Speed Control of DC Motors: the speed of a motor is given by the
relation
V I a Ra 60 A V I a Ra
N K where Ra= armature circuit resistance. It is
Z p
obvious that the speed can be controlled by varying
(i) Flux/pole, Φ (Flux Control)
(ii) Resistance Ra of armature circuit (Rheostatic Control) and
(iii) Applied voltage V (Voltage Control).
Speed Control of Shunt motor:
(i) Variation of Flux or Flux Control Method: By decreasing the flux, the
speed can be increased and vice versa. The flux of a dc motor can be
changed by changing Ish with help of a shunt field rheostat. Since Ish is
relatively small, shunt field rheostat has to carry only a small current,
which means I2shR loss is small, so that rheostat is small in size.
(ii) Armature or Rheostatic Control Method: This method is used when
speeds below the no-load speed are required. As the supply voltage is
normally constant, the voltage across the armature is varied by inserting a
variable rheostat in series with the armature circuit. As controller
resistance is increased, voltage across the armature is decreased, thereby
decreasing the armature speed. For a load constant torque, speed is
approximately proportional to the voltage across the armature. From the
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speed/armature current characteristic, it is seen that greater the resistance
in the armature circuit, greater is the fall in the speed.
(iii) Voltage Control Method:
(a) Multiple Voltage Control: In this method, the shunt field of the motor
is connected permanently to a fixed exciting voltage, but the armature is
supplied with different voltages by connecting it across one of the several
different voltages by means of suitable switchgear. The armature speed
will be approximately proportional to these different voltages. The
intermediate speeds can be obtained by adjusting the shunt field regulator.
(b) Ward-Leonard System: This system is used where an unusually wide
and very sensitive speed control is required as for colliery winders,
electric excavators, elevators and the main drives in steel mills and
blooming and paper mills. M1 is the main motor whose speed control is
required. The field of this motor is permanently connected across the dc
supply lines. By applying a variable voltage across its armature, any
desired speed can be obtained. This variable voltage is supplied by a
motor-generator set which consists of either a dc or an ac motor M2
directly coupled to generator G. The motor M2 runs at an approximately
constant speed. The output voltage of G is directly fed to the main motor
M1. The voltage of the generator can be varied from zero up to its
maximum value by means of its field regulator. By reversing the direction
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of the field current of G by means of the reversing switch RS, generated
voltage can be reversed and hence the direction of rotation of M1. It
should be remembered that motor generator set always runs in the same
direction.
Speed Control of Series Motors:
1. Flux Control Method: Variations in the flux of a series motor can be
brought about in any one of the following ways:
(a) Field Diverters: The series winding are shunted by a variable
resistance known as field diverter. Any desired amount of current can be
passed through the diverter by adjusting its resistance. Hence the flux can
be decreased and consequently, the speed of the motor increased.
(b) Armature Diverter: A diverter across the armature can be used for
giving speeds lower than the normal speed. For a given constant load
torque, if Ia is reduced due to armature diverter, the must increase
(∵Ta I a ) This results in an increase in current taken from the supply
(which increases the flux and a fall in speed (N I/ )). The variation in
speed can be controlled by varying the diverter resistance.
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(c) Trapped Field Control Field: This method is often used in electric
traction. The number of series filed turns in the circuit can be changed.
With full field, the motor runs at its minimum speed which can be raised
in steps by cutting out some of the series turns.
(d) Paralleling Field coils: this method used for fan motors, several
speeds can be obtained by regrouping the field coils. It is seen that for a
4-pole motor, three speeds can be obtained easily.
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2. Variable Resistance in Series with Motor: By increasing the resistance
in series with the armature the voltage applied across the armature
terminals can be decreased. With reduced voltage across the armature, the
speed is reduced. However, it will be noted that since full motor current
passes through this resistance, there is a considerable loss of power in it.
Problems 4
1- A 230 V dc shunt motor runs at 800 rpm and takes armature current of 50 A.
Find resistance to be added to the field circuit to increase speed to 1000 rpm
at an armature current of 80 A. Assume flux proportional to field current.
Armature resistance = 0.15 Ω and field winding resistance = 250 Ω.
