DC Machines
EEE 360
DC Machines
EEE 360
Chapter 10:
DC Machines
 The direct current (dc) machine can be operated
as a motor or as a generator. However, it is
most often used for a motor.
 The major advantages of dc machines are the
easy speed and torque regulation.
 However, their application is limited to mills,
mines and trains. As examples, trolleys and
underground subway cars may use dc motors.
 Earlier automobiles were equipped with dc
dynamos to charge their batteries.
DC Machines
 Even today, the starter for an internal
combustion engine is a series dc motor.
 In the past, dc generators were utilized to excite
synchronous generators.
 However, the recent development of power
electronics has reduced the use of dc motors
and generators.
 Electronically controlled ac drives are gradually
replacing the dc motor drives in factories.
 Nevertheless, a large number of dc motors are
still employed by industry and several thousand
are sold annually.
DC Machines
Construction
DC Machine Construction
Armature core
Commutating
field winding
Commutating
pole
Commutator
Pole
shoe
S
Main
pole
core
N
• +
+
•
+
•
+
• +
• +
+
+
Brush
+
+
+
+
– –
+ +
– –
+ +
Series field
winding
–
–
–
–
–
–
–
–
–
–
–
–
Shunt field
winding
•
•
•
•
•
S
Compensating
winding
N
Air gap
Stator yoke
Armature
(rotor) winding
Rotor shaft
Figure 10.1 General arrangement
of a dc motor.
 The stator of the dc motor
has poles, which are
excited by dc current to
produce magnetic fields.
 The poles may have two
dc excitation windings:
one connected in parallel
and another connected in
series with the rotor.
 In the neutral zone, in the
middle between the
poles, commutating poles
are placed to reduce
sparking of the
commutator.
Karady & Holbert
DC Machine Construction
Armature core
Commutating
field winding
Commutating
pole
Commutator
Pole
shoe
S
Main
pole
core
N
• +
+
•
+
•
+
• +
• +
+
+
Brush
+
+
+
+
– –
+ +
– –
+ +
Series field
winding
–
–
–
–
–
–
–
–
–
–
–
–
Shunt field
winding
•
•
•
•
•
S
Compensating
winding
N
Air gap
Stator yoke
Armature
(rotor) winding
Rotor shaft
Figure 10.1 General arrangement
of a dc motor.
 The commutator is a
rotating electrical
switch that reverses
current direction.
 The commutating
poles are also
supplied by dc
current.
 Compensating
windings are mounted
on the main poles.
1
DC Machines
EEE 360
DC Machine Construction
Commutating
pole
Armature core
Commutating
field winding
Commutator
Pole
shoe
S
Main
pole
core
N
•
•
•
•
•
+
+
+
+
+
+
+
+
+
+
+
+
Brush
– –
+ +
– –
+ +
Series field
winding
–
–
–
–
–
–
–
–
–
–
–
–
Shunt field
winding
•
•
•
•
•
S
Compensating
winding
N
Air gap
Stator yoke
Rotor shaft
Armature
(rotor) winding
Figure 10.1 General arrangement
of a dc motor.
 The rotor has a ringshaped laminated iron
core with slots.
 Coils with several turns
are placed in the slots.
 The distance between
the two legs of the coil
is about 180 electric
degrees.
 The ends of each coil
are connected to a
commutator segment.
DC Machine Construction
Commutating
field winding
Commutating
pole
Commutator
Pole
shoe
S
Main
pole
core
N
• +
+
•
+
•
+
• +
• +
+
+
+
+
+
+
Brush
– –
+ +
– –
+ +
Series field
winding
–
–
–
–
–
–
–
–
–
–
–
–
Shunt field
winding
•
•
•
•
•
S
Compensating
winding
N
Air gap
Stator yoke
Rotor shaft
Armature
(rotor) winding
Figure 10.1 General arrangement
of a dc motor.
 The commutator
segments connect the
coils in series.
 The commutator
consists of insulated
copper segments
mounted on an
insulated tube.
 Two brushes are
pressed to the
commutator to permit
current flow.
