--- title: "Single-phase full converter: rectification and inversion" subject: "Power Electronics" chapter: 2 key: "semester_7/power_electronics:full-converter-rect-inversion" source: https://www.bctengineer.com/q/semester_7/power_electronics/full-converter-rect-inversion site: BCT Engineering --- # Single-phase full converter: rectification and inversion Power Electronics · Chapter 2: Single phase AC to DC conversion - Asked in: 2071 Chaitra, 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-rect-inversion - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2 ## Question Explain the operation single phase full converter with highly inductive load. How it can be operated in rectification as well as in inversion mode? ## Short answer With highly inductive load the output current is continuous and Vdc = (2Vm/π) cosα. For α < 90°, Vdc > 0, power flows from ac to dc (rectification). For α > 90°, Vdc < 0 while current direction is unchanged; a dc source must force current, returning power to ac (inversion), with α limited below 180° minus overlap and turn-off margins. ## Answer A **single-phase full converter** is a fully controlled bridge of four thyristors. With a highly inductive load the load current $I_o$ is continuous and flows in one direction only, but the average output voltage can be positive or negative. It is therefore a **two-quadrant converter**: it works as a rectifier for $\alpha < 90^\circ$ and as a line-commutated inverter for $\alpha > 90^\circ$. ```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 with highly inductive load - **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 $$ V_{dc} = \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{2V_m}{\pi}\cos\alpha $$ #### Rectification mode ($0 < \alpha < 90^\circ$) - $\cos\alpha > 0$, so $V_{dc}$ is positive; $I_o$ is positive. - Power $P = V_{dc}I_o > 0$: energy flows from the **ac supply to the dc load** (first quadrant). - Example: motoring of a separately excited dc motor. #### Inversion mode ($90^\circ < \alpha < 180^\circ$) - $\cos\alpha < 0$, so $V_{dc}$ is negative while $I_o$ is still positive (thyristors conduct in one direction only). - $P = V_{dc}I_o < 0$: energy flows from the **dc side to the ac supply** (fourth quadrant). - Conditions needed: 1. A **dc source** in the load (e.g. back emf of a dc machine or a battery) with its polarity reversed relative to rectification, so that it can drive current through the thyristors against negative $V_{dc}$: $E > |V_{dc}|$. 2. Continuous current (large inductance). 3. A firing-angle limit: $\alpha_{max} \approx 180^\circ - (\mu + \gamma)$ where $\mu$ is the overlap angle and $\gamma$ the turn-off margin; otherwise the outgoing thyristor cannot regain blocking ability and **commutation failure** occurs. - Example: regenerative braking of a dc motor, HVDC inverter station. | Quantity | Rectification | Inversion | |---|---|---| | Firing angle | $0^\circ$ – $90^\circ$ | $90^\circ$ – $180^\circ$ | | $V_{dc}$ | positive | negative | | $I_o$ | positive | positive | | Power flow | ac → dc | dc → ac | | Quadrant | I | IV | | Load needs | R-L or R-L-E | R-L with dc source E | At $\alpha = 90^\circ$, $V_{dc} = 0$ and the average power is zero. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With a highly inductive load, why is the single-phase full converter called a two-quadrant converter? A. Vdc can be positive or negative while Io stays positive B. Io can reverse while Vdc stays positive C. Both Vdc and Io can reverse D. It only works with two thyristors at a time Answer: A. Thyristors conduct one way, so Io is always positive, but Vdc = (2Vm/pi)cos(alpha) changes sign with alpha, giving rectifier and inverter operation. (Key point: Two-quadrant operation) 2. What is the average output voltage of the full converter with continuous (highly inductive) load? A. (Vm/pi) cos(alpha) B. (2Vm/pi)(1+cos(alpha)) C. (2Vm/pi) sin(alpha) D. (2Vm/pi) cos(alpha) Answer: D. Integrating Vm sin(wt) from alpha to pi+alpha over pi gives (2Vm/pi)cos(alpha). (Key point: Average output voltage) 3. For which firing angles does the converter act as a line-commutated inverter? A. 0 to 90 degrees B. 90 to 180 degrees C. Only exactly 90 degrees D. Only below 30 degrees Answer: B. For alpha > 90 degrees cos(alpha) is negative, so Vdc is negative while Io is positive, so power flows dc to ac. (Key point: Mode by firing angle) 4. Which is a required condition for inversion operation? A. A resistive load with no inductance B. Firing angle kept at exactly 180 degrees C. A dc source (back emf) in the load with E > |Vdc| D. A freewheeling diode across the load Answer: C. A reversed-polarity dc source must drive current through the thyristors against the negative Vdc, along with continuous current and a limited alpha to avoid commutation failure. (Key point: Conditions for inversion) ## Related in Power Electronics - [Series full converters: rectification and inversion](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aseries-full-converters-rect-inversion) — 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× - [Series full converters in rectification mode](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aseries-full-converters-rectification) — Power Electronics, chapter 2 - [Three-phase full converter at 50% of maximum output](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-full-converter-50-percent) — Power Electronics, chapter 3 - [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× - [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× ## Related in other subjects - [Choppers; single-phase full bridge inverter](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Achoppers-full-bridge-inverter) — Advanced Electronics, chapter 7, asked 2× - [Single-phase fully controlled converter DC motor speed](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afull-converter-dc-motor-speed) — Utilization of Electrical Energy, chapter 3 - [Aluminium cable for 25 kW single-phase load](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aaluminium-cable-25kw) — Industrial Power Distribution and Illumination, chapter 5