DC Machines with explanation in detail of everything

Chapter 6
DC Machines
EET103/4

Introduction
• An electrical machine is link between an electrical
system and a mechanical system.
• Conversion from mechanical to electrical: generator
• Conversion from electrical to mechanical: motor

Introduction
Machines are called
• AC machines (generators or motors) if the electrical
system is AC.
• DC machines (generators or motors) if the
electrical system is DC.

DC machines can be divide by:
a) DC motor
b) DC Generator
DC Machines
DC Motor DC Generator

DC Machines Construction
cutaway view of a dc machine

Rotor of a dc machine

Stator of a dc machine

DC Machines Fundamentals
• Stator: is the stationary part of the machine. The
stator carries a field winding that is used to
produce the required magnetic field by DC
excitation.
• Rotor (Armature): is the rotating part of the
machine. The rotor carries a distributed winding,
and is the winding where the e.m.f. is induced.
• Field winding: Is wound on the stator poles to
produce magnetic field (flux) in the air gap.
• Armature winding: Is composed of coils placed in
the armature slots.
• Commutator: Is composed of copper bars,
insulated from each other. The armature winding is
connected to the commutator.
• Brush: Is placed against the commutator surface.
Brush is used to connect the armature winding to
external circuit through commutator

In DC machines, conversion of energy from electrical to mechanical
form or vice versa results from the following two electromagnetic
phenomena
1.When a conductor moves in a magnetic field, voltage is induced in
the conductor.
2. When a current carrying conductor is placed in magnetic field, the
conductor experiences a mechanical forces.

Generator action:
An e.m.f. (voltage) is induced in a conductor if it
moves through a magnetic field.
Motor action:
A force is induced in a conductor that has a current
going through it and placed in a magnetic field
•Any DC machine can act either as a generator or
as a motor.

DC Machines Equivalent Circuit
The equivalent circuit of DC machines has two
components:
Armature circuit:
• It can be represented by a voltage source and a
resistance connected in series (the armature
resistance). The armature winding has a
resistance, RA.
The field circuit:
• It is represented by a winding that generates the
magnetic field and a resistance connected in
series. The field winding has resistance RF.

• 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 of the
magnetic field and rotor
current generates a force that
drives the motor.
Basic Operation of DC Motor

• The magnetic field lines enter into
the rotor from the north pole (N)
and exit toward the south pole (S)
• The poles generate a magnetic
field that is perpendicular to the
current carrying conductors
• The interaction between the field
and the current produces a
Lorentz force
• The force is perpendicular to both
the magnetic field and conductor

• The generated force turns the
rotor until the coil reaches the
neutral point between the poles.
• At this point, the magnetic field
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 counter
clockwise rotation.

• Before reaching the neutral
zone, the current enters in
segment 1 and exits from
segment 2
• Therefore, current enters the coil
end at slot ‘a’ and exits from slot
‘b’ during this stage
• After passing the neutral zone,
the current enters segment 2 and
exits from segment 1,
• This reverses the current
direction through the rotor coil,
when the coil passes the neutral
zone
• The result of this current reversal
is the maintenance of the
rotation

Classification of DC Motor
1. Separately Excited DC Motor
• Field and armature windings are either connected
separate.
2. Shunt DC Motor
• Field and armature windings are either connected in
parallel.
3. Series DC Motor
• Field and armature windings are connected in
series.
4. Compound DC Motor
• Has both shunt and series field so it combines
features of series and shunt motors.

Important terms
• VT – supply voltage
• EA – internal generated voltage/back e.m.f.
• RA – armature resistance
• RF – field/shunt resistance
• RS – series resistance
• IL – load current
• IF – field current
• IA – armature current
• IL – load current
• n – speed

Generated or back e.m.f. of DC
Motor
• General form of back e.m.f.,
Φ = flux/pole (Weber)
Z = total number of armature conductors
= number of slots x number of conductor/slot
P = number of poles
A = number of parallel paths in armature
[A = 2 (for wave winding), A = P (for lap winding)]
N = armature rotation (rpm)
EA = back e.m.f.
A
P
ZN
EA 

60


Torque Equation of a DC Motor
• The armature torque of a DC motor is given by
P = number of poles
IA = armature current
Ta = armature torque
)
(
2
meter
Newton
A
P
ZI
T A
a 





