DC generator construction, working, emf equation

A DC generator converts mechanical energy into DC electrical energy by electromagnetic induction.

Construction

   +----------- Yoke (frame) ------------+
   |  [Pole core + shoe]    [Pole core]  |
   |   field winding         field wdg   |
   |        ( Armature core + winding )  |
   |        Commutator -- Brushes        |
   +---- Shaft, bearings, end covers ----+

1. Yoke (frame): cast iron or cast steel outer frame. It gives mechanical support and carries the return path of the magnetic flux.
2. Pole core and pole shoe: laminated steel bolted to the yoke. The shoe spreads the flux evenly over the air gap and holds the field coils.
3. Field winding: copper coils on the pole cores that produce the main flux when excited.
4. Armature core: cylindrical, made of thin laminated silicon steel (to reduce eddy-current loss) with slots on the outside to hold conductors.
5. Armature winding: insulated copper conductors in the slots, connected as lap or wave winding.
6. Commutator: cylinder of hard-drawn copper segments insulated by mica. It rectifies the armature emf and collects the current.
7. Brushes and brush gear: carbon brushes in holders press on the commutator and carry current to the external circuit.
8. Shaft, bearings and end covers: support and rotate the armature.

Working principle

When the armature is rotated by a prime mover in the field of the poles, its conductors cut flux and an emf $e = Blv$ is induced (Faraday's law). The emf in each coil is alternating; the commutator and brushes convert it to a unidirectional output at the terminals.

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, $E_g = \dfrac{ZP}{2\pi A}\Phi\omega = K_a\Phi\omega$.