Electromechanical systems - System Analysis and Control

Simplified DC motor models¶

Recall the electromechanical model of a DC motor from the previous section.

Figure 1:Model circuit for a brushed DC motor, showing electrical and mechanical components.

Also recall the equations of motion we derived.

\boxed{\begin{aligned} L \frac{\dd i}{\dd t} + R i + K \omega &= v_\textsf{in} \\ J \dot \omega + b\,\omega - K i &= - T_L \end{aligned}}

(1)

The inputs are the applied voltage $v_\textsf{in}$ and load torque $T_L$ , and the outputs are the motor current $i$ and angular velocity $\omega$ . The parameters are the motor inductance $L$ , resistance $R$ , motor constant $K$ , rotor inertia $J$ , and viscous friction coefficient $b$ .

Single ODE model¶

We can eliminate $i$ from Eq. (1) to obtain a single equation of motion in terms of $\omega$ . This is most easily done in the Laplace domain, so let’s start by taking the Laplace transform of Eq. (1) assuming zero initial conditions:

\begin{aligned} (L s + R) I + K \Omega &= V_\textsf{in} \\ (J s + b)\Omega - K I &= - T_L \end{aligned}

(2)

Solving the second equation for $I$ and substituting into the first gives

\bigl[(L s + R)(J s + b) + K^2\bigr] \Omega = K V_\textsf{in} - (Ls+R) T_L

(3)

We can expand this and transform back to the time domain to obtain

\boxed{J L\, \ddot\omega + (J R + b L)\, \dot\omega + (b R + K^2)\, \omega = K v_\textsf{in} - L\, \dot T_L - R\, T_L}

(4)

This is a second-order ODE in $\omega$ (just like a spring-mass damper or RLC circuit) that captures the full electromechanical dynamics of the DC motor.

Ideal motor model¶

For motors that are well-designed and well-built, the inductance $L$ and viscous friction $b$ are often quite small compared to the other parameters. Therefore, we can make the approximations $L \approx 0$ and $b \approx 0$ . We do not assume that the resistance $R$ is small, since even an ideal motor contains a large length of copper wire and resistance cannot be neglected. Setting $L=0$ and $b=0$ in Eq. (4) gives

\boxed{J R\, \dot\omega + K^2\, \omega = K v_\textsf{in} - R\, T_L}

(5)

This is a first-order ODE in $\omega$ that describes the dynamics of an ideal DC motor.

Motor performance analysis¶

With our models in hand, we can begin to answer questions about DC motor performance. In this section, we will look at the effect of motor parameters on speed, torque, and efficiency.

Steady-state speed¶

As we saw in the previous section, as the DC motor spins faster, it generates an increasing back EMF that opposes the applied voltage. So we expect the motor to reach a steady-state speed. We can find this speed by assuming constant inputs $v_\textsf{in}$ and $T_L$ and setting $\dot\omega = 0$ in our equations of motion (4). This produces

(b R + K^2)\, \omega = K v_\textsf{in} - R\, T_L

(6)

Solving for $\omega$ gives the steady-state speed. We can similarly solve for the steady-state current $i$ by setting $\frac{\dd i}{\dd t} = 0$ in Eq. (1). The results are:

\boxed{\omega_\textsf{ss} = \frac{K v_\textsf{in} - R\, T_L}{b R + K^2}, \quad i_\textsf{ss} = \frac{b v_\textsf{in}+K T_L}{b R+K^2}}

(7)

The steady-state speed changes in predictable as a function of the motor parameters:

Increasing the input voltage $v_\textsf{in}$ increases the speed and current.
Increasing the load torque $T_L$ decreases the speed and increases the current. This is known as droop.
Increasing the resistance $R$ or friction $b$ decreases the speed.

No-load speed and stall torque¶

Two important performance metrics are the no-load speed and the stall torque.