Commutator
S
Main
pole
core
N
Commutating
field winding
Commutating
pole
Commutator
Pole
shoe
S
Main
pole
core
N
• +
+
•
+
•
+
• +
• +
+
+
Brush
+
+
+
+
– –
+ +
– –
+ +
Series field
winding
–
–
–
–
–
–
–
–
–
–
–
–
Shunt field
winding
•
•
•
•
•
S
Compensating
winding
N
Air gap
Stator yoke
Armature
(rotor) winding
Rotor shaft
Figure 10.1 General arrangement
of a dc motor.
 The high voltage
produces flashover
and arcing between
the commutator
segment and the
brush.
 The arcing can be
lessened by
reducing the
magnetic field at the
brush site by the
commutating poles.
Karady & Holbert
•
•
•
•
•
+
+
+
+
+
+
+
+
+
+
+
+
Brush
Series field
winding
– –
+ +
–
–
– –
+ +
–
–
–
–
–
–
–
–
–
–
Shunt field
winding
•
•
•
•
•
S
Compensating
winding
N
Air gap
Stator yoke
Rotor shaft
Armature
(rotor) winding
Figure 10.1 General arrangement
of a dc motor.
 The compensating
winding is shortcircuited to damp rotor
oscillations.
 The poles are
positioned on an iron
core that provides a
closed magnetic circuit.
 The motor housing
supports the iron core,
the brushes and the
bearings.
DC Machine Construction
Armature core
Commutating
field winding
Commutating
pole
Commutator
Pole
shoe
S
Main
pole
core
N
• +
+
•
+
•
+
• +
• +
+
+
+
+
Brush
+
+
Series field
winding
– –
+ +
–
–
– –
+ +
–
–
–
–
–
–
–
–
–
–
Shunt field
winding
•
•
•
•
•
S
Compensating
winding
N
Air gap
Stator yoke
Rotor shaft
Armature
(rotor) winding
Figure 10.1 General arrangement
of a dc motor.
DC Machine Construction
Armature core
Commutating
field winding
Pole
shoe
DC Machine Construction
Armature core
Commutating
pole
Armature core
 The brushes are placed
in the neutral zone, where
the magnetic field is close
to zero, to reduce arcing.
 The commutator switches
the current from one rotor
coil to the adjacent coil,
which requires the
interruption of the coil
current.
 The sudden interruption
of an inductive current
generates high voltages.
DC Machine Construction
Rotation
Idc/2
Brush
Idc/2
Idc
Shaft
– – – – –
Pole
winding
– – – – –
|
1
N
8
2
6
4
3
7
S
5
+ + + + +
Insulation
Rotor
winding
+ + + + +
Idc
Copper
segment
Figure 10.2 Commutator with
the rotor coils connections for
a two-pole dc motor.
 Figure 10.2 depicts the
commutator and the rotor
coil connections for a
hypothetical rotor with
eight coils.
 The commutator switches
the location of the current
entry and exit as the rotor
turns.
 As an example, the motor
current (Idc) flows into the
rotor at segment 1 and
flows out of the rotor from
segment 5.
2
DC Machines
EEE 360
DC Machine Construction
Rotation
Idc/2
Brush
Idc/2
Idc
Pole
winding
Shaft
– – – – –
– – – – –
|
1
8
N
2
3
7
6
S
4
5
+ + + + +
Insulation
Rotor
winding
+ + + + +
Idc
Copper
segment
Figure 10.2 Commutator with
the rotor coils connections for
a two-pole dc motor.
 Further clockwise rotation
switches the entry and exit
points to segments 2 and
6, respectively, when the
next commutator segment
slides to the brush.
 The brushes divide the
rotor into two parallel
electrical paths.
 The current flows in
opposite directions in the
two halves of the rotor
separated by the brushes.
Small DC Machines
DC Machines
Photographs showing a small
DC motor construction
Small DC Machines
 The commutator
segments are made
out of copper.
 Mica insulation is
placed between the
segments.
Figure 10.3 Details of the
commutator of a dc motor.
 The coil endings are
welded to the copper
segments.
Small DC Machines
Figure 10.5 Rotor of a dc motor.
Figure 10.4 DC motor stator with
poles visible.
 The stator of a
small dc machine
with four main
poles is shown.
 An iron frame supports the iron core.
 The ring-shaped
iron core has poles
with an excitation
winding, which is
secured by tape.
Small DC Machines
 The rotor of a dc motor with the commutator is
seen in Figure 10.5.
