--- title: "Extinction angle control: current leads by β/2" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:extinction-angle-fundamental-leads" source: https://www.bctengineer.com/q/semester_7/power_electronics/extinction-angle-fundamental-leads site: BCT Engineering --- # Extinction angle control: current leads by β/2 Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2070 Chaitra, 8 marks - Video: [Power Factor Improvement Techniques | AC to DC Converter | Controlled Rectifier in Hindi](https://www.youtube.com/watch?v=-LqL-OTBIwY) — ElectroTech CC - 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-fundamental-leads - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Show that the fundamental component of input current leads input voltage by a phase angle of β/2 in case of extinction angle control method of power factor improvement assuming highly inductive load and β be the extinction angle. ## Short answer In extinction angle control with highly inductive load, the switch turns on at 0 and off at π–β each half cycle. The fundamental Fourier coefficients are a₁ = (2Ia/π) sinβ, b₁ = (2Ia/π)(1+cosβ). The phase angle φ₁ = arctan(a₁/b₁) = β/2, so the fundamental current leads the supply voltage by β/2, realizing leading displacement factor. ## Answer In **extinction angle control**, the forced-commutated switch of a single-phase converter is turned ON at $\omega t = 0$ and turned OFF at $\omega t = \pi-\beta$ ($\beta$ = extinction angle); the load current freewheels from $\pi-\beta$ to $\pi$. With a highly inductive load ($i_o = I_a$ constant) the input current is: $$ i_s(\omega t) = \begin{cases} +I_a, & 0 < \omega t < \pi-\beta \\ 0, & \pi-\beta < \omega t < \pi \\ -I_a, & \pi < \omega t < 2\pi-\beta \\ 0, & 2\pi-\beta < \omega t < 2\pi \end{cases} $$ ```text vs ___ / \ / ---/-----\-----------/---- \___/ is +Ia +------+ | | ----+ +--+ +--- | | -Ia +-------+ 0 180-b 180 360-b 360 (pulse centre at (180-b)/2, i.e. b/2 before the voltage peak at 90) ``` #### Fourier analysis Let $i_s = \frac{a_0}{2} + \sum (a_n\cos n\omega t + b_n\sin n\omega t)$. The waveform has half-wave symmetry, so $a_0 = 0$. $$ \begin{aligned} a_1 &= \frac{1}{\pi}\left[\int_0^{\pi-\beta}I_a\cos\omega t\,d(\omega t) - \int_{\pi}^{2\pi-\beta}I_a\cos\omega t\,d(\omega t)\right] \\ &= \frac{I_a}{\pi}\left[\sin(\pi-\beta) - \{\sin(2\pi-\beta) - \sin\pi\}\right] = \frac{I_a}{\pi}\left[\sin\beta + \sin\beta\right] = \frac{2I_a}{\pi}\sin\beta \end{aligned} $$ $$ \begin{aligned} b_1 &= \frac{1}{\pi}\left[\int_0^{\pi-\beta}I_a\sin\omega t\,d(\omega t) - \int_{\pi}^{2\pi-\beta}I_a\sin\omega t\,d(\omega t)\right] \\ &= \frac{I_a}{\pi}\left[\{1 - \cos(\pi-\beta)\} + \{\cos(2\pi-\beta) - \cos\pi\}\right] \\ &= \frac{I_a}{\pi}\left[(1+\cos\beta) + (\cos\beta + 1)\right] = \frac{2I_a}{\pi}(1+\cos\beta) \end{aligned} $$ #### Magnitude and phase of the fundamental $$ i_{s1} = a_1\cos\omega t + b_1\sin\omega t = I_{s1,m}\sin(\omega t + \phi_1) $$ with $I_{s1,m}\sin\phi_1 = a_1$ and $I_{s1,m}\cos\phi_1 = b_1$. Therefore $$ \begin{aligned} \tan\phi_1 &= \frac{a_1}{b_1} = \frac{\sin\beta}{1+\cos\beta} = \frac{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}{2\cos^2\frac{\beta}{2}} = \tan\frac{\beta}{2} \\ \phi_1 &= +\frac{\beta}{2} \end{aligned} $$ $$ \begin{aligned} I_{s1,m} &= \sqrt{a_1^2 + b_1^2} = \frac{2I_a}{\pi}\sqrt{\sin^2\beta + (1+\cos\beta)^2} = \frac{2I_a}{\pi}\sqrt{2(1+\cos\beta)} \\ &= \frac{4I_a}{\pi}\cos\frac{\beta}{2} \end{aligned} $$ So $$ i_{s1} = \frac{4I_a}{\pi}\cos\frac{\beta}{2}\,\sin\left(\omega t + \frac{\beta}{2}\right) $$ Since the supply voltage is $v_s = V_m\sin\omega t$, the positive phase angle $+\beta/2$ shows that the **fundamental input current leads the input voltage by $\beta/2$**. Hence the displacement factor is $\cos(\beta/2)$ leading, and the converter draws leading (capacitive) reactive power, which is the basis of power factor improvement by this method. Example: $\beta = 30^\circ$, $I_a = 20$ A gives $I_{s1} = \frac{2\sqrt2 \times 20}{\pi}\cos15^\circ = 17.39$ A rms, leading by $15^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In extinction angle control, how is the switch operated? A. Turned on at wt = beta and off at pi B. Turned on at wt = 0 and off at pi - beta C. Turned on at pi - beta and off at pi D. Turned on and off at several points in each half cycle Answer: B. The switch is ON from 0 to pi - beta; the load current freewheels from pi - beta to pi, so the input current is zero then. (Key point: Extinction angle control) 2. For the highly inductive load, what are a1 and b1 of the input current fundamental? A. a1 = (2Ia/pi)(1+cos beta), b1 = (2Ia/pi) sin beta B. a1 = (4Ia/pi) sin beta, b1 = 0 C. a1 = (2Ia/pi) sin beta, b1 = (2Ia/pi)(1+cos beta) D. a1 = 0, b1 = (4Ia/pi) Answer: C. Integrating gives a1 = (2Ia/pi) sin(beta) and b1 = (2Ia/pi)(1+cos(beta)). (Key point: Fourier coefficients) 3. How does the phase angle phi1 come out as beta/2? A. tan(phi1) = a1/b1 = sin(beta)/(1+cos(beta)) = tan(beta/2) B. tan(phi1) = b1/a1 = tan(beta) C. tan(phi1) = a1*b1 = sin(beta) D. cos(phi1) = a1/Ia = sin(beta) Answer: A. Using sin(beta) = 2 sin(beta/2)cos(beta/2) and 1+cos(beta) = 2cos^2(beta/2), the ratio reduces to tan(beta/2). (Key point: Phase angle derivation) 4. What does the positive phase angle +beta/2 mean for the converter? A. The fundamental current lags the voltage and draws lagging VARs B. The fundamental current is in phase with the voltage C. The fundamental current leads the voltage, giving displacement factor cos(beta/2) leading D. The displacement factor is cos(beta) lagging Answer: C. i_s1 = (4Ia/pi)cos(beta/2) sin(wt + beta/2), so current leads by beta/2, which offsets lagging reactive power and improves power factor. 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