Figure 1 shows the basic power-circuit for current commutated chopper. Here, $T_{1}$ is the main thyristor. The commutation circuit consists of auxiliary thyristor $T_{2}$ , capacitor $C,$ inductor $L,$ diodes $D_{1}$ and $D_{2} . D_{F}$ is the freewheeling diode and R is the charging resistor. The main $\mathrm{SCR} T_{1}$ is commutated by a current pulse generated in the commutation circuitry. The important feature of this type of chopper is that in at the reverse voltage across the device is applied through a diode connected in antiparallel to the SCR. Since this is limited to about one volt, the turn-off time of SCR increases in comparison with the voltage commutations.

As in voltage commutated chopper, here also the energy for current commutation comes from the energy stored in a capacitor. To start the circuit, the capacitor is charged to a voltage $E_{\text { dc }}$ through the path $E_{\mathrm{dc+}}-C-R-E_{\mathrm{dc-}}$ . The main thyristor $T_{1}$ is triggered at $t=t_{0},$ so that load voltage $e_{0}=E_{\mathrm{dc}}$ and load current $i_{0}=I_{0 m},$ up to $t=t_{i},$ as shown in Fig.2.

(i) Mode 1 Operation: Attime $t=t_{1},$ auxiliary $\mathrm{SCR} T_{2}$ is triggered to commutate the main thyristor $T_{1}$ . When thyristor $T_{2}$ is turned-on, an oscillatory current $\left(i_{c}=\frac{E_{\mathrm{dc}}}{\omega_{r} L} \sin \omega_{r} t\right)$ is set up in the circuit consisting of $C, T_{2},$ and $L .$ At $t_{2},$ the capacitor current $i_{c}$ reverses, therefore, $\mathrm{SCR}T_{2}$ gets turned-off due to natural commutation, and at $t_{2}, V_{c}=-E_{\mathrm{dc}} .$ In this mode, main thyristor $T_{1}$ remains unaffected and hence load voltage and load current remains $E_{\mathrm{dc}}$ and $I_{0 m}$ respectively.

(ii) Mode 2 Operation: Since $T_{2}$ is turned-off at $t_{2},$ oscillatory current $i_{c}$ flows through $C, L, D_{2}$ and $T_{1}$ . As shown in Fig.2, after $t_{2},$ current $i_{c}$ would flow through thyristor $T_{1}$ and not through $D_{1},$ because $D_{1}$ is reverse biased by a small voltage drop across conducting thyristor $T_{1} .$ Hence, after $t_{2}, i_{c}$ would flow through $T_{1}$ and not through $D_{1}$ . As current $i_{c}$ flows in the opposite direction in $T_{1},$ it decreases the current $i_{T_{1}} .$

At $t_{3}, i_{c}=i_{T_{1}}$ and so the net current through $T_{1}$ is zero and it turns-off. As the oscillatory current through $T_{1}$ turns it OFF, it is called as current commutated chopper. During this mode, load voltage remains $E_{\text { dc}}$ through $T_{1}$ .

(iii) Mode 3 Operation: Since $T_{1}$ is turned-off at $t_{3}, i_{c}$ becomes more than $i_{0} .$ After $t_{3}, i_{c}$ supplies load current $i_{0}$ and diode $D_{1}$ begins to conduct the current $\left(i_{c}-i_{0}\right)$ and the drop in $D_{1}$ due to this current keeps the thyristor $T_{1}$ reverse-biased for the time $t_{q}\left(=t_{4}-t_{3}\right)$

(iv) Mode 4 Operation: As shown in Fig.2 at $t_{4}, i_{c}=i_{0}$ and $i D_{1}=0$ therefore, diode $D_{1}$ is reverse-biased. After $t_{4},$ a constant current equal to $i_{0}$ flows through $E_{\mathrm{dc}}-C-L-D_{L}-$ load and therefore, capacitor C is charged linearly to source-voltage $E_{\mathrm{dc}}$ at $t_{5} .$ Therefore, during the period $\left(t_{4}-t_{5}\right), i_{c}=i_{0}$ .

As $D_{1}$ is turned-off at, $t_{4}, v_{c}=E_{t_{1}}$ . Now, the load voltage $e_{0}=E_{\mathrm{dc}}-V_{c}$ at $t_{4}$ and at $t_{5}, V_{c}=E_{\mathrm{dc}}$ . Hence, at $t_{5}$ load voltage become become $e_{0}=E_{\mathrm{dc}}-E_{\mathrm{dc}}=0 .$ During the interval $\left(t_{4}-t_{5}\right), V_{c}$ increases linearity and therefore, load voltage $e_{0}$ decreases to zero linearity, during this period.

(v) Mode 5 Operation: As shown in Fig. 2 at $t_{6},$ capacitor C is actually overcharged to a voltage somewhat greater than source voltage $E_{\mathrm{dc}} .$ Therefore at $t_{5},$ the freewheeling diode $D_{F}$ becomes forward biased and starts to conduct the load current $i_{0} .$ As $i_{c}$ is not zero at $t_{5},$ the capacitor C is still connected to load through source $E_{d c}-C-L$ and $D_{2},$ and as a result C is overcharged by the transfer of energy from L to C. Therefore, at $t_{6}, i_{c}=0$ and $V_{c}$ becomes more than $E_{\mathrm{dc}}$ .

During interval $t_{5}$ to $t_{6}, i_{0}=i_{c}+i_{F}$ and therefore, as $i_{c}$ decays, $i_{F}$ builds up. At $t_{6}, i_{F}=i_{0}$ and $i_{c}=0 .$ Commutation process is completed at $t_{6}$ and the commutation interval is $\left(t_{6}-t_{1}\right) .$

From $t_{6}$ onwards, load current freewheels through $D_{F}$ and decays. As $i_{c}$ is zero and $D_{2}$ is open circuited, capacitor voltage $V_{c}$ decays through R for the freewheeling period of the chopper.

At $t=T,$ the main thyristor $T_{1}$ is triggered again and the cycle repeats.

Advantages:

1.The capacitor always remains charged with the correct polarity.

2.Commutation is reliable as long as load current is less than the peak commutation current $I_{c p} .$

3.The auxiliary thyristor $T_{2}$ is naturally commutated as its current passes through zero value.