 The rotor has a laminated iron core with slots.
Coils are placed in the slots.
 Photograph shows
the interconnection
between the coil
endings, the
commutator, and the
ball bearing
supported shaft.
Figure 10.5 Rotor of a dc motor.
Karady & Holbert
3
DC Machines
Small DC Machines
EEE 360
Small DC Machines
Figure 10.6 Cutaway view of a dc motor.
 The rotor,
commutator and
brush assembly
are presented in
Figure 10.6.
 A fan attached to
the rotor shaft at
the right-hand
side cools this
motor.
Figure 10.6 Cutaway view of a dc motor.
Small DC Machines
Figure 10.6
 The graphite brush
is pushed to the
commutator by a
spring.
 The square-shaped
brush holder
attached to the
housing secures
the graphite brush.

This motor has two
Cutaway view of a dc motor.
brush assemblies.
DC Machine Operation
 In a dc motor, the stator poles are supplied
by dc excitation current, which produces a
dc magnetic field.
 The rotor is supplied by dc current through
the brushes, commutator and coils.
 The interaction between the stator
magnetic field and rotor current generates
a force that drives the motor.
DC Machines
Operating Principle
DC Machine Operation
 Figure 10.7 demonstrates that the magnetic field lines
enter into the rotor from the north pole (N) and exit
toward the south pole (S).
 The stator poles
generate a
magnetic field that
is more or less
perpendicular to
the current-carrying
conductors in the
rotor
(a) Rotor current flow from segment 1 to 2 (slot a to b)
Figure 10.7 Force generation and commutation of a simplified dc motor.
Karady & Holbert
4
DC Machines
EEE 360
DC Machine Operation
 The interaction between the magnetic field and the
rotor coil current produces a Lorentz force
 The Lorentz force
is perpendicular to
both the magnetic
field and conductor
current
 This force drives
the motor in the
clockwise direction
(a) Rotor current flow from segment 1 to 2 (slot a to b)
DC Machine Operation
Figure shows that the Lorentz force direction is the same
when current direction changes after commutation.
(b) Rotor current flow from segment
2 to 1 (slot b to a)
 In this figure the
current flows from
segment 2 to 1 (slot b
to a).
 The direction of the
generated Lorenz
force is the same
clockwise direction as
on the previous slide.
Figure 10.7 Force generation and commutation of a simplified dc motor.
Figure 10.7 Force generation and commutation of a simplified dc motor.
DC Machine Operation
DC Machine Operation
Figure 10.8 Fleming’s left
hand rule for motors.
 The force on an
individual conductor is
described by Fleming’s
left hand rule for motors
 In this figure the second
finger points in the
direction of current flow
and the index finger in
the magnetic field
direction.
 The thumb then specifies
the resultant force (and
rotational motion)
direction.
A summary of the force generation is
 The generated force turns the rotor clockwise until the
coil reaches the neutral point between the poles
 At this point, the magnetic field perpendicular to the rotor
current becomes practically zero together with the force
 However, inertia drives the motor beyond the neutral
zone where the direction of the magnetic field reverses
 To avoid the reversal of the force direction, the
commutator changes the current direction, which
maintains the clockwise rotation
 An actual dc machine has several rotor coils and
commutator segments, which assures smooth
continuous rotation
DC Machine Operation
Reversal of rotational direction
 The motor rotation direction can be reversed by
switching the polarity of the stator field winding
voltage
 This voltage polarity reversal swaps the north
and south poles
DC Machines
DC Generator
 Furthermore, the rotation direction can also be
changed by reversing the rotor current direction
(i.e., the rotor voltage polarity)
Karady & Holbert
5
DC Machines
EEE 360
DC Generator
DC Generator
1
1
 A turbine or other
machine drives
the generator rotor
 The N-S poles
produce a dc
magnetic field and
the rotor coil turns
in this field
2
2
(a) Rotor current flow from
segment 1 to 2 (slot a to b)
(a) Rotor current flow from
segment 1 to 2 (slot a to b)
Figure 10.12 DC generator operation.
Figure 10.12 DC generator operation.
DC Generator
1
2
(a) Rotor current flow from
segment 1 to 2 (slot a to b)
Figure 10.12 DC generator operation.
