--- title: "Three-phase single-way diode rectifier: average and rms" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:3ph-single-way-diode-avg-rms" source: https://www.bctengineer.com/q/semester_7/power_electronics/3ph-single-way-diode-avg-rms site: BCT Engineering --- # Three-phase single-way diode rectifier: average and rms Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2074 Asoj, 8 marks; 2070 Asar, 8 marks (asked 2 times) - 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/3ph-single-way-diode-avg-rms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3 ## Question Explain the operation of three-phase single way rectifier with diode with neat circuit diagram and waveforms. Derive the expression for average and rms values of the output voltage. ## Short answer Each diode conducts for 120° when its phase is most positive. Average output voltage Vdc = (3√3 Vm)/(2π) = 0.827 Vm; RMS output voltage Vrms = Vm sqrt(1/2 + 3√3/(8π)) = 0.8407 Vm. ## 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. ```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 and waveforms - 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 ``` #### Average output voltage The output consists of three identical pulses per cycle; taking D1's pulse ($v_a$ from $30^\circ$ to $150^\circ$): $$ \begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6}^{5\pi/6}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\frac{\pi}{6} - \cos\frac{5\pi}{6}\right] \\ &= \frac{3V_m}{2\pi}\left[\frac{\sqrt3}{2} + \frac{\sqrt3}{2}\right] = \frac{3\sqrt3\,V_m}{2\pi} = 0.827\,V_m \end{aligned} $$ #### RMS output voltage $$ \begin{aligned} V_{rms}^2 &= \frac{3}{2\pi}\int_{\pi/6}^{5\pi/6}V_m^2\sin^2\omega t\,d(\omega t) = \frac{3V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\pi/6}^{5\pi/6} \\ &= \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} - \frac{1}{2}\left(-\frac{\sqrt3}{2} - \frac{\sqrt3}{2}\right)\right] = \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} + \frac{\sqrt3}{2}\right] \\ &= V_m^2\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\right] \end{aligned} $$ $$ V_{rms} = V_m\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\right]^{1/2} = 0.8407\,V_m $$ Here $V_m$ is the peak phase voltage. Form factor $= 0.8407/0.8270 = 1.017$ and ripple factor $= \sqrt{FF^2-1} = 18.3\%$, much better than single-phase rectifiers (48% for full wave). Example: with 230 V per phase, $V_m = 325.27$ V, $V_{dc} = 268.99$ V and $V_{rms} = 273.45$ V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which diode conducts in the three-phase single-way rectifier? A. The one connected to the phase with the largest line voltage B. All three diodes together C. The one connected to the most positive phase, for 120 degrees D. The one connected to the phase passing through zero Answer: C. The diode on the most positive phase conducts for 120 degrees; the others are reverse biased. (Key point: Diode conduction) 2. What are the ripple frequency and the lowest output voltage (50 Hz supply)? A. 300 Hz; 0.866 Vm B. 100 Hz; 0 C. 50 Hz; Vm/2 D. 150 Hz; Vm/2 Answer: D. Three pulses per cycle give 3f = 150 Hz, and the envelope never falls below Vm/2 at crossover points. (Key point: Ripple and minimum voltage) 3. What is the average output voltage? A. 0.827 Vm = 3 sqrt(3) Vm/(2 pi) B. 0.637 Vm C. 1.654 Vm D. 0.955 Vm Answer: A. (3/2pi) integral of Vm sin from pi/6 to 5pi/6 gives 3 sqrt(3) Vm/(2 pi) = 0.827 Vm. (Key point: Average value) 4. Which pair is correct for this rectifier? A. Vrms = 0.8407 Vm; diode PIV = sqrt(3) Vm B. Vrms = 0.707 Vm; PIV = Vm C. Vrms = 0.8407 Vm; PIV = Vm D. Vrms = 1.6554 Vm; PIV = 2Vm Answer: A. Vrms = Vm[1/2 + 3 sqrt(3)/(8 pi)]^(1/2) = 0.8407 Vm, and each diode blocks the peak line voltage sqrt(3) Vm. 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