Back emf and its role in DC motor

Back emf is the emf induced in the armature conductors of a running DC motor. When the armature rotates in the magnetic field, its conductors cut the flux and an emf is induced (generator action). By Lenz's law this emf opposes the applied voltage, so it is called back emf or counter emf, $E_b$.

$$E_b = \frac{P \phi Z N}{60 A}, \qquad V = E_b + I_a R_a$$

   +----Ra----+
   |   Ia ->  |
   V         (Eb)  opposes V
   |          |
   +----------+

So the armature current is

$$I_a = \frac{V - E_b}{R_a}$$

Roles of back emf

1. Limits armature current: $R_a$ is very small (e.g. 0.5 Ω). Without $E_b$, a 220 V motor would draw $220/0.5 = 440$ A. With $E_b = 210$ V, $I_a = (220 - 210)/0.5 = 20$ A only. Back emf acts like a self-adjusting resistance.

2. Makes the motor self-regulating (adjusts current to load):

   • Load increases → speed falls slightly → $E_b$ falls → $I_a$ rises → torque rises to meet the load.
   • Load decreases → speed rises → $E_b$ rises → $I_a$ falls → torque falls. The motor automatically draws just the current needed by the load.

3. Makes energy conversion possible: multiplying $V = E_b + I_a R_a$ by $I_a$:

   $$V I_a = E_b I_a + I_a^2 R_a$$

   $E_b I_a$ is the gross mechanical power developed. Without back emf there would be no conversion; all input would be wasted as heat. Maximum mechanical power occurs when $E_b = V/2$.

4. Determines speed: $N \propto \dfrac{E_b}{\phi} = \dfrac{V - I_a R_a}{\phi}$. This is the basis of speed control by flux or armature voltage.

5. Need for a starter: at starting $N = 0$ so $E_b = 0$, and current would be very large. That is why a starter resistance is put in series until $E_b$ builds up.