 The conductors in
slot a are cutting the
magnetic field lines
entering into the rotor
from the north pole,
which induce positive
voltage.
 The conductors in
slot b are cutting the
field lines exiting from
the rotor to the south
pole, which induce
negative voltage.
DC Generator
(b) Rotor current flow from
segment 2 to 1 (slot b to a)
Figure 10.12 DC generator operation
(b) Rotor current flow from
segment 2 to 1 (slot b to a)
Figure 10.12 DC generator operation
Karady & Holbert
 However,
simultaneously the
commutator reverses
its terminals, which
assures that the
output voltage (Vdc)
polarity is unchanged
 However, when the coil
passes the neutral zone,
conductors in slot a are
then moving toward the
south pole and cut flux
lines exiting from the
rotor
 The conductors in slot b
cut the flux lines entering
the rotor from the north
pole
 This changes the
polarity of the induced
voltage in the coil
DC Generator
DC Generator
 With the polarity of
the induced voltage
reversed, the voltage
induced in a is now
positive, and in b is
negative
 The conductors in
the rotor slots cut
the magnetic flux
lines, which
induce voltage in
the rotor coils
 The coil has two
sides: one is
placed in slot a,
the other in slot b
 The cutting of the magnetic field lines generates
voltage in the conductors.
 The voltages generated in the two sides of the coil
are added.
 The induced voltage in the coil is proportional with
the magnetic flux derivative. The derivative can be
expressed with motor rotational speed
Nr
B
ℓg
v
E  2N B  v
g
r
g
is the number of turns in the rotor coil;
is the magnetic flux density in the gap;
is the length of the rotor; and
is the velocity of the rotor.
6
DC Machines
EEE 360
DC Generator Equivalent
Circuit
DC Machines
DC Generator Equivalent
Circuit
 In a generator, the magnetic field produced by
the stator poles induces a voltage in the rotor (or
armature) coils when the generator is rotated.
 The dc field current of the poles generates a
magnetic flux (Φag).
 This implies that the flux is proportional with the
field current (If) if the iron core is not saturated:
 ag  K 1 I f
where K1 is a proportionality constant
DC Generator Equivalent
Circuit
DC Generator Equivalent
Circuit
 The rotor conductors cut the field lines which
generates voltage in the coils. The generated
voltage is calculated by
 The combination of the equations gives a
practical expression for the induced voltage:
E g  2 Nr B  g v
 This formula is modified by substituting for the rotor
coil speed and the magnetic flux
v 
Dg
2
 ag  B  g Dg
where Dg is the rotor diameter and ω is the rotor
angular velocity
DC Generator Equivalent
Circuit
 When the generator is loaded, the load current
produces a voltage drop on the rotor winding
resistance
 In addition, there is a more or less constant
voltage drop of 1 V to 3 V on the brushes
 These two voltage drops reduce the terminal
voltage of the generator
 The terminal voltage can be calculated by using
KVL, which gives relation between the induced
and terminal voltage
Karady & Holbert
Eag  2 Nr B  g v  2 Nr B  g  Dg 2
 Nr B  g Dg    Nr  ag 
 For further simplify, we combine the number of
rotor turns and the constant K1 into a new
constant Km, which is the machine constant
Eag  Nr  ag   Nr K1 If   K m If 
DC Generator Equivalent
Circuit
The terminal voltage equation is
Eag  Vdc  Iag Ra  Vbrush
where
Vdc
Iag
Ra
is the generator terminal voltage;
is the load current;
is the generator armature (or rotor)
winding resistance;
Vbrush is the voltage drop on the brushes; and
Eag is the field current (or stator current)
generated induced voltage.
7
DC Machines
EEE 360
DC Generator Equivalent
Circuit
 This equation suggests that the dc generator can be
represented by a voltage source and a resistance
connected in series
 It must be noted that the flux is generated by
inductances
 The exact equivalent circuit includes the rotor and
stator inductances as well as the magnetizing
inductance
 However, for steady-state analysis, the inductances
can be neglected since the dc current does not
produce a voltage drop on an inductance
DC Generator Equivalent
Circuit
 In this diagram, a separate dc source (Vf) supplies
the stator excitation circuit
 The generator induced voltage (Eag) drives current
(Iag) through the load
 The induced
voltage is larger
than the terminal
or load voltage
 The generator
has to be rotated
by an engine or
Figure 10.14 Equivalent circuit of a
turbine
separately excited dc generator.
