--- title: "Full converter, 25 A, α = 30°, fundamental input current" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:full-converter-25a-30deg" source: https://www.bctengineer.com/q/semester_7/power_electronics/full-converter-25a-30deg site: BCT Engineering --- # Full converter, 25 A, α = 30°, fundamental input current Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2079 Bhadra, 8 marks - Video: [single phase full wave controlled rectifier with r load | single phase full wave bridge converter](https://www.youtube.com/watch?v=v4PfolYHB3s) — 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/full-converter-25a-30deg - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Figure shows single phase full converter circuit with highly inductive load so that load current is constant and equal to 25 A. Explain its operation for firing angle of 30°. Draw the input and output waveforms of voltages and currents. Calculate the fundamental component of input AC current. [Figure: single-phase full converter — thyristors T1, T3 (top) and T4, T4 [sic] (bottom) in a bridge fed from 230 V, 50 Hz with input current Is; output V0 across the LOAD with current I0] ## Short answer Thyristor pairs are fired at α=30° and conduct for 180°; output voltage includes a negative portion due to inductive load. Fundamental input current rms = (2√2/π)×25 = 22.51 A, lagging the supply voltage by the firing angle 30°. ## Answer A **single-phase full converter** uses four thyristors in a bridge. With a highly inductive load the current is continuous and constant ($I_o = 25$ A), so it can work as a rectifier ($\alpha < 90^\circ$) or inverter ($\alpha > 90^\circ$, with a dc source). The figure's bottom pair is taken as T4 (left) and T2 (right). ```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° Data: $V_s = 230$ V, $V_m = 325.27$ V, $I_o = 25$ A. - **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 ```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) ``` #### 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 $$ I_{s1} = \frac{4I_o}{\sqrt2\,\pi} = \frac{2\sqrt2}{\pi}I_o = 0.9003\,I_o, \quad \text{lagging } v_s \text{ by } \phi_1 = \alpha $$ $$ \begin{aligned} I_{s1} &= 0.9003 \times 25 = 22.51\ \text{A (rms)} \\ I_{s1,peak} &= \frac{4 \times 25}{\pi} = 31.83\ \text{A} \end{aligned} $$ So $i_{s1} = 31.83\sin(\omega t - 30^\circ)$ A. Also, $V_{dc} = \frac{2V_m}{\pi}\cos\alpha = 207.07 \times 0.866 = 179.33$ V. **Answer: Fundamental input current $I_{s1} = 22.51$ A (rms), lagging $v_s$ by $30^\circ$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why do T1, T2 continue to conduct after wt = pi when vs turns negative? A. The gate signal is held B. The large load inductance maintains the current C. A freewheeling diode forces them on D. Because the thyristors are reverse biased Answer: B. Inductance keeps Io flowing, so vo goes negative until T3, T4 are fired at pi + alpha. (Key point: Operation after pi) 2. What does the input current look like for alpha = 30°? A. Sinusoid lagging by 30° B. Square wave of +/-25 A lagging vs by 30° C. Pulses of width 150° only D. Dc of 25 A Answer: B. Each pair conducts 180° and the current is a square wave delayed by alpha. (Key point: Input current waveform) 3. What is the rms fundamental of is for Io = 25 A? A. 31.83 A B. 25 A C. 22.51 A D. 17.7 A Answer: C. Is1 = (2sqrt(2)/pi)Io = 0.9003 x 25 = 22.51 A rms (31.83 A is the peak). (Key point: Fundamental rms) 4. When does the full converter operate as an inverter? A. alpha < 90° B. alpha > 90° with a dc source in the load C. alpha = 0 D. Only with a resistive load Answer: B. Vdc = (2Vm/pi) cos alpha becomes negative for alpha > 90°, returning power when a dc source is present. 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