Chapter 8
The DC Motor
8.1 Commutation
The rotary actuator in Section 7.3 has a useful range of something less than 180◦ : It
produces a uniform torque while the coil is in the gap (in a “horizontal” plane), but the
torque goes to zero when the coil reaches a vertical plane. If we rotate it past this torque
null, the torque reverses direction, tending to restore the loop to the vertical plane. To
make this device into a useful, continuous rotation motor, we need to (a) keep the torque
pointing in the same direction, and (b) remove the region of zero torque.
Let’s consider problem (a) first. If we could reverse the direction of the current when the
coil crosses the vertical plane, then the reversal of the current would reverse the reversal
of torque and the non-zero torque would always be in the same direction. If the load is
sufficiently light, inertia will probably carry the coil through the torque null and we would
have a motor with continuous rotation.
We can do this with a device called a commutator
which is essentially a rotary switch connected to
and hence synchronized with the coil. The sta-
tionary parts of the switch are called brushes and
are usually made of carbon.
8-1
The problem of the torque null can be solved by
adding a second coil so that when the first coil has
left the gap, the second can continue to produce
torque.
This will work, but is in-
efficient: only half of the
wire is carrying current at
any one time, and wire not
carrying current is not pro-
ducing torque. We can
fix this by connecting the
coils together in series, as
shown here for a three coil
winding with a six segment
commutator.
8-2
A cross sectional view makes it easier to see how the com-
mutation works but makes it more difficult to see how the
coils are connected on the far end.
Y LUWXDOEUXVKHV
Note that although the brushes lie in a horizontal plane,
the current in a coil changes direction when it crosses
the vertical plane. For this reason it is often convenient
to draw the brushes as though they contacted the coils
directly at their periphery. In this case current switching
occurs in the same plane as these “virtual” brushes.
We can also put the magnet on the inside of the coil, with
a cylindrical iron shell on the outsite to complete the flux
path.
To find the total torque produced by a motor, we can sum the L
contributions due to each turn. Because the total armature L
current Ia is divided between the two halves of the winding, L
each conductor is carrying a current of Ia /2.
Each length l of this conductor in the gap experiences a force of F = BlIa /2 and contributes
a torque of T = rF = rBlIa /2. Since each turn of the coil has two segments of length l in
the gap, each turn contributes 2rBlIa /2 = rBlIa . With N total turns we get
T = N rBlIa = Kt Ia (8.1)
where Kt = N rlB is called the torque constant of the motor.
8-3
It is sometimes convenient to describe the torque in terms of
the total gap flux Φ = πrlB Substituting rlB = Φ π into 8.1
O
we get
Φ N U
SU
T = N RlBIa = N Ia = ΦIa = KΦ ΦIa = Kt Ia
π π
Where KΦ is called the motor constant and is often written
as simply K, i.e. T = KΦIa .
8.2 Moving Coil Motors
These cross sectional views hide the fact
that the volume where the cylindrical pole
piece or magnet must reside is completely
surrounded by the coil winding. This
makes construction difficult as the coil
must be wound with this already inside.
Nevertheless, motors are made using this
structure and are called ball wound mo-
tors due to the resemblance of the coil to
a ball of string.
One way to make construction eas-
ier is to change the path of the
rear portion of the coil, creating an
open, cylindrical winding rather
than a closed ball. This produces
what is called a cylindrical coreless
motor.
By going to the opposite extreme and compressing the cylindrical portion of the winding
and leaving only the ends, we get what is called a pancake motor. Because the armature
winding is often fabricated as a printed circuit rather than wound with wire, it is also often
called a printed circuit motor.
8-4
The class of motors where the moving member is made entierly of current carrying con-
ductors (usually copper) and their supporting structure, but no magnetic material is called
moving coil motors. Moving coil motors have a number of advantages: Their low moving
mass gives them a low moment of inertia and hence good dynamic response. The fact that
the coils do not enclose magnetic material gives them low inductance and low magnetic
losses. The homogeneity of the winding reduces variations of torque with angular position
(torque ripple).
