Single-phase inverter with AC motor load

A single-phase full-bridge inverter with square-wave output applies $+V_s$ for half a period and $-V_s$ for the other half to the motor. An ac motor is modelled per phase as a series R–L load (back EMF neglected or lumped in R–L), so the current is exponential and lags the voltage; feedback diodes carry current when it is opposite to the voltage.

Circuit

     +-------+-----------+
     |       |           |
     |   T1 D1       T3 D3
    Vs       |           |
     |       A--[R  L]---B
     |       |  motor    |
     |   T4 D4       T2 D2
     |       |           |
     +-------+-----------+

Operation (steady state)

• $0 < t < T/2$: T1, T2 are gated, $v_0 = +V_s$. At $t = 0$ the current is still negative ($-I_0$), so it first flows through D1, D2 back to the source. When it reaches zero, T1 and T2 take over and $i_0$ rises to $+I_0$.
• $T/2 < t < T$: T3, T4 are gated, $v_0 = -V_s$. The positive current first flows through D3, D4, falls to zero, then T3, T4 conduct and $i_0$ goes to $-I_0$.

 v0  +Vs|________
        |        |
      0 |--------+--------- t
        |        |________
     -Vs           T/2     T
 i0     |      ___
     +I0|    /     \
      0 |---/-------\------- t
     -I0|__/         \___/
        D1D2|T1T2|D3D4|T3T4

Current during the positive half cycle ($0 \le t \le T/2$)

$$\begin{aligned} V_s &= Ri_0 + L\frac{di_0}{dt}, \qquad i_0(0) = -I_0 \\ i_0(t) &= \frac{V_s}{R} - \left(\frac{V_s}{R} + I_0\right)e^{-t/\tau}, \qquad \tau = \frac{L}{R} \end{aligned}$$

At $t = T/2$, $i_0 = +I_0$ (half-wave symmetry):

$$\begin{aligned} I_0 &= \frac{V_s}{R} - \left(\frac{V_s}{R} + I_0\right)e^{-T/2\tau} \\ I_0 &= \frac{V_s}{R}\cdot\frac{1 - e^{-T/2\tau}}{1 + e^{-T/2\tau}} \end{aligned}$$

The diodes D1, D2 conduct until $i_0 = 0$, i.e. for

$$t_1 = \tau\ln\!\left[\frac{2}{1+e^{-T/2\tau}}\right]$$

Current during the negative half cycle ($T/2 \le t \le T$)

$$\begin{aligned} -V_s &= Ri_0 + L\frac{di_0}{dt}, \qquad i_0(T/2) = +I_0 \\ i_0(t) &= -\frac{V_s}{R} + \left(\frac{V_s}{R} + I_0\right)e^{-(t - T/2)/\tau} \end{aligned}$$

Remarks

• Output voltage: $V_{0,rms} = V_s$; fundamental $V_{1,rms} = \dfrac{4V_s}{\sqrt2\,\pi} = 0.9V_s$.
• The harmonic currents (3rd, 5th, …) cause extra copper loss and torque pulsation in the motor; their magnitudes fall roughly as $1/n^2$ (the harmonic voltage falls as $1/n$ and the reactance $nX_L$ rises as $n$), so the motor inductance filters them.
• Feedback diodes are essential with a motor load; without them the inductive current would have no path when the conducting switches turn off, and large voltage spikes would occur.