--- title: "Full controlled rectifier, 25 A, α = 30°, derivation" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:full-controlled-rectifier-25a-30deg-derive" source: https://www.bctengineer.com/q/semester_7/power_electronics/full-controlled-rectifier-25a-30deg-derive site: BCT Engineering --- # Full controlled rectifier, 25 A, α = 30°, derivation Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2080 Bhadra, 8 marks - Video: [Single Phase Full Wave Controlled Rectifier with RL Load | Power Electronics | Lecture 44](https://www.youtube.com/watch?v=CHl408osl2s) — 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/full-controlled-rectifier-25a-30deg-derive - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Draw the circuit diagram of a single phase full controlled rectifier having supply Input RMS voltage of 230 V, 50 Hz. with a highly inductive load so that load current is constant and equal to 25 Amp. i) Explain its operation for firing angle α = 30°. ii) Derive the expression for average and RMS values of the output voltage iii) Calculate the value of average output voltage. ## Short answer With α=30°, T1-T2 conduct from 30° to 210° and T3-T4 from 210° to 390°, maintaining constant 25 A. Average output voltage, found by integrating vs over a half cycle, is Vdc = (2Vm/π)cosα = 179.33 V; rms output equals 230 V. ## Answer A **single-phase full controlled rectifier (full converter)** uses four thyristors in a bridge. With a highly inductive load the load current is continuous and constant ($I_o = 25$ A), so each thyristor pair conducts for $180^\circ$. ```text P (+) +-------+-------+ | | T1 T3 | | a o----+ +----o b | | T4 T2 | | +-------+-------+ N (-) ac supply vs between a and b (is into a) Load (R-L, Io) between P and N ``` #### i) Operation for α = 30° Data: $V_s = 230$ V, $V_m = \sqrt2 \times 230 = 325.27$ V, $\alpha = 30^\circ$. | Interval ($\omega t$) | Conducting | $v_o$ | $i_s$ | |---|---|---|---| | $30^\circ$ – $180^\circ$ | T1, T2 | $+v_s$ (positive) | $+25$ A | | $180^\circ$ – $210^\circ$ | T1, T2 | $+v_s$ (negative) | $+25$ A | | $210^\circ$ – $360^\circ$ | T3, T4 | $-v_s$ (positive) | $-25$ A | | $360^\circ$ – $390^\circ$ | T3, T4 | $-v_s$ (negative) | $-25$ A | - At $\omega t = 30^\circ$, T1 and T2 are fired; the load is connected to $v_s$. - After $180^\circ$, $v_s$ becomes negative but T1, T2 keep conducting because the inductance keeps the current flowing; $v_o$ goes negative for $30^\circ$. - At $210^\circ$, T3 and T4 are fired. They apply reverse voltage to T1, T2, which turn off (**natural/line commutation**). The cycle repeats. - Input current $i_s$ is a square wave of $\pm 25$ A, lagging $v_s$ by $\alpha = 30^\circ$. ```text vs ___ ___ / \ / -------/-----\-----------/---- \___/ vo: follows vs from 30 to 210 deg, then -vs from 210 to 390 deg (goes below zero for 30 deg in each half cycle) io: constant 25 A is +Io +---------+ +--- | | | 0 ------+ | | +---------+ -Io 30 210 390 (deg) ``` #### ii) Average and RMS output voltage Output repeats every $\pi$; $v_o = V_m\sin\omega t$ from $\alpha$ to $\pi + \alpha$: $$ \begin{aligned} V_{dc} &= \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m \sin\omega t\, d(\omega t) = \frac{V_m}{\pi}\big[-\cos\omega t\big]_{\alpha}^{\pi+\alpha} \\ &= \frac{2V_m}{\pi}\cos\alpha \end{aligned} $$ $$ \begin{aligned} V_{rms} &= \left[\frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m^2 \sin^2\omega t\, d(\omega t)\right]^{1/2} = \left[\frac{V_m^2}{2\pi}\int_{\alpha}^{\pi+\alpha}(1-\cos 2\omega t)\, d(\omega t)\right]^{1/2} \\ &= \frac{V_m}{\sqrt2} = V_s \end{aligned} $$ So the rms output equals the supply rms (230 V) for any $\alpha$ in continuous conduction. #### iii) Average output voltage $$ \begin{aligned} V_{dc} &= \frac{2 \times 325.27}{\pi}\cos 30^\circ = 207.07 \times 0.866 \\ &= 179.33\ \text{V} \end{aligned} $$ **Answer: $V_{dc} = 179.33$ V, $V_{rms} = 230$ V.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At 210° when T3 and T4 are fired, what happens to T1 and T2? A. They keep conducting B. They turn off by line commutation C. They are turned off by gate pulse D. They fail due to overvoltage Answer: B. Firing T3, T4 while vs is negative reverse biases T1, T2 and turns them off naturally. (Key point: Thyristor turn-off) 2. Why does vo go negative from 180° to 210°? A. The supply has failed B. A freewheeling diode conducts C. The inductance keeps T1, T2 conducting after vs reverses D. T3 and T4 conduct early Answer: C. The large inductance keeps load current flowing through T1, T2, so vo = vs which is negative for alpha degrees. (Key point: Negative output interval) 3. What does integrating Vm sin wt from alpha to pi + alpha over period pi give? A. (Vm/pi)(1 + cos alpha) B. (2Vm/pi) cos alpha C. (2Vm/pi) sin alpha D. Vm/pi cos alpha Answer: B. (Vm/pi)[-cos wt] from alpha to pi + alpha = (2Vm/pi) cos alpha. (Key point: Average output voltage derivation) 4. What is the rms output voltage in continuous conduction, and does it depend on alpha? A. Vm/sqrt(2) = Vs, independent of alpha B. Vs cos alpha C. Vm/2 D. 2Vm/pi, independent of alpha Answer: A. The squared waveform sin^2 integrates over a half-period regardless of start angle, so Vrms = 230 V. 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