ECE 3010 ELEMENTS OF ELEC. MACHINES AND DIGITAL SYSTEMS
CHAPTER 4
Chapter 4 – DC Electric Machines
Objectives
1. To understand what is meant by the term “dc electric
machine.”
2. To understand the basic operation and construction of a
dc machine.
3. To understand how to analyze the input and output
characteristics of a given dc electric machine.
Introduction
• An electric machine can be considered to be any device
which is able to convert energy in one form, into
another. For our purposes an electric machine takes
electrical energy and converts it into mechanical energy,
or conversely, takes mechanical energy and converts it
into electrical energy. The mechanical energy can be in
either a rotational or linear form, likewise, the electrical
energy can be constant values, otherwise known as dc
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(direct current) values, or time varying values, otherwise
known as ac (alternating current) values.
• The electric machines which we will focus on consist of
rotating motors and generators which perform their
energy conversions through the interaction of magnetic
fields created using coils of wire which form
electromagnets.
Basic dc electric machine
• The basic dc machine uses a pair of magnetic fields
interacting with one another. The magnetic fields are
created using dc (constant / time-invariant) voltages and
currents applied to electromagnets. It is the interaction
of these magnetic fields which link the electrical and
mechanical parts of these machines.
• We can get one electromagnet to move by using a
second magnet to impart some mechanical energy
without having them mechanically coupled.
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• We can initially consider a simple representation for a dc
motor consisting of a fixed magnet, referred to as the
stator, and a moveable magnet, referred to as the rotor
N
S
N
Stationary
Magnet (Stator)
S
Moveable
Magnet (Rotor)
• The basic operation of a dc motor takes advantage of
the fact that if the rotor magnet faces the same polarity
on the stator, the poles will exert a repulsive force on
each other;
N
S
S
Stator
N
Rotor
this will result in the rotor moving in relation to the stator.
N
N
S
S
Stator
Rotor
As the rotor rotates, the opposite pole of the rotor will be
attracted to the stator, and the rotor will continue to
rotate and try to align the opposite poles.
N
S
Stator
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S
N
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The result is that, without the rotor and stator being
mechanically connected, we have imparted a
mechanical rotation on the rotor using the interaction of
the two magnetic fields.
• There are two problems if we use permanent magnets
for the rotor and stator: we need to manually rotate the
rotor to start with same polarity poles facing one
another; and the rotating action will cease once we get a
180° rotation of the rotor.
• We can solve this problem by using electromagnets,
rather than permanent magnets. We can electrically
control the polarity of an electromagnet to ensure we
start with matching poles in alignment, and reverse the
polarity of the rotor after 180°.
• Using electromagnets is fairly straight forward since an
electromagnet can be constructed by using a coil of wire
and a power source.
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N
S
S
CHAPTER 4
N
+–
–+
Stationary
Magnet (Stator)
Moveable
Magnet (Rotor)
Note that the polarity of the voltage sources determines
the direction of current flow, and therefore the polarity of
the individual electromagnets.
• Since we can electrically control the polarity of the
electromagnet, when the opposite poles are coming into
alignment we can disconnect the source from the rotor
and allow its inertia to take the rotor past the point of
alignment.
N
S
Stator
Rotor
and then re-apply the voltage source, with its polarity
reversed, which will result in a reversal of the magnetic
polarity on the rotor, and the rotation will continue, with
the two south poles once again repelling one another.
N
N
S
S
Stator
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• The polarity of the rotor can be reversed via mechanical
means using a system of brushes. The brushes provide
a stationary contact which changes its connection to the
rotor as the axle of the rotor rotates.
• Since the rotor is a moving part, in order to maintain a
complete circuit, the conductive ‘brushes’ ride on a split,
conductive, band on the axle of the rotor. The conductor
shown on the axle is not continuous, but switches sides
as brush travels the circumference of the axle.
Conductor
x
y
Brush
• Note that there is also a type of dc machine called a
brush-less dc motor which uses sensors and electronics
to perform the polarity switching.
