--- title: "Full converter, 20 A, α = 30°, fundamental input current" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:full-converter-20a-30deg-fundamental" source: https://www.bctengineer.com/q/semester_7/power_electronics/full-converter-20a-30deg-fundamental site: BCT Engineering --- # Full converter, 20 A, α = 30°, fundamental input current Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2081 Baishakh, 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-20a-30deg-fundamental - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Explain the operation of a single phase full converter circuit with highly inductive load with neat circuit diagram and associated waveforms. If the load current is constant and equal to 20Amp, draw the waveform input ac current is for firing angle of 30° and calculate the fundamental component of is. ## Short answer With highly inductive load, is is a square wave of ±20 A. Its fundamental component has peak value 25.46 A, rms value 18.01 A, and lags the supply voltage by 30°. ## 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 constant ($I_o$), 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) Highly inductive load (Io = 20 A) between P and N ``` #### Operation 1. **$\alpha$ to $\pi + \alpha$:** in the positive half cycle, $T_1$ and $T_2$ are fired at $\omega t = \alpha$. The load current flows $T_1 \to$ load $\to T_2$, so $v_o = v_s$ and $i_s = +I_o$. After $\omega t = \pi$, $v_s$ is negative, but $T_1, T_2$ keep conducting because the inductance maintains the current, so $v_o$ becomes negative. 2. **At $\pi + \alpha$:** $T_3$ and $T_4$ are fired. The supply voltage reverse biases $T_1, T_2$, which turn OFF (line commutation). Now $v_o = -v_s$ and $i_s = -I_o$. 3. The cycle repeats. The average output voltage is $$ V_o = \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m \sin\omega t \, d(\omega t) = \frac{2V_m}{\pi}\cos\alpha $$ For $\alpha < 90^\circ$ the converter rectifies; for $\alpha > 90^\circ$ (with a suitable d.c. source in the load) it inverts. #### Waveforms for $\alpha = 30^\circ$ ```text vs .-. .-. / \ / \ ----/-----\-----/-----\---> wt '-' '-' vo _ _ | '. | '. -----+---'._.---+---'._.--> wt a pi pi+a (negative 30 210 part) is +Io_________ | | -----+ | +---> wt |______| -Io T1,T2 ON T3,T4 ON ``` The input current $i_s$ is a square wave of amplitude $I_o$, lagging the supply voltage by $\alpha$. #### Fundamental component of $i_s$ for $I_o = 20$ A, $\alpha = 30^\circ$ From the Fourier series of a square wave of amplitude $I_o$, the peak value of the fundamental is $$ I_{s1(peak)} = \frac{4 I_o}{\pi} = \frac{4 \times 20}{\pi} = 25.46\ \text{A} $$ so its rms value is $$ I_{s1} = \frac{4 I_o}{\pi\sqrt{2}} = \frac{2\sqrt{2}}{\pi} I_o = 0.9003 \times 20 = 18.01\ \text{A} $$ The fundamental lags the supply voltage by $\phi_1 = \alpha = 30^\circ$: $$ i_{s1}(t) = 25.46\sin(\omega t - 30^\circ)\ \text{A} $$ (For reference: $I_s = 20$ A rms, displacement factor $\cos 30^\circ = 0.866$, input power factor $= 0.9003 \times 0.866 = 0.78$ lagging.) **Answer:** the fundamental of $i_s$ is 18.01 A rms (25.46 A peak), lagging $v_s$ by $30^\circ$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With a highly inductive load, what is the input ac current waveform of a single-phase full converter? A. A sine wave in phase with vs B. A square wave of amplitude Io, lagging vs by alpha C. A quasi-square wave with zero intervals D. A pulsed wave leading vs by alpha Answer: B. Load current is constant and switches between the pairs, so is = +Io and -Io alternately, 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 gate turn-off signal B. By load current falling to zero C. Supply voltage reverse biases them (line commutation) D. By a freewheeling diode Answer: C. At pi + alpha vs is negative, so firing T3, T4 applies reverse voltage across T1, T2 and turns them off naturally. (Key point: Commutation of thyristors) 3. For Io = 20 A, what is the rms fundamental of is? A. 20 A B. 25.46 A C. 14.14 A D. 18.01 A Answer: D. Fundamental peak is 4Io/pi = 25.46 A; rms is 0.9003 x 20 = 18.01 A. (Key point: Fundamental magnitude) 4. What is the phase of the fundamental of is relative to vs at alpha = 30°? A. Lagging by 30° B. Leading by 30° C. Lagging by 15° D. In phase Answer: A. The square wave is delayed by alpha, so its fundamental lags vs by phi1 = alpha = 30°. (Key point: Fundamental phase) ## Close variants set in other exams - [Full converter, 15 A, fundamental input current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-converter-15a-fundamental) — Power Electronics, chapter 2, asked 2× ## Related in Power Electronics - [Full converter, 25 A, α = 30°, fundamental input current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-converter-25a-30deg) — Power Electronics, chapter 2, asked 4× - [Single-phase full converter, 10 A, α = 30°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-converter-10a-30deg) — Power Electronics, chapter 2, asked 4× - [Full converter, 25 A, α = 30°, input power factor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-converter-25a-30deg-pf) — Power Electronics, chapter 2, asked 4× - [GTO full converter, extinction angle: RMS and input current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Agto-extinction-angle-rms) — Power Electronics, chapter 2, asked 2× - [GTO full converter, extinction angle 30°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Agto-extinction-angle-avg-rms) — Power Electronics, chapter 2, asked 2× - [Full converter, 10 A, α = 45°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-converter-10a-45deg) — Power Electronics, chapter 2, asked 4× ## Related in other subjects - [Input current, pf and torque of single phase IM](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asingle-phase-im-input-torque) — Electric Machines II, chapter 3