--- title: "Single-phase half-wave thyristor rectifier, R load" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:half-wave-thyristor-r-load-rms" source: https://www.bctengineer.com/q/semester_7/power_electronics/half-wave-thyristor-r-load-rms site: BCT Engineering --- # Single-phase half-wave thyristor rectifier, R load Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2076 Chaitra, 8 marks - Video: [Single-Phase Half-Wave Controlled Rectifier with R Load | Working | Power Electronics | SCR](https://www.youtube.com/watch?v=4zyTah3-P0k) — Mind of Engineer - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/half-wave-thyristor-r-load-rms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Explain the operation of single phase half wave rectification using thyristor with resistive load. Draw the waveform of output voltage and the voltage appearing across the thyristor. Derive the expression for RMS value of output voltage. ## Short answer The thyristor conducts from α to π in the positive half-cycle and blocks otherwise. RMS output voltage is derived as Vrms = (Vm/2)√[(1/π)(π-α+sin2α/2)]. ## Answer A **single-phase half-wave controlled rectifier** uses one thyristor in series with the load. The thyristor conducts only in the positive half cycle, from the firing angle $\alpha$ to $\pi$; the output is controlled by changing $\alpha$. ```text T (thyristor) +----->|---------+ | ig -> | (~) vs = R vo | Vm sin wt | +----------------+ ``` #### Operation (resistive load) - $0 < \omega t < \alpha$: thyristor is forward biased but not fired, so it **blocks**; $v_o = 0$, $v_T = v_s$ (positive). - At $\omega t = \alpha$ a gate pulse turns it ON; $v_o = v_s$, $i_o = v_s/R$, $v_T \approx 0$. - At $\omega t = \pi$, $v_s$ and hence $i_o$ fall to zero; the thyristor turns off naturally. - $\pi < \omega t < 2\pi$: thyristor is reverse biased; $v_o = 0$, $v_T = v_s$ (negative). The thyristor must withstand peak reverse voltage $V_m$ and peak forward voltage $V_m$ (if $\alpha \ge 90^\circ$). #### Waveforms ```text vs ___ / \ --/-----\---------/-- \_______/ vo _ | \ ------+--\----------+-- vT _ / | --/--+---+---------/-- \_______/ 0 a 180 360 (deg) vo: sine from a to 180; vT: vs from 0 to a and from 180 to 360 (reverse), zero while ON ``` #### Average output voltage $$ V_{dc} = \frac{1}{2\pi}\int_{\alpha}^{\pi} V_m\sin\omega t\, d(\omega t) = \frac{V_m}{2\pi}(1+\cos\alpha) $$ #### Derivation of RMS output voltage $$ \begin{aligned} V_{rms}^2 &= \frac{1}{2\pi}\int_{\alpha}^{\pi} V_m^2\sin^2\omega t\, d(\omega t) \\ &= \frac{V_m^2}{4\pi}\int_{\alpha}^{\pi}(1-\cos2\omega t)\, d(\omega t) \\ &= \frac{V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\alpha}^{\pi} \\ &= \frac{V_m^2}{4\pi}\left[\pi - \alpha + \frac{\sin2\alpha}{2}\right] \end{aligned} $$ $$ V_{rms} = \frac{V_m}{2}\left[\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin2\alpha}{2}\right)\right]^{1/2} $$ Check: at $\alpha = 0$, $V_{rms} = V_m/2$, the value for an uncontrolled half-wave rectifier. RMS load current is $I_{rms} = V_{rms}/R$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Between 0 and alpha, what is the state of the thyristor and the voltage across it? A. ON, vT = 0 B. OFF but forward biased, vT = vs C. Reverse biased, vT = -vs D. ON with vo = vs Answer: B. Without a gate pulse it blocks, so vo = 0 and the thyristor carries the full supply voltage. (Key point: Blocking before firing) 2. What is the correct starting expression for Vrms^2? A. (1/pi) integral of Vm^2 sin^2 from 0 to pi B. (1/2pi) integral of Vm^2 sin^2 from alpha to pi C. (1/2pi) integral of Vm sin from alpha to pi D. (1/pi) integral of Vm^2 sin^2 from alpha to 2pi Answer: B. The output conducts from alpha to pi in a period of 2pi, and rms uses the square of the voltage. (Key point: RMS derivation start) 3. What is Vrms for the half-wave thyristor circuit? A. Vm/2 [(1/pi)(pi - alpha + sin(2 alpha)/2)]^(1/2) B. Vm/sqrt(2) cos alpha C. (Vm/2pi)(1 + cos alpha) D. Vm/2 sin alpha Answer: A. Integrating (1 - cos 2wt)/2 gives the stated result, reducing to Vm/2 at alpha = 0. 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