--- title: "Full converter, 15 A, fundamental input current" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:full-converter-15a-fundamental" source: https://www.bctengineer.com/q/semester_7/power_electronics/full-converter-15a-fundamental site: BCT Engineering --- # Full converter, 15 A, fundamental input current Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2074 Chaitra, 8 marks - Video: [Fourier Analysis of Single Phase Full Wave Controlled Rectifier | Power Electronics | Lecture 47](https://www.youtube.com/watch?v=DuyI1L2g1SM) — 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-15a-fundamental - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Draw the circuit diagram of single phase full converter with highly inductive load and explain its operation. If the load current is constant at 15A, draw the waveform of input ac current and calculate the fundamental component of the input ac current. ## Short answer With constant 15 A load, input current is a ±15 A square wave lagging vs by α. Fundamental rms value Is1=(2√2/π)×15=13.50 A, lagging the supply voltage by α. ## Answer A **single-phase full converter** is a fully controlled bridge of four thyristors. With a highly inductive load the load current is continuous and ripple-free ($I_o = 15$ A), each thyristor pair conducts for $180^\circ$, and the output voltage can be positive or negative (two-quadrant converter). ```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 - **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 $\alpha$. Output voltage (continuous conduction): $$ V_{dc} = \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{2V_m}{\pi}\cos\alpha, \qquad V_{rms} = \frac{V_m}{\sqrt2} $$ - $0 < \alpha < 90^\circ$: $V_{dc} > 0$, rectifier mode (power ac to dc). - $90^\circ < \alpha < 180^\circ$: $V_{dc} < 0$, inverter mode (needs a dc source in the load). #### Input current waveform $i_s = +15$ A from $\alpha$ to $\pi+\alpha$ and $-15$ A from $\pi+\alpha$ to $2\pi+\alpha$: a square wave lagging $v_s$ by $\alpha$. ```text vs ___ ___ / \ / -----/-----\-----------/------ \___/ is +15 A +---------+ +-- | | | 0 ------+ | | +---------+ -15 A a 180+a 360+a ``` #### 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 $$ I_{s1} = \frac{4I_o}{\sqrt2\,\pi} = \frac{2\sqrt2}{\pi}I_o = 0.9003\,I_o, \quad \text{lagging } v_s \text{ by } \phi_1 = \alpha $$ $$ \begin{aligned} I_{s1,peak} &= \frac{4 \times 15}{\pi} = 19.10\ \text{A} \\ I_{s1} &= \frac{2\sqrt2}{\pi} \times 15 = 0.9003 \times 15 = 13.50\ \text{A (rms)} \end{aligned} $$ The rms value of the total input current is $I_s = I_o = 15$ A, so the harmonic factor is $\sqrt{(I_s/I_{s1})^2 - 1} = 48.3\%$, and the input power factor is $0.9003\cos\alpha$ (lagging). **Answer: $I_{s1} = 13.50$ A (rms), i.e. $i_{s1} = 19.10\sin(\omega t - \alpha)$ A, lagging the supply voltage by $\alpha$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why do T1 and T2 keep conducting after the supply reverses at wt = pi? A. The gate pulse is held on B. The large load inductance keeps the load current flowing C. The battery supplies the current D. T3 and T4 are already conducting Answer: B. The highly inductive load maintains Io, so T1 and T2 stay on and vo goes negative until T3 and T4 are fired. (Key point: Continued conduction) 2. How are T1 and T2 turned off when T3 and T4 are fired? A. By forced commutation with a separate circuit B. By a negative gate pulse C. By natural (line) commutation, since T3 and T4 apply reverse voltage across them D. By the load current falling to zero Answer: C. Firing T3, T4 while vs is negative puts the supply voltage in reverse across T1, T2 and turns them off. (Key point: Turn-off method) 3. What is the input current waveform for a constant load current Io? A. Square wave of amplitude Io, lagging vs by alpha B. Sine wave in phase with vs C. Quasi-square wave with zero intervals D. Square wave of amplitude Io in phase with vs Answer: A. Each pair carries +Io or -Io for 180 degrees, delayed by the firing angle alpha. (Key point: Input current waveform) 4. What is the rms fundamental of the input current for Io = 15 A? A. 19.10 A B. 15.00 A C. 13.50 A D. 10.60 A Answer: C. Is1 = (2*sqrt(2)/pi) Io = 0.9003 x 15 = 13.50 A rms; 19.10 A is its peak. 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