--- title: "Three-phase full bridge rectifier with R load" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:3ph-full-bridge-r-load-avg" source: https://www.bctengineer.com/q/semester_7/power_electronics/3ph-full-bridge-r-load-avg site: BCT Engineering --- # Three-phase full bridge rectifier with R load Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2078 Bhadra, 8 marks - Video: [Analysis of 3 Phase Full Wave Controlled Rectifier with R load | Power Electronics | Lecture 58](https://www.youtube.com/watch?v=oW82hMvHlAc) — 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-full-bridge-r-load-avg - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3 ## Question Draw and explain the output waveforms of three phase full bridge rectifier with R load. Find the average value of output voltage. ## Short answer The six-diode bridge connects the load to the largest line voltage; each diode conducts 120° with commutation every 60°, giving a six-pulse output between 0.866VmL and VmL. Integrating one line-voltage segment gives Vdc=(3/π)VmL=1.35VL=1.654Vm, e.g. 540V for a 400V line supply. ## Answer A **three-phase full bridge rectifier** with six diodes (or six thyristors at $\alpha = 0$) connects the load at every instant to the **largest line voltage**: the most positive phase through the upper group (D1, D3, D5) and the most negative phase through the lower group (D4, D6, D2). ```text P (+) o-----+------+------+-----+ | | | | D1 D3 D5 | | | | R a o----------------+ | | | b o-----------------------+ | | c o------------------------------+ | | | | | D4 D6 D2 | | | | | N (-) o-----+------+------+-----+ (a joins D1-D4, b joins D3-D6, c joins D5-D2) ``` #### Operation and output waveform (R load) | $\omega t$ (w.r.t. $v_a$) | Conducting | $v_o$ | |---|---|---| | $30^\circ$ – $90^\circ$ | 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 line voltages (peak VmL), six per cycle: vo vab vac vbc vba vca vcb _/\__/\__/\__/\__/\__/\_ max VmL min 0.866 VmL 30 90 150 210 270 330 390 (deg) io = vo / R (same shape) ``` - Each diode conducts for $120^\circ$; a commutation occurs every $60^\circ$. - Output is a **six-pulse** wave (ripple frequency $6f = 300$ Hz for 50 Hz), varying between $0.866V_{mL}$ and $V_{mL}$; with R load the current has the same shape. - Each phase current is two $120^\circ$ pulses (positive and negative) per cycle, so no dc flows in the supply. #### Average output voltage Taking one segment of $v_{ab} = V_{mL}\sin\theta$ from $\pi/3$ to $2\pi/3$ (period $\pi/3$): $$ \begin{aligned} V_{dc} &= \frac{3}{\pi}\int_{\pi/3}^{2\pi/3}V_{mL}\sin\theta\,d\theta = \frac{3V_{mL}}{\pi}\left[-\cos\theta\right]_{\pi/3}^{2\pi/3} = \frac{3V_{mL}}{\pi}\left[\frac{1}{2} + \frac{1}{2}\right] \\ &= \frac{3V_{mL}}{\pi} = \frac{3\sqrt2\,V_L}{\pi} = 1.35\,V_L = 1.654\,V_m \end{aligned} $$ If the bridge uses thyristors fired at $\alpha$ (R load), $V_{dc} = 1.35V_L\cos\alpha$ for $\alpha \le 60^\circ$ and $V_{dc} = 1.35V_L\left[1+\cos(\alpha+60^\circ)\right]$ for $60^\circ < \alpha \le 120^\circ$. Example: 400 V, 50 Hz supply gives $V_{dc} = 1.35 \times 400 = 540$ V, and $I_{dc} = 540/R$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In a three-phase full bridge rectifier, at any instant the load is connected to: A. The smallest phase voltage B. The largest line voltage, via the most positive and most negative phases C. A single phase only D. The average of the three phase voltages Answer: B. The most positive phase conducts through the upper group and the most negative phase through the lower group, so the largest line voltage appears across the load. (Key point: Which devices conduct) 2. In the six-diode bridge, how long does each diode conduct and how often does commutation occur? A. 60° each, commutation every 30° B. 180° each, commutation every 90° C. 120° each, commutation every 60° D. 120° each, commutation every 120° Answer: C. Each diode conducts for 120° and a new pair takes over every 60°, giving six pulses per cycle. (Key point: Conduction and commutation) 3. For a 50 Hz supply, the output of the three-phase full bridge has: A. Ripple frequency 300 Hz, varying between 0.866 VmL and VmL B. Ripple frequency 150 Hz, varying between 0 and VmL C. Ripple frequency 100 Hz, varying between 0.5 VmL and VmL D. Ripple frequency 50 Hz, varying between 0.866 VmL and VmL Answer: A. It is a six-pulse wave, so ripple is 6f = 300 Hz, and the line-voltage envelope varies between 0.866 VmL and VmL. (Key point: Output ripple) 4. The average output voltage of the three-phase full bridge with line rms voltage VL is: A. 1.17 VL B. 0.827 VL C. 1.35 VL D. 0.955 VL Answer: C. Vdc = 3 VmL / pi = 3 sqrt2 VL / pi = 1.35 VL, e.g. 540 V for a 400 V supply. 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