Single-phase semiconverter with freewheeling diode (purely resistive load) For a single-phase semiconverter (half-controlled bridge) feeding a purely resistive load and equipped with a freewheeling diode, if the firing angle is α, what are the conduction and freewheeling intervals within each half-cycle?

Introduction / Context:
Semiconverters are widely used controlled rectifiers. The presence of a freewheeling diode primarily affects inductive loads by providing a path for current continuity when the source reverses. For purely resistive loads, current is zero whenever the instantaneous source voltage is negative or the device is off.

Given Data / Assumptions:

• Single-phase semiconverter, resistive (R) load only.
• Freewheeling diode is present.
• Firing angle = α.

Concept / Approach:

With a purely resistive load, current follows voltage instantly (no energy storage). Therefore, once the input crosses zero or becomes negative, the current drops to zero and a freewheeling path is not required. The thyristor conducts only from the firing instant α to the end of that positive half-cycle at π.

Step-by-Step Solution:

Verification / Alternative check:

Waveforms confirm that current becomes zero as soon as the source crosses zero because there is no stored energy to sustain it. Hence the freewheeling diode does not conduct for a purely resistive load.

Why Other Options Are Wrong:

Options invoking a nonzero freewheeling interval (e.g., π − α and α) apply to RL loads. Extended conduction windows such as π + α are not applicable to an R-load with a semiconverter.

Common Pitfalls:

Assuming the mere presence of a freewheeling diode implies it must conduct; it only does so if energy storage (L) forces current continuity when the source reverses.

Final Answer:

π − α and 0