--- title: "Three-phase single-way controlled rectifier, α = π/6" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:three-phase-single-way-pi-6" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-single-way-pi-6 site: BCT Engineering --- # Three-phase single-way controlled rectifier, α = π/6 Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2071 Chaitra, 8 marks - Video: [3 Phase Half Wave Controlled Rectifier with R load | Power Electronics | Lecture 53](https://www.youtube.com/watch?v=zIns_hBFNWs) — Electronics Maddy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-single-way-pi-6 - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3 ## Question Explain the operation of three phase single way controlled rectifier for firing angle α = π/6 with associated waveforms and hence deduce the expression for average and root mean square value of the output voltage assuming highly inductive load. ## Short answer At alpha = pi/6, each thyristor conducts 120 degrees and each output pulse runs from 0.866 Vm to zero, never going negative. Average voltage is Vdc = 0.827 Vm cos(pi/6) = 0.716 Vm. RMS value is Vrms = Vm sqrt(1/2 + 3 sqrt3 cos(2 alpha)/(8 pi)) with alpha = pi/6. ## Answer A **three-phase single-way (half-wave, three-pulse) controlled rectifier** uses three thyristors, one per phase, connected to a star-connected supply (or transformer secondary) with the load returned to the neutral. The output voltage is controlled by delaying the firing of each thyristor by $\alpha$ from its natural commutation point. Here the load is highly inductive and $\alpha = \pi/6 = 30^\circ$. ```text T1 a o------>|-----+ T2 | b o------>|-----+------o P (+) T3 | | c o------>|-----+ Load | n o----------------------o N (-) a, b, c, n: star-connected secondary va = Vm sin wt, vb lags 120, vc lags 240 ``` #### Operation - Each thyristor's anode goes to one phase and the cathodes are joined to the positive output. A thyristor can conduct only when its phase is the most positive **and** it has been fired. - The firing angle $\alpha$ is measured from the **natural commutation point** (the instant a diode would start conducting), i.e. from $30^\circ$ after the zero crossing of its phase voltage. So T1 is fired at $\omega t = 30^\circ + \alpha$, T2 at $150^\circ + \alpha$, T3 at $270^\circ + \alpha$. - When T2 is fired, $v_b > v_a$, so T1 becomes reverse biased and turns off (**natural/line commutation**). - With a **highly inductive load**, the load current is continuous and constant, so each thyristor conducts for exactly $120^\circ$, even if its phase voltage becomes negative. The output follows the phase of the conducting thyristor. | Interval ($\omega t$) | Conducting | $v_o$ | |---|---|---| | $60^\circ$ – $180^\circ$ | T1 | $v_a$ | | $180^\circ$ – $300^\circ$ | T2 | $v_b$ | | $300^\circ$ – $420^\circ$ | T3 | $v_c$ | #### Waveforms for α = π/6 ```text phase: va vb vc fired at: 60 180 300 (deg) ends at: 180 300 420 each phase is used from 60 to 180 deg of its own sine; vo just touches zero at the end of each pulse (boundary of continuity) ``` ```text vo .-. .-. .-. / \ / \ / \ | \ | \ | \ -----+------+-----+------+-----+------+-- 60 180 180 300 300 420 each pulse: 0.866 Vm -> Vm (at 90) -> 0 ``` - Output pulses start at $V_m\sin60^\circ = 0.866V_m$ and fall to zero at the end of each $120^\circ$ interval; $v_o$ never becomes negative at $\alpha = 30^\circ$. - Each thyristor current is a $120^\circ$ rectangular pulse of height $I_o$; the load current is constant. #### Average output voltage Taking the conduction of T1 (the waveform repeats every $2\pi/3$): $$ \begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\left(\frac{\pi}{6}+\alpha\right) - \cos\left(\frac{5\pi}{6}+\alpha\right)\right] \\ &= \frac{3V_m}{2\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{3} = \frac{3V_m}{2\pi}\cdot\sqrt3\cos\alpha \end{aligned} $$ $$ V_{dc} = \frac{3\sqrt3\,V_m}{2\pi}\cos\alpha = 0.827\,V_m\cos\alpha $$ where $V_m$ is the peak phase voltage. Maximum value (at $\alpha = 0$) is $V_{dm} = 0.827V_m$, so $V_{dc} = V_{dm}\cos\alpha$. At $\alpha = \pi/6$: $V_{dc} = 0.827V_m \times 0.866 = 0.716\,V_m$. #### RMS output voltage $$ \begin{aligned} V_{rms}^2 &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m^2\sin^2\omega t\,d(\omega t) = \frac{3V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\pi/6+\alpha}^{5\pi/6+\alpha} \\ &= \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} - \frac{\sin\left(\frac{5\pi}{3}+2\alpha\right) - \sin\left(\frac{\pi}{3}+2\alpha\right)}{2}\right] = \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} + \frac{\sqrt3}{2}\cos2\alpha\right] \end{aligned} $$ (using $\sin\left(\frac{5\pi}{3}+2\alpha\right) - \sin\left(\frac{\pi}{3}+2\alpha\right) = 2\cos(\pi+2\alpha)\sin\frac{2\pi}{3} = -\sqrt3\cos2\alpha$) $$ V_{rms} = V_m\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\cos2\alpha\right]^{1/2} $$ At $\alpha = \pi/6$: $V_{rms} = V_m\left[0.5 + 0.2067 \times 0.5\right]^{1/2} = 0.7768\,V_m$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For alpha = pi/6, T1 is fired at: A. 30° B. 60° C. 90° D. 150° Answer: B. T1 fires at 30° + alpha = 60°; T2 at 180° and T3 at 300°. (Key point: Firing instant for alpha = 30°) 2. At alpha = 30° with highly inductive load, each output pulse: A. Starts at zero and rises to Vm B. Starts at 0.866 Vm, peaks at Vm and falls to zero without going negative C. Goes negative at its end D. Is constant at 0.827 Vm Answer: B. Pulse starts at Vm sin 60° = 0.866 Vm, peaks at 90° and reaches zero at 180°, the boundary of continuity. (Key point: Output pulse shape) 3. At alpha = pi/6, the average output voltage is: A. 0.827 Vm B. 0.716 Vm C. 0.7768 Vm D. 0.477 Vm Answer: B. Vdc = 0.827 Vm cos 30° = 0.716 Vm. (Key point: Average voltage) 4. At alpha = pi/6, the rms output voltage is: A. 0.716 Vm B. 0.827 Vm C. 0.7768 Vm D. 0.5 Vm Answer: C. Vrms = Vm [0.5 + 0.2067 x cos 60°]^(1/2) = 0.7768 Vm. 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