DC Generator Equivalent
Circuit
 A dc generator is represented by a voltage
source and a resistance connected in series
DC Machines
DC Motor Equivalent Circuit
Figure 10.15 Equivalent circuit of a
separately excited dc motor.
DC Motor Equivalent Circuit
Equivalent circuit of a separately excited dc
motor
 The excitation current (If) is supplied by a separate
DC source (Vf)
Vbrush
Rf
Ra
Electrical
power in
max
Vf If
Eam
Iam
Vdc
DC power
supply
Mechanical
power out
DC Motor Equivalent Circuit
 The motor terminal voltage is the sum of the
induced voltage, the voltage drop on the
armature resistance, and the brush voltage drop.
The motor-induced voltage equations are
Vdc  E am  Iam Ra  Vbrush
Eam  K m If 
 The rotor angular
frequency is related
to the shaft speed
(nm) by ω = 2 π nm
Figure 10.15 Equivalent circuit of a separately excited dc motor.
Karady & Holbert
8
DC Machines
EEE 360
DC Motor Equivalent Circuit
Shunt excited dc motor
 The shunt machine is the most frequently used motor.
 It requires only one power supply.
 A load change has little effect on the motor speed or the
generator voltage.
Figure 10.16 Equivalent
circuit of a shunt motor
connection.
DC Motor Equivalent Circuit
Series excited dc motor
 The field current and motor current are the same
 This motor has large starting torque, which is an
advantage for trolley and tramway drives
Figure 10.17 Equivalent circuit
of a series motor connection.
 The sudden loss of
mechanical load
increases the motor
speed rapidly and
can lead to
mechanical failure
and an accident.
DC Motor Equivalent Circuit
Compound connection
 This motor has two excitation windings on the stator poles
 One field coil is connected in series (Rfs) with the rotor
 The other field winding is in parallel (Rfp)
DC Machines
Operation Analysis
Figure 10.18
Compound motor
connection.
Operation Analysis
General equation for DC machine analysis
 Combining the machine constant and induced
voltage equations and neglecting the brushes
voltage drop gives
K m If ω  E am  Vdc  IamRm
 The mechanical output power and torque are
Pout  Eam Iam
T
Pout
 K m Iam If
ω
DC Machines
Separately Excited Motor
Operation
 The mechanical losses (ventilation and friction loss;
typical values 5%–10%) should be subtracted from
the output power calculated
Karady & Holbert
9
DC Machines
EEE 360
Separately Excited Motor
The rated values are
Pm  40 kW
Vdc  240 V
R a  0.25 
R f  120 
The motor constant is calculated using measured
values
Vm  Vdc
Im  8 A
If  2 A
nm  1000 rpm
The motor angular speed in test conditions
rev  1 min 
rad
 2  rad 
ω o  2  nm  
1000

  104.7
rev
min
60
sec
s




Separately Excited Motor
 With the motor constant determined, we now
select an arbitrary speed for analysis
 The motor speed and excitation (field) current
are
If  2 A
nm  1000 rpm
 The induced voltage for these conditions is
Separately Excited Motor
 The induced voltage is calculated by using the
loop voltage equation (KVL) for the rotor circuit
Eo  Vm  Im R a
 240 V  (8 A)(0.25 Ω)  238 V
 The motor constant is calculated from the
induced voltage
Km 
Eo
(238 V)
V s

 1.137
If ωo (2 A)(104.7 rad/sec)
A
Separately Excited Motor
 The motor current is calculated from the
equivalent circuit using the loop voltage equation
(KVL) for the rotor:
Iam 
Vdc  Eam
240 V  238 V

8 A
Ra
0.25 
Eam  K m If ωm
 (1.137 V  s A )(2 A)(104.7 rad/s)
 238 V
Separately Excited Motor
The motor starting current is calculated by
substituting a speed of zero into the current
equation, that is, Eam = 0 since ω = 0, such that
Istart  Iam (  0) 
Vm 240 V

 960 A
Ra 0.25 
The rated current is
Im_rated
P
40000 W
 m 
 166.7 A
Vdc
240 V
Karady & Holbert
Separately Excited Motor
The input electrical power, including the field
excitation, is
2
Pin  Vm Iam  If R f
 (240 V)(8 A)  (2 A)2 (120  )
 2400 W
The output mechanical power is
Pout  Eam Iam
 (8 V)(238 A)  1904 W
10
DC Machines
EEE 360
Separately Excited Motor
 The motor operation is stable to the right side of
the maximum, where an increase in load power
decreases the motor speed
nm  0.rpm 1rpm  2000rpm
The output power is
plotted using the
Mathcad program
60
Pout  1A nm
kW
Pout  2A nm 40
kW
Pout  4A nm
kW
20
Pm
kW
0
0
500
1000
1500
2000
Figure 10.22 Separately
excited motor output
power versus speed at
different field currents.
Separately Excited Motor
The torque of the motor is calculated by dividing
the output power with the angular speed:
Tm 
Pout
1904 W

 18.2 N  m
ωm 104.7 rad/s
An alternative equation for the torque is
Tm  K m If Iam
 (1.137 V  s A )(2 A )(8 A )
 18.2 N  m
nm
rpm
Separately Excited Motor
nm  0.rpm 1rpm  2000rpm
5000
Tm 1A n m 4000
N m
Tm 2A n m
3000
N m
Tm 4A n m
N m
2000
1000
0
0
500
1000
1500
2000
nm
rpm
Figure 10.24 Separately excited
motor torque versus speed at
different field currents.
 The motor
torque varies
linearly with the
speed for a
given field
current
 The field current
affects the motor
starting current
and the motor
operating range
 The motor data are
Vdc  240 V
Figure 10.25 Shunt motor
simplified equivalent circuit.
R a  0.25 Ω
R f  120 Ω
• In the case of a shunt
motor, the machine field
winding (Rf) is connected
in parallel with the supply
• The voltage drop across
the brushes and possible
saturation of the iron
core are neglected.
Karady & Holbert
Shunt Motor
Shunt Motor
Shunt Motor
Pm  40 kW
DC Machines
 First, we compute the motor constant from
measured values:
Im  8 A
nm  1000 rpm
Vm  Vdc
 The field current at the
steady-state test condition is
V
240 V
If  m 
2A
R f 120 
 Using KCL, the armature current in the test
condition is
Ia  Im  If  8  2  6 A
11
DC Machines
EEE 360
Shunt Motor
 The motor angular speed is
ωo  2 π nm 
2 π (1000 rpm)
 104.7 rad/s
60 s/min
Shunt Motor
 The calculation of the induced voltage requires
the angular speed: nm = 1000 rpm
ωm  2 π nm  104.7
 Applying KVL, the induced voltage is
Em  Vm  Ia Ra
 (240 V)  (6 A)(0.25  )
 238.5 V
 The shunt motor constant is
Km 
Em
238.5 V

 1.139 V  s/A
If ωo (2 A)(104.7 rad/sec)
Shunt Motor
 The induced voltage when nm = 1000 rpm is
Eam  K m If ωm
 (1.139 V  s A )(2 A)(104.7 rad/s)
 238.5 V
 The loop voltage equation applied to the rotor
circuit in permits the calculation of the armature
motor current, which is
V  Eam 240 V  238.5 V

6A
Iam  m
Ra
0.25 
Shunt Motor
 Using the Mathcad program the motor current
vs. speed is plotted
nm  0rpm 1rpm  1200rpm
1000
Im n m Vm
A
Im_rated
A
Motor Operation
500
0
 500
0
Generator Operation
200
400
600
800
1000
nm
rpm
Figure 10.26 Shunt motor current
versus motor speed at rated voltage.
