--- title: "Single-phase full converter, 10 A, α = 30°" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:full-converter-10a-30deg" source: https://www.bctengineer.com/q/semester_7/power_electronics/full-converter-10a-30deg site: BCT Engineering --- # Single-phase full converter, 10 A, α = 30° Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2082 Baishakh, 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-10a-30deg - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Figure below shows a single-phase full converter circuit. Explain its operation with highly inductive load so that the load current is constant and equal to 10 A. Draw the waveforms of input voltage Vs, output voltage vo and input AC current Is for firing angle of 30°. Calculate the average value of the output voltage. [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 A single-phase full converter with highly inductive load maintains constant 10 A current. At firing angle 30 degrees, average output voltage is (2Vm/pi)cos30 = 179.33 V. Input current is a square wave of amplitude 10 A lagging the supply by 30 degrees. ## 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 = 230 V, 50 Hz between a and b (is into a) Highly inductive load (Io = 10 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 $I_o = 10$ A, lagging the supply voltage by $\alpha$. #### Average output voltage at $\alpha = 30^\circ$ $$ \begin{aligned} V_m &= \sqrt{2} \times 230 = 325.27\ \text{V} \\ V_o &= \frac{2V_m}{\pi}\cos\alpha = \frac{2 \times 325.27}{\pi}\cos 30^\circ = 207.07 \times 0.866 = 179.33\ \text{V} \end{aligned} $$ Other values: with $I_o = 10$ A, the load power is $179.33 \times 10 = 1793.3$ W, and the rms input current is $I_s = I_o = 10$ A. **Answer:** $V_o = 179.33$ V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. After wt = pi, why do T1 and T2 keep conducting although vs is negative? A. Their gate pulses are still present B. The highly inductive load maintains the current, so vo becomes negative C. The supply is disconnected D. T3 and T4 are already ON Answer: B. The inductance keeps the current flowing, so T1/T2 conduct until T3/T4 are fired at pi + alpha. (Key point: Operation beyond pi) 2. What is the shape of the input current is for constant Io? A. A square wave of amplitude Io lagging the supply voltage by alpha B. A sine wave leading vs C. A square wave in phase with vs D. A pulse of amplitude Io only during 0 to alpha Answer: A. is = +Io while T1,T2 conduct and -Io while T3,T4 conduct, shifted by alpha. (Key point: Input current) 3. What is the average output voltage at alpha = 30 degrees for 230 V rms? A. 207.07 V B. 310.6 V C. 89.7 V D. 179.33 V Answer: D. Vo = (2Vm/pi) cos alpha = 207.07 x 0.866 = 179.33 V with Vm = 325.27 V. 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