Electric machine input and output
relationships
• Considering we want to use electromagnets in our
electric motor, what we are interested in are the inputs
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to the motor, in terms of voltages and currents, and the
outputs from the motor, in terms of torques and
rotational velocities. For an electric machine, the torque
is related to the current flowing in the rotor (the armature
current) via the following relationship:
T = Kφ ia
T = torque (Nm)
Kφ = machine const.
ia = armature current(Amps)
• Likewise, the rotational speed is related to the armature
voltage via the following:
Ea = Kφω m
Ea = armature voltage (V )
Kφ = machine const.
ω m = armature speed(radian / s)
• The Kϕ term, known as the machine constant, is
dependent on the physical construction of the electric
machine.
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• From these relationships we can see that to analyze
and understand the operation of any electric machine
requires us to apply some basic circuit analysis
techniques involving voltages and currents.
DC Electric Circuits Review
• Electric machines, though mechanically different from
the traditional static non-moving electric circuits made of
sources, resistors, etc, can be modelled and analyzed
using the same basic circuit analysis techniques
involving voltages and currents.
Basic electrical devices and electrical relationships
• We can analyze the basic operation of an electric
machine by modelling the machine using what are
known as passive components and then analyze the
model.
• Passive components are those devices whose
behaviour is governed by their material properties and,
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as a result, do not require an external source of energy
to function.
• The basic passive, electric circuit, elements are the
resistor, capacitor and inductor.
Resistor
Inductor
Capacitor
The basic active elements we will be considering are
voltage sources and current sources.
+–
Voltage
Source
Current
Source
• We can construct any electric circuit by connecting
various types of these circuit elements together. When
discussing a circuit we can define the following terms: a
node is the point where multiple items are connected
together; and a loop is a closed path where current can
flow within the circuit.
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• The parameters of interest when dealing with any circuit
are voltage and current.
• Voltage is defined as the work required, per unit charge,
to move the charge from point a to point b. It can also
be considered as the potential difference between two
points, the voltage at a given point b relative to a point a
is defined as:
wba
vba = vb − va =
Volts
q
wba = work to move charge from a to b (Joules)
q = unit charge (Coulombs)
• Current is defined as the rate of movement of electric
charge. Specifically, it has been defined as the measure
of net positive charge transferred per unit time:
dQ
Amperes
dt
Q = charge (Coulombs)
t = time (seconds)
i=
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• Combining the concept of passive components with
voltage and current results in: devices with two nodes; a
voltage difference between the two nodes; and a current
which enters one node and exits the other. Since
voltage and current are vector quantities it is important
to maintain the correct sign when performing any
calculations. We will use the passive sign convention to
label the parameters. The convention is that the current
is assumed to enter the node of higher potential.
i
+
Passive
Element
v
–
• The relationships between currents, and relationships
between voltages, within a given electric circuit are
governed Kirchhoff's Laws.
• Kirchhoff’s current law (KCL) requires that the algebraic
sum of currents at a node must be zero.
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N
∑ in = 0
n=1
• Kirchhoff’s voltage law (KVL) requires that the sum of
voltages around a loop (closed path) is zero.
N
∑ vn = 0
n=1
• Using Kirchhoff’s Laws we can define two specific types
of connections which are known as series and parallel.
• A series connection can be defined as having two or
more elements connected such that the same current
flows through all elements.
• A parallel connection can be defined as two or more
elements connected such that the voltages across them
are equal.
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• Being able to identify these two types of connections is
important since we can apply circuit simplification
techniques to these types of connections. Making circuit
representation simpler makes the analysis of their
parameters simpler. It is important to realize that the
concept of series and parallel do not refer to their
graphical (pictorial) representations, but refer to their
electrical configurations.
Example - Passive sign law with KVL and KCL
• Given the following circuit, with the known parameters
indicated, use KCL and KVL to determine the voltage
and current parameters for all of the components.
5A
!