However, they do have a few shortcomings: Since the torque producing windings must pass
through the air gap, a large gap is required to accommodate a large current. However, a
large gap requires a large, expensive magnet to produce a large B field. Because the torque
is produced directly on the wire of the coils, the winding must be strengthened to withstand
the forces and couple them to the shaft.
8.3 The Slotted Armature
The disadvantages noted for the moving coil motor were in fact insurmountable ones until
the mid 20th century; the necessary high energy permanent magnet materials and high
strength adhesives were simply not available. While waiting for these to be invented, makers
of DC motors used a structure called the slotted armature.
8-5
In this type of motor, the iron core
rotates along with the windings,
so it is sometimes referred to as a
moving iron motor. Rather than
being in the gap as with the mov-
ing coil motor, the coils are wound
in slots cut into the rotor. One
obvious advantage of this struc-
ture is that the air gap can now
be made as small as manufactur-
ing tolerances, thermal expansion,
and centrifugal force will allow.
One apparent disadvantage is that it seems that it should not work.
Because of the high permeability of the iron of the rotor, nearly all
of the flux is diverted around the winding and the B field at the
conductor is essentially zero.
The fact that millions of motors like this are sold every year suggests
that we might look a bit closer. The previous picture is somewhat
misleading, as it shows the B field in the vicinity of the conductor
when no current is flowing. If we look at the total field due to both
the permanent magnet and the coil when current is flowing we get a
slightly different picture.
In this case, the field is no longer symmetric; it is more intense at the upper edge of the
slot. If we use the same argument we used with the sheared plate electrostatic actuator,
i.e. that the force produced by the field works in the direction which would shorten the
field lines, then there would be a net downward force on the edge of the slot, Combined
with the corresponding upward force on the opposite side of the rotor, this would produce a
counterclockwise torque. Note that this torque is in the same direction as the torque which
8-6
would be produced directly on the coil if the iron rotor were absent.
We can use an energy method to quantify the torque. Assuming the energy conversion
process is lossless, dWe = dWf + dWm . In this case, the field energy is constant, so dWf = 0
and
Pin(elec) = Pout(mech)
ei = T ω
dλ dθ
i = T
dt dt
So we have that
dλ
T =i
dθ
For simplicity, assume the armature contains a single
coil of N turns. If the total flux in the gap is Φ, then Q
the portion of that flux linking the coil depends on
the angle of the coil. If the gap flux is uniform, this
will vary linearly from zero when θ = 0 to N Φ with T
)
the coil in a vertical plane.
λ
λ + 1Φ
To this we must add the flux linkage λ0 due to
λ 1Φ
the current in the coil, which doesn’t vary with
angle. λ − 1Φ
θ
π π
π π
Gλ
Gθ
Differentiating with respect to θ gives a torque which is 1Φ
π
constant in magnitude and changes sign when the coil θ
passes through the vertical plane. π π
The total flux is Φ = BA = Bπrl so
dλ 2N Bπrl
T =i =i = 2N Bilr
dθ π
which is the same as for the moving coil motor (8.1).
8-7
8.4 Back EMF and the Tachogenerator
The equations we have developed so far do not take into consideration the rotational velocity
of the motor. If we force a constant current through the armature, we get a constant torque.
However, as with the voice coil, when the conductors in the rotor move through the magnetic
field of the stator, an electromotive force or voltage is generated.
Each segment of length of length l in the gap produces a voltage E = Blu where u = ωr is
the tangential velocity of the conductor. Each turn of the coil has two segments of length l
in the gap, so each contributes a voltage of 2Blrω.
The total of N turns in the windings appears to the brushes
as two coils of N/2 turns in parallel. Thus the total voltage
at the brushes is
E = N Blrω = Ke ω
Where Ke = N Blr is called the back emf constant of the
motor, and has the same value as Kt .
