--- title: "Three-phase six-step (180° conduction) inverter" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:three-phase-inverter-180-six-step" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-inverter-180-six-step site: BCT Engineering --- # Three-phase six-step (180° conduction) inverter Power Electronics · Chapter 5: Inverter - Asked in: 2080 Bhadra, 8 marks; 2079 Bhadra, 8 marks; 2076 Asoj, 8 marks; 2073 Shrawan, 8 marks (asked 4 times) - Video: [Power Electronics - Six Step 180-Degree Inverter - Part 1](https://www.youtube.com/watch?v=ScL8umsZ-F0) — Power Electronics with Dr. K - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-inverter-180-six-step - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question With the help of suitable circuit diagram, explain the 180° conduction mode (six step output voltage) of three phase inverter. Also draw the output waveforms of instantaneous phase and line voltage for star-connected load. ## Short answer In the 180° conduction mode of a three-phase bridge inverter, each switch conducts for 180°, three switches are ON at any time, and they are gated at 60° intervals in the sequence S1-S6. The phase voltage is a six-step wave with steps of Vdc/3 and 2Vdc/3 every 60°, and the line voltage is a quasi-square wave of 120° width with amplitude Vdc. RMS values are Vph,rms = 0.4714 Vdc and VL,rms = 0.8165 Vdc. ## Answer In the 180° conduction mode of a three-phase bridge inverter each switch conducts for 180°, three switches are ON at any time (one from each leg), and switches are gated in the order S1, S2, …, S6 at 60° intervals. The output phase voltage is a six-step wave and the line voltage is a 120° quasi-square wave. #### Circuit ```text +----+--------+--------+ | | | | Vdc/2 S1 D1 S3 D3 S5 D5 --- | | | | O +-a +--b +--c | --- | | | | Vdc/2 S4 D4 S6 D6 S2 D2 | | | | +----+--------+--------+ a, b, c -> 3-phase load ``` Switches S1, S3, S5 connect phases a, b, c to the positive rail; S4, S6, S2 to the negative rail. S1 and S4 (same leg) are complementary. #### Conduction table and phase voltages (star resistive load, each phase $R$) | Mode | $\omega t$ | ON switches | $v_{an}$ | $v_{bn}$ | $v_{cn}$ | |---|---|---|---|---|---| | 1 | 0–60° | S5, S6, S1 | $V/3$ | $-2V/3$ | $V/3$ | | 2 | 60–120° | S6, S1, S2 | $2V/3$ | $-V/3$ | $-V/3$ | | 3 | 120–180° | S1, S2, S3 | $V/3$ | $V/3$ | $-2V/3$ | | 4 | 180–240° | S2, S3, S4 | $-V/3$ | $2V/3$ | $-V/3$ | | 5 | 240–300° | S3, S4, S5 | $-2V/3$ | $V/3$ | $V/3$ | | 6 | 300–360° | S4, S5, S6 | $-V/3$ | $-V/3$ | $2V/3$ | ($V = V_{dc}$.) Example, mode 1: phases a and c are on the positive rail, b on the negative. The equivalent circuit is $R/2$ (a, c in parallel) in series with $R$: current $i = \dfrac{V}{1.5R}$, so $v_{bn} = -iR = -2V/3$ and $v_{an} = v_{cn} = iR/2 = V/3$. #### Waveforms ```text gates: S1 |######......| (0-180) S2 ..|######...| (60-240) S3 ....|######.| (120-300) van 2V/3 __ V/3 __| |__ 0 -|--------|------------- -V/3 |__ __| -2V/3 |__| vab +V |_____| 0 -| |__| |__|-- -V |_____| 0 60 120 180 240 300 360 ``` - Phase voltage $v_{an}$: $V/3$, $2V/3$, $V/3$, $-V/3$, $-2V/3$, $-V/3$ in successive 60° steps. $v_{bn}$, $v_{cn}$ are the same wave shifted by 120° and 240°. - Line voltage $v_{ab} = v_{an} - v_{bn}$: $+V$ for 0–120°, 0 for 120–180°, $-V$ for 180–300°, 0 for 300–360°. $v_{bc}$, $v_{ca}$ are shifted by 120°, 240°. #### Important values $$ \begin{aligned} V_{ph,rms} &= \frac{\sqrt2}{3}V_{dc} = 0.4714\,V_{dc}, \qquad V_{L,rms} = \sqrt{\frac23}\,V_{dc} = 0.8165\,V_{dc} \\ v_{an} &= \sum_{n=6k\pm1}\frac{2V_{dc}}{n\pi}\sin n\omega t \end{aligned} $$ Triplen harmonics are absent. Because two switches of a leg must not conduct together, a small dead time is given between turning S1 off and S4 on. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the 180° conduction mode of a three-phase bridge inverter, how many switches conduct at any instant? A. Two, one upper and one lower B. Three, one from each leg C. Four, two from each of two legs D. Six, all switches Answer: B. Each switch conducts for 180°, so one switch in each of the three legs is ON at any time. (Key point: Switches ON in 180° mode) 2. For a star resistive load in the interval 0-60° (S5, S6, S1 ON), what is v_bn? A. V/3 B. -V/3 C. -2V/3 D. -V/2 Answer: C. Phase b is alone on the negative rail; the equivalent circuit is R/2 in series with R, so v_bn = -2V/3. (Key point: Phase voltage in mode 1) 3. What is the shape of the line voltage v_ab in the 180° mode? A. Six-step wave with levels V/3 and 2V/3 B. Quasi-square wave: +V for 120°, 0 for 60°, -V for 120°, 0 for 60° C. Pure square wave of ±V/2 D. Sinusoid Answer: B. v_ab = v_an - v_bn gives +V for 0-120°, 0 for 120-180°, -V for 180-300° and 0 for 300-360°. (Key point: Line voltage shape) 4. Why is a small dead time given between turning S1 off and S4 on? A. To remove triplen harmonics B. To raise the output voltage C. To make the line voltage quasi-square D. To avoid both switches of a leg conducting together and shorting the dc source Answer: D. Two switches of one leg conducting together would short the dc supply, so a dead time is inserted. 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