--- title: "Three-phase single-way uncontrolled rectifier" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:three-phase-single-way-diode-avg" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-single-way-diode-avg site: BCT Engineering --- # Three-phase single-way uncontrolled rectifier Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2079 Bhadra, 8 marks - Video: [27. Three phase uncontrolled half wave rectifier](https://www.youtube.com/watch?v=SBEf3yph52s) — Read Electric Vehicle - 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-diode-avg - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3 ## Question Explain the operation of 3-phase single way uncontrolled rectifier and describe expression for average output voltage. ## Short answer Three diodes, one per phase with common cathode output and neutral return, conduct in turn for 120° as the most positive phase changes. The output is the upper envelope of the phase voltages with ripple frequency 3f. Average DC voltage is (3√3/2π)Vm = 0.827Vm = 1.17Vph = 0.675Vline. ## Answer A **three-phase single-way (half-wave, three-pulse) uncontrolled rectifier** uses three diodes, one in each phase of a star-connected supply (or transformer secondary), with the load connected between the common cathode point and the neutral. It is the simplest three-phase rectifier and is also called the three-pulse midpoint (M3) rectifier. ```text D1 a o------>|-----+ D2 | b o------>|-----+------o P (+) D3 | | 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 diode's anode is connected to one phase; the cathodes are joined to form the positive output. The diode connected to the **most positive phase** conducts and the other two are reverse biased. - Measured from the zero crossing of $v_a$, $v_a$ is the highest from $30^\circ$ to $150^\circ$, $v_b$ from $150^\circ$ to $270^\circ$ and $v_c$ from $270^\circ$ to $390^\circ$. These crossover points ($30^\circ$, $150^\circ$, $270^\circ$) are the points of **natural commutation**. - Each diode conducts for $120^\circ$; the output is the **upper envelope** of the three phase voltages, with three pulses per cycle (ripple frequency $3f = 150$ Hz), never falling below $V_m/2$. - Load current returns through the neutral; each diode must block a peak inverse voltage of $\sqrt3 V_m$ (peak line voltage). | Interval ($\omega t$) | Conducting | $v_o$ | |---|---|---| | $30^\circ$ – $150^\circ$ | D1 | $v_a$ | | $150^\circ$ – $270^\circ$ | D2 | $v_b$ | | $270^\circ$ – $390^\circ$ | D3 | $v_c$ | ```text vo va vb vc _----_ _----_ _----_ / \__/ \__/ \ / D1 ^ D2 ^ D3 \ 30 150 270 390 (deg) vo = upper envelope; minimum Vm/2 at crossover points, peak Vm ``` #### Expression for average output voltage Since the three pulses are identical, the average over one pulse ($2\pi/3$) equals the average over a cycle: $$ \begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6}^{5\pi/6}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\frac{\pi}{6} - \cos\frac{5\pi}{6}\right] \\ &= \frac{3V_m}{2\pi}\left[\frac{\sqrt3}{2} + \frac{\sqrt3}{2}\right] = \frac{3\sqrt3\,V_m}{2\pi} = 0.827\,V_m \end{aligned} $$ In terms of rms values: $V_{dc} = \frac{3\sqrt3}{2\pi}\sqrt2\,V_{ph} = 1.17\,V_{ph} = 0.675\,V_L$. #### Remarks - Each diode carries the full load current for $120^\circ$: $I_{D,avg} = I_o/3$, $I_{D,rms} = I_o/\sqrt3$ (inductive load). - Peak inverse voltage per diode $= \sqrt3V_m$. - The transformer secondary carries dc current (unidirectional), which can saturate the core; a zig-zag secondary or a bridge circuit avoids this. - Example: $V_{ph} = 230$ V gives $V_{dc} = 0.827 \times 325.27 = 268.99$ V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Measured from the zero crossing of va, where do the diodes commutate? A. 0, 120 and 240 degrees B. 90, 210 and 330 degrees C. 60, 180 and 300 degrees D. 30, 150 and 270 degrees Answer: D. va is highest from 30 to 150 degrees, vb from 150 to 270, vc from 270 to 390, so these crossover points are natural commutation points. (Key point: Natural commutation points) 2. Which is the correct average output voltage? A. 0.827 Vm = 1.17 Vph B. 1.35 VL C. 0.637 Vm D. 1.654 Vm Answer: A. (3/2pi) integral of Vm sin from pi/6 to 5pi/6 = 3 sqrt(3) Vm/(2 pi) = 0.827 Vm = 1.17 Vph = 0.675 VL. (Key point: Average voltage expression) 3. With inductive load, what are the average current and PIV of each diode? A. Io/2; Vm B. Io/3; sqrt(3) Vm C. Io; 2Vm D. Io/sqrt(3); Vm Answer: B. Each diode carries Io for 120 degrees, so I_avg = Io/3, and it blocks the peak line voltage sqrt(3) Vm. (Key point: Diode ratings) 4. What is a drawback of this circuit for the transformer? A. The secondary carries unidirectional (dc) current, which can saturate the core B. The primary carries no current C. It needs a delta secondary only D. The neutral must carry zero current Answer: A. Each secondary winding carries current in one direction only, so dc magnetization can saturate the core; a zig-zag secondary or bridge avoids this. 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