Dc motors

ELECTRICAL MACHINES
DIRECT CURRENT MOTORS - WORKING PRINCIPLE,
TYPES AND CHARACTERSTICS

2
INTRODUCTION
• For special applications such as in steel mills, mines and electric trains, it is
advantageous to use d.c. motors.
• Speed/ torque characteristics of d.c. motors are much more superior to that of
a.c. motors.
• D.C. motors are of three types viz., shunt-wound, series-wound, and
compound-wound.
D.C. Motor Principle
• Operation of d.c. motor is based on the principle that when a current carrying
conductor is placed in a magnetic field, the conductor experiences a
mechanical force.
• The direction of the force is given by Fleming’s left hand rule and magnitude
is given by; F = BIL Newtons
Lecture Notes by Dr.R.M.Larik

3
WORKING OF D.C. MOTOR
(i) The field magnets are excited developing alternate N and S poles
(ii) All armature conductors under N-pole carry currents in one direction
while all the conductors under S-pole carry currents in the opposite
direction.
Applying Fleming’s left hand rule, it is
clear that force on each conductor is
tending to rotate the armature in
anticlockwise direction.
When the motor terminals are connected to an external source of d.c. supply:

4
WORKING OF D.C. MOTOR

5
BACK OR COUNTER E.M.F.
• When d.c. voltage V is applied in a shunt- wound motor, the field magnets are
excited and current flows in armature conductors.
• The driving torque acts on the armature which begins to rotate.
• When the armature of a d.c. motor rotates, the armature conductors move
through the magnetic field and electromotive force (e.m.f.) is induced in them
• The induced e.m.f. acts in opposite direction to the applied voltage V (Lenz’s
law), known as back or counter e.m.f. Eb which is always less than the applied
voltage V, when the motor is in normal operation.
• The applied voltage V has to force current through the armature against the
back e.m.f. Eb.

6
BACK OR COUNTER E.M.F. (Contd)
• The electric work done in overcoming and causing the current to flow against
Eb is converted into mechanical energy developed in the armature.
• Net voltage across the armature = V – Eb
• If Ra is the armature resistance, then,
armature current Ia =
V − Eb
Ra
• Since V and Ra are usually fixed, the value
of Eb will determine the current drawn by
the motor.
• If the speed of the motor is high, then back
e.m.f. Eb (= PΦZN/60 A) is large and
hence the motor will draw less armature
current and vice-versa.

7
VOLTAGE EQUATION OF D.C. MOTOR
Let in a d.c. motor
V = applied voltage
Eb = back e.m.f.
Ra = armature resistance
Ia = armature current
Since back e.m.f. Eb acts opposite to the
applied voltage V, the net voltage across the
armature circuit is V- Eb.
The armature current Ia is given by;
Ia =
V − Eb
Ra
or V = Eb + Ia Ra
This is known as Voltage Equation of the d.c. motor.

8
POWER EQUATION OF D.C. MOTOR
The voltage equation of d.c. motor is:
V = Eb + Ia Ra
On multiplying the above equation is by la, we get,
V Ia = Eb Ia + Ia
2 Ra
This is known as Power Equation of the d.c. motor.
VIa = electric power supplied to armature (armature input)
EbIa = power developed by armature (armature output)
Ia
2 Ra = electric power wasted in armature (armature Cu loss)
Out of the armature input, a small portion (about 5%) is wasted as Ia
2 Ra and the
remaining portion EbIa is converted into mechanical power within armature.

9
NUMERICAL PROBLEMS
Problem 1(a)
A 220-V d.c. shunt motor has an armature resistance of 0.5 . If the full-load
armature current is 20 A, find the back e.m.f.
Solution:
Supply voltage = 220 V DC, Ra = 0.5 , Ia = 20 A
Back e.m.f. ?
Eb = V – Ia Ra
= 220 – 20 x 0.5
= 210 V
Problem 1(b)
A 440-V, shunt motor has armature resistance of 0.8  and field resistance
of 200 . Determine the back e.m.f. when the motor is giving an output of
7.46 kW at 85 percent efficiency.

