DFRT and equivalent circuit of single phase IM

Double field revolving theory (DFRT) states that a pulsating magnetic field of amplitude $\phi_m$ can be resolved into two rotating fields, each of amplitude $\phi_m/2$, rotating at synchronous speed $N_s$ in opposite directions.

$$\phi_m\cos\omega t \cos\theta = \frac{\phi_m}{2}\cos(\theta - \omega t) + \frac{\phi_m}{2}\cos(\theta + \omega t)$$

Torque from the two fields

• If the rotor runs at speed $N$ in the forward direction, slip w.r.t. forward field: $s_f = s = \dfrac{N_s - N}{N_s}$.
• Slip w.r.t. backward field: $s_b = \dfrac{N_s + N}{N_s} = 2 - s$.
• Forward field gives torque $T_f$, backward field gives $T_b$ (opposite). Net torque $T = T_f - T_b$.
• At standstill $s = 1$, $s_b = 1$: $T_f = T_b$, net torque zero, so the motor is not self-starting. Once rotating, $T_f > T_b$ and it runs on.

 T  |      Tf
    |     /\         Tnet = Tf - Tb
    |    /  \
 ---+---------------------> N
    |  \  /
    |   \/   Tb

Equivalent circuit

Since each rotating field is half the total, the rotor can be represented by two half-rotors: one acted on by the forward field (slip $s$) and one by the backward field (slip $2-s$). Magnetising reactance and rotor impedance are each split into halves.

 I1   R1     X1
o--/\/\--mmm--+----------+
              |  0.5Xm   |  0.5R2'/s
              +--mmm--+--+--/\/\--mmm-- (Forward)
  V           |          |  0.5X2'
              +----------+
              |  0.5Xm   |  0.5R2'/(2-s)
              +--mmm--+--+--/\/\--mmm-- (Backward)
              |          |  0.5X2'
o-------------+----------+

Forward and backward impedances:

$$\begin{aligned} Z_F &= R_F + jX_F = \frac{(j0.5X_m)\left(\frac{0.5R_2'}{s} + j0.5X_2'\right)}{\frac{0.5R_2'}{s} + j0.5(X_m + X_2')} \\ Z_B &= R_B + jX_B = \frac{(j0.5X_m)\left(\frac{0.5R_2'}{2-s} + j0.5X_2'\right)}{\frac{0.5R_2'}{2-s} + j0.5(X_m + X_2')} \end{aligned}$$

Operation through the equivalent circuit

1. Input current: $I_1 = \dfrac{V}{R_1 + jX_1 + Z_F + Z_B}$.
2. Air-gap powers: $P_{gF} = I_1^2 R_F$, $P_{gB} = I_1^2 R_B$.
3. Torques: $T_f = \dfrac{P_{gF}}{\omega_s}$, $T_b = \dfrac{P_{gB}}{\omega_s}$, net $T = \dfrac{P_{gF} - P_{gB}}{\omega_s}$.
4. Mechanical power developed: $P_m = (1-s)(P_{gF} - P_{gB})$.
5. Rotor copper loss: $s P_{gF} + (2-s) P_{gB}$.

Interpretation:

• At $s = 1$: $Z_F = Z_B$, so $P_{gF} = P_{gB}$ and $T = 0$ (no starting torque).
• At normal slip (small $s$): $0.5R_2'/s$ is large, so $Z_F \gg Z_B$. Most of the voltage appears across the forward branch, the forward field is strong and the backward field is weak, giving positive net torque.
• The backward field adds extra rotor copper loss and a double-frequency torque pulsation, so a single-phase motor has lower efficiency and is noisier than a three-phase motor.