 The motor current
becomes zero near
1000 rpm
 Above this speed,
the motor would
operate as a
1200
generator, but the
generator
operation requires
an external driver
Karady & Holbert
rad
sec
 The field current is found from applying Ohm’s
Law at steady-state conditions:
If 
Vm 240 V

2A
R f 120 
Shunt Motor
 Using KCL, the motor input current is the sum of
the field current and armature current:
Im  If  Iam  2  6  8 A
 The rated motor current is
P
40000 W
Im_rated  m 
 166.7 A
Vdc
240 V
 The starting current is
V
240 V
Istart  m 
 960 A
Ra 0.25 
Shunt Motor
 The input electrical motor power at 1000 rpm is
Pin  V m Im  (240 V)(8 A)  1920 W
 The output mechanical power is computed from
Pout  Eam Iam  (238.5 V)(6 A)  1431 W
 The numerical values
at 1000 rpm are
Pin  1.92 kW
Pout  1.43 kW
12
DC Machines
EEE 360
Shunt Motor
 The input power
is a maximum
when the motor
starts
nm  0.rpm 1rpm  1000rpm
250
Pin n m Vm 200
kW
Pout  n m Vm
150
 The input power
declines with
increasing speed
kW
100
Pm
kW
50
0
0
200
400
600
800
1000
nm
rpm
Figure 10.27 Shunt motor input and output
powers versus speed at rated voltage.
 The bell-shaped
output power
curve reaches a
maximum value
around 500 rpm
Shunt Motor
 The motor torque is the output power divided by
the angular speed. The torque at 1000 rpm and
240 V is
Tm 
Pout
1430 W

 13.65 N  m
ωm 104.7 rad/s
 The alternative torque equation is
Tm  K m If Iam
Shunt Motor
 The torque linearly decreases with increasing
speed
nm  0.rpm 1rpm  1000rpm
3000
Tm nm Vm
2000
N m
Tm nPrated Vm
1000
N m
0
0
200
400
600
800
1000
nm
rpm
Figure 10.29 Shunt motor torque
versus speed at rated voltage.
The curve
demonstrates that
the starting torque
is very high,
which is a
desirable quality
for certain
applications.
Shunt Generator
 The generator operation will be analyzed when
the generator charges a battery.
Vbatt = 238 V Ibatt = 20 A Rf = 120.2 Ω Ra = 0.25 Ω
Ra
Iag
Eag
Pgen
max
Ibatt
Rf
Vbatt
Ifg
Figure 10.30 Equivalent circuit
of a dc shunt generator.
The goal here is the
calculation of the
required generator
speed to provide a
particular load/battery
current, assuming the
generator or battery
voltage is constant
Karady & Holbert
DC Machines
Shunt Generator
Shunt Generator
 Calculation of the generator field current from
Ohm’s Law
V
238 V
Ifg  batt 
 1.98 A
Rf
120.2 Ω
 Using KCL, the generator armature
current is the sum of the battery
current and the field current
Iag  Ifg  Ibatt
 1.98 A  20 A
 21.98 A
13
DC Machines
EEE 360
Shunt Generator
Shunt Generator
 The induced voltage is calculated from the rotor
circuit loop voltage equation (KVL):
Ibatt  0A 1A  200A
60
Pgen Ibatt 
Eag  Vbatt  R a Iag
kW
 238 V  (0.25 )(21.98 A)
Pm
kW
 243.5 V
40
20
0
 The required input (mechanical) power to the
generator is the product of the induced voltage
and the generator produced current:
0
50
100
150
200
Ibatt
A
Figure 10.31 Shunt generator input
power versus load (battery) current.
The curve
shows that the
required
generator
power
increasing
almost linearly
with the
battery current
Pgen  Iag Eag  (1.98 A )(243.5 V )  5352 W
Shunt Generator
Ibatt  0A 1A  200A
1300
1250
n n Ibatt 
rpm
1200
n n Irated
1150
rpm
1100
1050
1000
0
25
50
75
100
125
150
175
200
Ibatt
A
Figure 10.32 Shunt generator speed
versus load (battery) current.
 The graph shows
that the relation
between the
generator speed
and battery
current is linear
 The linear
relation permits
the accurate
control of the
load by speed
variation
Series Motor
Vdc  240 V
Figure 10.33 Series motor
simplified equivalent circuit.
Series Motor
Series Motor
 Motor constant calculation
The motor data are
Pm  20 kW
DC Machines
R a  0.25 Ω
R f  0.3 Ω
Figure 10.33 diagrams
the series motor
simplified equivalent
circuit if the voltage
drop across the
brushes and the ironcore saturation are
neglected
Karady & Holbert
 The motor current, speed and voltage are
measured to determine the motor constant
 At rated terminal voltage, the motor current is 8
A while running at 500 rpm:
Vm  Vdc  240 V
Im  8 A
nm  500 rpm
14
DC Machines
EEE 360
Series Motor
 Motor constant
 The motor angular speed
at the test condition is
2 π 500
60
 52.4 rad/s
ω o  2 π nm 
Figure 10.33 Series motor
simplified equivalent circuit.
