For a long time I wanted to simulate the electric and magnetic quantities in a Shaded Pole Motor and last week I finally found 10 minutes to dedicate to this activity. I find these motors fascinating for their simplicity and for how they work by using the laws of electromagnetism.

Shaded pole motors are a variety of induction motors. They are quite the cheap type of motors: you can usually find them in old washing machines (powering the water pump for example) or even in new white goods where they may be used for periodic (but not frequent) tasks, for instance every 30s for turning a wheel. It is extremely rare to find a shaded pole motor that runs continuously. In the picture below (from Wikipedia) you can find an example of such motors.



The advantages of these kind of motors reside mainly in that they are

• Simple (and hence reliable)
• Cheap
• Easy to build

However, the main disadvantages are that

• They are quite inefficient. Indeed they waste a lot of energy in the shaded poles (auxiliary) windings which get pretty hot.
• They are a bit bulky due to the need of having an iron core.
• They generally have low starting torque.
• The power factor is generally low contributing to higher copper losses and lower overall efficiency.

Before proceeding any further, it is important to lay down the hypothesis of my analysis. The analysis is performed in FEMM on an actual shaded pole motor. The main hypothesis are those that you can find in most electromagnetics problems, i.e.

• No saturation of the iron core
• Planar geometry
• Rotor is blocked

As far as the motor rated quantities, in general, the following should be given by the motor manufacturer although it is common for some motor data to be either missing or unknown

• Rated power $P_n = 16 W$
• Rated RMS current $I_n = 0.15 A$
• Rated RMS voltage $V_n = 230 V$
• Rated speed $\Omega_n = 246 \frac{rad}{s} = 2350 RPM$
• Rated frequency $f = 50 Hz$

It might be of interest to calculate also other quantities such as

• Rated torque $C_n = \frac{P_n}{\Omega_n}=0.065 Nm = 65 g cm$
• Rated power factor $\cos(\phi_n) = \frac{P_n}{V_n I_n} = 0.46$. Pretty bad (by the way).
• Rated peak current $\sqrt{2} * 0.15 = 0.21 A$

The geometry of the motor is illustrated in the following figure. The depth is 12 mm.

By inspecting the motor, it is possible to calculate the number of turns and other geometrical quantities that may be useful in later calculations

• Number of turns in the main winding $N = 4500$
• Number of turns in each auxiliary winding $N_1 = N_2 = 5$
• Rated magnetomotive force $fmm = N I_n \approx 951 A$ (peak value)
• Area of iron cross section $A = 22.4*12=268.8 mm^2 = 2.26 * 10^{-4} m^2$
• Area of each auxiliary windings cross section $A_1 = \frac{\pi D^2}{4} = 62.3 mm^2 = 0.623 * 10^{-4} m^2$

It can be noted that $\frac{A_1}{A} \approx \frac{1}{5}$. Since $V = \omega N \phi$ the flux through the iron core can be calculated easily assuming that no saturation occurs. Indeed
$\hat \phi = \frac{V}{\omega N} = \frac{\sqrt{2} V_n}{2 \pi f N} = \frac{\sqrt{2} * 230}{2 \pi 50 * 4500}=228 mWb$
as a consequence, we have that the peak induction (B) field value is $B_{peak} = \frac{\hat \phi}{A}=1.02 T$ which is a reasonable value for peak B field in iron (still below a typical saturation value of 1.2 T).
The peak value of the emf induced in the auxiliary windings is $V_{1 peak}= \sqrt{2} N_1 \omega \hat \phi = \sqrt{2} N_1 \omega B_{peak} A_1 = V_{2 peak} = 0.1 V$.

Given the winding resistance of $R_1 = R_2 = N_1 \rho \frac{l}{A_1} = 5*(1.68*10^{-8})\frac{6.28*6.3*10^{-3}}{4*10^{-6}} = 83*10^{-5} \Omega$ we can expect an induced current of about $I_{1 peak} = I_{2 peak}=\frac{V_{1 peak}}{R_1} = 120 A$

If we assume that no saturation occurs in the iron, then given a sinusoidal voltage applied to the main windings, the current will be sinusoidal as well. Since the magnetomotive force is sinusoidal, the flux is going to be sinusoidal too. Therefore, we have

• Current in the main winding $i(t) = \sqrt{2} I_n \cos(\omega t) $
• Flux in the main iron column $\phi (t) = \hat \phi \cos(\omega t)$
• Induced emf $v_1(t) = v_2(t) = – \frac{d}{dt} \phi _1 = V_{1 peak} \sin(\omega t)$

It is possible to calculate the value of the currents in the windings at different times. In the following table this is performed for some key time instants. The obtained values can then be inserted in the FEMM software to calculate the actual flux and induction field in the motor as well as the starting torque.

$\omega t$	$i(\omega t)$	$i_2(\omega t)$
0	$i_{max}$	0
$\frac{\pi}{4}$	$\frac{\sqrt{2}}{2} i_{max}$	$\frac{\sqrt{2}}{2} i_{2 max}$
$\frac{\pi}{2}$	0	$i_{2 max}$
$\frac{3 \pi}{4}$	$- \frac{1}{\sqrt{2}} i_{max}$	$\frac{1}{\sqrt{2}} i_{2 max}$
$\pi$	$-i_{max}$	0
$\frac{5 \pi}{4}$	$- \frac{1}{\sqrt{2}} i_{max}$	$- \frac{1}{\sqrt{2}} i_{2 max}$
$\frac{3 \pi}{2}$	0	$- i_{2 max}$
$\frac{7 \pi}{4}$	$\frac{1}{\sqrt{2}} i_{max}$	$- \frac{1}{\sqrt{2}} i_{2 max}$

The results of the FEMM simulation for an entire cycle are reported below. As you can see, a rotating magnetic field is generated in the rotor area and during a cycle the field makes a complete revolution. This behaviour can be observed in the GIF I made by using the different simulations accordin to the values in the table. The arrows represent the direction of the B field. Cyan coloured regions indicate 0 T while the light purple regions are close to 1 T (sometimes higher). As it can be seen this design has a tight iron spot (below the rightmost shaded pole) where saturation is likely to occur. A good remedy is to make the motor exactly simmetric with respect to the left shaded pole.