The no-load speed $\omega_\textsf{max}$ is the speed when there is no load torque ( $T_L = 0$ ). This is the fastest the motor can spin. From Eq. (7), we have
$\boxed{\omega_\textsf{max} = \frac{Kv_\textsf{in}}{b R + K^2} }$
(8)
The stall torque $T_\textsf{stall}$ is the load torque required to reduce the speed to zero ( $\omega = 0$ ). This is the maximum torque the motor can provide. This is also when current is maximized, so there is a corresponding stall current. From Eq. (7), we have
$\boxed{T_\textsf{stall} = \frac{Kv_\textsf{in}}{R},\quad i_\textsf{stall} = \frac{v_\textsf{in}}{R} }$
(9)

These calculations confirm the motor design trade-off we discussed in the previous section: if $bR \ll K^2$ , using a larger $K$ will increase the max torque but decrease the max speed, while using a smaller $K$ will increase the max speed but decrease the max torque.

Power balance¶

Returning to the full governing equations (1), we can multiply the first equation by $i$ and the second equation by $\omega$ to obtain power equations for the electrical and mechanical domains:

\begin{aligned} i \bigl(L \frac{\dd i}{\dd t} + R i + K \omega\bigr) &= v_\textsf{in} i \\ \omega \bigl(J \dot \omega + b\,\omega - K i\bigr) &= - T_L \,\omega \end{aligned}

(10)

Now use the product rule: $\omega \dot \omega = \frac{\dd}{\dd t}(\frac{1}{2}\omega^2)$ and similarly for $i$ . This gives

\begin{aligned} \frac{\dd}{\dd t} \left( \frac{1}{2} L i^2 \right) + R i^2 + K i \omega &= v_\textsf{in} i \\ \frac{\dd}{\dd t} \left( \frac{1}{2} J \omega^2 \right) + b\,\omega^2 - K i \omega &= - T_L \,\omega \end{aligned}

(11)

Add these equations together to obtain

\boxed{v_\textsf{in} i = T_L \,\omega + R i^2 + b\,\omega^2 + \frac{\dd}{\dd t} \left( \frac{1}{2} L i^2 + \frac{1}{2} J \omega^2 \right)}

(12)

This is an energy balance equation for the DC motor! The left-hand side is the electrical power supplied to the motor, while the right-hand side breaks down into five terms:

Power delivered to the load: $T_L \,\omega$
Power lost to electrical resistance: $R i^2$
Power lost to viscous friction: $b\,\omega^2$
Rate of change of stored magnetic energy in the inductor: $\frac{\dd}{\dd t} \left( \frac{1}{2} L i^2 \right)$
Rate of change of kinetic energy of the rotor: $\frac{\dd}{\dd t} \left( \frac{1}{2} J \omega^2 \right)$ In steady state operation, we can set the time derivatives to zero, and we obtain:

\underbrace{v_\textsf{in} i}_{P_\textsf{in}} \;=\; \underbrace{T_L \,\omega}_{P_\textsf{out}} \;+\; \underbrace{R i^2 + b\,\omega^2}_{P_\textsf{lost}}

(13)

We can say a lot more about power and efficiency of DC motors, but this is a good starting point. For a deeper dive, see the appendix on motor performance.

Example: Geared motor¶

In the problem below, we analyze a DC motor driving a load through a gearbox. Our task is to derive as set of ODEs relating the input voltage $v_\textsf{in}$ to the current $i$ , shaft angle $\theta_1$ , and load angle $\theta_3$ . We can assume our motor has parameters $K, b, L, R$ .

Solution: The system consists of one circuit (single loop) and three rotational mechanical components (the motor shaft/pinion, one gear, and the load). We also anticipate one kinematic constraint from the gearbox, since the gear and pinion move together.