Equivalent Circuit of DC Motor
F
T
F
R
V
I 
A
A
A
T R
I
E
V 

F
F
F
R
V
I 
A
A
A
T R
I
E
V 

A
L I
I 
Separately Excited DC Motor
Shunt DC Motor
F
A
L I
I
I 


)
( S
A
A
A
T R
R
I
E
V 


L
S
A I
I
I 

Series DC Motor
)
( S
A
A
A
T R
R
I
E
V 


F
T
F
R
V
I  F
L
A I
I
I 

Compound DC Motor

Speed of a DC Motor
• For shunt motor
• For series motor
1
2
1
2
1
2
2
1
1
2
1
2
,
A
A
A
A
E
E
n
n
then
If
E
E
n
n



 



2
1
1
2
2
1
1
2
1
2
A
A
A
A
A
A
I
I
E
E
E
E
n
n







Example 1
A 250 V, DC shunt motor takes a line
current of 20 A. Resistance of shunt
field winding is 200 Ω and resistance of
the armature is 0.3 Ω. Find the
armature current, IA and the back e.m.f.,
EA.

Solution
Given quantities:
• Terminal voltage, VT = 250 V
• Field resistance, RF = 200 Ω
• Armature resistance, RA = 0.3 Ω
• Line current, IL = 20 A
Figure 1

Solution (cont..)
the field current,
the armature current,
VT = EA + IARA
the back e.m.f.,
EA = VT – IARA = 250 V – (18.75)(0.3) = 244.375 V
A
25
.
1
200
V
250






F
T
F
F
A
L
R
V
I
I
I
I
18.75A
A
25
.
1
A
20




 F
L
A I
I
I

Example 2
A 50hp, 250 V, 1200 r/min dc shunt motor
with compensating windings has an
armature resistance (including the
brushes, compensating windings, and
interpoles) of 0.06 Ω. Its field circuit has a
total resistance Radj + RF of 50 Ω, which
produces a no-load speed of 1200 r/min.
There are 1200 turns per pole on the
shunt field winding.

Example 2 (cont..)
a) Find the speed of this motor when its
input current is 100 A.
b) Find the speed of this motor when its
c) Find the speed of this motor when its

Solution
Given quantities:
• Terminal voltage, VT = 250 V
• Field resistance, RF = 50 Ω
• Initial speed, n1 = 1200 r/min
Figure 2

Solution (cont..)
(a) When the input current is 100A, the armature
current in the motor is
Therefore, EA at the load will be
A
95
A
5
A
100
50
V
250
A
100










F
T
L
F
L
A
R
V
I
I
I
I
V
3
.
244
V
7
.
5
V
250
)
06
.
0
)(
A
95
(
V
250







 A
A
T
A R
I
V
E

Solution (cont..)
• The resulting speed of this motor is
min
/
r
1173
min
/
r
1200
250
3
.
244
1
1
2
2
1
2
1
2






V
V
n
E
E
n
E
E
n
n
A
A
A
A

Solution (cont..)
(b) When the input current is 200A, the armature
A
195
A
5
A
200
50
V
250
A
200










F
T
L
F
L
A
R
V
I
I
I
I
V
3
.
238
V
7
.
11
V
250
)
06
.
0
)(
195
(
V
250








A
R
I
V
E A
A
T
A

Solution (cont..)
min
/
r
1144
min
/
r
1200
250
3
.
238
1
1
2
2
1
2
1
2






V
V
n
E
E
n
E
E
n
n
A
A
A
A

Solution (cont..)
(c) When the input current is 300A, the armature
A
295
A
5
A
300
50
V
250
A
300










F
T
L
F
L
A
R
V
I
I
I
I
V
3
.
232
V
7
.
17
V
250
)
06
.
0
)(
295
(
V
250








A
R
I
V
E A
A
T
A

Solution (cont..)
min
/
r
1115
min
/
r
1200
V
250
V
3
.
232
1
1
2
2
1
2
1
2






n
E
E
n
E
E
n
n
A
A
A
A

Example 3
The motor in Example 2 is now connected in
separately excited circuit as shown in Figure 3.
The motor is initially running at speed, n = 1103
r/min with VA = 250 V and IA = 120 A, while
supplying a constant-torque load. If VA is reduced
to 200 V, determine
i). the internal generated voltage, EA
ii). the final speed of this motor, n2

Solution
Given quantities
• Initial line current, IL = IA = 120 A
• Initial armature voltage, VA = 250 V

Solution (cont..)
i) The internal generated voltage
EA = VT - IARA
= 250 V – (120 A)(0.06 Ω)
= 250 V – 7.2 V
= 242.8 V