10
NUMERICAL PROBLEMS
Problem 2(a)
A 250 V shunt motor takes a load of 20 A. The shunt field and armature
resistances are 200  and 0.3  respectively. Determine (i) value of back
e.m.f. (ii) mechanical power developed in the armature.
Solution:
Supply voltage = 250 V DC, IL = 20 A, Rsh = 200 , Ra = 0.3 Ω
Back e.m.f. ?, gross mechanical power of armature ?
Ish =
V
Rsh
=
250
200
= 1.25 A
Ia = IL – Ish = 20 – 1.25 = 18.75 A
Eb = V – Ia Ra
= 250 – 18.75 x 0.3
= 244.4 V

11
NUMERICAL PROBLEMS (Contd.)
Problem 2(a)
Solution (Contd.)
Mechanical power developed = Eb Ia
= 244.4 x 18.75
= 4582.5 W
Problem 2(b)
A 25-kW, 250-V, d.c. shunt motor has armature and field resistances of
0.06  and 100  respectively. Determine the total armature power
developed when the motor is taking 25 kW input.
Problem 2(c)
A 230 V motor has an armature circuit resistance of 0.6 . If the full load
current is 30 A and a no load armature current is 4 A, find the change in the
back e.m.f. from no load to full load.

12
Problem 3(a)
A d.c. motor takes an armature current of 110 A at 480 V. The armature
circuit resistance is 0.2 Ω. The motor has 6-poles and the armature is lap-
wound with 864 conductors. The flux per pole is 0.05 Wb. Calculate the
speed of motor.
Solution:
Supply voltage = 480 V DC, Ia = 110 A, Ra = 0. 2 , P = 6, A = P = 6 (for lap
wound), Z = 864, Φ = 0.05 Wb/pole
Speed of motor N ?,
Back e.m.f. Eb = V – Ia Ra
= 480 – 110 x 0.2 = 458 V
Also Eb =
Φ Z N
60
x
P
A
458 =
0.05 x 864 x N
60
x
6
6
N = 636 rpmLecture Notes by Dr.R.M.Larik

13
Problem 3(b)
A 4-pole, 500 V shunt motor has 720 wave-connected conductors in the
armature. The full load armature current is 60 A and the flux per pole is 0.03
Wb. The armature resistance is 0.2  and the contact drop is 1 V per brush.
Calculate full load speed of the motor.
Problem 4(a)
Find the no-load and full-load speeds for a four-pole, 220-V, and 20-kW, shunt
motor having the following data:
Field current = 5 A, armature resistance = 0.04 ohm, flux per pole = 0.04 Wb,
number of armature-conductors = 160, two - circuit wave-connection, full load
current = 95 A, No load current = 9 A
Solution:
Motor capacity = 20 kW, supply voltage = 220 V DC, Ish = 5 A, Ra = 0. 04 ,
Φ = 0.04 Wb/pole, Z = 160, P = 4, A = 2 (wave wound), IFL = 95 A, INL = 9 A
No load speed No ?, Full load speed of motor N ?,

14
Problem 4(a)
Solution (Contd.)
Ia(NL) = INL – Ish
= 9 – 5 = 4A
Ia(FL) = IFL – Ish
= 95 – 5 = 90 A
At no load
Back e.m.f. Ebo = V – Ia Ra
= 220 – 4 x 0.04 = 219.84 V
Also Eb =
∅ Z N 𝑜
60
x
P
A
No =
Eb x 60 x A
P ∅ Z
=
219.84 x 60 x 2
4 x 0.04 x 160
= 1030.5 r.p.m.

