--- title: "Bridge single-phase cycloconverter conduction sequence" subject: "Power Electronics" chapter: 6 key: "semester_7/power_electronics:bridge-cyclo-conduction-sequence" source: https://www.bctengineer.com/q/semester_7/power_electronics/bridge-cyclo-conduction-sequence site: BCT Engineering --- # Bridge single-phase cycloconverter conduction sequence Power Electronics · Chapter 6: AC voltage controller - Asked in: 2076 Asoj, 8 marks; 2075 Chaitra, 8 marks; 2072 Chaitra, 8 marks (asked 3 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/bridge-cyclo-conduction-sequence - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-6 ## Question Explain the bridge configuration of single phase (step-down) cycloconverter with necessary circuit diagram and waveform. Also tabulate the conduction sequence. ## Short answer Two back-to-back bridge converters (P and N) generate low-frequency AC. For f_o = f_i/2, P-converter conducts first two supply half-cycles (P1P2, P3P4) giving positive output, then N-converter conducts next two (N1N2, N3N4) giving negative output. Conduction sequence: P1P2→P3P4→N1N2→N3N4. ## Answer A single-phase bridge cycloconverter converts AC at frequency $f_i$ directly to AC at a lower frequency $f_o = f_i/n$, using two full-bridge thyristor converters connected back to back: the positive (P) converter supplies the positive half of the output and the negative (N) converter the negative half. #### Circuit ```text P-converter N-converter +--P1---P3--+ +--N1---N3--+ vs | | load | | ~ --+ +--[R]---+ +-- ~ | | | | +--P4---P2--+ +--N4---N2--+ ``` - P-converter: P1, P2 (conduct when supply positive) and P3, P4 (conduct when supply negative); output current flows A to B through the load (+ve). - N-converter: N1, N2 and N3, N4 connected in reverse, so load current flows B to A (−ve). - Only one converter is enabled at a time (circulating-current-free mode). #### Operation for $f_o = f_i/2$ (example, $\alpha = 0$, R load) For the first output half cycle (two input half cycles), the P-converter acts as a full-wave rectifier, so both input half cycles appear positive on the load. For the next two input half cycles the N-converter works, and both appear negative. ```text vs /\ /\ /\ /\ / \ / \ / \ / \ ---/----\/----\/----\/----\- (each hump is one half cycle) vo /\ /\ / \/ \ ---/--------\--------/----- \ /\ / \/ \/ |<--P--->|<--N--->| To = 2 Ti ``` So the output period is twice the input period, i.e. $f_o = f_i/2$ (25 Hz from 50 Hz). With firing angle $\alpha$, each hump starts at $\alpha$, reducing the rms output. #### Conduction sequence ($f_o = f_i/2$) | Input half cycle | Supply polarity | Converter | Thyristors ON | Output | |---|---|---|---|---| | 1 (0 to $\pi$) | + | P | P1, P2 | + | | 2 ($\pi$ to $2\pi$) | − | P | P3, P4 | + | | 3 ($2\pi$ to $3\pi$) | + | N | N1, N2 | − | | 4 ($3\pi$ to $4\pi$) | − | N | N3, N4 | − | For $f_o = f_i/3$, each converter works for three input half cycles (P1P2, P3P4, P1P2, then N3N4, N1N2, N3N4 style alternation), giving 3 humps per output half cycle. #### RMS output (R load) $$ V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} $$ **Uses:** low-speed, high-power AC drives (cement mills, ship propulsion), induction heating supplies. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does a single-phase bridge cycloconverter consist of? A. One thyristor bridge with a DC link B. Two full-bridge thyristor converters connected back to back, P for positive and N for negative output C. Two diode bridges in series D. A single thyristor in series with the load Answer: B. The P converter supplies the positive output half and the N converter the negative half. (Key point: Bridge cycloconverter structure) 2. In the mode described, how many converters are enabled at any time? A. Both, with an inter-group reactor B. Only one C. Neither, until the load current reverses D. Both, with alpha equal to 90 degrees Answer: B. Only one converter is enabled at a time, so no circulating current flows between P and N. (Key point: Circulating-current-free mode) 3. For fo = fi/2 with alpha = 0, which thyristors conduct in input half cycles 1 to 4? A. P1P2, P3P4, N1N2, N3N4 B. P1P2, N1N2, P3P4, N3N4 C. P1P2, P3P4, N3N4, N1N2 D. N1N2, N3N4, P1P2, P3P4 Answer: A. The P converter works for the first two half cycles (P1P2 then P3P4) and the N converter for the next two (N1N2 then N3N4). (Key point: Conduction sequence for fo = fi/2) 4. In the fo = fi/2 operation, why is the output frequency half the input frequency? A. Each thyristor is fired at twice the supply angle B. The load has an inductive time constant C. The output period equals two input periods D. The output period equals one input period Answer: C. Both input half cycles are positive in the P converter and negative in the N converter, giving To = 2Ti, so 50 Hz gives 25 Hz. (Key point: Output frequency) ## Related in Power Electronics - [Single-phase step-down cycloconverter](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asingle-phase-step-down-cyclo) — Power Electronics, chapter 6, asked 2× - [Step-up and step-down single-phase cycloconverter](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-up-step-down-cyclo) — Power Electronics, chapter 6 - [Step-down cycloconverter at one-fourth frequency](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Astep-down-cyclo-quarter-frequency) — Power Electronics, chapter 6 - [Single-phase cycloconverter operation](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asingle-phase-cycloconverter) — Power Electronics, chapter 6 - [Three-phase six-step (180° conduction) inverter](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-inverter-180-six-step) — Power Electronics, chapter 5, asked 4× - [Three-phase bridge inverter, 120° conduction](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-inverter-120-conduction) — Power Electronics, chapter 5 ## Related in other subjects - [Chopper definition; single-phase half bridge inverter](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Achopper-half-bridge-inverter) — Advanced Electronics, chapter 7 - [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× - [DFRT and equivalent circuit of single phase IM](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Adfrt-equivalent-circuit) — Electric Machines II, chapter 3 - [Step-down chopper with RL load](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Astep-down-chopper-rl) — Advanced Electronics, chapter 7