"
+ 2V –
#
– 3V +
$
–1A
Current and voltage relationships for specific
devices
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• The relationship between the voltage and current
associated with a resistor is governed by a linear
relationship known as Ohm’s Law.
v(t) = Ri(t)
v(t) = voltage (V)
i(t) = current (A)
R = resistance (Ω)
or for when dealing with dc values
V = RI
• Though both capacitors and inductors are passive
devices, they are referred to as reactive devices. For
reactive components their voltage and current
characteristics will behave as functions of time.
• The capacitor is a passive energy storage device, which
stores energy through the storage of electric charge.
The amount of charge stored (i.e. the voltage) is related
to the amount of time current (the flow of charge) is
flowing into the capacitor.
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Q = capacitance × voltage
Q
C
C = →1Farad =1
V
V
dq(t)
dv(t)
i(t) =
=C
Amps
dt
dt
1 t
v(t) = ∫ i(t)dt Volts
C −∞
Note that the current associated with a capacitor is
determined by the derivative of the voltage across the
capacitor. As a result, intuitively, the voltage function
cannot change instantaneously since this would result in
an infinite current which, in the real world, is impossible.
For a capacitor, an instantaneous change in current will
result in an exponential change in voltage.
• Note that the current through the capacitor is a function
of the change in voltage across the capacitor. Therefore,
if the voltage across the capacitor is constant (i.e. a dc
value),
dV
I =C
=0
dt
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• The inductor is another passive device which stores
energy. The inductor stores energy via magnetic flux.
Recall from basic physics, if a current flows through a
wire, a magnetic field will be formed externally around
the wire. Whenever a time varying magnetic flux passes
through a loop of wire, a voltage will be induced on the
loop.
• The basic inductor is a coil of wire which, by its nature,
concentrates the magnetic flux in the vicinity of the coil.
The magnetic field which is formed represents stored
potential which is a voltage. The voltage and current
associated with the inductor is related to the ‘capacity or
size’ (measured in Henrys) of the inductor
v(t) = L
di(t)
Volts
dt
1 t
i(t) = ∫ v(t)dt Amps
L −∞
The current in an inductor cannot change
instantaneously. For an inductor, an instantaneous
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change in voltage will result in an exponential change in
current.
• The voltage across the inductor is a function of the
change in current through the inductor. Therefore, if the
current through the inductor is constant (i.e. a dc value),
dI
V =L =0
dt
Power
• The goal of any electric machine is to convert power
between mechanical and electrical forms. Therefore we
must be able to calculate the power absorbed by the
electrical elements within our circuit. Power is defined
as the rate of transfer, or absorption, of energy per unit
time in a system.
p(t) = v(t)[units J / C] × i(t)[units C / s]
= v(t) × i(t) J / s
∴ p(t) = v(t) × i(t) W
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• The instantaneous power associated with a resistor is
always positive. This means that a resistor always
absorbs power and never supplies power.
p(t) = v(t) × i(t) W
combining with Ohm' s Law:
2
v(t)
v
(t)
p(t) = (i(t) × R) × i(t) = i 2 (t)R = v(t) ×
=
W
R
R
• The instantaneous power delivered to a capacitor is the
product of the voltage across, and the current into the
capacitor:
dv(t)
p(t) = v(t) × i(t) = Cv(t)
Watts
dt
• The capacitor does not absorb power but instead stores
the power delivered to it, and later releases that power.
By integrating the expression for the power associated
with the capacitor over time, the amount of energy
stored in the capacitor can be calculated as:
w(t) =
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t
1
p(τ )dτ = Cv 2 (t)Joules
2
−∞
∫
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• The inductor is another passive device which can store
energy. The instantaneous power associated with an
inductor can be calculated using:
p(t) = v(t) × i(t) = Li(t)
di(t)
Watts
dt
and the stored energy:
w(t) =
t
1 2
p(
τ
)d
τ
=
Li (t)Joules
∫
2
−∞
Example - Device specific behaviours and power
calculations
• Given the solved circuit from the previous example
determine what types of devices are represented by the
unknown boxes.
5A
!
