**1 Answer**

written 5.8 years ago by |

The Class C commutation circuit is shown in Fig.1. In this method, the main thyristor (SCR $T_1$) that is to be commutated is connected in series with the load. An additional thyristor (SCR $T_2$), called the complementary thyristor is connected in parallel with the main thyristor.

**Circuit Operation:**

**(a) Mode 0 [Initial-state of circuit] :** Initially, both the thyristors are OFF.

Therefore, the states of the devices are,

$T_{1} \longrightarrow \mathrm{OFF}, T_{2} \longrightarrow \mathrm{OFF}, \quad \therefore \quad E_{\mathrm{c}_{1}}=0$

**(b) Mode 1 :** When a triggering pulse is applied to the gate of $T_{1},$ the thyristor $T_{1}$
is triggered. Therefore, two circuit current, namely, load current $I_{L}$ and charging
current $I_{C}$ start flowing. Their paths are:

Load current $I_{L}$ ;

$$E_{\mathrm{dc+}}-R_{1}-T_{1}-E_{\mathrm{dc}-}$$

Charging current $I_{C}$;

$$E_{\mathrm{dc+}}-R_{2}-\mathrm{C}_{+}-\mathrm{C_-}-T_{1}-E_{\mathrm{dc-}}$$

Capacitor **C** will get charged by the supply votage $E_{\mathrm{dc}}$ with the polarity shown in
Fig.1 . The states of circuit components becomes

$$T_{1} \longrightarrow \mathrm{ON}, \quad T_{2} \longrightarrow \mathrm{OFF}, \quad E_{\mathrm{c1}}=E_{\mathrm{dc}}$$

**(c) Mode 2 :** When a triggering pulse is applied to the gate of $T_{2}, T_{2}$ will be
turned on. As soon as $T_{2}$ is ON, the negative polarity of the capacitor **C** is applied
to the anode of $T_{1}$ and simultaneously, the positive polarity of capacitor **C** is
applied to the cathode. This causes the reverse voltage across the main thyristor
$T_{1}$ and immediately turns it off.

Charging of capacitor **C** now takes place through the load and its polarity
becomes reverse. Therefore, charging path of capacitor **C** becomes,

$$E_{\mathrm{dc}+}-R_{1}-C_{+}-C_--T_{2(a-k)}-E_{\mathrm{dc}-}$$

Hence, at the end of Mode2 the states of the devices are

$$T_{1} \longrightarrow \mathrm{OFF}, \quad T_{2} \longrightarrow \mathrm{ON}, \quad E_{c1}=-E_{\mathrm{dc}}$$

**(d) Mode 3:** Now, when thyristor $T_{1}$ is triggered, the discharging current of
capacitor turns the complementary thyristor $T_{2}$ OFF. The state of the circuit at
the end of this Mode 3 becomes,

$T_{1} \longrightarrow \mathrm{ON}, \quad T_{2} \longrightarrow \mathrm{OFF}, \quad E_{\mathrm{c1}}=E_{\mathrm{dc}}$

Therefore, this Mode 3 operation is equivalent to Mode 1 operation.

The waveforms at the various points on the commutation circuit are shown in Fig.2. An example of this class of commutation is the well known McMurray-Bedford inverter . With the aid of certain accessories, this class is very useful at frequencies below about 1000 Hz. Sure and reliable . commutation is the other characteristic of this method.