15
Problem 4(a)
Solution (Contd.)
At full load
Back e.m.f. Eb = V – Ia Ra
= 220 – 90 x 0.04 = 216.4 V
Also Eb =
∅ Z N
60
x
P
A
N =
Eb x 60 x A
P ∅ Z
=
216.4 x 60 x 2
4 x 0.04 x 160
= 1014 r.p.m.
Example 4(b): The counter e.m.f. of a shunt motor is 227 V, the field
resistance is 160  and field current is 1.5 A. If the line current is 39 A, find
the armature resistance. Also find the armature current when the motor is
stationary. Lecture Notes by Dr.R.M.Larik

16
TYPES OF D.C. MOTORS
Shunt-Wound Motor
• Shunt field windings are designed to produce
the necessary Magnetomotive Force (mmf) by
means of a relatively large number of turns of
wire having high resistance.
• Shunt field current is relatively small as
compared to the armature current.
Series-Wound Motor
• The field windings is designed with much fewer
turns than shunt field windings for the same
mmf
• Series field winding has a relatively small
number of turns of thick wire and, therefore, will
possess a low resistance.

17
TYPES OF D.C. MOTORS (Contd.)
Compound-Wound Motor
• It has two field windings; one connected in parallel with the armature and the
other in series with it.
• When the shunt field winding is directly connected across the armature
terminals it is called short-shunt connection.
• When the shunt winding is connected such that it shunts the series
combination of armature and series field it is called long-shunt connection.

18
SPEED RELATIONS
If a d.c. motor has initial values of speed, flux per pole and back e.m.f. as N1, ϕ1
and Eb1 respectively and the corresponding final values are N2, ϕ2 and Eb2, then,
N1 
Eb1
Φ1
and N2 
Eb2
Φ2
N2
N1
=
Eb2
Eb1
x
Φ1
Φ2
(i) For a shunt motor, flux practically remains constant so that ϕ1 = ϕ2.
N2
N1
=
Eb2
Eb1
(ii) For a series motor, ϕ  Ia prior to saturation.
N2
N1
=
Eb2
Eb1
x
Ia1
Ia2
where Ia1 = initial armature current
Ia2 = final armature current

19
NUMERICAL PROBLEMS
Problem 5(a)
A 250-V shunt motor runs at 1000 r.p.m. at no-load and takes 8A. The total
armature and shunt field resistances are respectively 0.2  and 250 .
Calculate the speed when loaded and taking 50 A. Assume the flux to be
constant.
Solution:
Supply voltage = 250 V DC, No = 1000 r.p.m., Io = 8A, Ra = 0.2 ,
Rsh = 250 , I = 50 A
Speed N ? (when loaded and taking 50 A)
N
No
=
Eb
Ebo
x
Φ0
Φ
Since Φo = Φ
N
No
=
Eb
Ebo
Ish =
V
Rsh
=
250
250
= 1 A

20
Problem 5(a)
Solution (Contd.):
F.L. Armature current Ia = I – Ish = 50 – 1 = 49 A
N. L. armature current Iao = Io – Ish = 8 – 1 = 7 A
N. L. Back e.m.f. Ebo = V – Iao Ra
= 250 – 7 x 0.2 = 248.6 V
F. L. Back e.m.f. Eb = V – Ia Ra
= 250 – 49 x 0.2 = 240.2 V
N
1000
=
240.2
248.6
N = 966.2 r.p.m.
Problem 5(b): A d.c. series motor operates at 800 r.p.m. with a line current of
100 A from 230-V mains. Its armature circuit resistance is 0.15  and its field
resistance 0.1 . Find the speed at which the motor runs at a line current of
25 A, assuming that the flux at this current is 45 per cent of the flux at 100 A.

