**1 Answer**

written 5.8 years ago by |

Figure 1 shows the typical Class *D* commutation circuit. In this commutation method, an auxiliary thyristor $\left(T_{2}\right)$ is required to
commutate the main thyristor $\left(T_{1}\right),$ Assuming ideal thyristors and the lossless components, then the waveforms are as in Fig.2. Here, inductor *L* is necessary
to ensure the correct polarity on capacitor *C*.

Thyristor $T_{1}$ and load resistance $R_{L}$ form the power circuit, whereas $L, D$ and $T_{2}$ form the commutation circuit.

**Circuit operations:**

**(a) Mode 0 [Initial operation] :** When the battery $E_{\mathrm{dc}}$ is connected, no current
flows as both thyristors are OFF. Hence, initially, the state of the circuit components becomes,

$$T_{1} \longrightarrow \mathrm{OFF}, \quad T_{2} \longrightarrow \mathrm{OFF}, \quad E_{C}=0$$

**(b) Mode 1 :** Initially, SCR $T_{2}$ must be
triggered first in order to charge the
capacitor *C* with the polarity shown. This
capacitor *C* has the charging path $E_{\mathrm{dc}+}-C_{+}-C_{-}-T_{2}-R_{L}-E_{\mathrm{dc-}} .$ As soon as capacitor *C* is fully charged, SCR
$T_{2}$ turns-off. This is due to the fact that,
as the voltage across the capacitor
increases, the current through the
thyristor $T_{2}$ decreases since capacitor *C*
and thyristor $T_{2}$ form the series circuit.

Hence the state of circuit components at the end of Mode 1 becomes,

$T_{1} \longrightarrow \mathrm{OFF}, T_{2} \longrightarrow \mathrm{OFF}, E_{C}=E_{\mathrm{dc}}$

**(c) Mode 2:** When thyristor $T_{1}$ is
triggered, the current flows in two paths:

(a) Load current $I_{L}$ flows through $E_{\mathrm{dc+}}-T_{1}-R_{L}-E_{\mathrm{dc}-}$

(b) Commutation current (Capacitor-discharges through) flows through $C_{+}-T_{1}-L-D- C_{-}$

After the capacitor **C** has completely discharged, its polarity will be reversed,
i.e., its upper plate will acquire negative charge and the lower plate will acquire,
positive charge. Reverse discharge of capacitor *C* will not be possible due to the
blocking diode *D*.

Therefore, at the end of Mode 2 the state of the circuit components becomes,

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

**(d) Mode 3 :** When the thyristor $T_{2}$ is triggered, capacitor *C* starts discharging
through the path $C_{+}-T_{2_{\{A-K)}}-T_{1_{(k-A)}}-C_{-}$ . When this discharging current
becomes more than the load current $I_{L},$ thyristor $T_{1}$ gets OFF.

Therefore, at the end of Mode $3,$ the state of circuit component becomes,

$$T_{1} \longrightarrow \mathrm{OFF}, \quad T_{2} \longrightarrow \mathrm{ON}$$

Again, capacitor *C* will charge to the supply voltage with the polarity shown
and hence SCR $T_{2}$ gets OFF. Therefore, thyristors $T_{1}$ and $T_{2}$ both get OFF,
which is equivalent to Mode 0 operation.