--- title: "Three-phase diode bridge rectifier" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:three-phase-diode-bridge-avg-rms" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-diode-bridge-avg-rms site: BCT Engineering --- # Three-phase diode bridge rectifier Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2080 Baishakh, 8 marks; 2071 Chaitra, 8 marks; 2070 Chaitra, 8 marks (asked 3 times) - Video: [Three-Phase Full-Wave Rectifier Operation](https://www.youtube.com/watch?v=-XUsu08Farw) — Engineering with Prof. Kim - 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-diode-bridge-avg-rms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3 ## Question With the help of suitable circuit diagram and waveforms, explain the operation of three phase full wave bridge rectifier using diodes (6 diodes). Derive the expression for average value and rms value of the output voltage. ## Short answer The output is a six-pulse waveform from the line voltages. Average output voltage Vdc = (3√3 Vm)/π = 1.654 Vm = 1.35 VL. RMS output voltage Vrms = Vm sqrt(3/2 + 9√3/(4π)) ≈ 1.6554 Vm. ## Answer A **three-phase full-wave bridge rectifier** uses six diodes: D1, D3, D5 form the positive (common-cathode) group and D4, D6, D2 the negative (common-anode) group. At any instant the diode of the most positive phase in the top group and the diode of the most negative phase in the bottom group conduct, so the load gets the **highest line voltage**. ```text P (+) o-----+------+------+-----+ | | | | D1 D3 D5 | | | | Load a o----------------+ | | | b o-----------------------+ | | c o------------------------------+ | | | | | D4 D6 D2 | | | | | N (-) o-----+------+------+-----+ (phase a joins D1-D4, b joins D3-D6, c joins D5-D2) ``` #### Operation - Each diode conducts for $120^\circ$; a new pair takes over every $60^\circ$. Conduction sequence: D1D2, D2D3, D3D4, D4D5, D5D6, D6D1 (numbering = order of conduction). - Output voltage is the upper envelope of the six line voltages $v_{ab}, v_{ac}, v_{bc}, v_{ba}, v_{ca}, v_{cb}$: a **six-pulse** waveform with ripple frequency $6f = 300$ Hz. - Each segment lasts $60^\circ$, varying from $\frac{\sqrt3}{2}V_{mL}$ to $V_{mL}$ and back. | $\omega t$ (w.r.t. $v_a$) | Diodes | $v_o$ | |---|---|---| | $30^\circ$ – $90^\circ$ | D1, D6 | $v_{ab}$ | | $90^\circ$ – $150^\circ$ | D1, D2 | $v_{ac}$ | | $150^\circ$ – $210^\circ$ | D3, D2 | $v_{bc}$ | | $210^\circ$ – $270^\circ$ | D3, D4 | $v_{ba}$ | | $270^\circ$ – $330^\circ$ | D5, D4 | $v_{ca}$ | | $330^\circ$ – $390^\circ$ | D5, D6 | $v_{cb}$ | ```text vo vab vac vbc vba vca vcb _/\__/\__/\__/\__/\__/\_ <- peak VmL min 0.866 VmL 30 90 150 210 270 330 390 (deg) six 60-degree caps per cycle ``` #### Average output voltage Let $V_{mL} = \sqrt3 V_m$ be the peak line voltage. Taking one $60^\circ$ segment of a line voltage $V_{mL}\sin\theta$ from $\theta = \pi/3$ to $2\pi/3$: $$ \begin{aligned} V_{dc} &= \frac{6}{2\pi}\int_{\pi/3}^{2\pi/3}V_{mL}\sin\theta\,d\theta = \frac{3V_{mL}}{\pi}\left[\cos\frac{\pi}{3} - \cos\frac{2\pi}{3}\right] \\ &= \frac{3V_{mL}}{\pi} = \frac{3\sqrt3\,V_m}{\pi} = 1.654\,V_m = 1.35\,V_L \end{aligned} $$ #### RMS output voltage $$ \begin{aligned} V_{rms}^2 &= \frac{3}{\pi}\int_{\pi/3}^{2\pi/3}V_{mL}^2\sin^2\theta\,d\theta = \frac{3V_{mL}^2}{2\pi}\left[\theta - \frac{\sin2\theta}{2}\right]_{\pi/3}^{2\pi/3} \\ &= \frac{3V_{mL}^2}{2\pi}\left[\frac{\pi}{3} + \frac{\sqrt3}{2}\right] = 3V_m^2\left[\frac{1}{2} + \frac{3\sqrt3}{4\pi}\right] \end{aligned} $$ $$ V_{rms} = V_m\left[\frac{3}{2} + \frac{9\sqrt3}{4\pi}\right]^{1/2} = 1.6554\,V_m $$ The rms value is almost equal to the average (form factor $= 1.6554/1.6540 = 1.0009$, ripple factor about 4%), so the output is very smooth. Example: for a 400 V line supply, $V_{dc} = 1.35 \times 400 = 540$ V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the three-phase diode bridge, how long does each diode conduct and how often does the pair change? A. 180 degrees; every 90 degrees B. 120 degrees; every 60 degrees C. 120 degrees; every 120 degrees D. 60 degrees; every 30 degrees Answer: B. Each diode conducts 120 degrees and a new diode pair takes over every 60 degrees, giving six pulses per cycle. (Key point: Conduction pattern) 2. What does the output voltage of the bridge follow, and at what ripple frequency (50 Hz supply)? A. The highest phase voltage; 150 Hz B. The average of three phase voltages; 50 Hz C. The upper envelope of the line voltages; 300 Hz D. The lowest line voltage; 100 Hz Answer: C. The load sees the highest line voltage at each instant, a six-pulse waveform with 6f = 300 Hz ripple. (Key point: Output waveform) 3. What is Vdc in terms of the peak line voltage VmL? A. 3VmL/pi = 1.35 VL B. VmL/pi C. 3VmL/(2pi) D. 2VmL/pi Answer: A. Averaging VmL sin(theta) over a 60 degree segment from pi/3 to 2pi/3 gives 3VmL/pi = 1.35 VL. (Key point: Average voltage derivation) 4. Why is the output of this rectifier very smooth? A. Because rms is much larger than average B. Because of the filter inductor built in C. Because each diode conducts 180 degrees D. Because the rms is almost equal to the average (form factor about 1.001) Answer: D. Vrms = 1.6554 Vm vs Vdc = 1.654 Vm, so ripple is small (about 4%). 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