﻿2.1A DC Motor Model    0:00  # DC Motor Model - Differential Equations

## Physical Model A  DC motor is comprised of a housing that holds a permanent magnet, a rotor with an electromagnet, and a commutator. A disassembled motor is shown in the figure below.

Fig 2A.1 DC Motor

Fig 2A.1 shows a model of the motor. The rotor is modeled as a rotating mass with an applied torque and load. When voltage is applied to the motor, the rotor’s electromagnet (shown in red) is energized, generating a torque that causes the rotor to spin. If the voltage applied to the rotor is held constant, the rotor will stop rotating once it aligned with the permanent magnets in the housing (shown in greed). The commutator is a mechanical device that changes the polarity of the voltage applied to the rotor electromagnets as the rotor spins, forcing the motor to spin continuously.

Fig 2A.2 DC Motor Modeled

The average torque generated by the rotor is proportional to the current through the rotors windings and is given by the equation

T = KTi
2A.1

where KT is the motor's torque constant and i the winding current

The current, i, is determined by the voltage applied to the windings, the winding resistance and winding inductance. When the rotor is not moving the voltage on the windings is the same as the voltage source connected to the motor. However, as the rotor spin it acts as a generator, inducing a voltage on windings – call the "back EMF". This voltage is opposite in polarity to the applied voltage and so reduce the current and torque. The back EMF follows the equation:

VEMF=KEMF
_________dt
2A.2

When you apply a voltage to the motor, the created by the rotor causes it to spin. As the speed increases, the back EMF increase and eventually the motor reaches a speed where friction and load at the rotor shaft exactly balance the torque supplied by the rotor. This is the steady state speed of the motor and is proportional to the applied voltage.

A DC motor has a self-regulating feature. If you apply more load to the motor shaft, the rotor will slow down. However, slowing down the rotor reduces the back EMF and so increase the effective voltage across the windings. This in turn increases the current through the winding and the rotor torque, causing the rotor to speed up and compensate for the applied load at the shaft.

## Mathematical Model

A mathematical model of the the motor can be derived by separating the electrical and mechanical systems. This approximation models the rotor winding as an RL circuit, where the resistance and inductance come from the wire windings on the rotor. The back EMF is modeled as a voltage source following Eqn 1.1. The physical rotor is modeled as a rotating shaft with inertial J, and some viscous friction. The torque created by the rotor follows Eqn 1.0. The load comes from any externally applied torque on the motor's shaft.

Fig. 2A.3 Mathematical Model

The current through the RL cicruit is given by (remind me)

Use Kirkoff's voltage law and sum the voltage drops around the circuit.
Ri+L
di_________dt
= Vapplied - VEMF
2A.3

The angular position of the rotating mass is governed by (remind me)

Apply Newton's Law
∑Torque = Inertia
d2θ_________dt2
Tmotor + Tload - b
_________dt
= J
d2θ_________dt2
2A.4

where b is the coefficient of viscous friction.

Combine Eqn 2A.3 and 2A.2 to obtain

Ri + L
di_________dt
=Vapplied - Kω
_________dt
2A.5

Combine Eqn 1.3 and 1.1 to obtain

KT i + Tload - b
_________dt
=J
d2θ_________dt2
2A.6

Finally combine Eqn 1.3 and 1.1 to obtain

JL
d3θ_________dt3
+(Lb + RJ)
d2θ_________dt2
+(KTKω+Rb)
_________dt
= KTVapplied + L
2A.7

When L is small, as is the case for most motors, Eqn 1.4 simplifies to

RJ
d2θ_________dt2
+(KTKω+Rb)
_________dt
= KTVapplied + RTload
2A.8
where the following constants are typically provided by the motor manufacturer
R = rotor electrical resistance
J = rotor inertia
b = motor internal viscous friction constant
KT = motor torque constant
Kω = motor velocity constant