What is a Separately Excited DC Motor?

Separately Excited DC Motors are similar to all other DC motors but the difference is the excitation System. For the Production of Magnetic Flux, there is a separate field coil in the stator which is fed by an external DC source.

The Field system of this type of motor is similar to the field system of the shunt motor. Generally except for separately excited DC motors all other motor's excitation systems and armature windings are fed by a single source of DC supply. The field winding of this type of DC generator has a large number of turns of fine wire. Therefore, the length of such a winding is greater with a small cross-section. As a result, the resistance of this field winding is high so the current is limited. The Circuit Diagram of Separately Excited DC Motors is given Below

Separately Excited DC Motor

from the above-given circuit diagram, it is clear that the motor is fed by an external DC Source, and the excitation system is also fed by an external source of DC Supply. During the operating period of the motor, the armature Voltage may vary but the Excitation system Voltage should be kept Constant because the Speed of the DC Motor is a function of flux produced by the Excitation system.

Relation Between Back EMF and Load Current

When the armature of the DC motor rotates under the influence of the driving force, the armature of the conductors moves through the magnetic field and generates an electromotive force(emf) in them. The induced emf is in opposite direction to the externally applied voltage and this induced voltage is known as back emf and denoted by E. Emf induced in any DC motor is given by the formula

Where

N = Speed of DC motor

P = Number of pole

φ = magnetic flux

Z = Number of Conductor

A = Number of Parallel Path

for a dc motor Number of Pole(P), Number of conductors (Z), and Number Of Parallel Path (A) is constant hence we can replace this emf equation in a general form By Removing all Constant By a new Constant K then

If this motor is connected with a DC Source of terminal voltage V and a load Current I start to flow in the motor then due to internal armature resistance(R), a voltage will drop then We can write the KVL equation for this motor like this

V = E+IR

E=V - IR

k𝜙N =V - IR