– 1V +
+
–6A " 3V
–
+ 2V –
– 3V +
#
$
–1A
5A
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Equivalent circuits - Thevenin’s
theorem
• We want to be able to represent an electric machine as
a simple circuit. Thevenin’s theorem is a technique
which can be applied to any electric device/circuit where
we can analyze its external electrical behaviour from a
pair of terminals, and then represent the device/circuit
with a single voltage source and a single resistor.
• We can apply Thevenin’s theorem to any linear network
which contains linear circuit elements, such as resistors,
capacitors, inductors, and sources, to obtain a Thevenin
equivalent circuit. The Thevenin equivalent circuit
consists of an independent voltage source, Voc, and a
series resistor, RTH.
Mystery
Device
?
?
+–
?
?
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Thevenin Equiv.
=
Voc
+
–
Rth
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The value of the voltage source (Voc) is the open circuit
voltage which would be seen at the terminals of the
linear network.
Mystery
Device
?
?
+–
?
?
I=0A
+
Voc
–
Thevenin Equiv. I=0A
V
=
Voc
+
–
Rth
V
+
Voc
–
From the perspective of the voltmeter, which is
interacting with the two external terminals, the voltage
seen at the two terminals is electrically equivalent
regardless of whether the box is filled with a multitude of
devices or contains only a single Thevenin voltage
source.
• The value of the Thevenin resistance (Rth) can be
calculated using Ohm’s Law and determining the ‘shortcircuit’ current.
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Mystery
Device
?
?
+–
?
?
I!0A
+
0V
–
CHAPTER 4
Thevenin Equiv.
A
=
Voc
+
–
Rth
I!0A
A
+
0V
–
• At this point we have enough information to be able to
model a dc electric machine, and by applying some
basic circuit analysis techniques we can analyze the
terminal behaviour of the machine.
DC Machines
• DC machines use dc voltages and currents to establish
the required magnetic fields, and therefore accept or
provide dc values at its terminals.
• The actual magnetic fields are created using coils of
wire which form electromagnets. The normal
terminology when dealing with dc machines uses the
term field winding - which is the non-changing portion of
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the machine - for the stator winding/magnet, and the
armature winding for the rotor winding/magnet.
• When using a dc machine as a motor we will be
powering the armature winding to allow it to form an
electromagnet, and it will operate as previously
described. However, when using the same dc machine
as a generator we will mechanically move the rotor
through the magnetic field of the field winding and
induce a current to flow in the rotor.
• To understand how the dc machine acts as a generator,
recall from basic physics that when a current flows
along a wire a magnetic field is established around the
wire. The direction of current flow and the direction of
the magnetic field are related via the right hand rule,
where if the thumb of the right hand is in the direction of
current flow, the direction of the magnetic field will be
indicated by the direction of the four fingers.
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Magnetic
Field
i(t)
+
This phenomena also works in ‘reverse.’ If the wire is
moved through an existing magnetic field, a current will
be induced by the magnetic field to flow in the wire.
When using the electric machine as a generator we are
mechanically moving a coil of wire through a fixed
magnetic field and are able to draw electrical energy
from the moving armature.
Machine construction
• The basic electric machine consists of a stator and a
rotor. The stator is constructed as a coil of wire. We can
pass a current through two coils of wire, which will form
a magnetic field between them, where we can place the
rotor.
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S
N
CHAPTER 4
S
N
• The rotor also consists of a coil of wire which will move
through the magnetic field provided by the field winding
on the stator. To simplify the initial description of the
rotor, first consider a single loop of wire, rather than a
coil consisting of many loops.
S
a
N
+ einduced V –
S
N
b
This can be further simplified by only considering one
side of the loop - a single straight piece of wire passing
through the magnetic field. From basic physics we know
that moving a single conductor of length l at speed v
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with the wire ‘cutting’ the flux lines of a magnetic field of
flux density B will result in a voltage on the wire of eab =
Blv Volts.
B
b
l
S
N
S
N
a
v
ø
The implication of this is that regardless of whether we
are using the machine as a motor or generator, as long
as the rotor / armature winding is moving through the
magnetic field of the stator / field winding, we will have a
voltage on the armature coil.
