Introduction

Although a far greater percentage of the electrical machines in service are a.c. machines, the d.c. machines are of considerable industrial importance. The principal advantage of the d.c. machine, particularly the d.c. motor, is that it provides a fine control of speed. Such an advantage is not claimed by any a.c.

motor. However, d.c. generators are not as common as they used to be, because direct current, when required, is mainly obtained from an a.c. supply by the use of rectifiers. Nevertheless, an understanding of d.c. generator is important because it represents a logical introduction to the behaviour of d.c. motors.

Indeed many d.c. motors in industry actually operate as d.c. generators for a brief period. In this UNIT, we shall deal with various aspects of d.c. generators.

 D.C. Motor Characteristics

There are three principal types of d.c. motors viz., shunt motors, series motors

and compound motors. Both shunt and series types have only one field winding

wound on the core of each pole of the motor. The compound type has two

separate field windings wound on the core of each pole. The performance of a

d.c. motor can be judged from its characteristic curves known as motor

characteristics, following are the three important characteristics of a d.c. motor:

(i) Torque and Armature current characteristic (Ta/Ia)

It is the curve between armature torque Ta and armature current Ia of a d.c.

motor. It is also known as electrical characteristic of the motor.

(ii) Speed and armature current characteristic (N/ia)

It is the curve between speed N and armature current Ia of a d.c. motor. It is very important characteristic as it is often the deciding factor in the selection of the motor for a particular application.

(iii) Speed and torque characteristic (N/Ta)

It is the curve between speed N and armature torque Ta of a d.c. motor. It is also

known as mechanical characteristic.

Losses in a D.C. Motor

The losses occurring in a d.c. motor are the same as in a d.c. generator  

(i) copper losses

 (ii) Iron losses or magnetic losses 

(iii) mechanical losses 

These losses cause 

(a) an increase of machine temperature and 

(b) reduction in the efficiency of the d.c. motor.

The following points may be noted:

(i) Apart from armature Cu loss, field Cu loss and brush contact loss, Cu losses also occur in interpoles (commutating poles) and compensating windings. Since these windings carry armature current (I_a),

Loss in interpole winding = I_a ²× Resistance of interpole winding.
Loss in compensating winding = I_a ²× Resistance of compensating winding.

(ii) Since d.c. machines (generators or motors) are generally operated at constant flux density and constant speed, the iron losses are nearly constant.

(iii) The mechanical losses (i.e. friction and windage) vary as the cube of the speed of rotation of the d.c. machine (generator or motor). Since d.c. machines are generally operated at constant speed, mechanical losses are
considered to be constant.

COMMUTATION IN DC MOTOR

 Commutation in D.C. Motors

Since the armature of a motor is the same as that of a generator, the current from the supply line must divide and pass through the paths of the armature windings.

In order to produce unidirectional force (or torque) on the armature conductors of a motor, the conductors under any pole must carry the current in the same direction at all times. This is illustrated in Fig. In this case, the current flows away from the observer in the conductors under the N-pole and towards

the observer in the conductors under the S-pole. Therefore, 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. The function of the commutator and the brush gear in a d.c. motor is to cause the reversal of current in a conductor as it moves from one side of a brush to the other. For

good commutation, the following points may be noted:

(i) If a motor does not have commutating poles (compoles), the brushes must be given a negative lead i.e., they must be shifted from G.N.A. against the direction of rotation of, the motor.

(ii) By using interpoles, a d.c. motor can be operated with fixed brush positions for all conditions of load. For a d.c. motor, the commutating poles must have the same polarity as the main poles directly back of them. This is the opposite of the corresponding relation in a d.c. generator.

Note. A d.c. machine may be used as a motor or a generator without changing the commutating poles connections. When the operation of a d.c. machine changes from generator to motor, the direction of the armature current reverses.

Since commutating poles winding carries armature current, the polarity of  commutating pole reverses automatically to the correct polarity.

ARMATURE REACTION IN DC MOTOR

Armature Reaction in D.C. Motors

As in a d.c. generator, armature reaction also occurs in a d.c. motor. This is expected because when current flows through the armature conductors of a d.c. motor, it produces flux (armature flux) which lets on the flux produced by the main poles. For a motor with the same polarity and direction of rotation as is for generator, the direction of armature reaction field is reversed.

(i) In a generator, the armature current flows in the direction of the induced e.m.f. (i.e. generated e.m.f. Eg) whereas in a motor, the armature current flows against the induced e.m.f. (i.e. back e.m.f. E_g). Therefore, it should be expected that for the same direction of rotation and field polarity, the armature flux of the motor will be in the opposite direction to that of the generator. Hence instead of the main flux being distorted in the direction of rotation as in a generator, it is distorted opposite to the direction of rotation. We can conclude that:
Armature reaction in a d.c. generator weakens the jinx at leading pole tips and strengthens the flux at trailing pole tips while the armature reaction in a d. c. motor produces the opposite effect.

