ECE 320
Energy Conversion and Power Electronics
Dr. Tim Hogan
Chapter 4: Concepts of Electrical Machines: DC Motors
(Textbook Sections 3.1-3.4, and 4.1-4.2)
Chapter Objectives
DC machines have faded from use due to their relatively high cost and increased maintenance
requirements. Nevertheless, they remain good examples for electromechanical systems used for
control. We’ll study DC machines here, at a conceptual level, for two reasons:
1. DC machines although complex in construction, can be useful in establishing the concepts of
emf and torque development, and are described by simple equations.
2. The magnetic fields in them, along with the voltage and torque equations can be used easily to
develop the ideas of field orientation.
In doing so we will develop basic steady state equations, again starting from fundamentals of the
electromagnetic field. We are going to see the same equations in ‘Brushless DC’ motors, when we
discuss synchronous AC machines.
4.1
Geometry, Fields, Voltages, and Currents
The geometry shown in Figure 1 describes an outer iron frame (stator), through which (i.e. its
center part) a uniform magnetic flux, φˆ , is established. The flux could be established by a current in
a coil or by a permanent magnet for example.
In the center part of the frame there is a solid iron cylinder (called rotor), free to rotate around its
axis. A coil of one turn is wound diametrically around the cylinder, parallel to its axis, and as the
stator and its coil rotate, the flux through the coil changes. Figure 2 shows consecutive locations of
the rotor and we can see that the flux through the coil changes both in value and direction. The top
graph of Figure 3 shows how the flux linkages of the coil through the coil would change, if the rotor
were to rotate at a constant angular velocity, ω.
λ = φˆ cos(ωt )
(4.1)
Figure 1. Geometry of an elementary DC motor.
1- 1
1
2
3
4
5
Figure 2. Flux through the stator coil of the simple dc motor shown in Figure 1.
Since the flux linking the coil changes with time, than a voltage will be induced in this coil, vcoil, as
shown below:
dλ
vcoil =
= −φˆ sin (ωt ) (V)
(4.2)
dt
1.5
1
1
2
λ
coil
0.5
3
0
-0.5
4
5
-1
-1.5
0
100
200
300
400
500
600
700
500
600
700
Angle (º)
1.5
1
1
0
5
v
dcoil
0.5
-0.5
2
4
-1
3
-1.5
0
100
200
300
400
Angle (º)
Figure 3. Flux and voltage in a coil of the motor in Figure 2 with coil positions 1-5.
1- 2
The points marked on the cosine and sine waveforms of Figure 3 correspond to the positions of the
rotor as shown in Figure 2.
Torque on the coil follows the Lorentz Force Law: which gives the force, F, in newtons on a
charge, q, in coloumbs that is exposed to a vectoral electric field, Eˆ , in volts per meter and magnetic
flux density, B, as
F = q Eˆ + v × B
(4.3)
(
)
where v is the velocity of a charged particle q. The first term gives the force on the electron that
moves it through the wire or the voltage applied to the wire. The second term gives the force on the
charge due to the magnetic field in which the wire is placed. With many electrons contributing to the
electrical current, I, (as a vector) in the wire, the force on the wire is
F = I×B
(4.4)
The force on the full coil is then twice this value to account for each of the wires shown in Figure 2.
To maintain this torque in a direction that continues the rotary motion of the rotor, the ends of the
rotor coil are connected to metal electrode ring segments called a commutator as shown in Figure 4.
These ring segments are attached to the rotor so as to rotate with it, and are electrically contacted
using two stationary brushes typically made of carbon and copper. The brushes are spring loaded and
pushed against the commutator. As the rotor spins, the brushes make contact with the opposite
segments of the commutator and they switch between segments of the commutator just as the induced
voltage goes through zero and switches sign.
B
bru
sh
ω
or
tat
u
m
com
bru
sh
Figure 4. A coil of a DC motor and brushes that are pushed up against the rotating
commutator (wire of the coil is highlighted red and soldered to the
commutator segments).
Because the current switches direction through the coil each time the brushes rotate to the
opposite segment of the commutator (and that this occurs as the voltage goes through zero), the
1- 3
1.5
1
1
0.5
0.5
terminal
1.5
0
0
v
v
coil
voltage at the terminals of the brushes is a rectified version of the voltage induced in the coil as
shown in Figure 5.
-0.5
-0.5
-1
-1
-1.5
-1.5
Time (arb. units)
Time (arb. units)
Figure 5. Induced voltage in a coil and terminal voltage in an elementary DC machine.
If a number of coils are placed on the rotor, as shown in Figure 6, with each coil connected to a
different segment of the commutator, then the total induced voltage to the coils, E, will be:
E = kφˆω
(4.5)
where k is proportional to the number of coils.