2- A 250 V, dc shunt motor has an armature resistance of 0.5 Ω and a field
resistance of 250 Ω. When driving a load of constant torque at 600 rpm, the
armature current is 20 A. If it is desired to raise the speed from 600 to 800
rpm, what resistance should be inserted in the shunt field circuit? Assume that
the magnetic circuit is unsaturated.
3- A dc shunt motor takes an armature current of 20 A from a 220 V supply.
Armature circuit resistance is 0.5 ohm. For reducing the speed by 50%,
calculate the resistance required in the series, with the armature, if (a) the
load torque is constant (b) the load torque is proportional to the square of the
speed.
4- A 7.48 kW, 220 V, 990 rpm shunt motor has a full load efficiency of 88%, the
armature resistance is 0.08 ohm and shunt field current is 2 A. If the speed of
this motor is reduced to 450 rpm by inserting a resistance in the armature
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circuit, find the motor output, the armature current, external resistance to be
inserted in the armature circuit and overall efficiency. Assume the load torque
to remain constant.
5- A 250 V dc shunt motor has armature circuit resistance of 0.5 Ω and a field
circuit resistance of 125 Ω. It drives a load at 1000 rpm and takes 30 A. The
field circuit resistance is then slowly increased to 150 Ω. If the flux and field
current can be assumed to be proportional and if the load torque remains
constant, calculate the final speed and armature current. [1186 rpm. 33.6 A]
6- A shunt-wound motor has a field resistance of 400 Ω and an armature
resistance of 0.1 Ω and runs off 240 V supply. The armature current is 60 A
and the motor speed is 900 rpm; Assuming a straight line magnetization
curve, calculate (a) the additional resistance in the field to increase the speed
to 1000 rpm for the same armature current and (b) the speed with the original
field current of 200 A. [(a) 44.4 Ω (b) 842.5 rpm]
7- A 250-V shunt motor has an armature current of 20 A when running at 1000
rpm against full load torque. The armature resistance is 0.5 Ω. What
resistance must be inserted in series with the armature to reduce the speed to
500 rpm at the same torque and what will be the speed if the load torque is
halved with this resistance in the circuit? Assume the flux to remain constant
throughout and neglect brush contact drop.
8- A 7.46 kW, 220 V, 900 rpm shunt motor has a full-load efficiency of 88 per
cent, an armature resistance of 0.08 Ω and shunt field current of 2 A. If the
speed of this motor is reduced to 450 rpm by inserting a resistance in the
armature circuit, the load torque remaining constant, find the motor output,
the armature current, the external resistance and the overall efficiency.
9- A 200 V, dc series motor takes 40 A when running at 700 rpm. Calculate the
speed at which the motor will run and the current taken from the supply if the
field is shunted by a resistance equal to the field resistance and the load
torque is increased by 50%. Armature resistance = 0.15 Ω, field resistance =
0.1 Ω It may be assumed that flux per pole is proportional to the field.
10- A 4-pole, 250 V dc series motor takes 20 A and runs at 900 rpm each field
coil has resistance of 0.025 ohm and the resistance of the armature is 0.1 ohm.
At what speed will the motor run developing the same torque if : (i) a diverter
of 0.2 ohm is connected in parallel with the series field (ii) rearranging the
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Characteristics of DC shunt motor
Armature torque vs armature current Tavs Ia characteristics
For a shunt motor flux can be assumed practically constant (through at heavy loads, ф decreases
somewhat due to increased armature reaction), hence Ta ∝ Ia
Therefore electrical characteristic is shown below, is practically a straight line through the origin. Shaft
torque is shown as dotted line.
Torque Current Characteristic of DC shunt motor
Speed vs armature current Navs Ia characteristics
As flux ɸ is assumed to be constant, we can say N ∝ Eb. But, as back emf is also almost constant, the speed
should remain constant. But practically, ɸ as well as Eb decreases with increase in load. Back emf Eb decreases
slightly more than ɸ, therefore, the speed decreases slightly. Generally, the speed decreases only by 5 to 15% of
full load speed. Therefore, a shunt motor can be assumed as a constant speed motor. In speed vs. armature
current characteristic in the following figure, the straight horizontal line represents the ideal characteristic and
the actual characteristic is shown by the dotted line.
