--- title: "Three-phase single-way controlled rectifier" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:three-phase-single-way-controlled-avg" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-single-way-controlled-avg site: BCT Engineering --- # Three-phase single-way controlled rectifier Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2075 Asoj, 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-controlled-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 circuit (with thyristors) with necessary waveforms. Also derive the expression for average value of the output voltage. ## Short answer With continuous current, each thyristor conducts 120° and commutates when the next phase is fired at 30°+α; the output is a sequence of 120° phase-voltage segments, negative for α>30°. Vdc=(3√3Vm/2π)cosα=0.827Vm cosα=Vdm cosα; Vdc positive for α<90°, zero at 90°, negative for α>90° (inverter). ## 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 (continuous current) is assumed. ```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 | $v_o$ | |---|---|---| | $90^\circ$ – $210^\circ$ | T1 | $v_a$ | | $210^\circ$ – $330^\circ$ | T2 | $v_b$ | | $330^\circ$ – $450^\circ$ | T3 | $v_c$ | (Table and waveform for $\alpha = 60^\circ$.) ```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 ``` - Thyristor currents: three $120^\circ$ blocks of height $I_o$, displaced by $120^\circ$. - Voltage across T1 while T2 conducts is $v_a - v_b = v_{ab}$, and while T3 conducts is $v_{ac}$; so each thyristor must block $\sqrt3 V_m$. #### 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). For $\alpha = 60^\circ$, $V_{dc} = 0.4135V_m$. If the load is purely resistive, the current cannot reverse, so for $\alpha > 30^\circ$ conduction becomes discontinuous and $V_{dc} = \frac{3V_m}{2\pi}\left[1 + \cos\left(\alpha + \frac{\pi}{6}\right)\right]$ instead. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this rectifier, T1 is turned off when T2 is fired because: A. The gate current of T1 is removed B. v_b > v_a, so T1 becomes reverse biased (line commutation) C. The load current becomes zero D. A separate commutation circuit forces it off Answer: B. Once T2 conducts with v_b higher than v_a, T1 is reverse biased and commutates naturally. (Key point: Turn-off mechanism) 2. What happens to the average output voltage Vdc = 0.827 Vm cos(alpha) as alpha goes beyond 90°? A. It stays positive, rectifier mode B. It becomes negative, inverter mode (needs a dc source in the load) C. It remains zero D. It doubles Answer: B. For 90° < alpha < 180°, cos(alpha) is negative so Vdc < 0 and the converter acts as an inverter, requiring a dc source in the load. (Key point: Operating modes) 3. While T2 or T3 is conducting, T1 must block a peak voltage of: A. Vm B. sqrt3 Vm C. 2 Vm D. 0.5 Vm Answer: B. The voltage across T1 is a line voltage (v_ab or v_ac), whose peak is sqrt3 Vm. (Key point: Thyristor voltage stress) 4. If the load is purely resistive instead of highly inductive, for alpha > 30° the rectifier: A. Still follows 0.827 Vm cos(alpha) B. Has discontinuous conduction with Vdc = (3Vm/2pi)[1 + cos(alpha + pi/6)] C. Cannot operate D. Gives negative average voltage Answer: B. A resistive load cannot carry reversed current, so conduction becomes discontinuous and a different Vdc formula applies. (Key point: Resistive load difference) ## Close variants set in other exams - [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× ## 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× ## Related in other subjects - [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