DC Machines




Chapter 1: Theory Of Operation & Construction
Chapter 2: DC Machines On Load
Chapter 3: DC Generators
Chapter 4: DC Motors
1
References
S.J. Chapman, "Electric Machinery
McGraw Hill, 4th edition, 2005.
Fundamentals",
2
Chapter 1
Theory Of Operation & Construction
1.
2.
3.
4.
5.
6.
Introduction
Electromechanical Energy Conversion
DC Machines Theory of Operation
Construction
Classification of DC Machines
Armature Voltage and Developed Torque
3
Electric Machine
Mechanical
Input
Electrical
Output
Generator
Electrical
Input
Motor
Mechanical
Output
Electromechanical Energy Conversion
Electrical system
v
+
_
i
w
Ideal Electric Machine
Motor
Energy Flow
Generator
T Mechanical system
v i=T w
4
Principle of Operation
If a conductor of length l moves at a linear speed v in a
magnetic field the induced voltage in the conductor is
Faraday’s law or
flux cutting rule
e=Blv
e
N
v
v
B
S
B
Right hand rule
5
e
Principle of Operation
For the current carrying conductor of length l the
force known as Lorentz force produced is
i
F
i
N
F
F=Bli
B
S
B
Left hand rule
6

Induced e.m.f
e
e
N
v
v
B
e=Blv
S
B
Right hand rule

Electromagnetic Force
F
i
i
N
F
B
F=Bli
S
B
Lift hand rule
7
DC Machines theory of operation
e
N
e
w
S
e
N
N
w
w
S
S
and for multiturn machine
8
Generated Voltage (EMF)


For counter-clockwise motion the rotor terminal under
the N pole is always positive with respect to that
under the S pole.
The brush terminal B1 is always positive and B2 is
negative so e12 is unidirectional
9
DC Machines
DC currents are required for both its field winding
( stator ) and armature winding (Rotor)
DC machine mainly used as DC motors
DC motors is characterized by superior torque
and a wide range of speed
DC machine is more costly than comparable AC
machine and their maintenance cost is higher.
10
Construction
4 Pole DC Machine
11
4 Pole DC Machine
12
2 Pole DC Machine
Shaft
Armature
Commutator
Stator
pole
Field
coil
13
DC Machine
14
Construction Of DC Machine


Armature
The armature (rotor) is made up of cylindrical steel
structure. However, it cannot be fabricated from
solid steel, since eddy and hysteresis losses may
reach prohibitive values and damage the armature
windings insulation. Reduction of eddy current
losses can be achieved by fabricating armature core
of insulated laminated silicon steel. Sheets of silicon
steel of 0.35 mm thick are punched to form the
shape of armature slots and teeth, then insulated on
both sides and assembled to form armature core.
Hysteresis loss can be reduced by adjusting the
amount of silicon in the steel such that the
hysteresis loop area of the alloy is kept as low as
possible.
15
The Armature
Slots
Cooling ducts for
air circulation
Teeth
16
Eddy Currents
17
Eddy Currents (cont.)
Dividing the iron core into thin insulated
laminations minimizes eddy current loss.
18
Hysteresis Loop
19
Hysteresis Loss


The amount of heat released per cycle
expressed in J/m3 is equal to the area in
Tesla.A/m of the hysteresis loop.
To reduce hysteresis losses, we select
magnetic materials that have a narrow
hysteresis loop, such as the grainoriented silicon steel used in the cores
of alternating-current transformers.
20
Construction Of DC Machine (Cont.)


The Commutator
The commutator is made up from a number of
hard drawn copper segments forming a cylinder
rotating with the shaft. For small machines this
is achieved by mounting the segments on a
Bakelite ring fitted on the shaft. The segments
are insulated from each other and fixed on the
Bakelite ring using adhesives. For machines with
high ratings, the commutator is assembled from
commutator segments shaped as shown. The
segments are fitted into two guides as shown in
figure and insulated from each other and from
the guides using mica paper.
21
The Commutator
Commutator: is a mechanical rectifier, which converts the alternating voltage
generated in the armature winding into direct voltage across the brush. It is made of
copper segments insulated from each other by mica and mounted on the shaft of the
machine. The armature windings are connected to the commutator segments.
Commutator
22
Construction Of DC Machine (Cont.)






