not be uniform, and the individual winding radius is not the same. Therefore, coefficient \(K_τ\) often needs to be determined experimentally. The equations derived can be used as an initial calibration to determine coefficient \(K_τ\).

Now, we are ready to address the process of deriving the back EMF coefficient \(K_m\) due to the counter EMF (electromotive force) voltage. We will first introduce the Lenz law [21] below.

Theorem 5.2 [21]

The Lenz law states that the induced electromotive force with different polarities results in a current whose magnetic field opposes the change in magnetic flux through the loop. The reason is to ensure that the original flux is maintained through the loop when current flows in it.

The process of generating counter EMF voltage on the motor winding is shown in Figure 5.11. This process starts with applying an armature voltage to the DC-motor rotor winding that generates an armature current. This generates torque and rotor motion. Because the rotor winding wire cuts through the magnetic field, an opposing voltage to the applied armature voltage, called back EMF voltage, is generated. This voltage is proportional to the DC-motor rotor speed because the back EMF voltage is proportional to the wire moving speed relative to the magnetic field.

Based on Lenz’s law [21], after certain simplifications, the voltage generated by back EMF effect can be calculated by

\[V_m=2\cdot N\cdot L\cdot B\cdot R\cdot ω=K_mω \label{5.9}\]

where parameters \(N\), \(L\), \(B\), and \(R\) were defined above; \(K_m=2NLBR\) is the DC-motor back EMF voltage coefficient; and \(ω\) is motor rotor angular speed. Because EMF increases as the motor speed increases, the motor output torque reduces, making it easy to control motor speed.

Example 5.3

As discussed above, using international units, what is the unit for the DC-motor torque coefficient \(K_τ\)?

The torque coefficient converts the armature current \(i_a\) in \(A\) (ampere) to the motor-shaft output torque \(τ\) in Nm (Newton-meter). As a result, the unit for \(K_τ\) is shown below.

\[K_τ: Nm/A \label{5.10}\]

Next, we will provide a rough idea of how the DC motor works using a single pole (winding pair) rotor as an example. For a permanent magnet DC motor, its rotor assembly consists of a pair of fixed brushes (represented by black rectangular blocks) that supply DC power to the rotor windings, as well as the rotating commutator (represented by brown half circles) in the middle. In the right plot of Figure 5.12, the motor state consists of two permanent magnets denoted by N and S on the outskirt of the motor. In order to make the DC-motor rotor rotate, the power (armature voltage) applied to the rotor windings needs to be switched through the commutator in the middle.

Figure 5.13 illustrates this process for a two-pole rotor (pair of windings). Here, the battery is connected to the two fixed motor brushes. The two-pole commutator attached to the motor rotor rotates along with the rotor and windings. When the rotor rotates in order to keep the output torque remaining in the same direction, the current directions of both windings are changed by the commutator.

Figure 5.13 uses five steps for a \(180^\circ\) anticlockwise rotation with an initial angle of 270\({}^\circ\). Starting with Step #1, with rotational angle equal to \(270^\circ\) (see top-left plot), the power is supplied to the two windings through the commutator. It generates two forces \(F\) in the opposite directions due to the stator magnetic field. This leads to a generated torque of \(T=2RF\). This torque rotates the motor rotor in the anticlockwise direction, where \(R\) is the orthogonal distance from generated force \(F\) to the center of the axis. Note that you can find the generated torque at the #1 location on the top-left plot of Figure 5.14 with known armature current \(i_a\) and DC-motor torque coefficient \(K_τ\).

In Step #2 with rotational angle of \(225^\circ\) (see bottom-left plot of Figure 5.13), we see the power is still supplied to the two windings in the same direction as Step #1. It generates two forces \(F\) in the opposite directions due to the stator magnetic field. This leads to a generated torque of \(T=2RF\left| \sin 225^\circ \right|\) (see “#2” mark on the bottom-left plot of Figure 5.14), which rotates the motor rotor in the anticlockwise direction.

In Step #3 with rotational angle of \(180^\circ\) (see top-middle plot of Figure 5.13), the power is not supplied to the two windings because they are disconnected and generate no force. This leads to a generated torque of \(T=0\) (see “#3” mark on the top-middle plot of Figure 5.14). Even though the generated torque is zero, the motor rotor continues rotating due to its rotational moment.