Figure 6. Multiple coils on the rotor of a DC machine.
We stated in equation (2.39) that Pelec = Pmech ⇒ T ⋅ ωmech = e ⋅ i thus with the multiple coil DC
machine:
E ⋅ i = T ⋅ω
(4.6)
kφˆω ⋅ i = T ⋅ ω
(4.7)
T = kφˆi
(4.8)
1- 4
If the DC machine is connected to a load or a source as in Figure 7, then the induced voltage and
terminal voltage will be related by:
Vterminals = E − ig Rwdg
for a generator
(4.9)
for a motor
(4.10)
Vterminals = E + im Rwdg
im
ig
Rwdg
Load
or
E
Source
Figure 7. Circuit with a DC machine with the current reference directions defined as
indicated for DC machine used as a generator, ig, or as a motor, im.
Example 4.1.1
A DC motor, when connected to a 100 (V) source and with no load connected to the motor runs at
1200 (rpm). Its stator resistance is 2 (Ω). What should be the torque and current if it is fed from a
220 (V) supply and its speed is 1500 (rpm)? Assume the field is constant.
With no load connected to the motor, we will assume the torque is zero (assuming no friction in
bearings). With the torque equal to zero, the current is zero since: T = kφi = Ki . This means for this
operation:
V = E = kφω = Kω
and with ω = 1200 (rpm ) ⋅
2π (rad) 1 (min)
⋅
= 125.66 (rad/s) , then
revolution 60 (s)
100 (V) = K ⋅125.66
K = 0.796 (V ⋅ s)
Then at 1500 (rpm)
ωo = 1500 (rpm ) ⋅
2π (rad) 1 (min)
⋅
= 157.08 (rad/s)
revolution 60 (s)
E = Kω = 125 (V)
1- 5
For a motor:
V = E + IRwdg
220 (V) = 125 (V) + IRwdg
I = 47.5 (A)
T = KI = 37.81 (N·m)
4.2
Energy Considerations
Conservation of energy governs that the total energy supplied to an electromechanical system
equals the energy output. Some of the energy output could be mechanical, some could be lost as
heat, and some could be stored. This is summarized in equation (3.10) of your textbook as:
⎛ Energy input ⎞ ⎛ Mechanical ⎞ ⎛ Increase in energy ⎞ ⎛ Energy ⎞
⎟ + ⎜ converted ⎟
⎟ + ⎜ stored in the
⎜ from electric ⎟ = ⎜ energy
⎟ ⎜ into heat ⎟
⎟ ⎜ magnetic field
⎟ ⎜ output
⎜ sources
⎠
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎝
In a lossless system, the last term is zero. When the losses are negligible, a differential change in
the electrical energy input corresponds to the sum of a differential mechanical energy output and a
differential change in the energy stored in the magnetic field, or
dWelec = dWmech + dWfld
(4.11)
Mechanical energy is force times distance, and the time rate of change of electrical energy is
power or dWelec = ei dt and (4.11) can be written as:
ei dt = Ffld dx + dWfld
(4.12)
where Ffld is the force produced by the magnetic field. From equation (3.1) in the handout notes, we
dλ
Nφ λ
saw e =
. Using the linear inductor assumption of (2.19) which was L =
= , then (4.12)
dt
i
i
becomes:
(4.13)
dWfld = i dλ − Ffld dx
The energy in the field is given as the energy stored in an inductor
1 2 1 λ2
Wfld = Li =
2
2 L
(4.14)
The current and force produced by the magnetic field can then be found from (4.13) as:
i=
Ffld = −
∂Wfld (λ , x )
∂λ
x
∂Wfld (λ , x )
i 2 dL(x )
=
∂x
2 dx
λ
(4.15)
(4.16)
If the magnetic force results in a torque as in a rotating mechanical terminal, the mechanical energy
from the field is then replaced with torque and angular displacement, Ffld dx ⇒ Tfld dθ , as:
1- 6
dWfld = i dλ − Tfld dθ
thus leading to
Tfld =
i 2 dL(θ )
2 dθ
(4.17)
(4.18)
In a rotary machine with multiple windings, the inductance in equation (4.18) would contain self and
mutual inductances of all coils involved.
Chapter Notes:
• The field of a DC motor can be created either by a DC current or a permanent magnet.
• The two fields, the one coming from the stator and the one coming from the moving rotor, are both
stationary (despite rotation) and they are perpendicular to each other.
• If the directions of current in the stator and in the rotor reverse together, torque will remain in the
same direction. Hence if the same current flows in both windings, it could be AC and the motor
will not reverse.
1- 7