--- title: "Half-controlled bridge with R load, α = 45°" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:half-controlled-bridge-45deg" source: https://www.bctengineer.com/q/semester_7/power_electronics/half-controlled-bridge-45deg site: BCT Engineering --- # Half-controlled bridge with R load, α = 45° Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2073 Chaitra, 8 marks - Video: [Single Phase Semi-controlled Bridge Rectifier with R-Load](https://www.youtube.com/watch?v=4mquz6f6ukI) — Anet Jose - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/half-controlled-bridge-45deg - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Figure shows below a full wave controlled rectifier with resistive load. The load voltage is 230V, 50 Hz and it is operated at firing angle 45°. Draw the waveform of output voltage, output current and ac input current. Calculate average value of output voltage, output dc power and input power factor. [Figure: single-phase half-controlled bridge — thyristors T1, T2 on top and diodes D1, D2 on bottom, fed from Vs = 220V, 50 Hz, feeding R = 10 Ω] ## Short answer Average output voltage V_dc = 176.75 V; output DC power P_dc = V_dc²/R = 3124 W. Input power factor = V_rms/V_s = 0.954 lagging. ## Answer The circuit is a **single-phase half-controlled (semi-converter) bridge**: thyristors T1, T2 on top and diodes D1, D2 at the bottom, feeding $R = 10\ \Omega$. With a resistive load the output is a chopped sine wave from $\alpha$ to $\pi$ in each half cycle. *Assumption:* the supply is taken as 230 V, 50 Hz as stated in the text (the figure shows 220 V; results for 220 V are given at the end). ```text P (+) +-------+-------+ T1 T2 a o----+ +----o b D1 D2 +-------+-------+ N (-) R = 10 ohm between P and N + half: T1 (a->P), D2 (N->b); - half: T2, D1 ``` #### Waveforms | Interval | Conducting | $v_o$ | $i_o$ | $i_s$ | |---|---|---|---|---| | $0$ – $45^\circ$ | none | 0 | 0 | 0 | | $45^\circ$ – $180^\circ$ | T1, D2 | $v_s$ | $v_s/R$ | $+v_s/R$ | | $180^\circ$ – $225^\circ$ | none | 0 | 0 | 0 | | $225^\circ$ – $360^\circ$ | T2, D1 | $-v_s$ | $-v_s/R$ | $v_s/R$ (negative) | ```text vo, io __ __ | \ | \ ---------+---\---------+---\----- is __ | \ ---------+---\---------+---/----- |__/ 45 180 225 360 (deg) ``` #### Average output voltage $V_m = \sqrt2 \times 230 = 325.27$ V, $\alpha = 45^\circ$. $$ \begin{aligned} V_{dc} &= \frac{V_m}{\pi}(1+\cos\alpha) = \frac{325.27}{\pi}(1.7071) = 176.75\ \text{V} \\ I_{dc} &= \frac{V_{dc}}{R} = 17.67\ \text{A} \end{aligned} $$ #### Output dc power $$ P_{dc} = V_{dc}I_{dc} = \frac{V_{dc}^2}{R} = \frac{176.75^2}{10} = 3123.97\ \text{W} $$ #### Input power factor RMS output voltage: $$ \begin{aligned} V_{rms} &= V_s\left[\frac{1}{\pi}\left(\pi-\alpha+\frac{\sin2\alpha}{2}\right)\right]^{1/2} = 230\left[\frac{2.3562+0.5}{\pi}\right]^{1/2} \\ &= 230 \times 0.9535 = 219.30\ \text{V} \end{aligned} $$ With R load, $i_s$ has the same rms value as $i_o$: $I_s = V_{rms}/R = 21.93$ A. All real input power is consumed in R: $$ \begin{aligned} P_{ac} &= \frac{V_{rms}^2}{R} = \frac{219.30^2}{10} = 4809.43\ \text{W} \\ PF &= \frac{P_{ac}}{V_sI_s} = \frac{4809.43}{230 \times 21.93} = \frac{V_{rms}}{V_s} = 0.9535\ \text{(lagging)} \end{aligned} $$ **Answer (230 V): $V_{dc} = 176.75$ V, $P_{dc} = 3124$ W, input PF = 0.954 lagging.** With 220 V (figure value): $V_{dc} = 169.06$ V, $P_{dc} = 2858.2$ W, PF = 0.954 (PF does not depend on the supply voltage). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a half-controlled bridge with R load at alpha = 45 degrees, what is vo from 0 to 45 degrees? A. vo = vs B. Zero, since no thyristor is fired C. vo = -vs D. Constant at Vm Answer: B. With resistive load the output is a chopped sine from alpha to pi; before firing nothing conducts. (Key point: Output waveform) 2. Which formula gives Vdc for the semi-converter with R load? A. (2Vm/pi) cos(alpha) B. (Vm/pi) cos(alpha) C. (Vm/pi)(1 + cos(alpha)) D. Vm/(2*pi) Answer: C. Vdc = (Vm/pi)(1 + cos 45) = 176.75 V for Vm = 325.27 V. (Key point: Average output voltage) 3. What is the output dc power with R = 10 ohm? A. 4809 W B. 3124 W C. 1768 W D. 31.2 kW Answer: B. Pdc = Vdc^2/R = 176.75^2/10 = 3124 W. (Key point: DC power) 4. How is the input power factor found for a resistive load? A. PF = cos(alpha) = 0.707 B. PF = Vdc/Vs C. PF = Vrms/Vs = 0.954 lagging, because Is has the same rms as Io D. PF = 1 Answer: C. All power goes into R, so PF = Vrms^2/R divided by Vs*Vrms/R = Vrms/Vs; it is independent of the supply voltage. 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