Single-phase semiconverter with highly inductive load and freewheeling diode If the firing angle is α (radians), what is the total conduction duration of the freewheeling diode in one input cycle?

Introduction / Context:
Semiconverters (half-controlled bridges) feeding highly inductive loads commonly include a freewheeling diode (FWD) across the load. The FWD conducts whenever the supply cannot support the instantaneous polarity required by the decaying inductor current. Knowing the conduction intervals helps determine average output voltage, device currents, and ripple.

Given Data / Assumptions:

• Single-phase semiconverter (two SCRs and two diodes) with a highly inductive load.
• Freewheeling diode is connected across the load.
• Firing angle of SCRs is α (in radians).
• Continuous current is assumed due to high inductance.

Concept / Approach:
In each half-cycle, controlled devices conduct from the firing instant to the point where the source reverses insufficiently to maintain conduction; then the FWD takes over for the remainder up to the next firing instance. For a semiconverter with continuous current, the FWD typically conducts over an interval equal to α within each half-cycle.

Step-by-Step Solution:
In a positive half-cycle, the SCR is triggered at angle α and conducts until the source crosses zero.From ωt = π to ωt = π + α, the source is negative and cannot sustain the current; the FWD conducts (duration = α in this half-cycle).Similarly, in the negative half-cycle, the appropriate devices conduct from ωt = π + α to 2π, and the FWD then conducts from ωt = 2π to 2π + α (mapped to the next cycle), again for duration α.Thus, total FWD conduction per full cycle = α + α = 2α.

Verification / Alternative check:
Waveform analysis of load voltage shows clamping to near zero during freewheeling; the span equals the delay before the next device firing in each half-cycle, which is α, summing to 2α per cycle.

Why Other Options Are Wrong:
α (A): Only one half-cycle; full cycle requires doubling to 2α.
4α (C): Overcounts; there are only two freewheeling intervals per cycle.
0.5α (D): Underestimates by a factor of four for the full cycle.

Common Pitfalls:
Forgetting that semiconverters have asymmetric device conduction yet the FWD still freewheels once per half-cycle, leading to 2α total per cycle under continuous current.

Final Answer:
2 α