Back emf of DC Motor

BACK EMF

In a DC motor when the armature rotates, the conductors cut the lines of force of magnet field, so that an emf is induced in the armature. This induced emf acts in opposition to the current in the machine and the applied voltage so this emf is called back emf or counter emf. According to lenz's law, the direction of the back emf opposes the supply voltage the back emf is calculated from the equation of induced emf in the generator.

E_b = $ZNP / 60A

Where

ϕ - flux/pole in wb

P - Number of poles

Z - Total number of conductors in the armature

N - Speed in rpm

A - No of parallel paths

The equivalent circuit of a motor is shown in fig 3.25. Here, the armature circuit is equivalent to a source of emf E_b, in series with a resistance R_a and a DC supply is applied across, series connection of R_a and E_b. The voltage equation is

V = E_b + I_aR_a

From the above voltage equation

Armature current I_a = V – E_b / R_a

Where

V - Applied voltage

E_b - Back emf

I_a - Armature current

R_a - Armature resistance

V - E_b - Net voltage in the armature circuit

 (i) If the motor speed is high, back emf E_b is large and armature current is small.

(ii) If the motor speed is low, back emf E_b will be less and armature current is more.

Example: 7

A 4 pole DC motor takes an armature current of 150 A at 440 V. If its armature circuit has a resistance of 0.15Ω. What will be the back emf at this load?

Voltage equation of DC motor is

V = E_b + I_aR_a

E_b = V - I_aR_a = 440-(150 × 0.15)

= 440 - 22.5

E = 417.5 V