In Step #4 with rotational angle of \(135^\circ\) (see bottom-right plot of Figure 5.13), the power is now supplied to the two windings in the opposite direction to Step #1. It generates two forces \(F\) in opposite directions due to the stator magnetic field. This leads to a generated torque of \(T=2RF\left|\sin 135^\circ \right|\) (see “#4” mark on the bottom-right plot of Figure 5.14), which rotates the motor rotor in the anticlockwise direction.

Finally in Step #5 with rotational angle of \(90^\circ\) (see top-right plot of Figure 5.13), the power is now supplied to the two windings in the opposite direction to Step #1. It generates two forces \(F\) in the opposite directions due to the stator magnetic field. This leads to a generated torque of \(T=2RF\) (see “#5” mark on the top-right plot of Figure 5.14), which rotates the motor rotor in the anticlockwise direction.

In Figure 5.14, we can see from the top-right plot that the two-pole DC motor in the top-left plot has a very large torque ripple. The motor output torque is in a form of an absolute sinusoid function varying from zero to peak torque output of \(2RF\) (100% ripple).

To reduce the output torque ripple, we need to increase the number of motor poles (winding pairs). Again, in Figure 5.14, we can see the four-pole DC motor in the bottom-left plot with its torque output profile (red trace) in the bottom-right plot. The torque ripple reduces significantly. The blue-dashed and green dash-dot curves on the bottom-right represent the torque generated by each pair of rotor windings. Because the two torque curves complement each other, the torque ripple can be reduced. We will use Example 5.4 to demonstrate how to calculate the level of torque ripple.

Example 5.4

Consider the four-pole DC motor with two pairs of the winding shown in the bottom-left of Figure 5.14. The green and blue torque curves can be expressed using the following equations, assuming \(R=F=1\).

\[ \begin{aligned} T_{blue}\left( θ \right) &=\left|2 \sin θ \right| \\[4pt] T_{green}\left( θ \right) &= \left|2 \sin \left(θ+\dfrac{π}{2}\right) \right| \end{aligned} \label{5.11}\]

The first intersection point of the two curves for \(θ>0\) is that \(θ=π/4\). In this case, we have the following.

\[T_{blue}\left(\dfrac{π}{4}\right)=2{\sin \dfrac{π}{4}\ }=T_{green}\left(\dfrac{π}{4}\right)=2{\mathrm{sin} \left(\dfrac{π}{4}+\dfrac{π}{2}\right)=\sqrt{2} } \label{5.12}\]

Therefore, \(T_{max}=2\sqrt{2}\) and \(T_{min}=2.\) As a result, the percentage ripple can be calculated below.

\[R_{\%}=\dfrac{T_{max}-T_{min}}{T_{max}}\times 100=\dfrac{2\sqrt{2}-2}{2\sqrt{2}}\times 100=29.3\% \label{5.13}\]

Increasing from one pair of poles to two reduces the torque ripple from 100% down to 29.3%, a reduction of 70.7%.

Now, we will introduce characteristics of the DC motor such as average peak torque and speed for a given armature voltage level. Using equation \ref{5.5}, and assuming a small inductor value by setting \(L_a=0\), we have the following equation.

\[e_a=\dfrac{R_a}{K_τ}τ+K_mω \label{5.14}\]

This equation provides the relationship of the DC-motor output torque \(τ\) and resulting speed \(ω\) for a given armature voltage \(e_a\). We can see from equation \ref{5.14} the relationship of motor torque and speed for a given armature voltage is an affine function (a mathematical function of linear transformation with a nonzero offset). We will define the DC-motor stall torque and no-load speed as below.

Definition 5.2

The stall torque \(τ_{stall}\) of a DC motor for a given armature voltage \(e_a\) is defined as the motor output torque when the motor speed is zero, which is equal to \(\dfrac{K_τ}{R_a}e_a\). The no-load speed \(ω_{no-load}\) for a given armature voltage is defined as the motor output speed with zero load torque, which is equal to \(\dfrac{e_a}{K_m}\).

Remark 5.2

Based on equation \ref{5.14} and Definition 5.2, we can reach the following conclusions. With fixed load, motor shaft speed \(ω\) is proportional to control voltage. With fixed control voltage, shaft speed is negatively proportional to torque load. Shaft output torque is proportional to motor current, independent of motor voltage. Motor efficiency is mainly impacted by its electric resistance and mechanical friction.

Figure 5.15 shows the relationship of DC-motor output torque \(τ\) and speed \(ω\), where the relationship varies as the armature voltage \(e_a\) changes. As the armature voltage increases, both stall torque \(τ_{stall}\) and no-load speed \(ω_{no-load}\) (marked in purple on Figure 5.15) increase. For a given armature voltage, the motor torque and speed relationship is affine, represented by a straight line.

