--- title: "Full-wave thyristor rectifier, 220 V, 50 Ω, 45°" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:full-wave-thyristor-220v-50ohm-45" source: https://www.bctengineer.com/q/semester_7/power_electronics/full-wave-thyristor-220v-50ohm-45 site: BCT Engineering --- # Full-wave thyristor rectifier, 220 V, 50 Ω, 45° Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2078 Bhadra, 8 marks - Video: [Single Phase Full Wave Controlled Rectifier with R Load | Power Electronics | Lecture 43](https://www.youtube.com/watch?v=3CAdY-fnPxA) — 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-wave-thyristor-220v-50ohm-45 - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Draw the circuit diagram of single phase full wave rectifier with associated waveforms. A single phase full wave thyristor rectifier has an input voltage of 220V rms. The load is a resistance of 50 ohms and firing angle is 45 degree in each positive half cycle. Find average output voltage, rms output voltage. ## Short answer For a resistive load, conduction from α to π. Vdc = (Vm/π)(1+cos45°) = 169.06 V. Vrms = (Vm/√2)√[(1/π)(π-α+sin2α/2)] = 209.77 V. ## Answer A **single-phase full-wave thyristor rectifier** (fully controlled bridge) with a resistive load conducts from $\alpha$ to $\pi$ in each half cycle. With an R load the current becomes zero at $\pi$, so the thyristors turn off naturally and $v_o$ never goes negative. ```text P (+) +-------+-------+ | | T1 T3 | | a o----+ +----o b | | T4 T2 | | +-------+-------+ N (-) ac supply vs between a and b (is into a) Load R = 50 ohm between P and N ``` #### Operation and waveforms - $0 < \omega t < \alpha$: all thyristors OFF, $v_o = 0$. - $\alpha$ to $\pi$: T1, T2 fired, $v_o = v_s$, $i_o = v_s/R$. - $\pi$ to $\pi+\alpha$: all OFF, $v_o = 0$. - $\pi+\alpha$ to $2\pi$: T3, T4 fired, $v_o = -v_s$ (positive). - The load current has the same shape as $v_o$. ```text vs ___ / \ / ---/-----\-----------/--- \___/ vo __ __ | \ | \ ------+---\-------+---\--- 0 45 180 225 360 (deg) (chopped half-sine pulses, same for io) ``` #### Expressions $$ V_{dc} = \frac{1}{\pi}\int_{\alpha}^{\pi} V_m\sin\omega t\, d(\omega t) = \frac{V_m}{\pi}(1+\cos\alpha) $$ $$ V_{rms} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin2\alpha}{2}\right)\right]^{1/2} $$ #### Calculation $V_m = \sqrt2 \times 220 = 311.13$ V, $\alpha = 45^\circ = 0.7854$ rad. $$ \begin{aligned} V_{dc} &= \frac{311.13}{\pi}(1 + \cos45^\circ) = 99.04 \times 1.7071 = 169.06\ \text{V} \\ V_{rms} &= 220\left[\frac{1}{\pi}\left(\pi - 0.7854 + \frac{\sin 90^\circ}{2}\right)\right]^{1/2} \\ &= 220\left[\frac{2.8562}{\pi}\right]^{1/2} = 220 \times 0.9535 = 209.77\ \text{V} \end{aligned} $$ Load currents: $I_{dc} = 169.06/50 = 3.38$ A, $I_{rms} = 209.77/50 = 4.20$ A. **Answer: $V_{dc} = 169.06$ V, $V_{rms} = 209.77$ V.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does vo never go negative with a resistive load? A. A freewheeling diode is used B. Current becomes zero at pi, so thyristors turn off naturally C. The supply is rectified first D. The gate is turned off Answer: B. With R load, io = vs/R falls to zero at pi, turning the thyristors off before the supply reverses. (Key point: Resistive-load behaviour) 2. Which formula gives Vdc for a resistive load? A. (2Vm/pi) cos alpha B. (Vm/pi)(1 + cos alpha) C. (Vm/2pi)(1 + cos alpha) D. Vm/pi cos alpha Answer: B. Integrating Vm sin wt from alpha to pi over a period of pi gives (Vm/pi)(1 + cos alpha). (Key point: Average voltage formula) 3. What is Vdc for 220 V rms and alpha = 45°? A. 209.77 V B. 198.07 V C. 99.04 V D. 169.06 V Answer: D. Vm = 311.13 V; (311.13/pi)(1 + 0.7071) = 169.06 V. (Key point: Numerical Vdc) 4. Which gives Vrms? A. Vs[(1/pi)(pi - alpha + sin(2 alpha)/2)]^(1/2) = 209.77 V B. Vdc/sqrt(2) = 119.5 V C. Vs cos alpha = 155.6 V D. Vm/2 = 155.6 V Answer: A. Vrms = 220 x sqrt(2.8562/pi) = 209.77 V. 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