--- title: "AC voltage controller RL load extinction angle" subject: "Power Electronics" chapter: 6 key: "semester_7/power_electronics:acvc-rl-extinction-angle" source: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-rl-extinction-angle site: BCT Engineering --- # AC voltage controller RL load extinction angle Power Electronics · Chapter 6: AC voltage controller - Asked in: 2081 Bhadra, 8 marks - Video: [M5 PE L5 AC Voltage Controller RL Load](https://www.youtube.com/watch?v=WVxyTSLiRKg) — Babu Thomas - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-rl-extinction-angle - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-6 ## Question A single-phase AC voltage controller with 230 V input voltage is connected to RL load with R = 2 Ω and XL = 2 Ω is operated at firing angle of 45°. The Output of AC voltage controller is discontinuous. Calculate: i) Extinction Angle ii) Conduction Angle iii) RMS value of the output voltage iv) Draw its waveform ## Short answer With alpha equal to load angle 45°, extinction angle is 225°, conduction angle is 180°. RMS output voltage equals supply voltage, 230 V, because the controller gives full output at this boundary. ## Answer For an AC voltage controller with R-L load, current continues after the voltage zero until the extinction angle $\beta$, found from the load current equation. **Given:** $V_s = 230$ V, $R = 2\ \Omega$, $X_L = 2\ \Omega$, $\alpha = 45°$. $$ \begin{aligned} Z &= \sqrt{2^2 + 2^2} = 2.828\ \Omega \\ \phi &= \tan^{-1}\frac{X_L}{R} = \tan^{-1}1 = 45° \end{aligned} $$ #### Load current and extinction condition For $\alpha \le \omega t \le \beta$: $$ i(\omega t) = \frac{V_m}{Z}\left[\sin(\omega t - \phi) - \sin(\alpha - \phi)\,e^{-(\omega t - \alpha)/\tan\phi}\right] $$ At $\omega t = \beta$, $i = 0$: $$ \sin(\beta - \phi) = \sin(\alpha - \phi)\,e^{-(\beta - \alpha)/\tan\phi} $$ #### (i) Extinction angle Here $\alpha = \phi = 45°$, so $\sin(\alpha - \phi) = 0$ and the transient term vanishes: $$ \sin(\beta - 45°) = 0 \;\Rightarrow\; \beta = 180° + 45° = 225° $$ #### (ii) Conduction angle $$ \gamma = \beta - \alpha = 225° - 45° = 180° $$ #### (iii) RMS output voltage $$ \begin{aligned} V_o &= V_s\sqrt{\frac{1}{\pi}\left[\beta - \alpha + \frac{\sin 2\alpha - \sin 2\beta}{2}\right]} \\ &= 230\sqrt{\frac{1}{\pi}\left[\pi + \frac{\sin 90° - \sin 450°}{2}\right]} = 230\sqrt{1} = 230\ \text{V} \end{aligned} $$ Load current $I_o = 230/2.828 = 81.3$ A (pure sinusoid). **Note:** $\alpha = \phi$ is the boundary between discontinuous ($\alpha > \phi$) and continuous conduction. Each SCR conducts exactly 180°, the current is a pure sine wave lagging by 45°, and the controller gives full output; any $\alpha < 45°$ also gives full output and loses control. #### (iv) Waveforms ```text vs,vo /\ /\ / \ / \ -----/----\----/----\--- 0 45 \ / 225 io __ \/ / \ / -----/----\------/------ 45 225 405 T1: 45 -> 225 T2: 225 -> 405 ``` **Answer: $\beta = 225°$, $\gamma = 180°$, $V_o = 230$ V.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For R = 2 ohm and XL = 2 ohm, what is the load angle phi? A. 30 degrees B. 90 degrees C. 60 degrees D. 45 degrees Answer: D. phi = tan^-1(XL/R) = tan^-1(1) = 45 degrees. (Key point: Load angle) 2. Why is the extinction angle beta = 225 degrees when alpha = 45 degrees? A. Because alpha = phi, the transient term vanishes and sin(beta - phi) = 0, so beta = 180 + 45 B. Because beta = alpha + 90 degrees always C. Because beta = 2 alpha + 135 degrees D. Because the load is resistive Answer: A. With alpha = phi, sin(alpha - phi) = 0 so the current is a pure sine and ends at beta = 180 + phi = 225 degrees. (Key point: Extinction angle) 3. What is the conduction angle gamma? A. 225 degrees B. 270 degrees C. 45 degrees D. 180 degrees Answer: D. gamma = beta - alpha = 225 - 45 = 180 degrees. (Key point: Conduction angle) 4. What is the significance of alpha = phi? A. It gives zero output voltage B. It is the boundary between discontinuous and continuous conduction, giving full output (230 V) C. It makes the current DC D. It requires forced commutation Answer: B. Each SCR conducts 180 degrees, so Vo = Vs = 230 V; any alpha below phi also gives full output and loses control. 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