Class D- Auxiliary Commutation
1 Answer

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.

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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.

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