The Brushes
Brushes are fitted to collect or conduct
current from or into the armature. They
are made of graphite with high hardness
to ensure long life time during service.
Brushes are usually fitted in brush box
(holder) are pressed on commutator
segments by virtue of spring. The brush
holder is fixed in the machine frame and
insulated from it. Details of brush holder
are shown in figure.
1. Brush holder box
2. Brush
3. Pressure spring
4. live pigtail
23
The Brushes
The purpose of the brush is to ensure electrical connections between the
rotating commutator and stationary external load circuit. It is made of
carbon and rest on the commutator.
Commutator and Brushes
24
Construction Of DC Machine (Cont.)


Field System
The magnetic flux in a DC machine is established by
electromagnets shaped in the form of salient poles
attached to a cylindrical yoke. The pole itself is made
from two main parts, shank and shoe. The shank
may be of circular or rectangular cross section while
fabricated from cast steel, wrought iron or cast iron.
There is no need for laminating it since flux flow is
unidirectional. As for the shoe, it keeps field
windings in place and distributes flux lines over
larger area and thus avoids saturation of flux lines in
armature teeth. Pole shoe is essentially laminated to
keep iron loss in it to a minimum. Iron loss in pole
shoe is produced by tooth pulsations. Field windings
are made by concentric number of turns of insulated
copper wire wound on a former. The former is made
of Bakelite and is shaped as shown in figure.
25
Construction Of DC Machine (Cont.)


Armature Windings
Armature windings are made of the shape of
pre-wound coils as shown in figure. The
armature is wound in a double layer fashion
to avoid irregular shape of end connections.
Moreover, coil pitch should be as near as
possible to pole pitch to ensure voltage
summation around the coil. On the other
hand, to obtain a reasonable value of
collected voltage, the coils are connected in
series maintaining that their voltages are in
the same sense. This is achieved in two
alternative methods of windings; namely, Lap
and Wave types of windings.
26
Construction Of DC Machine (Cont.)

The turn, coil, and the winding are shown schematically as:
End connection
Conductors
Turn



Coil
Winding
A turn consists of two conductors connected to one end by an end connector.
A coil is formed by connecting several turns in series.
A winding is formed by connecting several coils in series.
27
Construction Of DC Machine (Cont.)



Lap Windings
In this type of windings, the successive coils of the
armature overlap each other as shown in figure.
Successive coils are connected in series with their ends
connected to successive commutator segments. The
pitch of the coil as viewed from commutator end is
termed as the front pitch while that measured from
other end is called the back pitch. These pitches should
not be equal to ensure winding progression.
If an armature is wound with C coils and having 2p
number of poles, then the back pitch should be equal to
the number of coils occupying one pole pitch, i.e.
Back pitch y b= C/2p
coils
and hence, Front pitch yf = yb ± 1 coils
Positive sign is used for winding retrogression while the
negative sign is for winding progression.
Commutator pitch is then given by : yc = 1
Number of Commutator Segments = C
28
Construction Of DC Machine (Cont.)

Example
Design a suitable armature winding for an
armature with 12 slots each containing two coil
sides. The winding is double layer and number
of poles is 4.

Solution
C = 12 x 2 / 2 = 12 coils
yb = C/2p = 12/4 = 3 coils
yf = 3 + 1 = 4 for retrogressive winding
or yf = 3 - 1 = 2 for progressive winding
yf is taken equal to 2 to ensure less end
connection length.

29
Construction Of DC Machine (Cont.)
30
Construction Of DC Machine (Cont.)
31
Construction Of DC Machine (Cont.)

In a lap winding, the number of parallel paths (2a) is always equal to
the number of poles and also to the number of brushes.
Bottom coil sides
Top coil sides
Commutator
N
S
1
2
3
Brush
Elements of Lap Winding
32
Construction Of DC Machine (Cont.)
33
Construction Of DC Machine (Cont.)



Wave Windings
In this type of windings, the coils connected in series
are either progressing or retrogressing all the way. The
figure show schematic representation for a partially
wound armature with wave windings. The distance
between successive coils sides occupying
nearly the same location under similar poles is termed
the resultant pitch “ y “ and is equal to;
y = yb + yf
where yb and yf are the back and front pitches of the
winding.
The resultant pitch “ y “ occupies double pole pitch,
therefore for a machine with “ 2p “ poles;
y.p=C±1
On the other hand, commutator pitch in terms of
commutator segments will be then given by;
yc = y
34
Construction Of DC Machine (Cont.)

