--- title: "Full converter, 10 A, α = 45°" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:full-converter-10a-45deg" source: https://www.bctengineer.com/q/semester_7/power_electronics/full-converter-10a-45deg site: BCT Engineering --- # Full converter, 10 A, α = 45° Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2069 Chaitra, 8 marks - Video: [Full converter (Basics, Working, Circuit & Waveforms) Explained in Power Electronics](https://www.youtube.com/watch?v=DBvCP-LL-mE) — Engineering Funda - 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-10a-45deg - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Figure below shows a single phase full converter circuit with highly inductive load so that load current is constant and equal to 10 amp. Explain its operation for firing angle of 45° and draw the waveforms of input voltage Vs, output voltage V0 and input current Is. Calculate the fundamental component of input current. [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 The input current is a square wave of amplitude 10 A lagging the voltage by 45°. Its fundamental rms value is Is1 = (2√2/π)Io = 0.9003×10 = 9.00 A (rms), lagging vs by 45°. ## Answer A **single-phase full converter** is a bridge of four thyristors (T1–T4). With a highly inductive load the current is continuous and constant ($I_o = 10$ A), and each thyristor pair conducts for $180^\circ$. Data: $V_s = 230$ V, 50 Hz, $V_m = 325.27$ V, $\alpha = 45^\circ$. ```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 = 45° - **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 | $v_o$ | $i_s$ | |---|---|---|---| | $45^\circ$ – $180^\circ$ | T1, T2 | $+v_s$ (positive) | $+10$ A | | $180^\circ$ – $225^\circ$ | T1, T2 | $+v_s$ (negative) | $+10$ A | | $225^\circ$ – $360^\circ$ | T3, T4 | $-v_s$ (positive) | $-10$ A | | $360^\circ$ – $405^\circ$ | T3, T4 | $-v_s$ (negative) | $-10$ A | #### Waveforms of $v_s$, $v_o$ and $i_s$ ```text vs ___ ___ / \ / -------/-----\-----------/---- \___/ vo: follows vs from 45 to 225 deg, then -vs from 225 to 405 deg (goes below zero for 45 deg in each half cycle) io: constant 10 A is +Io +---------+ +--- | | | 0 ------+ | | +---------+ -Io 45 225 405 (deg) ``` Average output voltage: $$ V_{dc} = \frac{2V_m}{\pi}\cos\alpha = \frac{2 \times 325.27}{\pi}\cos45^\circ = 207.07 \times 0.7071 = 146.42\ \text{V} $$ #### 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 $$ I_{s1} = \frac{4I_o}{\sqrt2\,\pi} = \frac{2\sqrt2}{\pi}I_o = 0.9003\,I_o, \quad \text{lagging } v_s \text{ by } \phi_1 = \alpha $$ $$ \begin{aligned} I_{s1,peak} &= \frac{4 \times 10}{\pi} = 12.73\ \text{A} \\ I_{s1} &= 0.9003 \times 10 = 9.00\ \text{A (rms)} \end{aligned} $$ So $i_{s1} = 12.73\sin(\omega t - 45^\circ)$ A. Displacement factor $= \cos45^\circ = 0.707$; input PF $= 0.9003 \times 0.707 = 0.637$ lagging. **Answer: $I_{s1} = 9.00$ A (rms), lagging $v_s$ by $45^\circ$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the full converter with constant 10 A load current and alpha = 45 degrees, what is the input current is? A. Sinusoidal, 10 A peak B. A square wave of amplitude 10 A lagging vs by 45 degrees C. A square wave of amplitude 10 A leading vs by 45 degrees D. Pulses of 10 A lasting only 135 degrees Answer: B. Each pair conducts 180 degrees carrying Io, so is is a +/-Io square wave delayed by alpha. (Key point: Input current shape) 2. How do T1 and T2 turn off when T3 and T4 are fired at pi + alpha? A. By a gate turn-off pulse B. By forced commutation circuit C. By natural line commutation as the negative supply reverse biases them D. By the load current falling to zero Answer: C. At pi + alpha vs is negative, so firing T3, T4 applies reverse voltage across T1, T2, turning them off naturally. (Key point: Commutation) 3. With Vm = 325.27 V and alpha = 45 degrees, what is Vdc? A. 146.4 V B. 207.1 V C. 230 V D. 103.5 V Answer: A. Vdc = (2Vm/pi)cos(alpha) = 207.07 x 0.7071 = 146.4 V. (Key point: Average output voltage) 4. What is the rms fundamental component of the input current? A. 10 A B. 12.73 A C. 7.07 A D. 9.00 A Answer: D. Peak fundamental = 4Io/pi = 12.73 A; rms = 12.73/sqrt(2) = 9.00 A, lagging vs by 45 degrees. 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