(ii) In case of a d.c. generator, with brushes along G.N.A. and no commutating poles used, the brushes must be shifted in the direction of rotation (forward lead) for satisfactory commutation. However, in case of a d.c. motor, the brushes are given a negative lead i.e., they are shifted against the direction of rotation.

With no commutating poles used, the brushes are given a forward lead in a d.c. generator and backward lead in a d.c. motor.

(iii) By using commutating poles (compoles), a d.c. machine can be operated with fixed brush positions for all conditions of load. Since commutating poles windings carry the armature current, then, when a machine changes from generator to motor (with consequent reversal of current), the polarities of commutating poles must be of opposite sign.

Therefore, in a d.c. motor, the commutating poles must have the same polarity as the main poles directly back of them. This is the opposite of the corresponding relation in a d.c. generator.

Torque and Speed of a D.C. Motor

For any motor, the torque and speed are very important factors. When the torque increases, the speed of a motor increases and vice-versa. We have seen that for a d.c. motor;

N = K (V- I_aR_a)/ Ф = K E_b/ Ф…………………………………………….(i)

T_a α ФI_a…………………………………………………………………………(ii)

If the flux decreases, from Eq.(i), the motor speed increases but from Eq.(ii) the motor torque decreases. This is not possible because the increase in motor speed must be the result of increased torque. Indeed, it is so in this case. When the flux decreases slightly, the armature current increases to a large value. As a result, in spite of the weakened field, the torque is momentarily increased to a high value and will exceed considerably the value corresponding to the load. The surplus torque available causes the motor to accelerate and back e.m.f  (E_a=PФZN/60A) to rise. Steady conditions of speed will ultimately be achieved when back e.m.f. has risen to such a value that armature current[I_a = (V- E_a)/ R_a]develops torque just sufficient to drive the load.

Illustration
Let us illustrate the above point with a numerical example. Suppose a 400 V
shunt motor is running at 600 r.p.m., taking an armature current of 50 A. The armature resistance is 0.28 Ω. Let us see the effect of sudden reduction of flux by 5% on the motor.

Initially (prior to weakening of field), we have,

E_a = V-I_aR_a= 400 – 50 × 0.28 = 386 volts

We know that E_b α Ф N. If the flux is reduced suddenly, E_b α Ф because inertia
of heavy armature prevents any rapid change in speed. It follows that when the flux is reduced by 5%, the generated e.m.f. must follow suit. Thus at the instant of reduction of flux, E’_b = 0.95 × 386 = 366.7 volts.

Instantaneous armature current is

I’_a=(V- E’_b)/ R_a =(400-366.7)/0.28=118.9A

Note that a sudden reduction of 5% in the flux has caused the armature current to increase about 2.5 times the initial value. This will result in the production of high value of torque. However, soon the steady conditions will prevail. This will depend on the system inertia; the more rapidly the motor can alter the speed, the sooner the e.m.f. rises and the armature current falls.

Speed of a D.C. Motor

E_b = V-I_aR_a

But E_b=PФZN/60A

PФZN/60A  = V- I_aR_a

Or  N = (V- I_aR_a)/ Ф ×  60A/ PZ

Or N = K (V- I_aR_a)/ Ф

But         V- I_aR_a = E_a

Therefore N= K E_b/ Ф

Or N α E_b/ Ф

Therefore, in a d.c. motor, speed is directly proportional to back e.m.f. E_b and inversely proportional to flux per pole Ф.

Speed Relations

If a d.c. motor has initial values of speed, flux per pole and back e.m.f. as N₁ ,Ф₁ and E_b1 respectively and the corresponding final values are N₂ ,Ф₂ and E_b2 then,

N₁ α E_b1/ Ф₁ and N₂ α E_b2/ Ф₂

Therefore N₂/ N_{1 =} (E_b2/ E_b1) ×( Ф₁ / Ф₂)

(i) For a shunt motor, flux practically remains constant so that Ф1 = Ф₂.

therefore  N₂/ N_{1 =} E_b2/ E_b1

(ii) For a series motor, Ф α I_a prior to saturation.

therefore N₂/ N_{1 =} (E_b2/ E_b1) × (I_a1/I_a2)

where I_a1 = initial armature current
I_a2 = final armature current

Speed Regulation

The speed regulation of a motor is the change in speed from full-load to no-load and is expressed as a percentage of the speed at full-load i.e.

% Speed regulation = [( N.L. speed - F.L.speed)/F.L.speed ] × 100

=[(N_o -N)/N] × 100

where N_o = No – load .speed
N = Full – load speed