Speed vs armature current characteristics of DC shunt motor
Speed vs armature torque Navs Ta characteristics
From Tavs Ia a n d Navs Ia with increase with torque the speed of DC shunt motor decreases. The nature of the
characteristics is drooping in nature shown in figure as given below.
Speed vs armature torque characteristics of DC shunt motor
Characteristics of DC series motor
Armature torque vs armature current characteristics
We know that Ta ∝ ф Ia. In case of DC series motor as field windings also carry the armature current
ф ∝ Ia
؞TI2
At light loads, Ia and hence is small. But as Ia increases Ta increases as the square of the current up-to
saturation. After saturation becomes constant, the characteristic becomes a straight line as shown in Figure
below. Therefore a series motor develops a torque proportional to the square of the armature current. This
characteristic is suited where huge starting torque is required for accelerating heavy masses.
Torque Current Characteristic of DC series motor
Speed vs armature current characteristics
N b/
In DC series motor Ia
Therefore N a
If Ia increases, speed decreases. This characteristic is shown in figure below. Therefore the speed is
inversely proportional to armature current Ia. When load is heavy Ia ia heavy thus speed is low. When load
is low Ia is low thus speed becomes dangerously high. Hence series motor should never start without load on
it.
Speed vs armature current characteristics of DC series motor
Speed vs armature torque characteristics
From Tavs Ia and Navs Ia characterist ics Speed is inversely proportional to torque. The characteristic is
shown in figure as given below.
Speed vs armature torque characteristics of DC series motor
Characteristics of DC compound motor
There are two different types of compound motors in common use, they are the cumulative compound motor
and the differential compound motor. In the cumulative compound motor, the field produced by the series
winding aids the field produced by the shunt winding. The speed of this motor falls more rapidly with
increasing current than does that of the shunt motor because the field increases. In the differential compound
motor, the flux from the series winding opposes the flux from the shunt winding. The field flux, therefore,
decreases with increasing load current. Because the flux decreases, the speed may increases with increasing
load. Depending on the ratio of the series-to-shunt field ampere-turns, the motor speed may increases very
rapidly.
Characteristics of DC compound motors
Example 1: A 4-pole wave wound dc series motor has 944 wave-connected armature conductors.
At a certain load, the flux per pole is 34.6 mWb and the total mechanical torque developed is 209
N-m. Calculate the line current taken by the motor and the speed at which it will run with an
applied voltage of 500 V. Total motor resistance is 3 ohm.
Solution:
Example 2:
(output power= 30 Hp= 30×746 =22,380 W)
Hence Il = 25720/230 = 111.82 A
Example2. A 250 volt DC shunt motor has armature resistance of 0.25 ohm on load it takes an
armature current of 50A and runs at 750rpm. If the flux of the motor is reduced by 10% without
changing the load torque, find the new speed of the motor.
Solution:
Given data V = 250
Ra = 0.25
Ia = 50
N1 = 750 rpm
Φ2 = 90%Φ1
For shunt motor
Eb1 = V-Ia1Ra = 250-(50x0.25) = 237.5V
Eb2 = V-Ia2Ra
Load torque is constant
Ta1 = Ta2
Or Φ1 Ia1 = Φ2 Ia2
Or Φ1 × 50 = 0.9 Φ1 Ia2 hence Ia2 = 55.55A
Eb2 = 250-55.55X0.25 = 236.12V
N2 = 828 rpm
Example 3. A 230-V d.c. shunt motor has an armature resistance of 0.5 Ω and field resistance of
115 Ω. 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.
Example 3.
Example 4. A dc motor takes an armature current of 110A at 480 V. The armature circuit
resistance is 0.2 ohm. The machine has 6 poles and the armature is lap connected with 864
conductors. The flux per pole is 0.05 Wb . Calculate (i) the speed and (ii) gross torque developed
by the armature?
Solution:
Example 5:
Figure (a) Figure (b)
Example 6:
Example 7:
Example 8:
