Figure 1 shows the power-diagram of the single-phase bridge inverter. The inverter uses two pairs of controlled switches $\left(S_{1} S_{2} \text { and } S_{3} S_{4}\right)$ and two pairs of diodes $\left(D_{1} D_{2} \text { and } D_{3} D_{4}\right) .$ The devices of one pair operate simultaneously. In order to develop a positive voltage $\left(+E_{0}\right)$ across the load, switches $S_{1}$ and $S_{2}$ are turned-on simultaneously whereas to have a negative voltage $\left(-E_{0}\right)$ across the load, we need to turn-on the switches $S_{3}$ and $S_{4} .$ Diodes $D_{1}, D_{2}, D_{3}$ and $D_{4}$ are known as the feedback diodes.

Operation with Resistive Load:

Voltage and current waveforms with resistive-load are shown in Fig.2. The bridge-inverter operates in two-modes in one-cycle of the output.

(i) $\text{Mode-I} (0 \lt t \lt T / 2) :$ In this mode, switches $S_{1}$ and $S_{2}$ conducts simultaneously. The load voltage is $+E_{\mathrm{dc}}$ and load current flows from P to Q. The equivalent circuit for $\text{Mode-I}$ is shown in Fig.3(a). At $t=T / 2, S_{1}$ and $S_{2}$ are turned-off and $S_{3}$ and $S_{4}$ are turned-on.

(ii) $\text{Mode-II} (T /2 \lt t \lt T) :$ At $t=T / 2,$ switches $S_{3}$ and $S_{4}$ are turned-on and $S_{1}$ and $S_{2}$ are turned-off. The load voltage is $-E_{\mathrm{dc}}$ and load current flows from Q to P . The equivalent circuit for $\text{Mode-II}$ is shown in Fig.3(b). At $t=T, S_{3}$ and $S_{4}$ are turned-off and $S_{1}$ and $S_{2}$ are turned-on again.

As the load is resistive, it does not store any energy. Therefore, feedback diodes are not effective here.