Example
Design a suitable armature winding for an armature with 14 slots each containing two coil
sides. The winding is double layer and number of poles is 6.

Solution
C = 14 x 2 / 2 = 14 coils
y = C ± 1 / p = 14 ± 1 / 3 = 5 coils progressive only
yb = 3 & yf = 2
or yb = 2 & yf = 3

35
Construction Of DC Machine (Cont.)
36
Construction Of DC Machine (Cont.)
37
Construction Of DC Machine (Cont.)

In a wave windings, the number of parallel paths (2a) is always two.
Top coil sides
N
1
S
N
Bottom coil sides
S
2
Brush
Elements of Wave Winding
38
Construction Of DC Machine (Cont.)





From the previous example, it is clear that:
a) There is only two parallel circuits whatever the number of poles is.
b) Two brushes arms can only be used for voltage collection ( A & B ). However,
extra brush arms are used to limit the current in each brush arm ( C, D, E & F ).
Therefore number of brush arms = number of poles as in the case of Lap windings.
c) Again brush location is decided relative to position of poles and best location is
such that brushes should be in contact with coil sides in the neutral zone.
D) Number of parallel circuits in this type of windings whatever the number of poles
is always equal to TWO.
i.e. For wave windings 2a = 2
39
Classifications of DC Machines
Field
Armature
Separately excited
Field
Self excited
1- Shunt
Field
Armature
Armature
Self excited
2- Series
40
Self Excited
3- Compound
ff
F2
F1
ff
fs
A1
D1
D2
F1
D1
F2
i- Short-shunt
Cumulative
F1
A2
ii-Long-shunt
Cumulative
ff
fs
A1
F2
D2
A2
A2
ff
fs
A1
D1
iii- Short-shunt
Differential
D2
F1
A1
F2
fs
D1
D2
A2
iv-Long-shunt
Differential
41
Armature Voltage
Let
Z = total number of armature conductors
N = total number of turns in the armature winding = Z/2
2p = number of poles
2a = number of parallel paths = 2 for wave winding
2 p for lap winding
f = flux per pole Weber
Nm= speed of the motor in the revolutions per minute,
Time of 1 revolution = 60 N seconds
m
wm= speed of the motor in radians per second
2 p
Flux / Re v .

 2 pN m / 60
Emf generated in each conductor =

Time / Re v .
(60 / N m )
Total emf between brushes =emf/conductor * Number of conductor/path
 (2 pNm / 60)(Z / 2a)
But, wm  2Nm / 60
Where K a 
, therefore
Z .2 p N .2 p

2 .2a  .2a
Ea  (2 pwm / 2 )(Z / 2a)  Kawm
(armature constant)
42
Developed (or Electromagnetic) Torque

Consider the turn shown in the following Figure.
Area per pole A =
2rl
2p
 2p 
Flux density B  
A 2 r l
Current / conductor is
Ic 
The force on a conductor is
Ia
2a
fc  B l
Ia
2a
I
 .2 p. I
a
a
r
The torque developed by a conductor is Tc  f c r  B l
2a
2 .2a
The total torque developed is
Te 
Z .2 p.I a
E I
 K a I a  a a
2 .2a
wm
43
Example 1

Determine the induced voltage induced in
the armature of a dc machine running at
1750 rpm and having four poles. The flux
per pole is 25 mWb, and the armature is
lap-wound with 728 conductors.
44
Solution
N m  1750rpm
2p  4
f  25mWb
2a  2 p  4
Z  728
Ea  K afwm
pZ

fwm
2 a
pZf nm

60a
728  25  103  1750

60
 530.83V
45
Example 2

A lap-wound armature has 576 conductors
and carries an armature current of 123.5A.
If the flux per-pole is 20 mWb, calculate
the electromagnetic torque.
46
Solution
2a  2 p
f  20mWb
Z  576
I a  123.5 A
Te  K afI a
2 p.Z

fI a
2 .2a
576  20 10 3 123.5

2
 226.43Nm
47
Magnetization Curve of a DC Machine

The magnetizing curve is obtained experimentally by rotating the dcmachine at a given speed and measuring the open-circuit armature terminal
voltage as the current in the field winding is changed.
Ea