Example 5.5

For the DC motor with \(R_a=0.1\ Ω\), \(K_τ=0.5\ Nm/A\), \(K_m=0.5\ V/(rad/s)\), and \(e_a=5\ V\), find the stall torque in Newton-meter (Nm) and no-load speed in rad/s. The stall torque in Nm can be calculated using the following equation.

\[T_{stall}=\dfrac{K_τ}{R_a}e_a=\dfrac{0.5}{0.1}\times 5=25\ Nm \label{5.15}\]

The no-load speed can be calculated using the equation below.

\[ω_{no-load}=\dfrac{e_a}{K_m}=\dfrac{5}{0.5}=10\ rad/s \label{5.16}\]

Therefore, the calculated stall torque is \(25\ Nm\) and the calculated no-load speed \(10\) rad/s.

In practice, a gearbox is often used to change the range of motor output torque and speed. Assuming that the gearbox does not have power loss (normally there are a few percents), we can state that the power input to and output of the gearbox remain the same. That is, \(P=τ\cdot ω=τ_o\cdot ω_o\), where \(τ\) and \(ω\) are input axis torque and speed and \(τ_o\) and \(ω_o\) are output axle torque and speed. Due to the assumption of no transmission loss, we have the following relationship.

\[η=\dfrac{N_o}{N_i}=\dfrac{ω}{ω_o}=\dfrac{τ_o}{τ} \label{5.17}\]

where \(η\) is gear ratio, \(N_o\) is the output shaft number of teeth, and \(N_i\) is the input shaft number of teeth. As a result, the gearbox output torque and speed are in an inverted relationship.

Figure 5.16 demonstrates the effect of a gearbox on the output torque and speed. In general, if \(η>1\), i.e., \(N_o>N_i\), at the gearbox output shaft, the stall torque increases, and no-load speed reduces. Thus, if one wants to increase the torque output, one needs to select a gearbox with \(η>1\), and vice versa.

Example 5.6

For a DC motor with its maximum stall torque of \(25\ Nm\) and no-load speed of \(10\ rad/s\), find a gearbox with its gear ratio of \(η\) so that the stall torque of gearbox output shaft is doubled with respect to the motor shaft. Also calculate the associated no-load speed at the gearbox output shaft. Using equation \ref{5.17}, we have

\[η=\dfrac{N_o}{N_i}=\dfrac{ω}{ω_o}=\dfrac{τ_o}{τ}=\dfrac{50}{25}=2 \label{5.18}\]

The no-load speed of the gearbox output shaft can be calculated using the equation below.

\[\dfrac{ω}{ω_o}=2 \to \ ω_o=\dfrac{ω}{2}=\dfrac{10}{2}=5\ rad/s \label{5.19}\]

Therefore, the calculated gear ratio is \(η=2\) and the calculated correspondent no-load speed is \(5\, rad/s\).

Below, we will discuss the transient characteristics of a DC motor. To be specific, we will investigate the acceleration process of a DC motor from zero speed to its steady-state speed with and without axle friction. Consider a DC motor described by equation \ref{5.5} with the motor parameters shown in equation \ref{5.20}.

\[R_a=0.1\ Ω, L_a=0, K_τ=0.5\ Nm/A K_m=0.5\ V/(rad/s), J=0.4\ Kgm^2, b_m=1\ Nm/(rad/s) \label{5.20}\]

First, let’s study the case with \(b_m=0\), that is, there is no axle friction torque. For the given parameters in equation \ref{5.20}, based upon dynamic equations in equation \ref{5.5}, we can derive the system transfer function from armature voltage \(E_a(s)\) to motor speed \(Ω\)(s) with \(b_m=0\) as below, drawing on equation \ref{2.120}.

\[\dfrac{Ω\mathrm{(s)}}{E_a(s)}=\dfrac{K}{Ts+1}=\dfrac{2}{0.16s+1} \label{5.21}\]

With the derived transfer function in equation \ref{5.21}, we can simulate the step armature voltage of \(E_a\left(s\right)=5/s\) (five times of unit step, 5 volts) in MATLAB using the command step. Figure 5.17 presents the simulation results. The blue line in the top plot shows a straight line between the stall torque (25 Nm) and the no-load speed (10 rad/s) that can be calculated by equations \ref{5.15} and \ref{5.16}. The blue curve in the bottom plot shows the time response of the DC-motor speed described by \ref{5.21}.

