--- title: "Extinction angle control for power factor" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:extinction-angle-control-derive" source: https://www.bctengineer.com/q/semester_7/power_electronics/extinction-angle-control-derive site: BCT Engineering --- # Extinction angle control for power factor Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2072 Chaitra, 8 marks - Video: [POWER FACTOR IMPROVEMENT: EXTINCTION ANGLE (β) CONTROL](https://www.youtube.com/watch?v=CZrHmUWnJxA) — KK CONCEPTS OF ELECTRICAL ENGINEERING - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/extinction-angle-control-derive - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question With the help of suitable circuit diagram and waveforms, explain the operation of extinction angle control for power factor improvement in rectifier circuit. Derive the expression for average value and rms value of output voltage. ## Short answer Switch turns on at zero crossing and off at π−β, so output voltage is zero for the final β interval and input current leads voltage by β/2. Average output is (Vm/π)(1+cosβ). RMS output is (Vm/√2)[(π−β+0.5 sin2β)/π]^1/2. ## Answer **Extinction angle control** is a power factor improvement method in which a forced-commutated switch is turned ON at the zero crossing of the supply voltage ($\omega t = 0$) and turned OFF at $\omega t = \pi-\beta$, where $\beta$ is the **extinction angle**. The output voltage is controlled by varying $\beta$. Because the current pulse is advanced, the fundamental input current **leads** the voltage by $\beta/2$. #### Circuit ```text P (+) +-------+-------+-------+ | | | | S1 S2 | | | | Dm Load a o----+ +--o b | (Ia) | | | | D1 D2 | | | | | | +-------+-------+-------+ N (-) S1, S2: forced-commutated switches (GTO/IGBT) Dm: freewheeling diode ``` #### Operation (highly inductive load, $i_o = I_a$ constant) 1. At $\omega t = 0$, S1 is turned ON; current path $a \to$ S1 $\to$ load $\to$ D2 $\to b$. $v_o = v_s$, $i_s = +I_a$. 2. At $\omega t = \pi-\beta$, S1 is turned OFF by its gate (forced commutation, since $v_s$ is still positive). Load current freewheels through Dm: $v_o = 0$, $i_s = 0$. 3. At $\omega t = \pi$, S2 is turned ON; path $b \to$ S2 $\to$ load $\to$ D1 $\to a$; $v_o = -v_s$, $i_s = -I_a$. 4. At $2\pi-\beta$, S2 is turned OFF and Dm conducts until $2\pi$. ```text vs ___ / \ / ---/-----\-----------/--- \___/ vo ___ ___ / | / | (sine from 0 to 180-b, --/---+-----/---+---- zero from 180-b to 180) is +Ia +------+ | | ----+ +--+ +--- | | -Ia +------+ 0 180-b 180 360-b 360 ``` #### Average output voltage $$ \begin{aligned} V_{dc} &= \frac{1}{\pi}\int_0^{\pi-\beta}V_m\sin\omega t\,d(\omega t) = \frac{V_m}{\pi}\big[-\cos\omega t\big]_0^{\pi-\beta} \\ &= \frac{V_m}{\pi}(1+\cos\beta) \end{aligned} $$ $V_{dc}$ varies from $2V_m/\pi$ ($\beta = 0$) to $0$ ($\beta = \pi$). #### RMS output voltage $$ \begin{aligned} V_{rms}^2 &= \frac{1}{\pi}\int_0^{\pi-\beta}V_m^2\sin^2\omega t\,d(\omega t) = \frac{V_m^2}{2\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_0^{\pi-\beta} \\ &= \frac{V_m^2}{2\pi}\left[\pi-\beta - \frac{\sin(2\pi-2\beta)}{2}\right] = \frac{V_m^2}{2\pi}\left[\pi-\beta + \frac{\sin2\beta}{2}\right] \end{aligned} $$ $$ V_{rms} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}\left(\pi-\beta+\frac{\sin2\beta}{2}\right)\right]^{1/2} $$ #### Power factor improvement - Fundamental input current: $I_{s1} = \frac{2\sqrt2 I_a}{\pi}\cos\frac{\beta}{2}$, **leading** $v_s$ by $\phi_1 = \beta/2$. - Displacement factor $= \cos(\beta/2)$ leading, compared with $\cos\alpha$ lagging for a phase-controlled converter. - The converter can appear as a capacitive load and can partly compensate the lagging reactive power of other loads. - Self-commutated devices are needed, which adds cost. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When is S1 turned on and off in extinction angle control? A. On at alpha, off at pi B. On at wt = 0, off at wt = pi - beta by forced commutation C. On at pi/2, off at pi D. On at wt = 0, off at pi by line commutation Answer: B. S1 starts at the zero crossing and is turned off beta before the next zero crossing. (Key point: Switching instants) 2. What is the average output voltage? A. (Vm/pi)(1 + cos(beta)) B. (2Vm/pi) cos(beta) C. (Vm/pi)(1 - cos(beta)) D. (Vm/pi) sin(beta) Answer: A. Integrating from 0 to pi - beta gives (Vm/pi)[-cos(wt)] = (Vm/pi)(1 + cos(beta)). (Key point: Average output voltage) 3. What is the fundamental input current relative to vs? A. Lagging by beta B. Leading by beta/2 C. Lagging by beta/2 D. In phase Answer: B. The pulse is advanced, so the fundamental leads by beta/2 with displacement factor cos(beta/2) leading. (Key point: Input current phase) 4. Why are GTO/IGBT type switches required? A. To block reverse voltage B. To reduce the rms voltage C. To raise the supply frequency D. To turn off while vs is still positive, which line commutation cannot do Answer: D. Turn-off at pi - beta happens while the supply is positive, so forced commutation is required; this adds cost. 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