Saturation
E a  K a wm
wm1> wm2
Field
Linear
Ea
wm2
If
Flux-mmf relation in
a dc machine
If Nf
If
p
Magnetization curve
 The magnetization curve is of great importance because it represents the
saturation level in the magnetic system of the d.c. machine.
48
Chapter 2
DC Machines on Load
1.
2.
3.
4.
Armature reaction
L di/dt Voltages
Commutation in DC Machines
Methods Of Improving Commutation
49
Armature Reaction
50
Armature Reaction (cont.)
51
Armature Reaction (cont.)
52
Armature Reaction (cont.)
53
Armature Reaction (cont.)
54
Armature Reaction (cont.)
55
Armature Reaction (cont.)
56
Armature Reaction (cont.)
57
Armature Reaction (cont.)
58
Armature Reaction (cont.)
59
Armature Reaction (cont.)
60
Armature Reaction (cont.)
61
Armature Reaction (cont.)
62
Armature Reaction (cont.)
63
Armature Reaction (cont.)
64
L di/dt Voltages
65
L di/dt Voltages (cont.)
66
L di/dt Voltages (cont.)
67
Commutation in DC Machines
68
Commutation in DC Machines (cont.)
69
Methods Of Improving Commutation
70
Methods Of Improving Commutation (cont.)
71
Methods Of Improving Commutation (cont.)
72
Methods Of Improving Commutation (cont.)
73
Methods Of Improving Commutation (cont.)
74
Methods Of Improving Commutation (cont.)
75
Methods Of Improving Commutation (cont.)
76
Methods Of Improving Commutation (cont.)
77
Methods Of Improving Commutation (cont.)
78
Methods Of Improving Commutation (cont.)
79
Methods Of Improving Commutation (cont.)
80
Methods Of Improving Commutation (cont.)
81
Methods Of Improving Commutation (cont.)
82
Methods Of Improving Commutation (cont.)
83
Armature Reaction
 It is the effect of armature ampere-turns upon the value and the distribution of
the magnetic flux in the air gap.
Fluxes added each other
Fluxes oppose each other
N
S
If
N
Fluxes added each other

Near one tip of a pole, the net flux density
increases while it decreases near the other pole
tip, as a result the zero flux density shifts from
the brush-axis.
If the increased flux density causes magnetic
saturation, the flux per pole decreases. This
magnetizing effect of armature current
.
.
++
+
+
+
+
+
S
Fluxes oppose each other
AT Load
At No-Load

.
...
.
Saturation
effect
increases as the armature current increases.
B (resultant)
Bf+ Ba
Bf
Ba
84
Compensating Winding




The armature mmf distorts the flux density distribution and also produce
demagnetizing effect as a result the zero flux density shifts from the
brush-axis, and this causes poor commutation leading to sparking.
Much of the armature mmf can be neutralized by using a compensating
winding, which is fitted in slots cut on the main pole faces
MMF produced by compensating winding opposes the armature mmf.
The compensating winding is connected in series with the armature
winding so that its mmf is proportional to armature mmf.
Compensating
windings
Field
Armature
85
Commutation

The purpose of the commutator and brushes is to reverse the current in the
conductor when it goes from one pole to the next.
x
N
x
y
.. .
..
.
.
++
+
+
+
++
+
S
Brush
Reactance voltage:
the coil undergoing commutation is in the
interpolar region. When this coil moves in this
region, a voltage, called reactance voltage, is
induced in the coil undergoing commutation.
vr = L(di/dt)
This high voltage naturally causes sparking at
the brushes of the machine.
Coil undergoing
commutation
y
S
N
Commutator
segments
N
Brush
N
S
+Icoil
t
-Icoil
86
Interpoles or Commutator Poles

To improve the commutation, a small poles, called
interpoles or commutator poles, is created. Its winding
carries the armature current in such a direction that its flux
opposes the armature reaction flux and produces an emf that
equal and opposite to the reactance voltage vr.
Interpole
Interpole
winding
i
Ia
N
.
.
.
..
..
++
+
+
+
++
Ia
S
Armature
winding
a
87
Chapter 3
DC Generators
1.
2.
3.
4.
5.
Basic Principles
Classification of DC Generators
Voltage Build Up
Generator Characteristics
Efficiency
88
Principle of Operation
If a conductor of length l moves at a linear speed v in a
magnetic field the induced voltage in the conductor is
Faraday’s law or
flux cutting rule
e=Blv
e
N
v
v
B
S
B
Right hand rule
89
e
Classification Of DC Generators

