--- title: "Three-phase single-way thyristor rectifier: average" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:three-phase-single-way-thyristor-avg" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-single-way-thyristor-avg site: BCT Engineering --- # Three-phase single-way thyristor rectifier: average Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2069 Chaitra, 8 marks - Video: [3 phase half wave controlled rectifier working & waveforms in Power Electronics by Engineering Funda](https://www.youtube.com/watch?v=alFjAYCu3yo) — Engineering Funda - 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-thyristor-avg - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3 ## Question Explain the operation of three-phase single way controlled rectifier with thyristor with neat circuit diagram and waveforms. Derive the expression for average value of the output voltage. ## Short answer Three thyristors connected to star supply, each conducts for 120° after firing at α from its natural commutation point (30° after phase voltage zero crossing). With continuous load current, average output voltage V_dc = (3√3 V_m)/(2π) cosα = 0.827 V_m cosα, or 1.17 V_ph cosα. ## 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. A highly inductive load is assumed so that the load current is continuous. ```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. #### Conduction table | Thyristor | Fired at | Conducts till | $v_o$ | |---|---|---|---| | T1 | $30^\circ+\alpha$ | $150^\circ+\alpha$ | $v_a$ | | T2 | $150^\circ+\alpha$ | $270^\circ+\alpha$ | $v_b$ | | T3 | $270^\circ+\alpha$ | $390^\circ+\alpha$ | $v_c$ | #### Waveforms For $\alpha = 30^\circ$ the output just reaches zero at the end of each pulse: ```text vo .-. .-. .-. / \ / \ / \ | \ | \ | \ -----+------+-----+------+-----+------+-- 60 180 180 300 300 420 each pulse: 0.866 Vm -> Vm (at 90) -> 0 ``` For $\alpha = 60^\circ$ part of each pulse is negative: ```text vo |`-. |`-. |`-. | `-. | `-. | `-. -----+------`-.-----+------`-.-----+------`-. `-. | `-. | `-. 90 180 210 330 450 each pulse: Vm (at 90) -> 0 (at 180) -> -0.5 Vm (at 210), then jumps to next phase ``` #### 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). In terms of rms phase voltage ($V_m = \sqrt2V_{ph}$), $V_{dc} = 1.17V_{ph}\cos\alpha$ (more exactly $1.1695V_{ph}\cos\alpha$). For example, with $V_{ph} = 230$ V and $\alpha = 30^\circ$, $V_{dc} = 1.1695 \times 230 \times 0.866 = 232.96$ V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At alpha = 60° with highly inductive load, part of each output pulse is negative because: A. The thyristor keeps conducting for its 120° even after its phase voltage goes below zero B. The next thyristor is fired early C. The supply polarity reverses D. The load acts as a source of rectification Answer: A. The constant inductive current holds the thyristor on for 120°, so vo follows the phase voltage even into its negative region. (Key point: Conduction when phase goes negative) 2. How does Vdc = 0.827 Vm cos(alpha) vary with alpha for 0 to 90°? A. Increases from zero to maximum B. Decreases from 0.827 Vm to zero C. Remains constant D. Decreases from 1.17 Vm to 0.827 Vm Answer: B. Cos(alpha) falls from 1 to 0, so Vdc decreases from the maximum 0.827 Vm to zero at 90°. (Key point: Dependence on alpha) 3. In terms of rms phase voltage, the average output voltage is: A. 1.35 Vph cos(alpha) B. 0.827 Vph cos(alpha) C. 1.17 Vph cos(alpha) D. 2.34 Vph cos(alpha) Answer: C. Since Vm = sqrt2 Vph, 0.827 Vm = 1.17 Vph, e.g. 232.96 V for 230 V at alpha = 30°. (Key point: In terms of rms phase voltage) 4. Operation with 90° < alpha < 180° is called: A. Rectifier mode with higher voltage B. Inverter mode, needing a dc source in the load C. Discontinuous mode D. Free-wheeling mode Answer: B. Vdc is negative, so power flows from the dc side to the ac side, which needs a dc source such as a motor emf. (Key point: Inverter mode) ## Close variants set in other exams - [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 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× ## Related in Power Electronics - [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× - [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, α = 90°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-90deg) — Power Electronics, chapter 3, asked 4× - [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 = 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 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×