Figure 1 shows the basic power circuit diagram of voltage commutated chopper. This commutation circuit comprises an auxiliary $\operatorname{SCR} T_{2},$ a diode $D,$ inductor $L,$ and capacitor $C ;$ the complete chopper circuitry has been outlined with a dotted box. The main power switch is $\mathrm{SCR} T_{1}$ .

To start the circuit, capacitor C is initially charged with the polarity shown, by triggering the auxiliary $\mathrm{SCR} T_{2}$ . Capacitor C gets charged through the path $E_{\mathrm{dc}+}-C_{+}-C_{-}-T_{2}-L_{1}$ load $-E_{\mathrm{dc}-}$ As the charging current decays to zero, thyristor $T_{2}$ will be turned OFF. Figure 2 shows the associated voltage and current waveforms. For convenience, the chopper operation is divided into certain modes and is explained as under:

(i) Mode 1 Operation $\left(0\lt t \lt t_{2}\right) :$ The main $\mathrm{SCR} T_{1}$ is triggered at $t_{0}$ . Source current flows in two paths, load current $i_{o}$ constitutes one path and commutation current $i_{c},$ the other path. With the triggering of SCR $T_{1},$ load gets connected to the supply and the load voltage, $e_{0}=E_{\mathrm{dc}}$ .

Load current $i_{0}$ flows through the path, $E_{\mathrm{dc}+}-T_{1}-$ load $-E_{\mathrm{dc}-}$ , whereas the commutating current $i_{c}$ flows through the path, $C_{+}-T_{1}-L-D-C_{-}$ . The capacitor current first rises from zero to a maximum value when voltage across capacitor C is zero at $t=t_{1} / 2 .$ As $i_c$ , decreases to zero, capacitor is charged to a reverse voltage $\left(-E_{\mathrm{dc}}\right)$ at $t=t_{1}$ as shown in Fig.2. This reverse voltage on capacitor is held constant by diode D.

At $t=0,$ voltage across auxiliary thyristor $T_{2}$ is $\left(-E_{\mathrm{dc}}\right),$ whereas it is zero at $t_{1} / 2$ and $E_{\mathrm{dc}}$ at $t_{1} .$ This voltage variation is shown as cosine wave in Fig.2. Therefore, at $t=t_{1}, i_{T_{1}}=I_{0}, V_{c}=-E_{\mathrm{dc}}, E_{T_{2}}=E_{\mathrm{dc}}, e_{0}=E_{\mathrm{dc}},$ as shown in Fig.2. These conditions continue upto the period $t_{2}$ .

(ii) Mode 2 Operation $\left(t_{2} \lt t\lt t_{3}\right) :$ At a desired instant $t_{2},$ the auxiliary SCR $T_{2}$ is to be triggered for turning-off the main $\mathrm{SCR} T_{1} .$ With the turning-on of $T_{2}$ reverse capacitor voltage $\left(-E_{\mathrm{dc}}\right)$ appears across $T_{1},$ which reverse-biases it, and turns it OFF. Since the capacititor voltage commutates the main $\mathrm{SCR} T_{1},$ the given circuit of Fig.1 is called as voltage commutated chopper. Current $i_{T_{1}}$ becomes zero at $t_{2}$ .

After the SCR $T_{1}$ is turned-off, the capacitor C and $\operatorname{SCR} T_{2}$ provide the path for load current $i_{0}$ through $E_{\mathrm{dc}}-C-T_{2}-$ load. The load voltage is the sum of source voltage and the voltage across the capacitor. Hence, at instant $t_{2},$ load voltage is $e_{0}=E_{\mathrm{dc}}+E_{\mathrm{dc}}=2 E_{\mathrm{dc}},$ and it decreases linearly as the voltage across the capacitor decreases. During this mode, $V_{c}=E_{T_{1}},$ since the capacitor is directly connected across $T_{1}$ through $T_{2}$ . As the capacitor discharges through the load, $V_{c}$ and $E_{T_{1}}$ change from $\left(-E_{\mathrm{dc}}\right)$ to zero at $\left(t_{2}+t_{q}\right)$ Load-voltage $e_{0}$ changes from 2$E_{\text { dc }}$ at $t_{2}$ to $E_{\text { dc }}$ at $\left(t_{2}+t_{q}\right) .$ After $\left(t_{2}+t_{q}\right), V_{c}$ and $E_{T_{1}}$ start rising from zero towards $E_{\mathrm{dc}}$,whereas $e_{0}$ starts falling towards zero. Hence, in this mode, $V_{c}$ and $E_{T_{1}}$ change linearly from $\left(-E_{\mathrm{dc}}\right)$ at $t_{2}$ to $E_{\mathrm{dc}}$ at $t_{3},$ since load current $i_{0}$ is assumed constant. Similarly, $e_{0}$ changes linearity from 2$E_{\mathrm{dc}}$ at $t_{2}$ to zero at $t_{3}$

(ii) Mode 3 Operation $\left(t_{3} \lt t\lt t_{4}\right) :$ For this mode, $t_{3} \lt t \lt T$. At $t_{3}$,$V_{c}=E_{T 1}=E_{dc}, e_{0}=0$ and capacitor current decays to zero, therefore, SCR $T_{2}$ turns-off naturally. At $t_{3},$ since capacitor is slightly over- charged, freewheeling diode gets forward-biased. The load current after $t_{3}$ freewheels through this diode $D_{F}$ . Note that during freewheeling period from $t_{3}$ to $T, E_{T}$ is slightly negative as capacitor C is somewhat overcharged.

Thus, during this mode, $i_{c}=0, i_{T_{1}}=0, i_{f}=I_{0 m}, E_{T_{1}}=E_{\mathrm{dc}}$, $e_{0}=0, i_{T_{2}}=0$

Now, at $t=T,$ the main thyristor $T_{1}$ is triggered again and the cycle as described from $t=0$ to $t=T$ repeats.

Disadvantage:

1.A starting circuit is required, and the starting circuit should be such that it triggers auxiliary SCR $T_{2}$ first.

2.Load voltage jumps to twice the supply voltage when the commutation is initiated.

3.The circuit imposed turn-off time is load dependent. At very low load currents, the capacitor takes longer time to discharge, thus limiting the frequency of the chopper.

4.The main thyristor $T_{1}$ has to carry the load current as well as the commutation current.