1- Separately Excited DC Generator
Ia
IL
V f  ( R fw  R fc ) I f  R f I f
+
ra
+
wm
Rfw
Rfc
If
Vt
Ea


+ Vf 
RL
Ea  Vt  I a ra
E a  K a  wm
Vt  I L RL
Ia  IL
90
External Characteristic Curve
 It is a curve between terminal voltage and load current
at constant field current and constant rotor speed
Vt
RaIa
Ea
Terminal characteristic
with no armature reaction
% rated voltage
100
DVAR
80
60
Terminal characteristic
with armature reaction
40
20
0
20
40
60
80
100
% rated current
It
External characteristic of a separately excited dc generator
91
2- Self-Excited DC Generators

1- Shunt generator (Voltage build-up)
If
Rfc
Rfw
wm
IL
+
Ia
ra
+

Vt
Ea
Ea
E a  K a w m
Vt  I L RL
Ia  IL  I f
P
RL


V f  R f I f  Vt
E a  Vt  I a ra
Operating point
(emf due to
residual flux(
Field resistance line
IfRf versus If
Ea2
Ea1
Ear
If
0
If1
If2
Voltage build-up
92
Shunt generator (Cont.)
Critical field circuit resistance
Rf3
Ea
Rf2
Rf1
Conditions for a successful voltage buildup
Rf4
 Residual magnetism must be present in the
magnetic system
 Field winding mmf should aid the residual
magnetism.
 Field circuit resistance should be less than
critical field circuit resistance.
Vt4
0
If
Effect of field resistance
93
External Characteristic Curve

(without Armature Reaction effect)
The external characteristics of the self excited shunt generator
can be obtained from the magnetization curve and the field
resistance line, as illustrated in the following Figure.
94
2- Series Generator
IL
Ia
Vt  Ea  I a (ra  Rs )
+
ra
+
Rs
Vt
Ea
RL

IL  Ia  I f
Ea  K a  sw m

Ea
Vt
Magnetization Curve
Vt
External Characteristic
Slope =(Ra+Rs)
Ia(Ra+Rs)
Ia=If =IL
IL
95
3- Compound DC Generator
If
If
IL
+
Rfc
Rs
+
Vt
Ea
Rfw


Short Shunt
Vt  Ea  I a Ra  I L Rs
If 
Ea  I a Ra
R fw  R fc
Ea  K a ( sh   s )w m
Ea

Rfw
Ea  K a  sh   s  wm
IL  Ia  I f

Cumulative
+
Ia
Ra
+
Ia
Ra
Rfc
IL

Differential
Rs
Vt

Long Shunt
Vt  Ea  I a Ra  Rs 
IL  Ia  I f
Vt
If 
R fw  R fc
Ea  K a ( sh   s )wm
96
External Characteristic Curve
Vt
Over compound
Vt(rated)
Flat compound
Under compound
Differential
(Useful as a welding generator)
Ia
Ia(rated)
Feff  Fsh  Fs  FA
N f I f (eff )  N f I f  N s I fs  FA
I f (eff )  I f 
Ns
F
I fs  A
Nf
Nf
97
Graphical relationships for shunt generator without armature reaction
98
Graphical relationships for shunt generator with armature reaction
99
Graphical relationships for a series generator
100
Graphical relationships for a cumulative generator
101
Example: A 172-kW, 430-V, 400-A, 1800-rpm compounded dc generator
shown in figure. The magnetization curve is given at 1800-rpm. The
generator has compensating windings which eliminate armature reaction.




The machine is being driven at
1800-rpm, and Radj is currently
set to 55-Ω.
(a) What is the no-load terminal
voltage?
(b) If this is connected
cumulatively compounded, what
is its terminal voltage at full-load
(400-A)? What will its voltage
regulation be?
(c) If this generator did not have
compensating windings and
instead had an armature reaction
of 500 A . turns at full-load, what
would its terminal voltage be at
full-load (400-A)? What would its
voltage regulation be?

