Back Emf in DC Motor
Consider a shunt wound dc motor as shown.
When a dc voltage V is applied across the motor
terminals, the armature starts rotating due to the torque developed in it.
As the armature rotates, armature conductors cut the
pole magnetic field, therefore, as per law of electromagnetic induction, an emf
called back emf is induced in them.
The back emf (also called counter emf) is given by
where, P=number of poles of dc motor
Φ= flux per pole
Z=total number of
armature conductors
N=armature speed
A=number of parallel paths
in armature winding
As all other parameters are constant, therefore, Eb ∝ N
As per
Lenz's law, "the induced emf always
opposes the cause of its
production" . Here, the cause of generation of back emf is the rotation of
armature. Rotation of armature is due to armature torque. Torque is due
to armature current and armature current is due to supply dc voltage V.
Therefore, the ultimate cause of production of Eb is the supply voltage V.
production" . Here, the cause of generation of back emf is the rotation of
armature. Rotation of armature is due to armature torque. Torque is due
to armature current and armature current is due to supply dc voltage V.
Therefore, the ultimate cause of production of Eb is the supply voltage V.
Therefore, back emf is always directed opposite to supply voltage V.
Significance of back emf in dc motor
(1) As
the back emf opposes supply voltage V, therefore, supply voltage has to force
current through the armature against the back emf, to keep armature rotating. The
electric work done in overcoming and causing the current to flow against the
back emf is converted into mechanical energy developed in the armature.
It
follows, therefore, that energy conversion in a dc motor is only possible due
to the production of back emf.
Mechanical
power developed in the armature = EbIa
(2) Back
emf makes dc motor a self-regulating motor i.e Eb makes motor to
adjust Ia automatically as per the load torque requirement. Lets see
how.
From the motor figure,
V
and Ra are fixed, therefore, armature current Ia dpends
on back emf, which in turn depends on speed of the motor.
(a)
when the motor is running at no-load, small torque ( Ta=KIa ) is required
by the motor to overcome friction and windage. Therefore, a small current is
drawn by the motor armature and the back
emf is almost equal to the supply voltage.
(b) If the motor is suddenly
loaded, the load torque beomes greater than the armature torque and the motor
starts to slow down. As motor speed decreases, back emf decreases and
therefore, armature current starts
increasing. With increasing Ia , armature torque increases and at
some point it becomes equal to the load torque. At that moment, motor stops slowing
down and keeps running at this new speed.
(c)
If the load on the motor is suddenly reduced, the driving
torque becomes more than the load torque and the motor starts accelerating. As
the motor speed increases, back emf increases and therefore, armature current
decreases. Due to this reducing armature current, armature developed torque
decreases and at some point becomes equal to the load torque. That point
onwards, motor will stop accelerating and will start rotating uniformly at this
new slightly increased speed.
So, this shows how important is back emf in dc motor.
Without back emf, the electromagnetic energy conversion would not have been
possible at the first place.
