Some of the limitations of basic series inverter circuit of can be overcome by making some modifications in the basic form. One such modification is to replace the normal inductance by a mutually-coupled inductance as shown in Fig.1. Inductors $L_{1}$ and $L_{2}$ have the same inductance and are closely coupled. It can be seen that even if SCR $T_{2}$ is triggered a little before SCR $T_{1}$ is turned-off, there will not be any possibility of short circuit at the d.c input source.

Let us suppose that SCR $T_{2}$ is triggered shortly before $\mathrm{SCR} T_{1}$ is turned-off. At the instant when $\mathrm{SCR} T_{2}$ is triggered, the voltage across the capacitor will be slightly less than $\left(E_{c}+E_{\mathrm{dc}}\right)$ and the load voltage and current will be closed to zero. Hence, a voltage equal to the voltage across the capacitor minus the load voltage will appear across $L_{2}$ . Since $L_{1}$ is closely coupled to $L_{2},$ the same voltage will appear across $L_{1}$ . The voltage across $L_{1}$ will tend to increase the cathode potential of $\mathrm{SCR} T_{1}$ more than its anode potential and therefore, $\mathrm{SCR} T_{1}$ will be reverse-biased and turn-off. Thus, even if $\mathrm{SCR} T_{2}$ is turned-on before $\mathrm{SCR} T_{1}$ is switched-off, it will not result into short circuiting of the d.c. source. A similar operation will take place if $\mathrm{SCR} T_{1}$ is triggered before $\mathrm{SCR} T_{2}$ is turned-off.

Hence, we can increase the limit of output frequency to more than the resonance frequency of the $R-L-C$ resonant circuit.

Further, the drawback of high pulsed current from the d.c. supply can be overcome in a half bridge configuration as shown in Fig.2 where $L_{1}=L_{2}$ and $C_{1}=C_{2}$ . The power is drawn from the d.c. source during both half-cycles of output voltage. One-half of the load current is supplied by capacitor $C_{1}$ or $C_{2}$ and the other-half by the d.c. source.

Initially, capacitor $C_{2}$ is assumed to be charged to voltage $E_{c}$ with upper plate negative and lower plate positive. As the capacitors $C_{1}$ and $C_{2}$ together are connected across the battery, the total voltage across $C_{1}$ and $C_{2}$ should be equal to $E_{\mathrm{dc}}$ . Therefore, $C_{1}$ will be charged to $\left(E_{\mathrm{dc}}\right.$ $+E_{c} )$ value with upper plate positive. The various voltage and current waveforms of this circuit are shown in Fig.3.

As shown in Fig.3 , thyristor $T_{1}$ is triggered at instant $^{\prime} 0^{\prime}$ . With this, the two currents flow through the thyristor $T_{1}$ and load $R .$ Current $i_{1}$ flows through the path $E_{\mathrm{dc+}}-$ $T_{1}-L_{1}-R-C_{2+}-C_{2-}-E_{\mathrm{dc-}}$ thus charging $C_{2}$ . Capacitor $C_{1}$ which is already charged at this instant will provide the second current $i_{2} .$ This discharge current $i_{2}$ has the path $C_{1+}-T_{1}-L_{1}-R-$ $C_{1-}$.

Since the driving voltage $(E_{dc}+E_c)$ the capacitors $C_{1}, C_{2}$ and the initial conditions are identical for both these paths, the two currents $i_{1}$ and $i_{2}$ will always be equal. Hence, 50$\%$ of the load current is drawn from the input source and 50$\%$ from the discharge of the capacitor.

At the end of the half-cycle , when the load current becomes zero, $SCR T_{1}$ will be turned-off and the voltage across the capacitors reversed. In the steady state, capacitor $C_{2}$ will be charged to voltage $\left(E_{\mathrm{dc}}+E_{c}\right)$ in the opposite direction and capacitor $C_1$ to $E_c$. When $SCRT_2$ is triggered at instant Q,identical operation will takes place in the following negative half cycle.

Again, 50$\%$ of the load current is obtained from the d.c. input source and the rest 50$\%$ from the discharge of capacitor $C_{2}$ . Hence, the input d.c supply no more remains intermittent in nature and the ripples are reduced to the minimum.