The compounded dc generator
102
Solution


(a): RF + Radj = 75-Ω, From the
magnetization curve the no-load voltage
is VT = 445-V.
(b): If the load current IL = 400-A, then
IA ~ 400-A too, and IA(RA + RS) = 24-V.
And the effect of the series field mmf in
terms of equivalent shunt field amperes
is :

(c): With armature reaction present in
this case, the equivalent shunt field
ampere is given by:
The load triangle thus has a horizontal
length of 0.7-A and a vertical length of
24-V. From figure, the output terminal
voltage will be: VT = 425-V
and the voltage regulation will be:
The load triangle thus has a horizontal
length of 1.2-A and a vertical length of
24-V. From figure, the output terminal
voltage will be: VT = 440-V
and the voltage regulation will be:
103
Solution
104
Example: If the generator in the previous example has compensating
windings and is connected differentially compounded, what will its
voltage be when it is supplying 200-A?


Solution
In this circumstances , the equivalent shunt current due to the series windings
and without armature reaction is:
The resistive voltage drop is IA(RA + RS) = 200 x 0.06 = 12-V.
The output voltage of the generator can be found
The load triangle thus has a horizontal length of 0.6-A and a vertical length
of 12-V. From figure, the output terminal voltage will be: VT = 390-V.
105
Power Flow and Efficiency
DC Generators
IL
If
+
Ia
Rfc
Ra
+
Rs
Vt
Ea

Rfw
Pinput=
Pmech =
Pshaft
Ea I a
Rotaional losses
Va I a
I a2 Ra
Va I L
Vt I L
I 2f R f I L2 Rs
Poutput=
Pelectrical


Poutput

Vt I L
Vt I L   I 2 R  RotationalLosses

Vt I L
Ea I a  RotationalLosses
Pinput

Poutput
Poutput  Losses
106
Chapter 4
DC Motors
1.
2.
3.
4.
5.
6.
Basic Principles
Classification of DC Motors
Motor Characteristics
Efficiency
Speed Control
Starting
107
Principle of Operation
For the current carrying conductor of length l the
force known as Lorentz force produced is
i
F
i
N
F
F=Bli
B
S
B
Left hand rule
108
D.C. Motor Characteristics
Torque-Speed Characteristics
1- Separately excited & Shunt motors
Ia
(f is independent of the load torque )
Vt  Ea  I a ra
Ea  K a  w m
Vt  I a ra
wm 
Ka
T  Ka I a
Therefore ,
Vt
ra
wm 

T
2
K a  ( K a )
wm
Vt
Ka
ra
Slope ( K ) 2
a
T
109
Torque-Speed Characteristics
2- Series motors
Ea  Vt  I a ( Ra  Rs )
Ea  K af wm
Neglecting saturation
f  K1 I f  K1 I a
Ea  K a K1 I awm  K s I awm
wm 
Vt
R  Rs
 a
Ks Ia
Ks
But T  K af I a  K a K1 I a2  K s I a2
w m 
Ra  Rs
Vt

Ks
Ks T
110
Torque-Speed Characteristics
2- Compound motors
Cumulative Compound
ATt  ATshunt  ATseries
Differential Compound
ft  fshunt  fseries
Shunt motor
Vt
ra
wm 

T
2
K aft ( K aft )
111
Example 3

A 250V shunt motor has an armature
resistance of 0.25W and a field resistance
of 125W. At no-load the motor takes a line
current of 5.0A while running at 1200
rpm. If the line current at full-load is
52.0A, what is the full-load speed?
112
Solution
If
At no-load:
It  5 A
It
Ia
Ra
Rfc
n  1200 rpm  w 
m
m
1200  2
Rfw
 125 .66 rad / sec
60
+
wm
+
Vt


Vt 250
If 

 2 A , I a _ NL  I t _ NL  I f  5  2  3 A
R f 125
Ea _ NL  Vt  I a _ NL Ra  250  3  0.25  249.25V
K af 
Ea _ NL
wm _ NL
249.25

 1.984 V.sec/rad
125.66
113
At full-load:
I L  52 A
I a _ FL  It _ FL  I f  52  2  50 A
Ea FL  Vt  I a  FL Ra  250  50  0.25  237.5V
Ea _ FL  K afw m _ FL
Ea _ FL
237.5
 wm _ FL 

 119.71 rad/sec
K af 1.984
wm  60
nm _ FL 
 1142.4 rpm
2
114
Power Flow and Efficiency
DC Motors
IL
If
+
Ia
Rfc
Ra
Rs
+
Vt
Ea
Rfw



Pinput = Vt I L
Pelectrical
Va I L
I L2 Rs
Va I a
I 2f R f
Ea I a
I a2 Ra Rotational
losses
Poutput=
Pmech=
Pshaft
Poutput
Pinput

Pinput  Losses
Pinput
Vt I L   I 2 R  RotationalLosses

Vt I L

E a I a  RotationalLosses
Vt I L
115
Example 4
The field and armature resistance of a
220 V series motor are 0.2Ω and 0.1Ω,
respectively. The motor takes 30 A of
current while running at 700 rpm. If
the rotational losses are 350W,
determine the motor efficiency.
116
Solution
Ia
IL
+
Pin  Vt I t  220  30  6600W
Ra
Pf  I 2f R f  30 2  0.2  180W
+
Pa  I a2 Ra  30 2  0.1  90W

Vt
Ea

Rotational losses  350W
Pout  Pin 
Rs

I 2 R Rotational losses
Pout  6600  (180  90)  350  5980W
Pout 5980


 0.9061
Pin
6600
117
SPEED CONTROL
Since Vt  Ea  I a ra
Ea  K a  w m
Vt  I a ra
wm 
K af
So the speed of the d.c .motor can be controlled by controlling
Vt , ra , or 
1- Armature Voltage Control
 In this method ra and If (i.e.f) are kept constant,
and Vt is varied to change the speed.
 The motor must be separately excited to use
armature voltage control.
 Armature voltage control can control the speed
of the motor for speeds below rated speed but
not for speed above rated speed.
 This method is expensive because it requires a
variable d.c. supply for the armature circuit.
118
SPEED CONTROL (Cont.)
2- Field current control
In this method Vt and ra remain fixed and the speed is controlled by varying
If . This is normally achieved by using a field rheostat as shown in the
following Figure for shunt d.c. motor. Field control can control the speed of
the motor for speeds above base speed but not for speeds below base speed.
This method is simple to implement and less expensive, because the control
is at the low power level of the field circuit.
119
SPEED CONTROL (Cont.)
3- Armature resistance control
In this method, the armature terminal
voltage Vt and the field current If are kept
constant at their rated values. The speed
is controlled by changing the resistance in
the armature circuit. Armature resistance
control is simple to implement. However,
this method is less efficient because of
losses in the resistance. This method can
be used in all types of d.c. motors.
Armature resistance control can control
the speed of the motor for speeds below
base speed.
Shunt Motor.
Series Motor.
120
SPEED CONTROL TECHNIQUES
1- Ward-Leonard System
(Classical method)
The system uses a motor-generator (M-G) set to control the speed of the DC drive motor.
The motor of the M-G set runs at constant speed. By varying the generator field
current, the generator voltage changes, which in turn changes the speed of the DC
drive motor. The system is operated in two control modes.
Vt Control: the speed is changed from zero to the base speed by keeping If constant at
rated value and changing the terminal voltage. The torque can be maintained constant
during operation in this range of speed.
If Control: The field current control is used to obtain speed above the base speed. In this
mode, Vt remains constant and If is decreased to obtain higher speeds. The armature
current can be kept constant, thereby operating the motor in a constant horsepower
mode.
121
SPEED CONTROL TECHNIQUES
2- Solid-State Control
In Recent years, solid-state control have been used as a replacement of rotating
(M-G) set. Both armature control and field control can be achieved using
controlled rectifier or choppers.

Controlled Rectifiers
If the supply is ac, controlled rectifiers
can be used to convert it to a variable
voltage dc supply by changing the firing
angle  of the rectifier thyristors.
122
SPEED CONTROL TECHNIQUES

Chopper
A chopper converts a fixed-voltage dc supply
into a variable voltage dc supply.
The switch S can be a thyristor , a GTO or a
power transistor.
When the switching device S is on , Vt = V
(supply voltage) and motor current increases.
When S is off Vt = 0 and motor current decays
through the diode.
t on
V

V
The average Vt is
t
T
123
Starting

If a d.c. motor is directly connected
to a d.c. power supply, the starting
current will be dangerously high.
V  Ea
at starting w  0  Ea  0
Ia  t
ra
Ia

Starting
Vt
ra
Since ra is small, the starting current is
very large.
The starting current can be limited by
the following methods:
1- Use a variable-voltage supply.
2- Insert an external resistance at start,
as shown in the Figure.
124