Figure 1 shows the basic configuration of single-phase half-bridge inverter.

Switches $S_{1}$ and $S_{2}$ are the gate- commutated devices such as power BJTs, MOSFETs, GTO, IGBT, MCT, etc. When closed, these switches conducts and current flows in the direction of arrow.

Operation with Resistive Load:

The operation of the circuit can be divided into two periods:

$\begin{array}{l}{\text { (i) Period-I, where switch } S_{1} \text { is conducting from }0 \leq 1 \leq T/2 \text { and }} \\ {\text { (ii) Period-Il, where switch } S_{2} \text { is conducting from } T / 2 \leq t \leq T}\end{array}$

where $T=1 / f$ and f is the frequency of the output voltage waveform. Figure 2 shows the waveforms for the output voltage and switch currents for a resistive- load.

Switch $S_{1}$ is closed for half-time period $(T / 2)$ of the desired ac output. It connects point p of the d.c source to point A and the output voltage $e_{0}$ becomes equal to $+E_{d_{c}} / 2 .$

At $t=T / 2,$ gating signal is removed from $S_{1}$ and it turns-off. For the next half- time period $(T / 2\lt t \lt T),$ the gating signal is given to $S_{2} .$ It connects point N of the d.c source to point A and the output voltage reverses. Thus, by closing $S_{1}$ and $S_{2}$ alternately, for half-time periods, a square-wave ac voltage is obtained at the output. With resistive load, waveshape of load current is identical to that of output voltage. Simply by controlling the time periods of the gate-drive signals, the frequency can be varied. Here diodes $D_{1}$ and $D_{2}$ do not play any role. The voltage across the switch when it is OFF is $E_{\mathrm{dc}}$ . Gating circuit should be designed such that switches $S_{1}$ and $S_{2}$ should not turn-on at the same time.