--- title: "Symmetrical angle control for power factor" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:symmetrical-angle-control" source: https://www.bctengineer.com/q/semester_7/power_electronics/symmetrical-angle-control site: BCT Engineering --- # Symmetrical angle control for power factor Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2080 Bhadra, 8 marks - Video: [Power Factor Improvement of Controlled Rectifiers using Symmetric Angle Control (SAC)](https://www.youtube.com/watch?v=E3cf7yYVvsw) — kiran navale - 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 - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Explain the symmetrical angle control method for improving input power factor in single phase controlled rectifier. ## Short answer Forced-commutated switches are turned ON at (π-β)/2 and OFF at (π+β)/2, placing conduction symmetrically about the voltage peak. This makes the fundamental input current in phase with the supply voltage, giving unity displacement factor, but requires GTOs or IGBTs. ## Answer **Symmetrical angle control** is a power factor improvement method in which a forced-commutated switch is turned ON at $\alpha = (\pi-\beta)/2$ and turned OFF at $(\pi+\beta)/2$, so that the conduction interval $\beta$ is placed symmetrically about the peak of the supply voltage ($\omega t = \pi/2$). As a result, the fundamental input current is **in phase** with the supply voltage and the displacement factor is unity. #### Why it is needed In a phase-controlled (thyristor) converter the input current always lags the voltage by $\alpha$ (or $\alpha/2$ in a semiconverter). At large $\alpha$ the input power factor becomes very poor. Thyristors cannot be turned off at will, but GTOs, IGBTs or power transistors can, so the position of the current pulse can be chosen freely. #### 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, current $I_a$ constant) 1. Positive half cycle: S1 is turned ON at $\omega t = (\pi-\beta)/2$. Current flows $a \to$ S1 $\to$ load $\to$ D2 $\to b$; $v_o = v_s$, $i_s = +I_a$. 2. S1 is turned OFF at $(\pi+\beta)/2$ by a gate signal. The load current transfers to the freewheeling diode Dm; $v_o = 0$, $i_s = 0$. 3. Negative half cycle: S2 is turned ON at $(3\pi-\beta)/2$ and OFF at $(3\pi+\beta)/2$; current flows $b \to$ S2 $\to$ load $\to$ D1 $\to a$; $v_o = -v_s$ (positive), $i_s = -I_a$. 4. For the rest of the time the load current freewheels through Dm. ```text is +Ia +-----+ | | 0 ------+ +-------+ +------ | | -Ia +-----+ (90-b/2)(90+b/2) (270-b/2)(270+b/2) b = beta; vs is positive 0-180, negative 180-360 ``` #### Output voltage $$ V_{dc} = \frac{2}{2\pi}\int_{(\pi-\beta)/2}^{(\pi+\beta)/2} V_m\sin\omega t\, d(\omega t) = \frac{2V_m}{\pi}\sin\frac{\beta}{2} $$ $$ V_{rms} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}(\beta + \sin\beta)\right]^{1/2} $$ #### Input current and power factor - Input current: quasi-square wave of $\pm I_a$, width $\beta$, centred on the voltage peaks. - RMS input current: $I_s = I_a\sqrt{\beta/\pi}$ - Fundamental (rms): $I_{s1} = \frac{2\sqrt2 I_a}{\pi}\sin\frac{\beta}{2}$, with displacement angle $\phi_1 = 0$. - Displacement factor $= \cos 0 = 1$. - Power factor: $$ PF = \frac{I_{s1}}{I_s}\cos\phi_1 = \frac{2\sqrt2\,\sin(\beta/2)}{\sqrt{\pi\beta}} $$ #### Merits and demerits - Unity displacement factor at all output levels; only the harmonic (distortion) factor reduces the PF. - Less reactive power is drawn from the supply. - Demerit: it needs self-commutated devices and gate turn-off circuitry, so the cost is higher. ## 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 turned ON and OFF? A. ON at alpha = 0, OFF at pi B. ON at (pi - beta)/2, OFF at (pi + beta)/2, centred on voltage peak C. ON at pi - beta, OFF at pi D. ON at pi/2 and OFF when current reaches zero Answer: B. The conduction interval beta is placed symmetrically about wt = pi/2 so the current pulse is centred on the voltage peak. (Key point: Switching instants) 2. Why must GTOs/IGBTs be used instead of thyristors? A. They block higher voltage B. They are cheaper C. They need no gate drive D. Turn-off must be done at will while the current is still flowing Answer: D. Thyristors cannot be turned off by the gate, so the current pulse could not be positioned freely. (Key point: Why forced commutation) 3. What is the effect on the fundamental input current? A. It lags by beta/2 B. It is in phase with vs, giving unity displacement factor C. It leads by beta D. It becomes zero Answer: B. A pulse symmetrical about the voltage peak has fundamental in phase with vs, so DF = 1. (Key point: Displacement factor) 4. Why is the power factor still below 1? A. Because of the freewheeling diode losses B. Because displacement factor is poor C. Because input current is distorted (harmonics), PF = 2sqrt(2) sin(beta/2)/sqrt(pi beta) D. Because the supply is not sinusoidal Answer: C. Only the distortion factor reduces PF, since DF = 1. 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