The fact that Ke = Kt is due to the consistent units we have been using and requires that
when we state that T = Kt Ia or E = Ke ω we measure torque in newton-meters and angular
velocity in radians/second. Most product data sheets for real world motors will use different
units such as inch-ounces and rpm, in which case the values given for the torque and back
emf constants will not be equal. However, given one of these constants in any set of units,
it is always possible to find the other by converting it to the units we have been using.
The fact that a voltage is produced across the motor terminals should not be surprising.
For one thing it is necessary so that the electrical input power (Pe = Ia E) matches the
mechanical output power (Pm = T ω). Also, we saw the same behavior with the voice coil,
where a loudspeaker and dynamic microphone are essentially the same structure.
The DC motor is a reciprocal energy conversion device: we can put electrical power in
and mechanical power out, using it as a motor, or we can mechanically turn the shaft to
convert mechanical power to electrical power in which case it becomes a DC generator. If
we are interested in information, rather than power, we can use just the voltage produced,
E = Ke ω, without drawing significant current. This gives us a rotational velocity sensor
where the output (voltage) is linearly proportional to the velocity. When used for this
purpose it is called a tachogenerator.
8-8
8.5 Permanent Magnet DC Motors
In the motors we have seen so far, the magnetic field in the gap has been produced by one
or more permanent magnets. This produces what in many ways is an ideal motor. They are
small and require no inputs other than the driving input. The linear relationships between
torque and current and between steady state speed and voltage make them easy to use in
control systems. In small sizes DC motors are very inexpensive, but at higher power levels,
the increasing amount of permanent magnet material rapidly raises the cost.
8.5.1 PM DC Motor Characteristics
Up to now, we have implicitly been driving the DC motor with a current source, producing
an output torque linearly proportional to the input current. However, the in a voltage
oriented world, we need to know what happens when we connect a motor to a voltage
source.
First we need a schematic diagram symbol for the DC motor. This
consists of a stylized representation of the commutator and brushes.
However, this is only half of the picture, because the motor has
mechanical as well as electrical terminals. We must consider
these, along with the coupling equations, T = Kt ia and e = Ke ω,
whenever we use a motor in a circuit.
Connecting this to a voltage source seems a bad idea: in the absence of any current lim-
iting mechanism, a huge current would flow, producing a huge torque that would quickly
accelerate the load to dangerous speeds.
Fortunately, two things conspire to prevent this. First
of all, any real motor will have some non zero resistance
Ra in the armature winding. This will at least limit the
armature current to a finite value. More significantly, as
the angular velocity of the motor (and its load) increases,
so does the back emf. At some point a speed will be
reached where the back emf balances the applied voltage and no further acceleration occurs.
If the only load torque is inertial, i.e. if there is no frictional or torsional load, then in the
steady state no torque will be required to sustain the speed. In this case, ia = T /Kt = 0,
e = v and ω = v/Ke . In other words, the steady state speed is proportional to the applied
voltage.
8-9
When we combine the effects of the armature resistance and the back emf, we get the
following relationships among voltage, current, torque, and velocity.s Using KVL
v = Ra ia + Ke ω
From Ohm’s law
ia = (v − e)/Ra = (v − Ke ω)/Ra
Then the torque is
Kt v Kt Ke
T = Kt ia = (v − Ke ω) = Kt − ω (8.2)
Ra Ra Ra
v/Ra is the current which would flow if ω = 0, i.e. if the motor were stalled, and is called
the stall current. The torque produced with the motor stalled, T = Kt Rva , is called the stall
torque.
If we solve 8.2 for ω we get
Ra T = Kt (v − Ke ω)
Ra T
= v − Ke ω
Kt
v Ra T
ω = − (8.3)
Ke Kt Ke
v/Ke is the speed that would result if the load torque were zero and is called the no load
speed.
If we plot the torque vs. speed for a
fixed applied voltage, we get a straight
line which crosses the T axis at the stall
torque, and crosses the ω axis at the no
load speed. If we change the applied volt-
age, the curve moves parallel to the origi-
nal curve.
