--- title: "Three-phase fully controlled bridge with R load" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:3ph-full-bridge-controlled-r-load" source: https://www.bctengineer.com/q/semester_7/power_electronics/3ph-full-bridge-controlled-r-load site: BCT Engineering --- # Three-phase fully controlled bridge with R load Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2082 Baishakh, 3+3+2 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-controlled-r-load - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3 ## Question Explain the operation of a three phase fully controlled bridge rectifier with resistive load. Derive the expression of average and RMS value of output voltage. Draw the waveform of output voltage for α = 0° and α = 30°. ## Short answer A six-thyristor bridge fires T1-T6 in order, two conducting at a time, so the load receives line-to-line voltage segments. For resistive load conduction is continuous for alpha <=60° and discontinuous above. Integrating one 60° segment gives Vdc=(3√3/π)Vm cosα and Vrms=√3Vm[1/2+(3√3/4π)cos2α]^1/2 in continuous mode. ## Answer A **three-phase fully controlled bridge rectifier** (six-pulse full converter) uses six thyristors. With a resistive load the output current follows the output voltage, so conduction is continuous for $\alpha \le 60^\circ$ and discontinuous for $\alpha > 60^\circ$ (the line voltage reaches zero before the next thyristor is fired). ```text P (+) o-----+------+------+-----+ | | | | T1 T3 T5 | | | | R a o----------------+ | | | b o-----------------------+ | | c o------------------------------+ | | | | | T4 T6 T2 | | | | | N (-) o-----+------+------+-----+ (a joins T1-T4, b joins T3-T6, c joins T5-T2) ``` #### Operation (3 marks) - T1, T3, T5 form the positive group and T4, T6, T2 the negative group. Thyristors are fired in the sequence T1, T2, T3, T4, T5, T6 at $60^\circ$ intervals; two thyristors (one from each group) conduct at a time, so the load sees a line voltage. - $\alpha$ is measured from the natural commutation point, $30^\circ$ after the zero crossing of the phase voltage: T1 is fired at $\omega t = 30^\circ + \alpha$ (with respect to $v_a$). - Each thyristor needs a double pulse (or a $60^\circ$+ wide pulse) so that the incoming thyristor and its partner are both on at start-up. | $\omega t$ (w.r.t. $v_a$) | Conducting | $v_o$ | |---|---|---| | $30^\circ+\alpha$ – $90^\circ+\alpha$ | T6, T1 | $v_{ab}$ | | $90^\circ+\alpha$ – $150^\circ+\alpha$ | T1, T2 | $v_{ac}$ | | $150^\circ+\alpha$ – $210^\circ+\alpha$ | T2, T3 | $v_{bc}$ | | $210^\circ+\alpha$ – $270^\circ+\alpha$ | T3, T4 | $v_{ba}$ | | $270^\circ+\alpha$ – $330^\circ+\alpha$ | T4, T5 | $v_{ca}$ | | $330^\circ+\alpha$ – $390^\circ+\alpha$ | T5, T6 | $v_{cb}$ | - With R load, for $\alpha \le 60^\circ$ each segment of $v_o$ stays positive and each thyristor conducts $120^\circ$. - For $\alpha > 60^\circ$, the line voltage becomes zero inside the segment, the current drops to zero and the thyristors turn off; $v_o = 0$ until the next pair is fired. #### Average and RMS output voltage (3 marks) **For $0 \le \alpha \le 60^\circ$ (continuous):** Using a line voltage $v_{ab} = V_{mL}\sin\theta$ ($V_{mL} = \sqrt3V_m$); the $60^\circ$ segment runs from $\theta = \pi/3 + \alpha$ to $2\pi/3 + \alpha$: $$ \begin{aligned} V_{dc} &= \frac{3}{\pi}\int_{\pi/3+\alpha}^{2\pi/3+\alpha}V_{mL}\sin\theta\,d\theta = \frac{3V_{mL}}{\pi}\left[\cos\left(\frac{\pi}{3}+\alpha\right) - \cos\left(\frac{2\pi}{3}+\alpha\right)\right] \\ &= \frac{3V_{mL}}{\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{6} = \frac{3V_{mL}}{\pi}\cos\alpha = \frac{3\sqrt3V_m}{\pi}\cos\alpha \end{aligned} $$ $$ \begin{aligned} V_{rms}^2 &= \frac{3}{\pi}\int_{\pi/3+\alpha}^{2\pi/3+\alpha}V_{mL}^2\sin^2\theta\,d\theta = \frac{3V_{mL}^2}{2\pi}\left[\frac{\pi}{3} - \frac{\sin\left(\frac{4\pi}{3}+2\alpha\right) - \sin\left(\frac{2\pi}{3}+2\alpha\right)}{2}\right] \\ &= \frac{3V_{mL}^2}{2\pi}\left[\frac{\pi}{3} + \frac{\sqrt3}{2}\cos2\alpha\right] \end{aligned} $$ $$ V_{rms} = V_{mL}\left[\frac{1}{2} + \frac{3\sqrt3}{4\pi}\cos2\alpha\right]^{1/2} = \sqrt3V_m\left[\frac{1}{2} + \frac{3\sqrt3}{4\pi}\cos2\alpha\right]^{1/2} $$ **For $60^\circ < \alpha \le 120^\circ$ (discontinuous):** each segment runs from $\pi/3 + \alpha$ to $\pi$: $$ V_{dc} = \frac{3V_{mL}}{\pi}\left[1 + \cos\left(\alpha + \frac{\pi}{3}\right)\right], \qquad V_{rms} = V_{mL}\left[\frac{3}{2\pi}\left(\frac{2\pi}{3} - \alpha + \frac{1}{2}\sin\left(2\alpha + \frac{2\pi}{3}\right)\right)\right]^{1/2} $$ Output becomes zero at $\alpha = 120^\circ$. #### Waveforms for α = 0° and α = 30° (2 marks) ```text alpha = 0 (same as diode bridge) vo vab vac vbc vba vca vcb _/\__/\__/\__/\__/\__/\_ peak VmL, 30 90 150 210 270 330 min 0.866 VmL alpha = 30 vo each 60-deg segment starts at VmL (peak of the line voltage) and falls to 0.5 VmL: |`-.|`-.|`-.|`-.|`-.|`-. 60 120 180 240 300 360 (deg) ``` - $\alpha = 0^\circ$: $V_{dc} = \frac{3V_{mL}}{\pi} = 1.35V_L$ (same as diode bridge); six pulses per cycle, ripple at 300 Hz. - $\alpha = 30^\circ$: each pulse is the falling part of a line voltage from its peak $V_{mL}$ to $0.5V_{mL}$; $V_{dc} = 1.35V_L\cos30^\circ = 1.169V_L$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the three-phase fully controlled bridge, what is the firing sequence and spacing? A. T1, T3, T5, T2, T4, T6 at 120 degree intervals B. T1 to T6 in order at 60 degree intervals C. T1 to T6 in order at 30 degree intervals D. All six together every 60 degrees Answer: B. Thyristors are fired in the order T1, T2, ... T6 at 60 degree intervals, with two (one from each group) conducting at a time. (Key point: Firing sequence) 2. With resistive load, for which alpha is the output continuous? A. alpha up to 30 degrees B. alpha up to 90 degrees C. alpha up to 60 degrees D. Any alpha up to 120 degrees Answer: C. For alpha <= 60 degrees the line voltage stays positive; beyond that it reaches zero and the current stops, so conduction is discontinuous. (Key point: Continuous vs discontinuous) 3. For 0 <= alpha <= 60 degrees, what is Vdc? A. (3 VmL/pi) cos(alpha) = 1.35 VL cos(alpha) B. (3 VmL/(2pi)) cos(alpha) C. (3 VmL/pi)(1 + cos alpha) D. 1.35 VL sin(alpha) Answer: A. Integration over the 60 degree line-voltage segment from pi/3 + alpha to 2pi/3 + alpha gives (3 VmL/pi) cos(alpha). (Key point: Average output voltage) 4. What does each pulse look like for alpha = 30 degrees, and what is Vdc? A. It begins at 0.5 VmL and rises to VmL; Vdc = 1.35 VL B. It follows the peak of the line voltage; Vdc = 1.35 VL cos 60 C. It is zero for part of each segment; Vdc = 0 D. It starts at VmL and falls to 0.5 VmL; Vdc = 1.169 VL Answer: D. At alpha = 30 each segment is the falling part of a line voltage from peak to 0.5 VmL, and Vdc = 1.35 VL cos 30 = 1.169 VL. 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