--- title: "Three-phase single-way controlled rectifier, R load" subject: "Power Electronics" chapter: 3 key: "semester_7/power_electronics:three-phase-single-way-r-load" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-single-way-r-load site: BCT Engineering --- # Three-phase single-way controlled rectifier, R load Power Electronics · Chapter 3: Three phase AC to DC conversion - Asked in: 2079 Baishakh, 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-single-way-r-load - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3 ## Question Explain the operation of three phase single way controlled rectifier with purely resistive load. Derive the expression of average and RMS value of output voltage. Draw the waveforms of output voltage for α = 30° and 60° and calculate average value of output voltage. ## Short answer With R load, for α≤30° conduction is continuous; for α>30° it stops at zero until next firing. Vdc=0.827Vm cosα for α≤30° and (3Vm/2π)[1+cos(α+30°)] for 30°<α≤150°; Vrms=Vm[1/2+3√3cos2α/(8π)]^{1/2} for α≤30°, with another expression for discontinuous mode. Assuming 400V line, at α=30° Vdc≈234V; at α=60° Vdc≈156V. ## Answer A **three-phase single-way controlled rectifier** has three thyristors, one per phase of a star-connected supply, with the load between the common cathode and the neutral. With a **purely resistive load** the current has the same shape as the voltage, so a thyristor turns off as soon as its phase voltage reaches zero. Two cases arise: continuous conduction for $\alpha \le 30^\circ$ and discontinuous conduction for $\alpha > 30^\circ$. ```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 - $\alpha$ is measured from the natural commutation point ($30^\circ$ after the phase zero crossing): T1 is fired at $30^\circ+\alpha$, T2 at $150^\circ+\alpha$, T3 at $270^\circ+\alpha$. - **$\alpha \le 30^\circ$:** when T2 is fired, $v_a$ is still positive but $v_b > v_a$, so T1 is commutated. Each thyristor conducts for $120^\circ$ and $v_o$ stays positive (continuous). - **$\alpha > 30^\circ$:** $v_a$ reaches zero at $180^\circ$ before T2 is fired; the current falls to zero and T1 turns off. $v_o = 0$ until the next firing (discontinuous). Each thyristor conducts from $30^\circ + \alpha$ to $180^\circ$, i.e. for $(150^\circ - \alpha)$. #### Waveforms ```text vo (alpha = 30, R load) .-. .-. .-. / \ / \ / \ | \ | \ | \ -----+------+-----+------+-----+------+-- 60 180 180 300 300 420 each pulse: 0.866 Vm -> Vm (at 90) -> 0 ``` ```text vo (alpha = 60, R load) |`-. |`-. |`-. | `-. | `-. | `-. -----+------`+------+------`+------+------`+ 90 180 210 300 330 420 each pulse: Vm (at 90) -> 0 (at 180), then vo = 0 until the next thyristor is fired ``` #### Average and RMS output voltage **Case 1: $0 \le \alpha \le 30^\circ$** (T1 conducts from $\pi/6+\alpha$ to $5\pi/6+\alpha$): $$ \begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\left(\frac{\pi}{6}+\alpha\right) - \cos\left(\frac{5\pi}{6}+\alpha\right)\right] \\ &= \frac{3V_m}{2\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{3} = \frac{3V_m}{2\pi}\cdot\sqrt3\cos\alpha \end{aligned} $$ $$ V_{dc} = \frac{3\sqrt3\,V_m}{2\pi}\cos\alpha = 0.827\,V_m\cos\alpha $$ $$ V_{rms} = V_m\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\cos2\alpha\right]^{1/2} $$ **Case 2: $30^\circ < \alpha \le 150^\circ$** (T1 conducts from $\pi/6+\alpha$ to $\pi$): $$ \begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{\pi}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[1 + \cos\left(\alpha + \frac{\pi}{6}\right)\right] \end{aligned} $$ $$ \begin{aligned} V_{rms}^2 &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{\pi}V_m^2\sin^2\omega t\,d(\omega t) = \frac{3V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\pi/6+\alpha}^{\pi} \\ &= \frac{3V_m^2}{4\pi}\left[\frac{5\pi}{6} - \alpha + \frac{1}{2}\sin\left(\frac{\pi}{3} + 2\alpha\right)\right] \end{aligned} $$ $$ V_{rms} = V_m\left[\frac{3}{4\pi}\left(\frac{5\pi}{6} - \alpha + \frac{1}{2}\sin\left(\frac{\pi}{3}+2\alpha\right)\right)\right]^{1/2} $$ #### Numerical values *Assumption:* no voltage is given, so a 400 V line supply is taken, i.e. $V_{ph} = 230$ V, $V_m = 325.27$ V. $$ \begin{aligned} \alpha = 30^\circ: \quad V_{dc} &= 0.827 \times 325.27 \times \cos30^\circ = 232.96\ \text{V} \\ V_{rms} &= 325.27\left[0.5 + 0.2067 \times 0.5\right]^{1/2} = 252.66\ \text{V} \\ \alpha = 60^\circ: \quad V_{dc} &= \frac{3 \times 325.27}{2\pi}\left[1 + \cos90^\circ\right] = 155.30\ \text{V} \\ V_{rms} &= 325.27\left[\frac{3}{4\pi}\left(\frac{5\pi}{6} - \frac{\pi}{3} + \frac{1}{2}\sin\pi\right)\right]^{1/2} = 325.27\sqrt{0.375} = 199.19\ \text{V} \end{aligned} $$ **Answer: $V_{dc} = 0.716V_m$ (232.96 V) at $\alpha = 30^\circ$ and $0.477V_m$ (155.30 V) at $\alpha = 60^\circ$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With a purely resistive load, why does a thyristor in the three-phase single-way rectifier turn off when its phase voltage reaches zero? A. The load current has the same shape as the voltage, so it falls to zero with the phase voltage B. The next thyristor always fires exactly at that instant C. The inductor discharges completely at that instant D. A commutation capacitor forces it off Answer: A. A resistive load has no energy storage, so the current follows the voltage and reaches zero when the phase voltage reaches zero, turning the thyristor off. (Key point: Turn-off with R load) 2. For the resistive-load three-phase single-way rectifier, conduction is discontinuous when: A. alpha < 30° B. alpha > 30° C. alpha > 90° D. alpha = 0° Answer: B. For alpha <= 30° each thyristor is commutated by the next one; for alpha > 30° the current falls to zero first, so output is zero for a while. (Key point: Continuous vs discontinuous conduction) 3. For 30° < alpha <= 150° with R load, T1 conducts from (pi/6 + alpha) to pi. The average output voltage is: A. (3 sqrt3 Vm / 2pi) cos(alpha) B. (3Vm / 2pi) [1 + cos(alpha + pi/6)] C. (3Vm / pi) [1 + cos(alpha)] D. (3Vm / 2pi) [1 - cos(alpha + pi/6)] Answer: B. Integrating Vm sin(wt) from pi/6+alpha to pi and multiplying by 3/2pi gives (3Vm/2pi)[1 + cos(alpha + pi/6)]. (Key point: Average voltage for alpha > 30°) 4. For R load at alpha = 60°, what is the average output voltage in terms of Vm? A. 0.827 Vm B. 0.716 Vm C. 0.477 Vm D. 0.414 Vm Answer: C. Vdc = (3Vm/2pi)[1 + cos 90°] = 0.477 Vm (155.30 V for Vm = 325.27 V). (Key point: Result at alpha = 60°) ## Close variants set in other exams - [Three-phase single-way controlled rectifier, α = 30°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-30deg) — Power Electronics, chapter 3, asked 4× - [Three-phase single-way thyristor rectifier: average and RMS](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3A3ph-single-way-thyristor-avg-rms) — Power Electronics, chapter 3, asked 5× ## Related in Power Electronics - [Three-phase single-way controlled rectifier, α = 60°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-60deg) — Power Electronics, chapter 3, asked 3× - [Three-phase single-way diode rectifier: average and rms](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3A3ph-single-way-diode-avg-rms) — Power Electronics, chapter 3, asked 2× - [Three-phase single-way controlled rectifier](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-controlled-avg) — Power Electronics, chapter 3, asked 3× - [Three-phase single-way controlled rectifier, α = π/6](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-pi-6) — Power Electronics, chapter 3 - [Three-phase single-way controlled rectifier, α = 90°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-90deg) — Power Electronics, chapter 3, asked 4× - [Three-phase single-way thyristor rectifier: average](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-single-way-thyristor-avg) — Power Electronics, chapter 3, asked 3×