Unidirectional torque in dc machine

Production of unidirectional torque

When both the armature (through the brushes) and the field winding are supplied with direct current, the armature conductors carry current in a magnetic field, so each conductor experiences a force $F = BIl$ (Fleming's left-hand rule).

• Conductors under the N pole carry current in one direction (into the page); those under the S pole carry it in the opposite direction (out of the page).
• Because the field direction under the two poles is also opposite, the forces on the two sets of conductors act in the same rotational sense, so a net torque turns the armature.

When the armature turns, a conductor moves from under the N pole to under the S pole. If its current stayed the same, the force on it would reverse. The commutator prevents this: its segments reverse the connection of the conductor to the brushes at the moment it crosses the neutral axis, so the current in the conductor reverses at the same time as the field. The product $B\times I$ keeps the same sign, so the torque stays in the same direction (unidirectional).

   Under N pole        Under S pole
   current  (x)  --->  current  (.)  after commutation
   force  -> up        force  -> up   (same sense of rotation)

Factors affecting the torque

From $T = \dfrac{ZP}{2\pi A}\Phi I_a$, the torque depends on:

1. Flux per pole $\Phi$ (field current and magnetic circuit).
2. Armature current $I_a$ (load and supply voltage).
3. Number of armature conductors $Z$.
4. Number of poles $P$.
5. Number of parallel paths $A$ (type of winding: lap $A = P$, wave $A = 2$).

For a given machine $Z$, $P$, $A$ are fixed, so $T\propto\Phi I_a$. For a shunt motor $T\propto I_a$; for an unsaturated series motor $T\propto I_a^2$. Armature reaction (which weakens the flux) and the brush position also change the torque.