Commutator function and DC generator emf equation

Function of commutator segments

The commutator is a cylinder made of wedge-shaped, hard-drawn copper segments insulated from each other by mica, mounted on the shaft. Each segment is connected to the end of an armature coil, and carbon brushes press on its surface.

The emf induced in each armature conductor is alternating, because a conductor passes under N and S poles in turn. The commutator, together with the brushes, works as a mechanical rectifier:

1. Rectification: as a coil side moves from under one pole to the next, the coil connection to the brush is reversed at the same instant the emf reverses. So the external circuit always gets current in one direction (unidirectional DC).
2. Collecting current: it provides a sliding contact so current can be taken from the rotating armature to the stationary external circuit through the brushes.
3. Connecting coils in series/parallel paths: the segments join the coils so that their emfs add up between brushes (lap or wave paths).
4. Commutation: it reverses the current in a coil while the coil is short-circuited by the brush, keeping the output steady with many segments (low ripple).

 Coil emf (AC)      After commutator (DC)
   /\    /\           /\  /\  /\
  /  \  /  \         /  \/  \/  \
 /    \/    \       /            \
       (reverses)    (always positive)

EMF equation of a DC generator

Let

• $P$ = number of poles, $\phi$ = flux per pole (Wb)
• $Z$ = total number of armature conductors
• $A$ = number of parallel paths ($A = P$ for lap, $A = 2$ for wave)
• $N$ = speed (rpm)

By Faraday's law, average emf in one conductor = flux cut per revolution ÷ time for one revolution.

Flux cut by one conductor in one revolution:

$$d\phi = P\phi\ \text{Wb}$$

Time for one revolution:

$$dt = \frac{60}{N}\ \text{s}$$

So emf per conductor:

$$e = \frac{d\phi}{dt} = \frac{P\phi N}{60}\ \text{V}$$

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

$$\begin{aligned} E_g &= \frac{P\phi N}{60} \times \frac{Z}{A} \\ &= \frac{P \phi Z N}{60 A}\ \text{V} \end{aligned}$$

• Lap winding ($A = P$): $E_g = \dfrac{\phi Z N}{60}$
• Wave winding ($A = 2$): $E_g = \dfrac{P \phi Z N}{120}$

For a given machine $P$, $Z$, $A$ are fixed, so $E_g = k \phi N$ (or $E_g = k\phi\omega$), i.e. emf is proportional to flux and speed.

Example: $P = 4$, lap, $Z = 400$, $\phi = 0.02$ Wb, $N = 1500$ rpm gives $E_g = 0.02 \times 400 \times 1500 / 60 = 200$ V.