Figure 1 shows the parallel inverter circuit. Two diodes are connected as shown in Fig.1 to feedback the stored load energy during those periods when the load current reverses relative to the voltage. Figure 2 shows the load voltage and current waveform of parallel inverter.

Thyristors $T_{1}$ and $T_{2}$ are the main load carrying thyristors. Inductor L and capacitor C are the commutating components. Diodes $D_{1}$ and $D_{2}$ are the feedback diodes, which permit the load reactive power to be fed back to the d.c. supply. The circuit operation can be divided into different operating modes.

Mode $1 :$ During this mode, thyristor $T_{1}$ is triggered at instant $R,$ as shown in Fig.2 . Battery voltage now forces the current to the primary section $C A$ through the path $E_{dc+}-C-A-T_{1}-L-E_{\text { dc- }} .$ Thus, neglecting the small voltage drop across $L,$ the supply voltage $E_{\mathrm{dc}}$ will appear across the left-half of the transformer primary winding $C A .$ Terminal C is positive with respect to terminal $A.$ The flux produced due to this current induces the voltage in all sections of transformer winding as number of turns in section $C A$ are assumed to be equal to number of turn in secondary. The load voltage is nearly equal to $E_{\mathrm{dc}}$ and is in such a direction so as to force a current into the dot at terminal $P.$ This is because, the current in primary leaves the dot at terminal $A.$

Now, due to autotransformer action, voltage $E_{\mathrm{dc}}$ is induced in $C E$ section of primary winding. Therefore, terminal E will be at a potential of $2E_{\mathrm{dc}}$ with respect to $A .$ Thus, capacitor C will get charged to twice the supply voltage with the polarity as shown in Fig.1. Since the load is inductive, load current increases gradually from instant $R,$ as shown in Fig.2.

Mode 2 : This mode begins when thyristor $T_{2}$ is switched ON at instant $S,$ as shown in Fig.2. When $T_{2}$ is turned-on, capacitor C will immediately apply a reverse voltage of $2E_{dc}$ across $SCR T_1$ and turns it OFF. When SCR $T_{1}$ is turned-off, the capacitor will discharge through $\mathrm{SCR} T_{2},$ inductor $L,$ diode $D_{1},$ and a portion of a transformer primary winding $B A$ . Thus, the energy stored in the capacitor will be fed back to the load through the transformer coupling of windings $B A$ and $P Q .$ During this period, the potential of point B will be fixed by the d.c. input supply and the load voltage will still be positive but more than $E_{\mathrm{dc}}$ .

The load current, which earlier was flowing through $\operatorname{SCR} T_{1}$ , will now flow through $C B$ and diode $D_{1}$ to the negative input terminal. This can happen only if diode $D_{1}$ is forward biased and the capacitor discharge current is more than the load current. As the potential of point B increases sufficiently to reverse bias diode $D_{1}$ , the capacitor will no longer discharge through $\mathrm{SCR} T_{2}$ and point B will not get connected to the supply negative terminal. The current through inductor in L will now flow through diode $D_{2}, D E$ and $\mathrm{SCR} T_{2},$ and the trapped energy in inductor L will be fed back to the load.

The load current $T_{1}$ , which earlier was flowing through $C B,$ will now flow from D to C through diode $D_{2},$ and the load reactive energy will be returned to the d.c. supply. Since point D is now connected to the negative supply terminal, the load voltage polarity will be reversed and more than $E_{\text { dc }}$ .Also, the capacitor C will be charged in the opposite direction to slightly more than twice the supply voltage. Thyristor $T_{2}$ will stop conducting after all the energy in the commutating inductor L has been completely dissipated in the load. Immediately following the commutation of $\mathrm{SCR} T_{1},$ energy is transferred from the capacitor and inductor to the load. During this period, high frequency oscillations will be superimposed on normal rectangular waveform of the load voltage. After this transient period, only diode $D_{2}$ will continue to conduct. This will cause application of a reverse voltage across $\mathrm{SCR} T_{2},$ and thereby help in turning it OFF.

Mode 3: This mode begins when the load current becomes zero, diode $D_{2}$ will be blocked and $\mathrm{SCR} T_{2}$ will have to be triggered again at instant U to reverse the direction of the load current. When thyristor $T_{2}$ starts conducting, the load voltage will again become equal to $E_{\mathrm{dc}} .$

In Fig.2 , the transient waveforms during the commutation period are shown by the dashed lines. In Fig.2 $ \mathrm{SCR} T_{1}$ will conduct during the period $R S,$ when both load voltage and load currentare positive. At $S, \mathrm{SCR} T_{2}$ is triggered to commutate $\mathrm{SCR} T_{1}$ . After the commutation transient, the load voltage will be reversed and the current will continue to flow through diode $D_{2}$ in the same direction as before. SCR $T_{2}$ will be turned-off because of the reverse bias applied by $D_{2}$ . At $U,$ the load current will become zero when all the inductive energy is dissipated and SCR $T_{2}$ will be triggered again. During the period $U W,$ both load voltage and load current are reversed. At $W, \mathrm{SCR} T_{1}$ is triggered to turn-off $\mathrm{SCR} T_{2}$.