8-10
T
If we draw these lines for several different volt-
ages, we get a family of curves describing how the T ∝ω
speed/torque relationship varies with voltage. If V = V 2 > V1
on this same plot we draw the curve representing
T ∝ ω2
the speed/torque relationship of the load, then
V = V1
the intersection of this load line with one of the
curves gives the steady state speed for the corre- T = constant
sponding voltage, and the locus of the intersec-
tions with curves for different voltages gives the
variation of speed with voltage for that load.
ω
Constant Voltage Performance
Curves We can also plot a number
of other variables as a function of
speed for a fixed voltage. Since cur-
rent is proportional to torque and
electrical input power is proportional
to current, these both give straight
lines like the torque plot. More in-
teresting are the plots of output power
and efficiency (η). Pout , being the
product of T and ω is a parabola
with maximum value at one half
the no load speed. Although this
is the operating point which gives the maximum power, it may exceed the rated power for
the motor. If we want the maximum efficiency, important for battery powered equipment,
then the motor must at nearly its unloaded speed where unfortunately, the output power
is much less than its maximum value.
8.6 Wound Field Motors
Up to now the stator field has been provided by a permanent magnet, but it could also
be provided by an electromagnet. Figure 8.1 shows a number of ways this might be done,
along with a few “equivalent” permanent magnet structures. For obvious reasons, motors
using an electromagnet to provide the stator field are called wound field motors.
8-11
Figure 8.1: Some field configurations for PM and wound field DC motors.
8-12
In such a motor, the stator field strength will depend on the value of if , the current in the
field winding. In terms of the total airgap flux we have Φ = Φ(if ), or for the linear region
of the stator material Φ = Kf if .
Ra Rf
Ia If
We can modify our circuit diagram to indicate the pres- + + +
ence of the field winding and the associated flux. V E Φ field Vf
- - -
There are several possible sources for the field current:
1. It could be connected to a source which is independent of the armature supply (sep-
arate excitation).
2. It could be connected in parallel with the armature winding (shunt wound).
3. It could be connected in series with the armature winding (series wound).
Each of these gives a motor with different speed/torque characteristics.
For a fixed value of If (or Vf ) the field flux Φ will be fixed so a separately excited motor
with a fixed field supply behaves like a permanent magnet motor.
8.6.1 Shunt Wound Motor
With the field winding connected in parallel with the armature, the field voltage Vf must
be equal to the armature voltage V . Substituting KΦ for Ke in 8.3 we have
V Ra T V Ra T
ω= − = −
Ke Kt Ke KΦ (KΦ)2
Then the no load speed (ωnl ) is
V V V
ωnl = = = V
KΦ KKf If KKf Rff
Rf
But, since Vf = V , we have ωnl = KKf independent of the armature voltage V .
If the load torque is non zero,
Ra Rf 2 Ra T
ω= − 2 2 2 ω
KKf K Kf V
V=V2>V1
Unlike the family of speed/torque curves for the permanent mag-
V=V1
net (or separately excited) DC motor, which is a series of parallel
lines, the curves for the shunt wound motor are a set of lines pass- T
ing through the ω axis at ωnl having slope which varies as the
square of the armature voltage.
8-13
This makes controlling the speed of a shunt ω
wound motor a bit different from control- Ra=0
Ra Rf
ling a PM motor. On the one hand, if +
Ra=med
what we want is constant speed with a vari- V Ra=high
able supply voltage, the shunt wound mo- field
tor provides good speed regulation with a T
light load. On the other hand, with a larger load we can vary the speed by increasing the
effective armature resistance with a series resistor. This changes the intersection of the load
line with the speed torque curve by varying the slope of the latter.
ω
To vary the speed with a light load and
constant supply voltage we can vary the +
Ra Rf Rf=high
Rf=med
no load speed by varying the effective field V
Rf=low
resistance with a series resistor. field
Note that in this case, somewhat unintuitively, the speed increases as we increase the field
resistance. This suggests the question of what might happen if we let Rf = ∞ and the
answer is often “bad things.” If the field current is removed once the motor is running the
speed will increase until limited by the residual magnetism in the field iron, the friction in
the bearings, or the strength of the rotor assembly under centrifugal force.
