--- title: "Symmetrical angle control: average and RMS output" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:symmetrical-angle-control-derive" source: https://www.bctengineer.com/q/semester_7/power_electronics/symmetrical-angle-control-derive site: BCT Engineering --- # Symmetrical angle control: average and RMS output Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2073 Chaitra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/symmetrical-angle-control-derive - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Explain the operation of symmetrical angle control method for power factor improvement. Derive the expression for average and RMS value of the output voltage. ## Short answer Switch conducts symmetrically about voltage peak, from (π−β)/2 to (π+β)/2, making fundamental input current in phase with voltage. Average output is 2Vm/π sin(β/2). RMS output is (Vm/√2)[(β+sinβ)/π]^1/2. ## Answer **Symmetrical angle control** is a power factor improvement technique for single-phase converters in which a self-commutated switch (GTO, IGBT, power transistor) conducts for an angle $\beta$ placed **symmetrically about the peak** of each half cycle of supply voltage: it is turned ON at $(\pi-\beta)/2$ and OFF at $(\pi+\beta)/2$. The fundamental input current then lies in phase with the voltage (displacement factor = 1). #### 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) | Interval ($\omega t$) | Conducting | $v_o$ | $i_s$ | |---|---|---|---| | $(\pi-\beta)/2$ – $(\pi+\beta)/2$ | S1, D2 | $v_s$ | $+I_a$ | | $(\pi+\beta)/2$ – $(3\pi-\beta)/2$ | Dm | $0$ | $0$ | | $(3\pi-\beta)/2$ – $(3\pi+\beta)/2$ | S2, D1 | $-v_s$ | $-I_a$ | | $(3\pi+\beta)/2$ – $(5\pi-\beta)/2$ | Dm | $0$ | $0$ | - S1 is turned OFF by a negative gate pulse (forced turn-off), not by the supply; the load current then transfers to Dm. - The output is a series of "caps" of the sine wave centred at $90^\circ$ and $270^\circ$. ```text vo ___ ___ | | | | caps of |vs|, width b ---------+---+--------+---+------ is +---+ +Ia | | ---------+ +--------+ +------ -Ia +---+ (90-b/2)(90+b/2)(270-b/2)(270+b/2) ``` #### Average output voltage The output repeats every $\pi$: $$ \begin{aligned} V_{dc} &= \frac{1}{\pi}\int_{(\pi-\beta)/2}^{(\pi+\beta)/2}V_m\sin\omega t\,d(\omega t) = \frac{V_m}{\pi}\left[\cos\frac{\pi-\beta}{2} - \cos\frac{\pi+\beta}{2}\right] \\ &= \frac{V_m}{\pi}\left[\sin\frac{\beta}{2} + \sin\frac{\beta}{2}\right] = \frac{2V_m}{\pi}\sin\frac{\beta}{2} \end{aligned} $$ $V_{dc}$ varies from $0$ ($\beta = 0$) to $2V_m/\pi$ ($\beta = \pi$). #### RMS output voltage $$ \begin{aligned} V_{rms}^2 &= \frac{1}{\pi}\int_{(\pi-\beta)/2}^{(\pi+\beta)/2}V_m^2\sin^2\omega t\,d(\omega t) = \frac{V_m^2}{2\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{(\pi-\beta)/2}^{(\pi+\beta)/2} \\ &= \frac{V_m^2}{2\pi}\left[\beta - \frac{\sin(\pi+\beta) - \sin(\pi-\beta)}{2}\right] = \frac{V_m^2}{2\pi}\left[\beta + \sin\beta\right] \end{aligned} $$ $$ V_{rms} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}(\beta + \sin\beta)\right]^{1/2} $$ #### Power factor - $I_s = I_a\sqrt{\beta/\pi}$; fundamental $I_{s1} = \frac{2\sqrt2 I_a}{\pi}\sin\frac{\beta}{2}$ with $\phi_1 = 0$. - $PF = \dfrac{I_{s1}}{I_s} = \dfrac{2\sqrt2\sin(\beta/2)}{\sqrt{\pi\beta}}$; e.g. at $\beta = 90^\circ$, $PF = 0.90$, whereas a full converter giving the same output ($\cos\alpha = 0.707$) has $PF = 0.64$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In symmetrical angle control, when is the switch on? A. From 0 to beta B. From alpha to pi C. From pi - beta to pi D. From (pi-beta)/2 to (pi+beta)/2, symmetric about the voltage peak Answer: D. The conduction interval of width beta is centred on 90 degrees (and 270 degrees). (Key point: Switching instants) 2. What is the effect on the fundamental input current? A. It leads vs by beta/2 B. It lags vs by beta/2 C. It is in phase with vs (displacement factor 1) D. It lags by beta Answer: C. The current pulse is symmetric about the voltage peak, so phi1 = 0. (Key point: Power factor benefit) 3. What is the average output voltage? A. (Vm/pi)(1 + cos(beta)) B. (2Vm/pi) sin(beta/2) C. (2Vm/pi) cos(beta/2) D. (Vm/pi) sin(beta) Answer: B. Integrating Vm sin(wt) over (pi-beta)/2 to (pi+beta)/2 gives (Vm/pi)[2 sin(beta/2)]. (Key point: Average output voltage) 4. How does the load current continue after S1 turns off? A. It stops immediately B. It flows through S2 C. It transfers to the freewheeling diode Dm, with vo = 0 D. It flows back to the source Answer: C. S1 is forced off by its gate and the inductive load current moves to Dm. (Key point: Freewheeling and turn-off) ## Related in Power Electronics - [Extinction angle control for power factor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aextinction-angle-control-derive) — Power Electronics, chapter 2 - [Symmetrical angle control for power factor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asymmetrical-angle-control) — Power Electronics, chapter 2 - [Step-up chopper average and rms output](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-up-chopper-avg-rms) — Power Electronics, chapter 4 - [Series full converters: average and rms output](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aseries-full-converters-avg-rms) — Power Electronics, chapter 2 - [Three-phase single-way thyristor rectifier: average and RMS](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3A3ph-single-way-thyristor-avg-rms) — Power Electronics, chapter 3, asked 5× - [Step-up chopper operation and output voltage](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-up-chopper-principle) — Power Electronics, chapter 4, asked 8× ## Related in other subjects - [Average, rms, form and peak factor of sawtooth](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Asawtooth-average-rms-form-peak) — Basic Electrical Engineering, chapter 5 - [Define RMS and average value](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_and_electronics_engineering%3Adefine-rms-average) — Basic Electrical and Electronics Engineering, chapter 1 - [Importance and improvement of power factor](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Apower-factor-importance-improvement) — Basic Electrical Engineering, chapter 7 - [Causes of poor power factor and improvement](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Apoor-pf-causes-improvement) — Industrial Power Distribution and Illumination, chapter 10