Operation with RL Load:

Voltage and current waveforms for single-phase bridge inverter with RL load are shown in Figure 2 The operation of the circuit is explained in four-modes.

(i) $\text{Mode-I} \left(t_{1} \lt t \lt t_{2}\right) :$ At instant $t_{1},$ the switch $S_{1}$ and $S_{2}$ are turned-on.Switches are assumed to be ideal switches. Point P gets connected to positive point of d.c. Source $E_{\mathrm{dc}}$ and point Q gets connected to negative point of input supply. The output voltage, $e_{0}=+E_{\mathrm{dc}},$ Fig.3(a). The load current starts increasing exponentially due to the inductive nature of the load. The instantaneous current through $S_{1}$ and $S_{2}$ is equal to the instantaneous load current. During this interval, energy is stored in inductive load.

(ii) $\text{Mode-II} \left(t_{2} \lt t \lt t_{3}\right) :$ Both the switches $Q_{1}$ and $Q_{2}$ are turned-off at instant $t_{2} .$ Due to the inductive nature of the load, the load current does not reduce to zero instantaneously. There is a self-induced voltage across the load which maintains the flow of current in the same-direction. The polarity of this voltage is exactly opposite to that in mode-1, The output voltage becomes $-E_{\mathrm{dc}}$ but the load current continues to flow in the same direction, through $D_{3}$ and $D_{4}$ as shown in Fig.3(b). Thus, in this mode, the stored energy in the load inductance is returned back to the source. Load current decreases exponentially and goes to 0 at instant $t_{3}$ when all the energy stored in the load is returned back to supply. $D_{3}$ and $D_{4}$ are turned-off at $t_{3}$

(iii) $\text{Mode-III} \left(t_{3} \lt t \lt t_{4}\right) :$ Switches $S_{3}$ and $S_{4}$ are turned-on simultaneously at instant $t_{3}$ . Load voltage remains negative $\left(-E_{\mathrm{dc}}\right)$ but the direction of load current will reverse. The current increases exponentially in the other direction and the load again stores the energy.

(iv) $\text{Mode-IV} \left(t_{1} \lt t \lt t_{2}\right) :$ Switches $S_{3}$ and $S_{4}$ are turned-off at instant $t_{0}$ (or $t_4$).The load inductance tries to maintain the load current in the same direction by inducing the positive-load voltage. This will forward-bias the diodes $D_{1}$ and $D_{2}$. The load energy is returned back to the input d.c supply. The load voltage becomes $e_{0}=+E_{d c}$ but the load current remains negative and decreases exponentially towards $0 .$ At $t_{1}\left(\text { or } t_{5}\right),$ the load current goes to zero and switches $S_{1}$ and $S_{2}$ can be turned-on again. The conduction period with a very highly inductive load, will be $T / 4$ or $90^{\circ}$ for all the switches as well as the diodes.The conduction period of switches will increase towards $T / 2$ or $180^{\circ}$ with increase in the load power- factor.