--- title: "SCR half-wave rectifier numerical (200 V peak)" subject: "Advanced Electronics" chapter: 7 key: "semester_4/advanced_electronics:scr-hwr-1ma-200v" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/scr-hwr-1ma-200v site: BCT Engineering --- # SCR half-wave rectifier numerical (200 V peak) Advanced Electronics · Chapter 7: Introduction to Power Electronics - Asked in: 2074 Asoj, 2+2+2+2 marks - Video: [Problems on Single Phase Half Wave Controlled Rectifiers | Power Electronics | Lecture 42](https://www.youtube.com/watch?v=LqCPk-1m2UA) — Electronics Maddy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/scr-hwr-1ma-200v - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-7 ## Question A half wave rectifier circuit employing an SCR is adjusted to have a gate current of 1 mA. The forward breakdown voltage of SCR is 100V for Ig = 1mA. If sinusoidal voltage of 200V peak is applied, find: i) Firing angle ii) Conduction angle iii) Average current, assume load resistance = 100Ω iv) In this circuit if holding current IH = 200 mA, find average current in this case. ## Short answer Firing angle α = arcsin(100/200) = 30°, conduction angle = 150°. Average load current with no holding current is (200/(2π×100))(1+cos30°) = 0.594 A. With holding current 200 mA, off angle 174.26°, conduction 144.26°, average current ≈ 0.592 A. ## Answer Given: Vm = 200 V, forward breakover voltage VBO = 100 V at IG = 1 mA, RL = 100 Ω. The SCR fires when the instantaneous supply reaches VBO: Vm sin α = VBO. #### i) Firing angle ```text sin α = VBO/Vm = 100/200 = 0.5 α = 30° ``` #### ii) Conduction angle With zero holding current, the SCR conducts from α to 180°: ```text θc = 180° − α = 180° − 30° = 150° ``` #### iii) Average current (RL = 100 Ω) ```text Iav = (1/2π)∫α^π (Vm/RL) sin θ dθ = (Vm / 2πRL)(1 + cos α) = (200 / (2π × 100))(1 + cos 30°) = 0.3183 × 1.8660 Iav = 0.594 A ``` #### iv) Average current with IH = 200 mA Now the SCR turns OFF when the load current falls to IH before the end of the half cycle: ```text i = (Vm/RL) sin θ = 2 sin θ A 2 sin θ = 0.2 → sin θ = 0.1 θ_off = 180° − 5.74° = 174.26° ``` (At turn-on, i = 2 sin 30° = 1 A > IH, so the SCR latches normally.) ```text Iav = (1/2π)∫α^θoff 2 sin θ dθ = (2/2π)[cos 30° − cos 174.26°] = (1/π)(0.8660 + 0.9950) Iav = 0.592 A ``` New conduction angle = 174.26° − 30° = 144.26°. **Answer: α = 30°, conduction angle = 150°, Iav = 0.594 A (IH = 0) and Iav ≈ 0.592 A (IH = 200 mA, conduction 144.26°)** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When does the SCR fire in this circuit? A. At the peak of the supply B. When Vm sin α reaches the breakover voltage 100 V for IG = 1 mA C. When the supply crosses zero D. When the load current reaches IH Answer: B. Firing occurs when instantaneous supply equals VBO: 200 sin α = 100. (Key point: Firing condition) 2. What are the firing and conduction angles? A. α = 60°, conduction = 120° B. α = 30°, conduction = 30° C. α = 30°, conduction = 150° D. α = 150°, conduction = 30° Answer: C. sin α = 0.5 gives 30°; conduction runs from 30° to 180°, i.e. 150°. (Key point: Firing angle) 3. Which expression gives the average current? A. Iav = (Vm/(πRL))(1 − cos α) B. Iav = Vm/RL = 2 A C. Iav = (Vm/(2πRL))(1 − cos α) D. Iav = (Vm/(2πRL))(1 + cos α) = 0.594 A Answer: D. Integrating from α to π gives (1 + cos α), so Iav = 0.3183 × 1.866 = 0.594 A. (Key point: Average current formula) 4. With IH = 200 mA, why does the average current change only slightly? A. The SCR turns off at sin θ = 0.1 (174.26°), just before 180° B. The SCR never turns on C. The firing angle moves to 90° D. Conduction ends at the supply peak Answer: A. 2 sin θ = 0.2 gives θoff = 174.26°, so conduction shortens by only about 5.7° and Iav ≈ 0.592 A. 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