--- title: "Three-phase AC-DC with three thyristors, inductive load" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:three-phase-three-thyristor-waveforms" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-three-thyristor-waveforms site: BCT Engineering --- # Three-phase AC-DC with three thyristors, inductive load Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2074 Asoj, 8 marks - 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/three-phase-three-thyristor-waveforms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3 ## Question Explain the operation of three phase AC to DC conversion using three Thyristors. Draw the input and output voltage waveform and find average and rms value of output voltage expression from the obtained waveform. Assume highly inductive load. ## Short answer Three-phase single-way three-pulse controlled rectifier uses three thyristors from a star supply to neutral; with highly inductive load each conducts 120 degrees. Firing angle alpha is measured from the natural commutation point. Vdc = 3 sqrt3 Vm cos(alpha)/(2 pi) = 0.827 Vm cos(alpha), and Vrms = Vm sqrt(1/2 + 3 sqrt3 cos(2 alpha)/(8 pi)), where Vm is peak phase voltage. ## 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. The load is highly inductive, so the load current $I_o$ is constant and each thyristor conducts for $120^\circ$. ```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. #### Input (phase) and output voltage waveforms Taking $\alpha = 30^\circ$ as an example: | Interval ($\omega t$) | Conducting | $v_o$ | |---|---|---| | $60^\circ$ – $180^\circ$ | T1 | $v_a$ | | $180^\circ$ – $300^\circ$ | T2 | $v_b$ | | $300^\circ$ – $420^\circ$ | T3 | $v_c$ | ```text Inputs: va = Vm sin wt, vb = Vm sin(wt - 120), vc = Vm sin(wt - 240) Natural commutation points (where two phase voltages cross): 30, 150, 270 deg Firing points for alpha = 30: 60, 180, 300 deg ``` ```text vo .-. .-. .-. / \ / \ / \ | \ | \ | \ -----+------+-----+------+-----+------+-- 60 180 180 300 300 420 each pulse: 0.866 Vm -> Vm (at 90) -> 0 ``` - For $\alpha = 0$ the output is the upper envelope of the phase voltages. - For $0 < \alpha \le 30^\circ$ the output stays positive. - For $\alpha > 30^\circ$ the end of each pulse goes negative (the inductance keeps the thyristor on); at $\alpha = 90^\circ$ the average becomes zero. - The thyristor current is a $120^\circ$ block of height $I_o$; the phase (input) current of each winding equals its thyristor current. #### Average value 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 $V_{dc} = V_{dm}\cos\alpha$. #### RMS value $$ \begin{aligned} V_{rms}^2 &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m^2\sin^2\omega t\,d(\omega t) = \frac{3V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\pi/6+\alpha}^{5\pi/6+\alpha} \\ &= \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} - \frac{\sin\left(\frac{5\pi}{3}+2\alpha\right) - \sin\left(\frac{\pi}{3}+2\alpha\right)}{2}\right] = \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} + \frac{\sqrt3}{2}\cos2\alpha\right] \end{aligned} $$ (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 $\alpha = 30^\circ$: $V_{dc} = 0.716V_m$, $V_{rms} = 0.7768V_m$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For alpha = 0 with the three-thyristor converter, the output voltage is: A. The lower envelope of the phase voltages B. The upper envelope of the phase voltages C. A pure dc of 0.827 Vm D. The line voltage Answer: B. At alpha = 0 each thyristor conducts when its phase is the most positive, so vo is the upper envelope. (Key point: Output follows conducting phase) 2. With inductive load, what happens to vo when alpha exceeds 30°? A. The end of each pulse goes negative, since the inductance keeps the thyristor on B. The output becomes zero between pulses C. The conduction angle reduces below 120° D. The ripple frequency doubles Answer: A. The inductance keeps the thyristor conducting past its phase zero crossing, so part of each pulse is negative; the average falls to zero at alpha = 90°. (Key point: Output behaviour vs alpha) 3. The rms output voltage for the inductive-load case is: A. Vm [1/2 + (3 sqrt3 / 8pi) cos(2 alpha)]^(1/2) B. Vm [1/2 + (3 sqrt3 / 8pi) cos(alpha)]^(1/2) C. 0.827 Vm cos(alpha) D. Vm [1/2 - (3 sqrt3 / 8pi) cos(2 alpha)]^(1/2) Answer: A. Integrating Vm^2 sin^2(wt) over 120° and multiplying by 3/2pi leads to the cos(2 alpha) term. (Key point: RMS formula) 4. At alpha = 30° the average and rms output voltages are: A. 0.827 Vm and 0.716 Vm B. 0.716 Vm and 0.7768 Vm C. 0.7768 Vm and 0.716 Vm D. 0.477 Vm and 0.7768 Vm Answer: B. Vdc = 0.827 Vm cos 30° = 0.716 Vm; the rms is always larger, 0.7768 Vm. 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