Minimum width of stator teeth

The minimum width of a stator tooth is the narrowest tooth that can carry the flux without its flux density exceeding about 1.7 Wb/m². Above this value the teeth saturate, and the magnetising current and iron loss rise sharply.

Derivation

• Flux per pole: $\Phi = B_{av}\tau L$.
• This flux passes through the teeth under one pole, and the number of teeth is $S_s/P$.
• Iron area of these teeth $= \dfrac{S_s}{P} \times W_{ts} \times L_i$, where $L_i = k_i(L - n_dw_d)$ is the net iron length ($k_i \approx 0.9$).
• Setting $B_t \le 1.7$ Wb/m²:

$$\frac{\Phi}{\frac{S_s}{P}W_{ts}L_i} \le 1.7 \;\Rightarrow\; W_{ts(min)} = \frac{\Phi}{1.7 \times \frac{S_s}{P} \times L_i}$$

Example

$\Phi = 4.51 \times 10^{-3}$ Wb, $S_s = 24$, $P = 4$, $L = 0.124$ m, $L_i = 0.9 \times 0.124 = 0.1116$ m:

$$W_{ts(min)} = \frac{4.51\times10^{-3}}{1.7 \times 6 \times 0.1116} = 3.96\ \text{mm}$$

The actual tooth width is the slot pitch minus the slot width. Here the slot pitch is $13.78$ mm and the slot width is $6.75$ mm, so the tooth is $13.78 - 6.75 = 7.0$ mm wide. This is larger than the minimum, so the design is safe.