Types of DC Motors

TYPES OF DC MOTORS

The classification of DC motors is similar to that of the DC generator. The classification is based on the connection of field winding in relation to the armature. The types of DC motor:

1. Separately excited DC Motor

2. Self excited DC Motor

a. Series Motor

b. Shunt Motor

c. Compound Motor

(i) Long Shunt Compound Motor

(ii) Short Shunt Compound Motor

Separately Excited DC Motor

The field winding and armature are separated. The field winding is excited by a separate DC source. That is why it is called separately excited DC motor:

From this diagram:

Armature current I_a = Line current I_L

Back emf E_b = V – I_a R_a - V_brush

V_brush is very small and therefore is neglected.

Self Excited DC Motor

DC Series Motor

DC series motor means, the field winding is connected in series with armature. The field winding should have less number of turns of thick wire. R_se is the resistance of the series field winding normally R_se value is very small.

In a DC series motor,

I_L = line current drawn from the supply

Armature current I_a = Series field current I_se = I_L

I_a = I_se = I_L

The voltage equation is given by:

V = E_b +I_a R_a + I_se R_se +V_brush.

I_a = I_se

V = E_b +I_a (R_a+ R_se) +V_brush

V_brush - Voltage drop in the brush.

Normally it is neglected

V =E_b +I_a (R_a+R_se)

In a DC series motor, full armature current flows through the series field winding, therefore flux produced is directly proportional to the armature current.

φ α I_se α I_a

DC Shunt Motor

In a DC shunt motor, the field winding is connected across the armature.

Here, the shunt field winding has more number of turn with less cross-sectional area. R_sh is the shunt field winding resistance. The value of R_a is very small and R_sh is quite large. The input voltage V is equal to the voltage across the armature and field winding.

It is the line current drawn from the supply. The line current is divided into two paths one through the field winding and second through the armature:

I_L =I_a + I_sh

I_a = armature current,

I_sh = shunt field current.

I_sh = V / R_sh

Voltage equation of DC shunt motor

V = E_b +I_a R_a + V_brush

In shunt motor, flux produced by field winding is proportional to the field current

φ α I_sh

Here, the input voltage is constant and so that flux is also constant. Therefore DC shunt motor is also called a constant flux motor or constant speed motor.

DC Compound Motor

A DC compound motor consists of both series and shunt field windings.

(a) Long Shunt Compound Motor:

In this motor, the shunt field winding is connected across both armature and series field winding.

I_L = I_se + I_sh

I_se = I_a

I_L = I_a + I_sh

I_sh = V / R_sh

The voltage equation of this motor is given by:

V = E_b + I_a R_a + I_se R_se + V_brush

I_a = I_se

V = E_b + I_a (R_a + R_se) + V_brush

(b) Short Shunt Compound Motor:

In this type of motor, the shunt field winding is across the armature and series field winding is connected in series with this combination.

I_L = I_se

IL = Ia + Ish

I_L = I_se = I_a + I_sh

The voltage across the shunt field winding can be found out from the voltage equation:

V = E_b + I_a R_a + I_se R_se + V_brush

I_se = I_L

V = E_b + I_a R_a + I_L R_se + V_brush

Voltage drop across the shunt field winding = V – I_L R_se

V_sh = E_b + I_a R_a + V_brush

I_sh = (V - I_L R_se) / R_sh

The compound motor again can be classified into two types:

1. Cummulative Compound Motor.

2. Differential Compound Motor.

Cummulative Compound Motor

In this type of motor, the two field winding flux aid each other i.e., flux due to the series field winding strengthens the flux due the field winding. The winding connection of diagram is shown in Figure 2.27.

Differential Compound Motor

In this type of motor, the two field winding flux oppose each other i.e., flux due to series field winding weakens the field due to shunt field winding.

Example 2.1:

The armature of a DC machine has a resistance of 0.1 Ω and is connected to a 230 V supply. Calculate the back emf when it is running (i) as a generator giving 80 A (ii) as a motor taking 80 A.

Given data:

R_a = 0.1 Ω

Line voltage V = 230 V.

To find:

Back emf = ?

Solution:

(i) As a generator giving 80 A

Induced emf E_g = V + I_a R_a = 230 + 80 × 0.1

E_g = 238 V.

(ii) As a motor taking 80 A

Back emf E_b = V - I_a R_a = 230 - 80 x 0.1

E_b = 222 V.

Example 2.2:

A 250 V dc shunt motor takes 41 A at full load resistance of motor armature and shunt field winding are 0.1 Ω and 250 Ω respectively. Find the back emf on full load.

Given data:

V = 250 V

IL = 41 A

R_a = 0.1 Ω

R_sh = 250 Ω.

To find: Back emf on full load (E_b).

Solution:

Shunt Field Current

I_sh = V / R_sh = 250 / 250 = 1 A

Armature Current

I_a = I_L - I_sh = 41 - 1 = 40 A

Back emf E_b = V - I_a R_a = 250 - 40 × 0.1 = 246 V

E_b = 246 V.

Example 2.3:

A 4 pole DC motor takes an armature current of 6 A. The armature has 480 lap connected conductor. The flux per pole is 20 mwb. Calculate the gross torque developed by the motor.

Given data:

Number of poles P = 4

Armature current I_a = 6 A

Number of conductor Z = 480

Flux per pole φ = 20 mwb

For lap connection A = P.

To find:

Gross torque (T_a).

Solution:

Example 2.4:

A 200 V, 2000 rpm, 10 A separately excited de motor has an armature resistance of 2 Ω. Rated de voltage is applied to both the armature and field winding of the motor, if the armature draws 5 A from the source, calculate the torque developed by the motor.

Given data:

Supply Voltage V = 200 V

Motor Speed N = 2000 rpm

Armature resistance R_a = 2 Ω

Armature Current I_a2 = 5 A

 Armature Current I_a1 = 10 A

To find:

Torque developed by the motor.

Solution:

Back emf

E_b2 = V - I_a2 R_a = 200 - 5 x 2 = 190 V.

Back emf

E_bl = V - I_a1 R_a = 200 - 10 x 2 = 180 V.

Let the motor speed be (N₂) at armature current of 5 A.

Example 2.5:

A 4 pole de motor is lap wound with 500 conductors. The pole shoe is 22 cm long and the average flux density over one pole pitch is 0.4 T, the armature diameter being 30 cm. Find the torque and gross mechanical power developed when the motor is drawing 26 A and running at 1500 rpm.

Given Data:

No. of poles P = 4

No. of conductors Z = 500

Armature current I_a = 26 A

Motor speed N = 1500 rpm

For lap connection A = P.

To find:

(i) Gross mechanical power.

(ii) Torque developed.

Solution:

Example 2.6:

A 4 pole DC motor takes an armature current of 50 amps. The armature has 480 lap connected conductor. The flux per pole is 20 mwb. Calculate the gross torque developed by the motor.

Given data:

No. of poles P = 4

Armature current I_a = 50 A

No. of conductor Z = 480

Flux per pole φ = 20 × 10^-3

For lap connection A = P = 4.

To find:

Gross torque developed by the motor (T_a).

Solution: