--- title: "Three-phase controlled rectifier with three thyristors" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:three-phase-three-thyristor-avg-rms" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-three-thyristor-avg-rms site: BCT Engineering --- # Three-phase controlled rectifier with three thyristors Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2081 Bhadra, 8 marks - Video: [3 phase half wave controlled rectifier working & waveforms in Power Electronics by Engineering Funda](https://www.youtube.com/watch?v=alFjAYCu3yo) — Engineering Funda - 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-three-thyristor-avg-rms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3 ## Question Explain the operation of three-phase controlled rectifier with 3 numbers of thyristors. Derive the expression for Average and RMS values of the output voltage. ## Short answer Three thyristors connected to the three phases, cathodes common to load, neutral return; highly inductive load keeps current continuous, each conducts 120°. Average voltage is (3√3/2π)Vm cosα = 0.827Vm cosα; RMS voltage is Vm√(1/2 + (3√3/8π)cos2α). ## Answer A **three-phase single-way (half-wave, three-pulse) controlled rectifier** uses three thyristors, one per phase, connected to a star-connected supply (or transformer secondary) with the load returned to the neutral. The output voltage is controlled by delaying the firing of each thyristor by $\alpha$ from its natural commutation point. Because it uses three thyristors it is also called a three-pulse midpoint converter. The load is assumed highly inductive so that the current is continuous. ```text T1 a o------>|-----+ T2 | b o------>|-----+------o P (+) T3 | | 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 thyristor's anode goes to one phase and the cathodes are joined to the positive output. A thyristor can conduct only when its phase is the most positive **and** it has been fired. - The firing angle $\alpha$ is measured from the **natural commutation point** (the instant a diode would start conducting), i.e. from $30^\circ$ after the zero crossing of its phase voltage. So T1 is fired at $\omega t = 30^\circ + \alpha$, T2 at $150^\circ + \alpha$, T3 at $270^\circ + \alpha$. - When T2 is fired, $v_b > v_a$, so T1 becomes reverse biased and turns off (**natural/line commutation**). - With a **highly inductive load**, the load current is continuous and constant, so each thyristor conducts for exactly $120^\circ$, even if its phase voltage becomes negative. The output follows the phase of the conducting thyristor. | Interval ($\omega t$) | Conducting | $v_o$ | |---|---|---| | $30^\circ+\alpha$ – $150^\circ+\alpha$ | T1 | $v_a$ | | $150^\circ+\alpha$ – $270^\circ+\alpha$ | T2 | $v_b$ | | $270^\circ+\alpha$ – $390^\circ+\alpha$ | T3 | $v_c$ | ```text vo (alpha = 30 deg shown) .-. .-. .-. / \ / \ / \ | \ | \ | \ -----+------+-----+------+-----+------+-- 60 180 180 300 300 420 iT1 +------+ (120 deg block of Io) For alpha > 30 the end of each pulse dips below zero; at alpha = 90 the average is 0 ``` #### Average output voltage Taking the conduction of T1 (the waveform repeats every $2\pi/3$): $$ \begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\left(\frac{\pi}{6}+\alpha\right) - \cos\left(\frac{5\pi}{6}+\alpha\right)\right] \\ &= \frac{3V_m}{2\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{3} = \frac{3V_m}{2\pi}\cdot\sqrt3\cos\alpha \end{aligned} $$ $$ V_{dc} = \frac{3\sqrt3\,V_m}{2\pi}\cos\alpha = 0.827\,V_m\cos\alpha $$ where $V_m$ is the peak phase voltage. Maximum value (at $\alpha = 0$) is $V_{dm} = 0.827V_m$, so $V_{dc} = V_{dm}\cos\alpha$. - $0 \le \alpha < 90^\circ$: $V_{dc} > 0$, rectifier mode. - $\alpha = 90^\circ$: $V_{dc} = 0$. - $90^\circ < \alpha < 180^\circ$: $V_{dc} < 0$, inverter mode (needs a dc source in the load). #### RMS output voltage $$ \begin{aligned} V_{rms}^2 &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m^2\sin^2\omega t\,d(\omega t) = \frac{3V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\pi/6+\alpha}^{5\pi/6+\alpha} \\ &= \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} - \frac{\sin\left(\frac{5\pi}{3}+2\alpha\right) - \sin\left(\frac{\pi}{3}+2\alpha\right)}{2}\right] = \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} + \frac{\sqrt3}{2}\cos2\alpha\right] \end{aligned} $$ (using $\sin\left(\frac{5\pi}{3}+2\alpha\right) - \sin\left(\frac{\pi}{3}+2\alpha\right) = 2\cos(\pi+2\alpha)\sin\frac{2\pi}{3} = -\sqrt3\cos2\alpha$) $$ V_{rms} = V_m\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\cos2\alpha\right]^{1/2} $$ At $\alpha = 0$ these reduce to the diode rectifier values $0.827V_m$ and $0.8407V_m$. #### Thyristor ratings Average thyristor current $= I_o/3$, rms $= I_o/\sqrt3$, peak reverse (and forward) voltage $= \sqrt3V_m$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the three-thyristor controlled rectifier with highly inductive load, how long does each thyristor conduct? A. 60 degrees B. 120 degrees, regardless of alpha C. 180 degrees D. 120 - alpha degrees Answer: B. The current is continuous, so each thyristor conducts exactly 120 degrees until the next phase thyristor is fired. (Key point: Thyristor conduction) 2. What turns off T1 when T2 is fired? A. vb > va reverse biases T1 (natural commutation) B. The gate current of T1 is removed C. A commutating capacitor D. The load current falling to zero Answer: A. Once T2 conducts, vb exceeds va, so T1 is reverse biased and turns off by line commutation. (Key point: Turn-off mechanism) 3. What is Vdc? A. 1.17 Vm cos(alpha) B. 0.827 Vm cos(alpha) C. 0.827 Vm (independent of alpha) D. 0.675 Vm sin(alpha) Answer: B. Vdc = (3 sqrt(3) Vm/(2 pi)) cos(alpha) = 0.827 Vm cos(alpha). (Key point: Average voltage) 4. What is the rms output voltage, and what happens at alpha = 0? A. Vm[1/2 + (3 sqrt(3)/(8 pi)) cos(2 alpha)]^(1/2); it equals 0.8407 Vm B. Vm[1/2 + (3 sqrt(3)/(8 pi)) cos(alpha)]^(1/2); it equals 0.827 Vm C. 0.827 Vm cos(2 alpha); it equals 0.827 Vm D. Vm/sqrt(2) independent of alpha; it equals 0.707 Vm Answer: A. The rms depends on cos(2 alpha) and reduces to the diode value 0.8407 Vm at alpha = 0. 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