We will skip free-body diagrams and write the equations of motion directly. Let’s assume a contact force $F$ between the pinion and gear teeth.

\begin{aligned} L \frac{\dd i}{\dd t} + R i + v_b &= v_\textsf{in} && \textsf{(motor RL circuit)}\\ v_b &= K \dot\theta_1 && \textsf{(back EMF)}\\ T &= K i && \textsf{(torque constant)}\\ J_1 \ddot\theta_1 + b \dot\theta_1 &= T - r_1 F && \textsf{(motor shaft dynamics)}\\ J_2 \ddot\theta_2 + k(\theta_2-\theta_3) &= r_2 F && \textsf{(gear dynamics)}\\ J_3 \ddot\theta_3 + c \dot\theta_3 +k(\theta_3-\theta_2) &= 0 && \textsf{(load dynamics)}\\ r_1\theta_1 &= r_2 \theta_2 && \textsf{(kinematic constraint)} \end{aligned}

(14)

We have 7 equations in 7 unknowns: $i, v_b, T, F, \theta_1, \theta_2, \theta_3$ . This is a complete set of equations of motion for the system. We start by eliminating $v_b$ and $T$ using the second and third equations.

\begin{aligned} L \frac{\dd i}{\dd t} + R i + K \dot\theta_1 &= v_\textsf{in} \\ J_1 \ddot\theta_1 + b \dot\theta_1 - K i &= - r_1 F \\ J_2 \ddot\theta_2 + k(\theta_2-\theta_3) &= r_2 F \\ J_3 \ddot\theta_3 + c \dot\theta_3 +k(\theta_3-\theta_2) &= 0 \\ r_1\theta_1 &= r_2 \theta_2 \end{aligned}

(15)

Now we’re down to 5 equations in 5 unknowns: $i, F, \theta_1, \theta_2, \theta_3$ . Next, we can eliminate $\theta_2$ using the kinematic constraint, $\theta_2 = \frac{r_1}{r_2} \theta_1$ . This yields

\begin{aligned} L \frac{\dd i}{\dd t} + R i + K \dot\theta_1 &= v_\textsf{in} \\ J_1 \ddot\theta_1 + b \dot\theta_1 - K i &= - r_1 F \\ \frac{r_1}{r_2} J_2 \ddot\theta_1 + k\left(\frac{r_1}{r_2}\theta_1-\theta_3\right) &= r_2 F \\ J_3 \ddot\theta_3 + c \dot\theta_3 +k\left(\theta_3-\frac{r_1}{r_2}\theta_1\right) &= 0 \end{aligned}

(16)

Next, eliminate $F$ using the second and third equations:

\boxed{\begin{aligned} L \frac{\dd i}{\dd t} + R i + K \dot\theta_1 &= v_\textsf{in} \\ \left(\frac{r_2}{r_1}J_1 + \frac{r_1}{r_2}J_2\right) \ddot\theta_1 + \frac{r_2}{r_1}b \dot\theta_1 - \frac{r_2}{r_1} K i + k\left(\frac{r_1}{r_2}\theta_1-\theta_3\right) &= 0 \\ J_3 \ddot\theta_3 + c \dot\theta_3 +k\left(\theta_3-\frac{r_1}{r_2}\theta_1\right) &= 0 \end{aligned}}

(17)

Of course, this could be simplified further by eliminating $i$ as well (easier in the Laplace domain!), but we’ll stop here. We have derived a set of three ODEs relating the input voltage $v_\textsf{in}$ to the current $i$ , shaft angle $\theta_1$ , and load angle $\theta_3$ .

Generators¶

A generator is the reverse of a motor: it converts mechanical power into electrical power. However, the same basic principles apply. In fact, the exact same equations (1) govern the dynamics of a DC generator, but they need to be interpreted differently.

Now, the load torque becomes an input torque: $T_L = -T_\textsf{in}$ , and the input voltage becomes a load voltage $v_\textsf{in} = -v_L$ . We switched the signs in our convention so that the current and angular velocity in our original diagram point in the same direction as before when the generator is running. The equations of motion therefore become:

\boxed{\begin{aligned} L \frac{\dd i}{\dd t} + R i + K \omega &= -v_L \\ J \dot \omega + b\,\omega - K i &= T_\textsf{in} \end{aligned}}

(18)

If the generator is being used to drive some load $R_L$ , then we can substitute $v_L = R_L i$ . This is analogous to what would happen if a motor were driving some speed-dependent load $T_L = b_L \omega$ such as a fan or a drill. The generator follows a similar causal chain of events as the motor (see Causal chain of events in a DC motor):