 The induced voltage is calculated by the loop
voltage equation using the circuit in Figure 10.33
Em  Vm  Im (Ra  R f )
Series Motor
 The motor constant is
E
(235.6 A)
Km  m 
 0.562 V  s/A
Im ωo (8 A)(52.4 rad/s)
 The series motor can be controlled by regulating
the supply voltage
 This voltage is selected as the first independent
variable; the second variable is the motor speed
 The motor speed and voltage are
Vm = Vdc = 240 V
nm = 100 rpm
 (240 V)  (8 A)(0.55 Ω)  235.6 V
Series Motor
Series Motor
 Calculation of the angular speed, which is
m  2  nm 
2  100
 10.47 rad/sec
60
 The field current is equal with the motor current.
The field current is replaced by the motor current
in induced voltage equation
Eam  K m Im ωm
 The loop voltage equation for the equivalent circuit
gives the motor voltage–current relation:
Im (Ra  R f )  Vm  Eam
 The induced voltage is substituted in this equation:
Im (Ra  R f )  Vm  K m Im ωm
 The motor current from the equation
Im 

Vm
R a  R f  K m ωm
(240 V)
 3 7 .3 A
(0.25  0.3)Ω  (0.562 V  s/A)(10.47 rad/s)
Series Motor
 The numerical value of the motor constant is
Eam  K m Im ωm  (0.562 V  s/A )(37.3 A )(10.47 rad/s )
 219.5 V
 The motor rated current is
P
20,000 W
Im_rated  m 
 83.3 A
Vdc
240 V
 The starting current is calculated by substituting
a speed of zero into the motor current formula
Vm
240 V

 436 A
Istart 
Ra  R f (0.25  0.3) Ω
Karady & Holbert
Series Motor
 The input electrical and output mechanical
powers are the product of the motor current and
appropriate voltage
Pin  Vm Im
 (240 V)(37.3 A)
 8952 W
Pout  Eam Im
 (219.5 V)(37.3 A)  8187 W
15
DC Machines
EEE 360
Series Motor
 The torque of the motor is
P
8187 W
Tm  out 
 782 N  m
ωm 10.47 rad/sec
 The alternative torque equation is
Tm  K m Im2  (0.562 V  s/A )(37.3 A )2
 782 N  m
Series Motor
 Figure 10.34 demonstrates that the motor current
declines rapidly with increasing speed
Im nm Vm
400
A
300
Im_rated
A
200
100
0
 Using the Mathcad program, the series motor
current, power and torque are plotted
0
20
40
Past the output power
peak value, increasing
the speed decreases
both powers
nm  0.rpm 1rpm  1000rpm
125
Pin  n m Vm 100
kW
Pout  n m Vm 75
kW
Pm
kW
50
25
0
Operating point
0
20
40
60
80
100
nm
rpm
Figure 10.35 Series motor input and
output powers versus speed.
The stable operating
point, which is the
right-hand-side intersection of the Pout and
Pm curves at ~30 rpm,
Karady & Holbert
60
80
100
nm
rpm
Figure 10.34 Series motor current
versus speed.
Series Motor
 Figure 10.35 shows input electrical power is very
high during the starting conditions, when the
output mechanical power is practically zero
The characteristics
show that this
motor operates
well at low speed
In this example,
the speed at the
rated current is
only about 40 rpm
as seen in the plot
nm  0rpm 1rpm  500rpm
500
Series Motor
 Figure 10.37 demonstrates that the starting torque
is very high, but the torque declines rapidly as the
motor gains speed.
The torque curve
nm  0rpm 1rpm  1000rpm
5
1.5 10
Tm n m Vm
Nm
Tm n Prated Vm
N m
5
1 10
4
5 10
0
0
20
40
60
nm
rpm
Figure 10.37 Series motor
torque versus speed.
80
100
indicates that the
decrease of the
torque increases
the speed
This is dangerous
when the motor
suddenly loses its
load and the
speed increases
to an unsafe value
16