DC generator principle and emf equation

A DC generator converts mechanical energy into electrical energy using Faraday's law of electromagnetic induction: when a conductor cuts magnetic flux, an emf is induced in it, $e = Blv$ volts (direction by Fleming's right-hand rule).

Working:

1. The field poles produce a flux $\Phi$ across the air gap.
2. The prime mover rotates the armature, so the armature conductors cut this flux and an emf is induced in each conductor.
3. In a single coil the induced emf is alternating, because each coil side passes under a north pole and then a south pole.
4. The commutator and brushes act as a mechanical rectifier. Each time the coil emf reverses, the segments reverse their connection to the brushes, so the brush terminals give a unidirectional (DC) voltage.
5. With a load connected, current flows out through the positive brush.

        N pole            S pole
      +---------+       +---------+
      |    ^    |       |         |
      |  B -->  |       |         |
      |  ______ |_______|______   |
      | / loop (armature coil) \  |
      +-+-----------------------+-+
        | split-ring commutator |
     brush A ----[ load ]---- brush B

The figure shows a simple loop generator: one coil rotating between the N and S poles, with a split-ring commutator (two segments) and two brushes.

Derivation of the emf equation

Let

• $\Phi$ = flux per pole (Wb), $P$ = number of poles
• $Z$ = total armature conductors, $A$ = number of parallel paths
• $N$ = speed in rpm

In one revolution a conductor cuts the flux of all poles, $P\Phi$ weber, in time $60/N$ second.

$$\text{average emf per conductor} = \frac{d\phi}{dt} = \frac{P\Phi}{60/N} = \frac{P\Phi N}{60}\ \text{V}$$

The $Z$ conductors form $A$ parallel paths, so each path has $Z/A$ conductors in series. The generated emf equals the emf of one path:

$$E_g = \frac{P\Phi N}{60}\times\frac{Z}{A} = \frac{\Phi Z N P}{60A}\ \text{V}$$

• Lap winding: $A = P$, so $E_g = \dfrac{\Phi Z N}{60}$
• Wave winding: $A = 2$, so $E_g = \dfrac{\Phi Z N P}{120}$

In terms of angular speed, $E_g = \dfrac{ZP}{2\pi A}\Phi\omega = K_a\Phi\omega$.