Single-phase half-wave controlled rectifier with free-wheeling diode (R–L load): for firing angle α, identify the conduction intervals of the SCR and the free-wheeling diode over one cycle.

Difficulty: Medium

Correct Answer: α < ωt < π and π < ωt < 2π

Explanation:


Introduction / Context:
This question examines current continuity and commutation in a half-wave controlled rectifier feeding an R–L load with a free-wheeling diode (FWD). Understanding when the SCR and the diode conduct is essential for predicting output voltage and current waveforms.


Given Data / Assumptions:

  • Source: single-phase sine voltage.
  • Converter: half-wave controlled rectifier (one SCR) with a free-wheeling diode across the load.
  • Load: R–L (inductive) so current tends to continue when source reverses.
  • Firing angle: α in each positive half-cycle.


Concept / Approach:
During the positive half-cycle, after firing at ωt = α, the SCR conducts until the source crosses zero at ωt = π. When the source becomes negative (ωt > π), the inductor tries to keep current flowing; the FWD provides a path, clamping load voltage near zero and preventing negative current through the SCR. Thus, the diode conducts from π to 2π (until the source becomes positive again and the next firing occurs).


Step-by-Step Solution:

SCR interval: from ωt = α (trigger) to ωt = π (natural commutation at zero crossing).FWD interval: from ωt = π to ωt = 2π (freewheeling during negative half-cycle).Therefore: SCR → α < ωt < π; FWD → π < ωt < 2π.


Verification / Alternative check:

Waveforms show load voltage equals source during SCR conduction and near zero during freewheeling; load current is continuous if L is adequate.


Why Other Options Are Wrong:

Extending beyond 2π or trimming by α in the negative half-cycle is inconsistent with freewheeling action.


Common Pitfalls:

Forgetting that the FWD takes over at π; assuming SCR continues into negative half-cycle (it does not).


Final Answer:

α < ωt < π and π < ωt < 2π

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