--- title: "Full converter, 25 A, α = 30°, input power factor" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:full-converter-25a-30deg-pf" source: https://www.bctengineer.com/q/semester_7/power_electronics/full-converter-25a-30deg-pf site: BCT Engineering --- # Full converter, 25 A, α = 30°, input power factor Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2074 Asoj, 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-converter-25a-30deg-pf - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Figure shows a single phase full converter circuit with highly inductive load so that load current is constant and equal to 25 amp. Explain its operation for firing angle = 30°. Draw the waveforms of input voltage Vs, output voltage V0 and input current is. Calculate the fundamental component of input current and Input power factor. [Figure: single-phase full converter — four thyristors T1, T3 (top) and T4, T2 (bottom) in a bridge fed from Vs = 230 V, 50 Hz with input current is; output Vo across the load with load current Io] ## Short answer With constant load current 25 A and α=30°, input current is a square wave. Its fundamental rms component is I_s1 = (2√2/π)×25 = 22.51 A, lagging vs by 30°. The input power factor is (I_s1/I_s)cos30° = 0.780 lagging. ## Answer A **single-phase full converter** (four thyristors) with a highly inductive load has a constant load current, here $I_o = 25$ A. Data: $V_s = 230$ V, 50 Hz, $V_m = 325.27$ V, $\alpha = 30^\circ$. ```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 ``` #### Operation for α = 30° - **Positive half cycle:** T1 and T2 are forward biased and are fired at $\omega t = \alpha$. Load is connected to the supply, $v_o = v_s$, $i_s = +I_o$. - At $\omega t = \pi$, $v_s$ reverses, but the large inductance keeps $I_o$ flowing, so T1, T2 continue to conduct and $v_o$ becomes negative from $\pi$ to $\pi+\alpha$ (energy returns from the inductance to the supply). - At $\omega t = \pi+\alpha$, T3 and T4 are fired. Since $v_s < 0$, they apply reverse voltage across T1, T2, which turn off by **natural (line) commutation**. Now $v_o = -v_s$, $i_s = -I_o$. - At $2\pi + \alpha$ T1, T2 are fired again and the cycle repeats. - Each pair conducts for $180^\circ$; the input current is a square wave of amplitude $I_o$ lagging the voltage by $\alpha$. | 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 | #### Waveforms of $v_s$, $v_o$ and $i_s$ ```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) ``` Average output: $V_{dc} = \frac{2V_m}{\pi}\cos\alpha = 207.07 \times 0.866 = 179.33$ V. #### Fundamental component of input current The input current is a square wave of amplitude $I_o$, displaced by $\alpha$ from $v_s$. Its Fourier series is $$ i_s(t) = \sum_{n=1,3,5,\ldots}\frac{4I_o}{n\pi}\sin(n\omega t - n\alpha) $$ so the fundamental has peak $\frac{4I_o}{\pi}$ and rms value $$ I_{s1} = \frac{4I_o}{\sqrt2\,\pi} = \frac{2\sqrt2}{\pi}I_o = 0.9003\,I_o, \quad \text{lagging } v_s \text{ by } \phi_1 = \alpha $$ $$ I_{s1} = 0.9003 \times 25 = 22.51\ \text{A (rms)}, \qquad \phi_1 = \alpha = 30^\circ\ \text{lagging} $$ #### Input power factor RMS input current $I_s = I_o = 25$ A (square wave). $$ \begin{aligned} \text{DF} &= \cos\phi_1 = \cos30^\circ = 0.866 \\ \text{PF} &= \frac{I_{s1}}{I_s}\cos\phi_1 = \frac{22.51}{25} \times 0.866 = 0.9003 \times 0.866 = 0.780\ \text{(lagging)} \end{aligned} $$ Check by power balance: $P = V_{dc}I_o = 179.33 \times 25 = 4483.3$ W; $V_sI_s = 230 \times 25 = 5750$ VA; $PF = 4483.3/5750 = 0.780$. **Answer: $I_{s1} = 22.51$ A (rms) lagging by $30^\circ$; input PF = 0.780 lagging.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For alpha = 30 degrees, what happens to vo between 180 and 210 degrees? A. vo is zero because all thyristors are off B. vo follows vs, so it is negative while T1, T2 still conduct C. vo equals +325 V D. vo equals the battery voltage Answer: B. Inductance keeps T1, T2 on after vs reverses, so vo = vs is negative for alpha degrees each half cycle. (Key point: Output voltage during 180-210 degrees) 2. What is the average output voltage for Vm = 325.27 V and alpha = 30 degrees? A. 207.07 V B. 230.00 V C. 103.5 V D. 179.33 V Answer: D. Vdc = (2Vm/pi) cos(alpha) = 207.07 x 0.866 = 179.33 V. (Key point: Average output voltage) 3. What is the fundamental rms input current for Io = 25 A and its phase? A. 25 A lagging by 30 degrees B. 22.51 A lagging vs by 30 degrees C. 22.51 A leading vs by 30 degrees D. 28.0 A lagging by 60 degrees Answer: B. Is1 = 0.9003 x 25 = 22.51 A, and the fundamental lags the voltage by alpha. (Key point: Fundamental input current) 4. How is the input power factor found, and what is it? A. cos(30) = 0.866, ignoring distortion B. (Is1/Is) cos(alpha) = 0.9003 x 0.866 = 0.780 lagging C. Is1/Is = 0.9003 D. cos(60) = 0.5 Answer: B. PF = distortion factor x displacement factor. Check: P/VA = 4483.3/5750 = 0.780. 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