There is also a nice analogy between the motor and generator extreme use cases:

Motor (Electrical to Mechanical)	Generator (Mechanical to Electrical)
No-load ( $T_L=0$ ): min $i$ , max $\omega$ , max $v_b$	Open circuit ( $i=0$ , $T=0$ ): max $\omega$ , max $v_b$
Stall ( $\omega=0$ , $v_b=0$ ): max $i$ , max $T$	Short circuit ( $v_L=0$ ): min $\omega$ , max $i$ , max $T$

The first row is the “no-load” case: nothing connected to the motor shaft, no load connected to the output terminals of the generator. These cases result in the largest speed (no load torque for the motor; no current for the generator) and therefore the largest back EMF $v_b$ .
The second row is the “stall/short-circuit” extreme-loading case: the motor shaft is locked (stall) and the generator terminals are short-circuited. These cases result in the largest current (stall current for the motor; short-circuit current for the generator) and therefore the largest motor torque $T$ and the most heat dissipated in the windings.

Both extremes are inefficient and ideally we would want both machines to operate somewhere in between these two extremes. For more discussion on power and efficiency, see the appendix on motor performance. A similar analysis can be carried out for generators.

Test your knowledge¶

Exercise 1

Consider the spring-return solenoid in Figure 3 (left). A solenoid is a motor that has been ``unwrapped’’ to form a linear actuator. The plunger, with mass $m$ , is attached to a linear compression spring with spring constant $k$ . A positive voltage $v_\text{in}$ applied to the solenoid will retract the plunger, compressing the spring, yielding a positive retraction displacement $x$ .

The solenoid operates according to electromagnetic principles similar to those of a DC motor. The retraction force applied to the plunger by the solenoid is proportional to the current $i$ generated in the solenoid: $F_s = K_s\,i$ . This current $i$ is the loop current shown in Figure 3 (right). The motion of the solenoid in the magnetic field causes a back EMF given by the voltage $v_b = K_b\,\dot x$ , where $K_b$ is a constant. This voltage, as shown in the circuit diagram, will affect the current in the solenoid.

Derive a single differential equation relating the input voltage $v_\text{in}$ to the plunger displacement $x$ . Your equation should not contain $i$ or $v_b$ or $F_s$ ; it should only depend on the problem parameters $m$ , $k$ , $R$ , $L$ , $K_b$ , $K_s$ . Neglect any frictional forces in the plunger.

Figure 3:Left: Spring-return solenoid. Right: Equivalent circuit.

Solution to Exercise 1 #

The external forces acting on the plunger are the solenoid force $F_s$ and the spring force $F_k = - k x$ . Applying Newton’s second law gives

m \ddot x = F_s + F_k = K_s i - k x

(19)

The solenoid circuit equation the same as with a DC motor:

v_\text{in} = L \frac{\dd i}{\dd t} + R i + v_b = L \frac{\dd i}{\dd t} + R i + K_b \dot x

(20)

Therefore, our equations of motion are:

\begin{aligned} m \ddot x + k x - K_s i &= 0 \\ L \frac{\dd i}{\dd t} + R i + K_b \dot x &= v_\text{in} \end{aligned}

(21)

We can eliminate $i$ by solving the first equation for $i$ and substituting into the second:

\begin{gathered} i = \frac{m \ddot x + k x}{K_s} \\ L \frac{\dd}{\dd t} \left( \frac{m \ddot x + k x}{K_s} \right) + R \left( \frac{m \ddot x + k x}{K_s} \right) + K_b \dot x = v_\text{in} \end{gathered}

(22)

Now expand and simplify to obtain the final ODE relating the input voltage $v_\text{in}$ to the plunger displacement $x$ :

\boxed{ \frac{m L}{K_s} \dot{\ddot{x}} + \frac{R m}{K_s} \ddot x + \left( \frac{k L}{K_s} + K_b \right) \dot x + \frac{R k}{K_s} x = v_\text{in} }

(23)