Solution (cont..)
ii) Use KVL to find EA2
EA2 = VT - IA2RA
Since the torque is constant ant he flux is constant,
IA is constant. This yields a voltage of
EA2 = 200 V – (120 A)(0.06 Ω)
= 200 V – 7.2 V
= 192.8 V

Solution (cont..)
• The final speed of this motor
min
/
r
876
min
/
r
1103
V
8
.
242
V
8
.
192
1
1
2
2
1
2
1
2






n
E
E
n
E
E
n
n
A
A
A
A

Example 4
A DC series motor is running with a speed
of 800 r/min while taking a current of 20 A
from the supply. If the load is changed
such that the current drawn by the motor
is increased to 50 A, calculate the speed
of the motor on new load. The armature
and series field winding resistances are
0.2 Ω and 0.3 Ω respectively. Assume the
flux produced is proportional to the
current. Assume supply voltage as 250 V.

Solution
Given quantities
• Supply voltage, VT = 250 V
• Series resistance, RS = 0.3 Ω
• Initial armature current, Ia1 = IL1 = 20 A
Figure 4

Solution (cont..)
For initial load, the armature current, Ia1 = 20 A and
the speed n1 = 800 r/min
V = EA1 + Ia1 (RA + RS)
The back e.m.f. at initial speed
EA1 = V - Ia1 (RA + RS)
= 250 – 20(0.2 + 0.3)
= 240 V

Solution (cont..)
When the armature current increased, Ia2 = 50 A, the
back emf
EA2 = V – Ia2 (RA + RS)
= 250 – 50(0.2 + 0.3)
= 225 V
min
/
r
300
50
20
240
225
800
2
1
1
2
1
2
2
1
1
2
1
2
2
1
1
2
1
2











I
I
E
E
n
n
I
I
E
E
n
n
E
E
n
n
A
A
A
A
A
A


The speed of the motor on new load

Generating of an AC Voltage
• The voltage generated
in any dc generator
inherently alternating
and only becomes dc
after it has been rectified
by the commutator

Armature windings
• The armature windings are usually former-
wound. This are first wound in the form of flat
rectangular coils and are then puller.
• Various conductors of the coils are insulated
each other. The conductors are placed in the
armature slots which are lined with tough
insulating material.
• This slot insulation is folded over above the
armature conductors placed in the slot and is
secured in place by special hard wooden or fiber
wedges.

Lap and wave Windings
There are two types of windings mostly
employed:
• Lap winding
• Wave winding
The difference between the two is merely
due to the different arrangement of the end
connection at the front or commutator end of
armature.

Generated or back e.m.f. of DC
Generator
• General form of generated e.m.f.,
P = number of poles
[A = 2 (for wave winding), A = P (for lap winding)]
N = armature rotation (rpm)
E = e.m.f. induced in any parallel path in armature
A
P
ZN
E 

60


Classification of DC Generator
1. Separately Excited DC Generator
• Field and armature windings are either connected
separate.
2. Shunt DC Generator
• Field and armature windings are either connected in
parallel.
3. Series DC Generator
• Field and armature windings are connected in
series.
4. Compound DC Generator
• Has both shunt and series field so it combines
features of series and shunt motors.

Equivalent circuit of DC generator
F
A
L I
I
I 

A
L I
I 
Separately excited DC generator
F
F
F
R
V
I 
A
A
A
T R
I
E
V 

Shunt DC generator
F
T
F
R
V
I 
A
A
A
T R
I
E
V 


F
A
L I
I
I 

A
S
L I
I
I 

Series DC generator
)
( S
A
A
A
T R
R
I
E
V 


Compound DC generator
F
T
F
R
V
I 
A
A
A
T R
I
E
V 


Example
• A DC shunt generator
has shunt field winding
resistance of 100Ω. It
is supplying a load of
5kW at a voltage of
250V. If its armature
resistance is 0.02Ω,
calculate the induced
e.m.f. of the generator.

Solution
Given quantities
• Terminal voltage, VT = 250V
• Field resistance, RF = 100Ω
• Armature resistance, RA = 0.22Ω
• Power at the load, P = 5kW

Solution (cont..)
A
5
.
2
100
V
250






F
T
F
F
L
A
R
V
I
I
I
I
The field current,
A
20
V
250
W
5000



T
L
V
P
I
The load current,
The armature current, IA = IL + IF = 20A + 2.5A = 22.5A
The induced e.m.f.,
EA = VT + IA RA = 250V + (22.5)(0.22) = 254.95V

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