Choice of number of poles in DC machine

The number of poles of a dc machine is chosen after the output equation gives $D^2L$. More poles reduce some material but increase other costs and losses, so a compromise is made.

1. Frequency of flux reversal

$$f = \frac{Pn}{2}$$

The armature iron sees alternating flux at this frequency. Iron loss rises with $f$, so $f$ is normally kept between 25 and 50 Hz. This limits the number of poles at high speed.

2. Weight of iron

For the same total flux, the flux per pole $\Phi \propto 1/P$. So the yoke and armature core depth (which carry $\Phi/2$) become thinner, and the iron weight falls as $P$ increases.

3. Weight of copper

• Armature copper: the end connections are shorter for a smaller pole pitch, so armature copper decreases with more poles.
• Field copper: each pole needs its own coil. The total field copper stays roughly constant or falls a little.

4. Armature reaction (ampere-turns per pole)

$$AT_a = \frac{ac\,\pi D}{2P}$$

More poles lower the armature mmf per pole, so the distortion and the field mmf needed to overcome armature reaction are less. A typical limit is armature AT per pole below about 5000–10000, depending on output.

5. Current per brush arm / per path

For a lap winding, the current per path $= I_a/P$. It should not exceed about 200 A (and about 400 A per brush arm), otherwise the commutator becomes long and commutation becomes difficult. Large-current machines therefore need more poles.

6. Commutator length and brushes

More poles mean more brush arms with a smaller current in each, so the commutator is shorter, but there is more brush gear.

7. Overall size, labour and cost

More poles give a smaller overall diameter and less material, but they need more coils, more parts and more labour. Frequency and losses also rise.

8. Voltage between commutator segments (flashover)

For a lap winding, the terminal voltage $V$ is spread over $N_{seg}/P$ segments in series, so the average voltage between adjacent segments is $e_s = VP/N_{seg}$. This must be kept below about 20 V to avoid flashover. For the same terminal voltage, more poles therefore need more commutator segments, which makes the commutator larger and more costly.

Typical choice

Example: a 37 kW, 1400 rpm motor with 4 poles has $f = 4 \times 1400/120 = 46.7$ Hz and a current per path of about 45 A, so 4 poles is suitable. Six poles would give 70 Hz, which is too high.