As the single side of a coil moves through the centre of
the stator the velocity v of the wire will be positive half
the time and negative for the remaining time which
means that the voltage induced on the wire will look like
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eab V
ø
The problem with this is that the goal of the dc machine
acting as a generator is to have dc values at its
terminals not a constant magnitude with an alternating
sign. This problem can be solved by using a brushcommutator such that the terminal voltage taken from a
generator, or applied to a motor, are constant.
S
N
S
N
+ einduced V –
The resulting voltage measured at the two terminals will
appear as
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eab V
ø
The points of zero volts can be virtually eliminated
through optimizing the construction of the machine.
Expressions for the induced voltage and developed
torque
• The torque and rotational speed associated with an
electric machine can be described via:
T = Kφ ia
T = torque (Nm)
Kφ = machine const.
ia = armature current(Amps)
Ea = Kφω m
Ea = armature voltage (V )
Kφ = machine const.
ω m = armature speed(radian / s)
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• The strength of the stator magnetic field can be adjusted
by changing the current passing through its windings
(the field current), however this magnetic field can also
be generated by using permanent magnets in which
case the field will be a fixed value.
• If an electromagnet is used it is important to be aware
that the strength of the magnetic field is not linear. There
will be a small amount of residual magnetism without
any current flowing, and there is an upper limit where
the electromagnet will reach saturation. Generally this
nonlinear nature is described via a no-load saturation
curve which relates the armature voltage to the field
current.
Ea V
If A
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Configuration of a dc machine’s windings
• There are three different configurations available:
1. Separately excited dc machine, where the field circuit
and the armature circuit are independent of one
another. Note that the armature winding symbol
represents the lumped, rotating components of the dc
machine with the brushes on the top and bottom.
Field winding
1.
Armature
winding
Series connected dc machine, where field circuit and
armature circuit are connected electrically in series
with one another.
Field winding
Armature
winding
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CHAPTER 4
Shunt connected dc machine, where the field circuit
and armature circuit are electrically connected in
parallel.
Field winding
Armature
winding
DC machine equivalent circuit model
• By applying Thevenin’s theorem we can simplify the
rotor circuit into its Thevenin equivalent voltage source
and resistor. By representing the resistance of the wire
in the stator windings as resistors we have the following:
Ia
Rf
+
If
–
+
Vf
Ra
Ea
+
Vt
–
–
• In normal operations we want to control the speed and
torque characteristics of the electric machine. This can
be done by controlling Ea which is a function of If, and Ia
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which is in turn controlled by Ea. As a result, we normally
add a potentiometer in the field circuit to control the
current in the field winding.
DC generators
• The first machine we can describe, in terms of our
circuit model, is a separately excited generator where
the field circuit and armature circuit are completely
independent of one another.
• Since a generator is normally used to provide power to
an electrical load, in this model the resistor RL is
included as the electrical load.
Ia
Rf
+
If
–
Radj
+
+
Ra
Ea
RL
Vt
–
Vf –
We can apply KVL to the model to yield
V f − I f Radj − I f R f = 0
Ea − I a Ra − Vt = Ea − I a Ra − I a RL = 0
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We are able to adjust the output voltage, Vt, by
controlling the magnetic field of the field circuit, or via
the speed of rotation of the armature, the ωm term.
• The dc generator can also be configured in a shunt
configuration, and the circuit described using KVL, and
KCL
Radj
Rf +
+
IL
Ra
Vf
Ia
+
If
–
RL
Vt
Ea
–
–
Ea − I a Ra − Vt = 0
Ea = Kφω m
Vt − I f Radj − I f R f = 0
Ia − I f − I L = 0
If we consider the no-load configuration, and apply KVL
to the loop, we have
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ECE 3010 ELEMENTS OF ELEC. MACHINES AND DIGITAL SYSTEMS
Radj
CHAPTER 4
Rf +
IL
+
Ra
Vf
Ia
+
If
–
Vt
Ea
–
–
V f − I f Radj − I f R f = 0
Ea − I a Ra − I f Radj − I f R f = 0
however I a = I f , and if Ra << (Radj + R f )
Ea = I f (Radj + R f )
We are left with an equation where Ea=f (If), and if we
plot this line on the magnetization curve, which also
represents Ea=f (If), the intersection of the two lines
determines the no-load operating point of the machine.
