Half-bridge square-wave inverter at 120 Hz

In a half-bridge square-wave inverter, S1 and S2 are switched alternately, each for exactly half of the output period, so the load sees $+V_s/2$ and $-V_s/2$ in turn.

ON and OFF periods of S1 and S2

$$\begin{aligned} T &= \frac{1}{f} = \frac{1}{120} = 8.333\ \text{ms} \\ T_{ON} &= T_{OFF} = \frac{T}{2} = 4.167\ \text{ms} \end{aligned}$$

• S1: ON for $0 < t < 4.167$ ms, OFF for $4.167 < t < 8.333$ ms.
• S2: OFF for $0 < t < 4.167$ ms, ON for $4.167 < t < 8.333$ ms.

(In practice a small dead time is left between turning one switch off and the other on, to avoid shoot-through.)

 S1 gate  ____        ____
         |    |      |    |
 ________|    |______|    |___
 S2 gate       ______      ___
 _____________|      |____|
         0   T/2    T
 vo  +Vs/2 ____        ____
          |    |      |    |
 -Vs/2    |    |______|    |__

Answer: ON period = OFF period = 4.167 ms for each switch.

Load current in the first positive half cycle

In steady state the current at $t = 0$ is negative, $i(0) = -I_0$ (it flows through D1 until it reverses, then through S1). For $0 \le t \le T/2$, $v_o = +V_s/2$ and the R-L load gives

$$L\frac{di}{dt} + Ri = \frac{V_s}{2}$$

The solution is the steady-state part plus a decaying part, with $\tau = L/R$:

$$i(t) = \frac{V_s}{2R} + A\,e^{-t/\tau}$$

Using $i(0) = -I_0$: $A = -I_0 - \dfrac{V_s}{2R}$, so

$$i(t) = \frac{V_s}{2R}\left(1 - e^{-t/\tau}\right) - I_0\,e^{-t/\tau}, \qquad 0 \le t \le \frac{T}{2}$$

At $t = T/2$ the current reaches $+I_0$ (half-wave symmetry). Putting this in:

$$I_0 = \frac{V_s}{2R}\left(1 - e^{-T/2\tau}\right) - I_0 e^{-T/2\tau} \;\Rightarrow\; I_0 = \frac{V_s}{2R}\cdot\frac{1 - e^{-T/2\tau}}{1 + e^{-T/2\tau}}$$

With $T/2 = 4.167$ ms:

$$i(t) = \frac{V_s}{2R} - \frac{V_s}{2R}\cdot\frac{2}{1 + e^{-0.004167/\tau}}\,e^{-t/\tau}, \qquad 0 \le t \le 4.167\ \text{ms}$$

Conduction sequence in this half cycle: from $t = 0$ the current is negative, so D1 conducts and energy returns to the upper source. At $t_1 = \tau \ln\!\left(1 + \dfrac{2R I_0}{V_s}\right)$ the current crosses zero, and S1 then carries the positive current until $T/2$. In the next half cycle D2 and then S2 conduct in the same way with the signs reversed.