DC motor working principle and torque equation

A DC motor converts DC electrical energy into mechanical energy. It works on the principle that a current-carrying conductor placed in a magnetic field experiences a mechanical force, $F = B I l$ newtons, whose direction is given by Fleming's left-hand rule.

Working principle

1. The field winding produces a steady flux between N and S poles.
2. DC supply is given to the armature through brushes and commutator, so current flows in the armature conductors.
3. Each conductor carries current in the field and experiences a force $F = BIl$. Conductors under the N pole and under the S pole carry opposite currents (because of the commutator), so their forces act in the same rotational sense and produce a torque.
4. When a conductor moves from one pole to the next, the commutator reverses its current, so the torque stays in one direction and the rotation is continuous.
5. As the armature rotates, a back emf $E_b$ is induced which opposes the supply: $V = E_b + I_a R_a$.

        N pole
   ---------------
    (x) (x)  -> F      x = current into page
       armature        . = current out of page
    (.) (.)  <- F
   ---------------
        S pole
  Forces form a couple -> rotation

Torque equation

Method 1: from force on conductors

Let $r$ = armature radius (m), $l$ = conductor length (m), $I_a/A$ = current per conductor. Average flux density:

$$B = \frac{P\phi}{2\pi r l}$$

Force per conductor $= B \dfrac{I_a}{A} l$. Torque per conductor $= B \dfrac{I_a}{A} l\, r$.

Total torque for $Z$ conductors:

$$\begin{aligned} T_a &= Z \cdot \frac{P\phi}{2\pi r l} \cdot \frac{I_a}{A} \cdot l \cdot r \\ &= \frac{P \phi Z I_a}{2\pi A}\ \text{N·m} \end{aligned}$$

Method 2: from power balance

Mechanical power developed $= E_b I_a$, and also $= T_a \omega = T_a \dfrac{2\pi N}{60}$. So

$$T_a \cdot \frac{2\pi N}{60} = E_b I_a = \frac{P\phi Z N}{60 A} I_a$$ $$T_a = \frac{P \phi Z I_a}{2\pi A} = 0.159\, \frac{P \phi Z I_a}{A}\ \text{N·m}$$

Conclusions

• For a given machine, $T_a = k \phi I_a$, i.e. $T_a \propto \phi I_a$.
• Shunt motor ($\phi$ nearly constant): $T_a \propto I_a$.
• Series motor ($\phi \propto I_a$ before saturation): $T_a \propto I_a^2$.
• In terms of speed: $T_a = 9.55 \dfrac{E_b I_a}{N}$ N·m. The useful shaft torque is $T_{sh} = 9.55 \dfrac{P_{out}}{N}$, which is less than $T_a$ by the torque lost to iron and friction losses.