Back EMF of DC Motor

BACK EMF

Fig. 2.29 Equivalent Clrcult of dc Motor

When the armature of a DC motor rotates in the magnetic field, the armature conductor cuts the magnetic flux. Hence emf will be induced in the conductor according to Faraday's law of electromagnetic induction. The induced emf acts in opposite direction to the applied voltage (As per Lenz's Law) and it is referred as the back emf or counter emf E_b.

The back emf is given by,

E_b = ( ϕΖΝ / 60 ) × P/A   Volts

which is same as that of emf induced in a DC generator.

The relationship between the current, back emf and the applied voltage for a DC shunt motor is given by E_b = V ‒ l_aRa.

where

E_b = back emf in Volts

V = Supply Voltage in Volts

l_a = armature current in amperes

R_a = armature resistance in ohms.

The back emf E_b is always less than that of the applied voltage and hence the current is flowing against the direction of back emf.

1. Significance of back emf

The back emf acts like a governor. It makes the DC motor a self regulating machine. i.e., it makes the motor to draw as much armature current, which is just sufficient to develop the torque required by the load.

Net voltage across the armature circuit = V‒E_b

If R_a is the armature circuit resistance, then,

 I_b = [ V‒E_b ] / R_a

Since V and R_a are usually fixed, the value of E_b will determine the current drawn by the motor.

E_b = ( ϕΖΝ / 60 ) × P/A   Volt

Back emf depends, among other factors, upon armature speed. If speed is high, E_b is large. Hence armature current l_a seen from the above equation is small. If the speed is less, the E_b is less, hence more current flows which develops motor torque.

2. Voltage Equation of a Motor

The voltage V applied across the motor armature has to

(i) Overcome the back emf E_b and

(ii) Supply the armature ohmic drop I_aR_a

∴ V = E_b+l_aR_a

This is known as voltage equation of a motor.

 Vl_a= E_bI_a + I²_aR_a

This is known as power equation of the DC motor

VI_a = electrical input to the armature

E_bI_a = electrical equivalent of mechanical power developed in the armature (P_m)

 I²_aR_a = Cu loss in the armature

Hence, out of the armature input, some is wasted in I²R loss and the rest is converted into mechanical power within the armature.

3. Condition for Maximum Power

The mechanical power developed by the motor is P_m = E_bl_a

Now, P_m=VI_a ‒ I²_aR_a

Since, V and R_a are fixed, power developed by the motor depends upon armature current.

For maximum power, dp_m / dl_a should be zero.

 dp_m / dI_a = V ‒ 2I_aR_a = 0

or

I_aR_a = V/2

Now,

V= E_b+ l_aR_a

 V = E_b + V/2

 R_a= V/2

Hence mechanical power developed by the motor is maximum when back emf is equal to half the applied voltage. This condition is however, not realised in practice, because in that case current would be much beyond the normal current of the motor. Moreover half the input power wasted in the form of heat and taking other losses (mechanical and magnetic) into consideration, the motor efficiency will be well below 50%.