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300
CHAPTER 4
Ea=(Rf2+Radj2)If
(Rf2+Radj2)>(Rf1+Radj1)
200
Ea=(Rf1+Radj1)If
Ea V
100
0
0
0.4
0.8
1.2
1.6
2
If A
We can extend this analysis to cover the case with an
attached load if we consider that we have
Ea − I a Ra − Vt = 0
Vt − I f Radj − I f R f = 0
∴ Ea − I a Ra − I f Radj − I f R f = 0
Ea = (Radj + R f )I f + I a Ra
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CHAPTER 4
Ea=(Rf1+Radj1)If+RaIa
Actual
operating point
300
No-load
operating point
200
Ea=(Rf1+Radj1)If
Ea V
100
0
0
0.4
0.8
If A
1.2
1.6
2
Figure 4.34 – Operating point with load.
Example – DC machine
• Given a dc machine with the following parameters:
• rated operating parameters: 10kW, 250V, 1000rpm
• electrical parameters: Ra = 0.2Ω, Rf = 133Ω
• magnetization curve (@1000rpm):
If (A)
0
0.1
0.2
0.3
0.4
0.5
0.75
1.0
1.5
2.0
Ea(V)
10
40
80
120
150
170
200
220
245
263
If the machine is connected in a shunt configuration and
operated as a generator at the rated speed, determine:
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CHAPTER 4
• What is the no-load output voltage if there is no field
current?
Table 1
I
0.1
0.2
0.3
0.4
0.5
0.75
is needed0 for the
machine
to have
a no-load
• What
a
191
194.6
198.2
201.8
205.4
209
218
227
80
120
150
170
200
220
output
voltage 10of 245V?
E
40
a
• What is needed for the machine to have a no-load
output voltage of 250V?
Ea
300
Ea
225
150
75
0
0
0.5
1
1.5
2
If
263 − 245
x + 191 = y ⇒ 36I f + 191 = E a
2.0 −1.5
250 −191
for E a = 250V = 36I f + 191 ⇒ I f =
= 1.63A
36
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ECE 3010 ELEMENTS OF ELEC. MACHINES AND DIGITAL SYSTEMS
CHAPTER 4
V
250
⇒ Radj + R f = t ⇒ Radj =
−133 = 20.37Ω
If
1.63
• What is the no-load output voltage if the machine is
rotated at 800rpm without a field control resistor (Radj
= 0Ω)?
300
1000rpm
Magnetization
curve
200
800rpm
Magnetization
curve
Ea V
100
Ea=133If
0
0
0.4
0.8
1.2
1.6
2
If A
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CHAPTER 4
• What rotational speed is required to have a no-load
output voltage of 200V without a control resistor
(Radj=0Ω)?
1000
2π
60
xrpm
200 = Kφ
2π
60
∴ xrpm = 816.32rpm
245 = Kφ
DC motor
• DC motors follow a similar form to the dc generators.
They can be configured as separately excited, shunt,
and series implementations.
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CHAPTER 4
• If we begin by considering a separately excited dc
motor, we have the following equivalent circuit and
equations which describe it:
Ia
Rf
+
If
–
Ra
Ea
+ Vt
–
+–
Vf
Vt − I a Ra − Ea = 0
Ea = Kφω m
V
I R
∴ω m = t − a a
Kφ Kφ
T
T = KφI a ⇒ I a =
Kφ
V
TRa
∴ω m = t −
Kφ (Kφ )2
Note that if Ra is a small value, ωm will essentially be
constant regardless of load.
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• For a separately excited dc motor we can control the
speed by controlling Vt, or controlling the Kϕ terms
which are controlled via Vf or the resistance of the field
circuit (Rf+Radj).
• Series dc motor
Ia
+
Mechanical
Load
–
Ra
Rf
+ Vt
–
Ea
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CHAPTER 4
Note that the speed is inversely proportional to the
torque. This means that at low speeds we have a very
large torque.