21
Problem 6
A 230-V d.c. shunt motor has an armature resistance of 0.5  and field
resistance of 115 . At no load, the speed is 1,200 r.p.m. and the armature
current 2.5 A. On application of rated load, the speed drops to 1,120 r.p.m.
Determine the line current and power input when the motor delivers rated load.
Solution:
Supply = 230 V DC, Ra = 115 , Rsh = 0.5 , No = 1200 rpm, Iao = 2.5 A,
N = 1120 rpm,
Line current ?, Input power ? (at rated load)
Ish =
V
Rsh
=
230
115
= 2 A
Back emf Ebo = V – Iao Ra
= 230 – 2.5 x 0.5 = 228.75 V
Back emf Eb = V – IaRa
= 230 – Ia x 0.5

22
Problem 6
Solution (Contd.)
N
No
=
Eb
Ebo
1120
1200
=
230− Ia x 0.5
228.75
Ia = 33 A
IL = Ia + Ish
= 33 + 2 = 35 A
Input power = V x IL
= 230 x 35 = 8050 W

23
ARMATURE REACTION IN D.C. MOTORS
• At no load the small armature current does not
significantly affect the pole flux Φ1.
• At full load the armature current creates a flux Φ2.
• By superimposing Φ1 and Φ2 resulting flux Φ3
occurs.
• As a result the flux density increases under the left
half of the pole and decreases under right half
producing two effects:
a) Neutral zone shifts towards left resulting poor
commutation with sparking at the brushes.
b) Due to higher flux density at pole tip A
saturation sets in.

24
ARMATURE REACTION IN D.C. MOTORS (Contd.)
Correcting Measures
• A set of commutating poles are place
between the main poles which develop
mmf equal and opposite to armature mmf
• In the case of large d.c. motors
undergoing rapid multiple operations the
commutating poles do not adequately
neutralize the armature mmf
• To eliminate the problem special
compensating winding are connected in
series with the armature to produce mmf
equal and opposite to armature mmf (on
point to point basis).

25
COMMUTATION IN D.C. MOTORS
• To produce unidirectional force (or torque) on the armature conductors, the
conductors under any pole must carry the current in the same direction at all
times.
• When a conductor moves from the
influence of N-pole to that of S-pole,
the direction of current in the
conductor must be reversed. This is
termed as commutation.
• Commutator and the brush gear in a
d.c. motor causes the reversal of
current in a conductor as it moves
from one side of a brush to the other.

26
LOSSES IN A D.C. MOTOR
Mechanical losses
• The mechanical losses (friction and windage) vary as the cube of the
speed of rotation of the d.c. motor
• Since d.c. motors are generally operated at constant speed,
mechanical losses are considered to be constant.
Copper losses
Armature Cu loss, field Cu loss, brush contact loss, Cu losses in interpoles
(commutating poles) and compensating windings
Iron losses or magnetic losses
Since d.c. motors are generally operated at constant flux density and
constant speed, the iron losses are nearly constant.

27
D.C. MOTOR CHARACTERISTICS
Torque and Armature Current (Ta/Ia)
It is the curve between armature torque Ta and armature current Ia of a d.c.
motor and is known as electrical characteristic of the motor.
Speed and Torque (N/Ta)
It is the curve between speed N and armature torque Ta of a d.c. motor and is
known as mechanical characteristic.
Speed and Armature Current (N/Ia)
It is the curve between speed N and armature current Ia of a d.c. motor and is
very important characteristic as it is the deciding factor in the selection of the
motor for a particular application.

28
CHARACTERISTICS OF SHUNT MOTORS
➢ Ta/Ia Characteristic
• In a d.c. motor, Ta  ϕ Ia
• Since supply voltage of motor is constant,
Ish is constant & flux Φ is also constant
(neglecting armature reaction) Ta  Ia
• Hence Ta/Ia characteristic is a straight line
passing through the origin
• The shaft torque (Tsh) is less than Ta and is
shown by a dotted line.
• Very large current is required to start a
heavy load, hence, a shunt motor should
not be started on heavy load.

29
CHARACTERISTICS OF SHUNT MOTORS (Contd.)
➢ N/Ia Characteristic
• The speed N of a d.c. motor is given by; N 
Eb
Φ
• The flux Φ and back e.m.f. Eb in a shunt motor are almost constant,
hence, the speed of a shunt motor remains constant with the variation of
armature current (line AB)
• When load increases, Eb (= V- IaRa) and Φ
decrease due to increase in armature
resistance voltage drop and armature
reaction
• Eb decreases slightly more than Φ so the
speed of the motor decreases slightly with
load (line AC).

