--- title: "Three-phase single-way controlled rectifier, α = 90°" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:three-phase-single-way-90deg" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-single-way-90deg site: BCT Engineering --- # Three-phase single-way controlled rectifier, α = 90° Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2081 Baishakh, 8 marks - Video: [3 Phase Half Wave Controlled Rectifier with RL load | Power Electronics | Lecture 55](https://www.youtube.com/watch?v=hAm_lquzunc) — 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-90deg - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3 ## Question Explain the operation of a three phase single way controlled rectifier circuit with highly inductive load. Derive the expression for Average value of the output voltage. Also draw the waveform of output voltage for firing angle of 90°. ## Short answer A three-pulse half-wave rectifier uses one thyristor per phase, star-connected with load between cathodes and neutral. With highly inductive load each thyristor conducts exactly 120°; average DC voltage is (3√3/2π)Vm cosα = 0.827Vm cosα. At α=90° Vdc is zero, with equal positive and negative areas in each 120° output pulse. ## 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. ```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 | $v_o$ | |---|---|---| | $120^\circ$ – $240^\circ$ | T1 | $v_a$ | | $240^\circ$ – $360^\circ$ | T2 | $v_b$ | | $360^\circ$ – $480^\circ$ | T3 | $v_c$ | #### 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 $V_{dc} = V_{dm}\cos\alpha$. - $0 \le \alpha < 90^\circ$: $V_{dc} > 0$, rectifier mode. - $\alpha = 90^\circ$: $V_{dc} = 0$. - $90^\circ < \alpha < 180^\circ$: $V_{dc} < 0$, inverter mode (needs a dc source in the load). #### Waveform for α = 90° ```text phase: va vb vc fired at: 120 240 360 (deg) ends at: 240 360 480 each phase is used from 120 to 240 deg of its own sine; positive and negative areas are equal, so Vdc = 0 ``` ```text vo |`-. |`-. |`-. -----+---`-.-----+---+---`-.-----+---+---`-. `-. | `-. | 120 180 240 360 each pulse: +0.866 Vm -> 0 (at 180) -> -0.866 Vm (at 240); equal areas ``` At $\alpha = 90^\circ$: $V_{dc} = 0.827V_m\cos90^\circ = 0$. The output is a train of $120^\circ$ pulses, each going from $+0.866V_m$ to $0$ to $-0.866V_m$; the positive and negative areas cancel, so the average output voltage and the average power are zero, although the load current still flows (it is sustained by the inductance and, in practice, requires a dc source in the load). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Vdc at alpha = 90 degrees for this rectifier? A. 0.827 Vm B. 0.4135 Vm C. -0.827 Vm D. Zero Answer: D. Vdc = 0.827 Vm cos 90 = 0, because the positive and negative areas are equal. (Key point: Average voltage at 90 degrees) 2. At alpha = 90 degrees, when is T1 fired (relative to va zero crossing)? A. 30 degrees B. 90 degrees C. 120 degrees D. 150 degrees Answer: C. T1 is fired at 30 degrees + alpha = 120 degrees, and conducts until 240 degrees. (Key point: Firing instants) 3. What does each 120 degree output pulse look like at alpha = 90 degrees? A. From +0.866 Vm to 0 and down to -0.866 Vm B. Constant at Vm C. From 0 up to Vm and back to 0 D. From -Vm to +Vm Answer: A. A phase is used from 120 to 240 degrees of its sine, going from +0.866 Vm through 0 at 180 to -0.866 Vm. (Key point: Pulse shape) 4. At alpha = 90 degrees the average voltage is zero. What about the load current? A. It is zero too B. It still flows, sustained by the inductance C. It becomes discontinuous D. It reverses direction Answer: B. The inductor keeps the current continuous and constant, though the average power is zero. (Key point: Current with zero average voltage) ## Close variants set in other exams - [Three-phase single-way controlled rectifier, α = 60°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-60deg) — Power Electronics, chapter 3, asked 3× - [Three-phase single-way controlled rectifier, α = 30°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-30deg) — Power Electronics, chapter 3, asked 4× ## Related in Power Electronics - [Three-phase single-way controlled rectifier, α = π/6](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-pi-6) — Power Electronics, chapter 3 - [Three-phase single-way controlled rectifier, R load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-r-load) — Power Electronics, chapter 3, asked 3× - [Three-phase single-way controlled rectifier](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-controlled-avg) — Power Electronics, chapter 3, asked 3× - [Three-phase single-way thyristor rectifier: average](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-thyristor-avg) — Power Electronics, chapter 3, asked 3× - [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× - [Three-phase single-way diode rectifier: average and rms](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3A3ph-single-way-diode-avg-rms) — Power Electronics, chapter 3, asked 2× ## Related in other subjects - [3-phase half controlled bridge: no-load speed, firing angle](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ahalf-bridge-motor-no-load-speed) — Utilization of Electrical Energy, chapter 3, asked 3× - [Single-phase fully controlled converter DC motor speed](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afull-converter-dc-motor-speed) — Utilization of Electrical Energy, chapter 3 - [3-phase thyristor bridge: no-load speed and firing angle](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afull-bridge-motor-no-load-speed) — Utilization of Electrical Energy, chapter 3, asked 3× - [Voltage regulation expression for inductive load](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Avoltage-regulation-inductive-load) — Electrical Machines, chapter 2