8.6.2 Series Wound Motor
Ra+Rf field
If we connect the field winding in series with the arma- i
ture, then the field current if will be equal to the arma- + +
ture current ia , i.e. if = ia = i. V E
- -
8-14
If we are in the linear region of the core material, KΦ =
KKf if = KKf i so that
T = KΦi = KKf i2
s ω
T
i =
KKf
Then the speed is
V RT
ω = − 2 2
KΦ K Φ
V RKKf i2 T
= − 2 22 2 KK
KKf i K Kf i V f
R2
V R
= p −
KKf T KKf
where R = Ra + Rf .
8.7 Brushless Motors
The key element of the DC motor is the commutator, which keeps the current in the rotor
windings always in the same direction with respect to the stator field. Another way of
looking at the function of the commutator is that it always keeps the field of the rotor at
right angles to that of the stator, the condition of maximum torque.
Consider the following experiment: (1) glue a battery to the back
of a permanent magnet DC motor, (2) connect the battery and
the motor, (3) hold the shaft. We still have a rotary actuator, but
now the outer part (containing the permanent magnets) rather
than the inner part (containing the coils) is rotating.
Previously when we ran a DC motor, the commutator % U
switched the current in the armature windings so that % V
the field of the rotor remained perpendicular to the field
of the stator, resulting in maximum torque as the fields % U
tried to align.
% V
Holding the shaft rather than the frame of the motor doesn’t change what % U
the commutator is doing: the two fields are still kept perpendicular, but
now they are both rotating. However, the rotor is now the part with the Z
permanent magnets and the stator is the part with the coils. % V
8-15
There are two problems with this arangement: (1) in most applications we want the shaft
of the motor to rotate, not the frame; (2) having to rotate the power supply along with the
load is undesirable.
We can fix problem 1 by rearranging the motor so that the magnets are on the inside and
the coils are on the outside. This puts the rotor and stator back in the places we’re used
to.
We solve problem 2 by realizing that since
+V
the coils are now on the outside and are
stationary, we don’t need a mechanical 1 2 3
commutator to switch the current through
them, but instead can use transistors. A B C
Here’s a circuit that will do this (for a sta- 4 5 6
tor winding with three coils): -V
The trick is to know when to switch the
$FWLYDWH% $FWLYDWH&
currents. Recall that what we want to do $
is to keep the fields produced by the rotor
and stator perpendicular (or nearly per- & %
pendicular). We can do this by watching
$FWLYDWH$ $FWLYDWH$
the North pole of the rotor and switch-
ing current to the appropriate winding de- % &
pending on where it is pointing. This is
usually done with a Hall sensor which we $
$FWLYDWH& $FWLYDWH%
will study later.
8-16
8.8 Summary
Here’s a device scorecard for the different types of DC motors.
Classification
Class Name DC Motor
Subclass Name Permanent Magnet Wound Field Brushless
Sensor or Actuator Actuator
Type of Motion Rotational
Intended Output torque or velocity
Behavior
Controlling input current or voltage
Defining equations T = Kt ia T = KΦia T = Kt ia
v Ra T Ra Rf 2 Ra T v Ra T
ω= Ke − Kt Ke ω = KK f
− K 2 Kf 2 V 2
(shunt) ω= Ke − Kt Ke
V Ra
ω= √ − KK (series)
KKf T f
Capability
Range θ: continuous, ω: 0 to 20k rpm
Load Rating < 1 kW > 100 W < 1 kW
Interface
Voltage < 100 V < 100 V
Dynamics
Mass low (moving coil)
Physical, Economic
Size small small
Form Factor cylindrical
Cost low moderate moderate
Efficiency good
Reliability good excellent
Maintenance brushes none
Environmental
Safety sparking
Miscellaneous
Integratability fair good
Figure 8.2: DC Motor Scorecard
8-17