30
CHARACTERISTICS OF SHUNT MOTORS (Contd.)
➢ N/Ta Characteristic
• The curve is obtained by plotting the
values of N and Ta for various armature
currents
• The speed falls somewhat as the load
torque increases.
Conclusions
(i) The shunt motor is a constant-speed motor as there is slight change in its
speed from no-load to full load
(ii) The starting torque is not high because Ta  Ia

31
CHARACTERISTICS OF SERIES MOTORS
• The armature current increases with the
increase in the mechanical load
• The flux in the field also increases
• Upto magnetic saturation, ϕ  Ia so that Ta  Ia
2
• After magnetic saturation, Φ is constant so that Ta  Ia
• Thus upto magnetic saturation, the armature torque is
directly proportional to the square of armature current.
• In a d.c. motor, Ta  Φ Ia

32
CHARACTERISTICS OF SERIES MOTORS (Contd.)
• The speed N is given by;
N 
Eb
ϕ
where Eb = V – Ia (Ra + Rse)
• When the armature current increases, the
back e.m.f. Eb decreases while the flux ϕ
increases.
• Neglecting the Ia(Ra + Rsc) drop, which is
quite small under normal conditions,
N 
1
ϕ
N 
1
Ia
upto magnetic saturation

33
CHARACTERISTICS OF SERIES MOTORS (Contd.)
• Series motor develops high torque at low
speed and vice-versa because an
increase in torque requires an increase in
armature current, which is also the field
current.
• The result is that flux is strengthened and
hence the speed drops (as N 
1
ϕ
)
Conclusions
• Series motor has a high starting torque
• It is a variable speed motor and automatically adjusts the speed as the load
changes.
• It should never be started on no-load

34
CHARACTERISTICS OF COMPOUND MOTORS
• As the load increases, the series field
increases but shunt field strength
remains constant.
• Consequently, total flux is increased
and hence the armature torque
• Torque of a cumulative-compound
motor is greater than that of shunt
motor for a given armature current
due to series field

35
CHARACTERISTICS OF COMPOUND MOTORS (Contd.)
• The flux per pole increases with the
increase in load
• Consequently, the speed of the motor
decreases as the load increases
• When the load is added, the
increased amount of flux causes the
speed to decrease more than does
the speed of a shunt motor.

36
CHARACTERISTICS OF COMPOUND MOTORS (Contd.)
For a given armature current, the torque
of a cumulative compound motor is more
than that of a shunt motor but less than
that of a series motor.
Conclusions
• Due to the presence of shunt field, the motor is prevented from
running away at no-load.
• Due to the presence of series field, the starting torque is increased.

37
COMPARISON OF THREE TYPES OF MOTORS
• The speed regulation of a shunt motor is better than that of a series motor.
• For a given armature current, the starting torque of a series motor is more
than that of a shunt motor.
• The starting torque of a cumulative compound motor lies between series
and shunt motors
• Both shunt and cumulative compound motors have definite no-load speed.
• However, a series motor has dangerously high speed at no-load.Lecture Notes by Dr.R.M.Larik

38
APPLICATIONS OF D.C. MOTORS
✓Shunt Motors
(i) the speed is required to remain almost constant from no-load to full-load
(ii) the load is to be driven at a number of speeds and any one of which is
required to remain nearly constant
Industrial use: Lathes, drills, boring mills, shapers, spinning and weaving
machines etc.
✓Series Motors
(i) large starting torque is required
(ii) the load is subjected to heavy fluctuations and the speed is automatically
required to reduce at high torques
Industrial use: Electric traction, cranes, elevators, air compressors, vacuum
cleaners, hair drier, sewing machines etc.
Shunt motors are used where:
Series motors are used where:

39
APPLICATIONS OF D.C. MOTORS
✓Compound Motors
Compound motors are used where a fairly constant speed is required with
irregular loads or suddenly applied heavy loads.
Industrial use: Presses, shears, reciprocating machines etc.

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