• The final configuration of a shunt motor can be analyzed
using techniques similar to the shunt generator analysis
Magnetic
which was previously examined.
Field
i(t)
+
Example – DC motor
• A 220V shunt motor has an armature resistance of
0.2Ω, and a total field resistance of 110Ω. At no-load the
motor draws 5A and rotates at 1500 rpm. If the motor
draws 52A at rated voltage and rated load, calculate the
rated speed and rated shaft torque. The mechanical
losses are the same at no-load and at full load.
Inl=5A
+
Ra=0.2Ω
+
– Vt=220V
Rf+adj=110Ω
+
–
1500rpm with no
mechanical load
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Ea
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ECE 3010 ELEMENTS OF ELEC. MACHINES AND DIGITAL SYSTEMS
• Rated speed
CHAPTER 4
need to find Ea under rated
conditions;
• Rated shaft torque
need to determine power
delivered to a mechanical load ∴ power absorbed by
the armature under rated conditions – losses in
rotating the armature.
• Finding the no-load Ea and no-load Ia will allow us to
find the power absorbed by the armature under noload conditions. This can also be found by
considering:
∑P = 0
Applying KCL:
If =
V f 220
=
= 2A
R f 110
5 − 2 − I a = 0 ⇒ I a = 3A
Pin − Pelecloss − Pmechloss = 0
220 × 5 − ( 32 × 0.2 + 2 2 × 110 ) = Pmechloss
⇒ Pmechloss = 658.2W
Alternatively we can apply KVL:
220 − 3 × 0.2 − Ea = 0 ⇒ Ea = 219.4V
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Magnetic
ECE 3010 ELEMENTS OF ELEC. MACHINES AND DIGITAL SYSTEMS
CHAPTER 4
Field
i(t)
The power absorbed by the armature (while not
+
driving a mechanical load) are:
Pa = Ea I a = 219.4 × 3 = 658.2W
This is the power required just Ito
turn the
+ armature
nl=5A
(i.e. overcome the mechanical
losses in the
Ra=0.2Ω
armature). The same result as obtained
previously.
+
•
– Vt=220V
Rf+adj=110Ω
Finding the rated-load Ea and rated-load
Ia will allow
+
Ea
us to find the power absorbed –by the armature
under
–
1500rpm with no
mechanical loadThis can also be found by
full-load conditions.
considering:
Iratedload=52A
+
Ra=0.2Ω
+
– Vt=220V
Rf+adj=110Ω
+
Ea
–
? rpm with
mechanical load
–
V f 220
If =
=
= 2A
R f 110
52 − 2 − I a = 0 ⇒ I a = 50A
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Pin − Pelecloss − Parmature = 0
220 × 52 − ( 50 2 × 0.2 + 2 2 × 110 ) = Parmature
⇒ Parmature = 10500W
Pload = Parmature − Pmechloss = 10500 − 658.2 = 9841.8W
• Finding the rated-torque requires finding the rated
speed. This can also be found by considering:
220 − 50 × 0.2 − Ea( fullload ) = 0 ⇒ Ea( fullload ) = 210V
1500
Ea(noload ) 219.4 Kφω m(noload )
2π
=
=
= 60
Ea( fullload )
210
Kφω m( fullload ) ω m( fullload )
T=
⇒ ω m( fullload ) = 150.35rad / s
P 9148.8
=
= 65.46Nm
ω 150.35
Conclusion
This chapter provided a review of basic circuit concepts
and how they can be applied to the study of dc machines.
The following objectives were met:
1.
Understand the concepts of basic circuit elements,
parameters, and topologies.
2.
Be able to apply KVL, KCL, and Ohm’s Law to
analyze a circuit to solve for any voltage and current
parameters.
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3.
CHAPTER 4
Be able to determine the power associated with a
circuit element.
4.
Understand the basic operation and configuration of a
dc machine.
5.
Be able to analyze a dc electric machine and
determine its external parameters both electrical and
mechanical.
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