Because the system time constant \(T=0.16\), the 2% settling time (see Chapter 3) is \(t_s\left(2\%\right)=4\times 0.16=0.64\) second. That is, the speed response will reach 98% of the steady-state speed (in this case, no-load speed) in 0.64 second. The red dots in both the top and bottom plots show the migration process of the speed from zero to 10 rad/s corresponding to those in the bottom chart. Because there is no axle friction (\(b_m=0\)), the no-load speed of 10 rad/s is achieved at the steady state.

For the parameters given by equation \ref{5.20} with \(b_m=1\), and based upon the dynamic equations in \ref{5.5}, we can derive the system transfer function from armature voltage \(E_a(s)\) to motor speed \(Ω\)(s), as below.

\[\dfrac{Ω\mathrm{(s)}}{E_a(s)}=\dfrac{K}{Ts+1}=\dfrac{1.4286}{0.1143s+1} \label{5.22}\]

We can see that both DC gain \(K=1.4286\) and time constant \(T=0.1143\) are reduced, comparing with the no friction case. This result indicates the time response will be faster than the case without friction and the achieved steady-state speed will be reduced.

With the derived transfer function in equation \ref{5.22}, we can simulate the step armature voltage of \(E_a\left(s\right)=5/s\) (five times of unit step) in MATLAB using the command Step. Figure 5.18 presents the simulation results. The blue line in the top plot of Figure 5.18 shows a straight line between the stall torque (25 Nm) and no-load speed (10 rad/s) that is the same as these for the no friction case. The red line is the friction torque as a linear function of motor speed (\(bω\)). The orange line shows the available torque for accelerating the rotor. Note that the available torque for acceleration is significantly reduced from the case without friction torque.

The solid blue curve in the bottom plot of Figure 5.18 shows the time response of the motor system described by \ref{5.21} (no friction case); the red dash-dot line represents the time response of the motor system described by \ref{5.22} (friction case). Because the system time constant is \(T=0.1143\) for the friction case, the 2% settling time (see chapter 3) is \(t_s\left(2\%\right)=4\times 0.1143=0.4572\) second. That is, the speed response will reach 98% of the steady-state speed (in this case, \(ω=5\times K=7.143\ \)rad/s) in 0.4572 second, which is faster than the no friction case of 0.64 second.

The magenta dots in the top and bottom plots of Figure 5.18 show the migration process of the speed from zero to its steady state value of 7.143 rad/s. Due to the existence of axle friction (\(b_m=1\)), the no-load speed cannot be reached, and the steady-state speed becomes 7.143 rad/s (less than the no-load speed of 10 rad/s). Even though the case with friction torque converges to its steady state faster than the case without friction torque, the rotor speed without friction torque accelerates faster than the case with friction torque.

Example 5.7

For the DC motor with \(R_a=0.1\ Ω\), \(K_τ=0.5\ Nm/A\), \(K_m=0.5\ V/(rad/s)\), \(J=0.4\ Kgm^2\), and \(b_m=1\ Nm/(rad/s),\ \)derive the system transfer function from \(E_a(s)\) to output speed \(Ω\left(\mathrm{s}\right)\). Calculate the steady-state motor speed with \(e_a=5\ V\). From equation \ref{2.120}, we have the following.

\[K=\dfrac{K_τ}{R_ab_m+K_τK_m}=\dfrac{0.5}{0.1\times 1+{0.5}^2}=1.4286 \label{5.23}\]

The time constant can be calculated by the following equation.

\[T=\dfrac{JR_a}{R_ab_m+K_τK_m}=\dfrac{0.4\times 0.1}{0.1\times 1+{0.5}^2}=0.1143\ (sec) \label{5.24}\]

Therefore, the calculated \(K=1.4286\) \(rad/s\) and the calculated time constant \(T=0.1143\) second. The achieved steady-state speed in this case can be calculated by the following equation.

\[ω_{ss}=K\times e_a=1.4285\times 5=7.143\ rad/s \label{5.25}\]

In short, basic characteristics of DC motors are a) with a fixed load, the motor steady-state shaft speed is proportional to armature voltage; b) with fixed armature voltage, the motor steady-state shaft speed is negatively proportional to torque load; c) the shaft torque is proportional to the motor armature current (independent of motor armature voltage); and d) motor efficiency is mainly impacted by its armature electric resistance and mechanical shaft friction (generating heat).

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