--- title: "180° three-phase inverter and fundamental phase voltage" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:inverter-180-fundamental-per-phase" source: https://www.bctengineer.com/q/semester_7/power_electronics/inverter-180-fundamental-per-phase site: BCT Engineering --- # 180° three-phase inverter and fundamental phase voltage Power Electronics · Chapter 5: Inverter - Asked in: 2072 Chaitra, 8 marks - Video: [Lec 46(b) Fourier Series Expression of Three Phase Inverter | 180 Mode | Power Electronics](https://www.youtube.com/watch?v=K3-nz86hxV0) — Gatematic by Sohail Sir - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/inverter-180-fundamental-per-phase - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question Explain the operation of a three phase inverter for 180 degree conduction with neat circuit diagram and waveforms. How the fundamental component of output (per phase) voltage can be calculated? ## Short answer In 180° conduction, each switch conducts for half a cycle, producing a six-step phase voltage with levels ±Vdc/3 and ±2Vdc/3. The fundamental component per phase is obtained by Fourier analysis: peak value = 2Vdc/π, rms = √2 Vdc/π. ## Answer In 180° conduction each switch of the six-switch bridge conducts for half a cycle, three switches are ON at a time (one per leg), and firing is in the order S1 to S6 at 60° intervals. With a star resistive load this gives a six-step phase voltage. #### Circuit ```text +-----+--------+--------+ | | | | | S1 D1 S3 D3 S5 D5 Vdc | | | | +--a +--b +--c | | | | | S4 D4 S6 D6 S2 D2 | | | | +-----+--------+--------+ a, b, c -> star load (R per phase) ``` #### Operation | 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 | $-V/3$ | $-V/3$ | $2V/3$ | In mode 1, a and c are on the positive bus and b on the negative bus: total resistance $R/2 + R = 1.5R$, so $v_{bn} = -2V/3$ and $v_{an} = v_{cn} = V/3$. Other modes follow similarly. ```text van 2V/3 ___ V/3 __| |__ 0 ----------+----------- wt -V/3 |__ __ -2V/3 |___| vab V _______ 0 |___ ___ -V |_____| 0 60 120 180 240 300 360 ``` Line voltage $v_{ab}$: $+V$ for 0–120°, 0 for 120–180°, $-V$ for 180–300°, 0 for 300–360°. #### Fundamental component of phase voltage Method 1 — Fourier analysis of the six-step wave (odd, half-wave symmetric): $$ \begin{aligned} b_n &= \frac{2}{\pi}\int_0^{\pi}v_{an}\sin n\theta\,d\theta = \frac{2V}{3n\pi}\Big[2 + \cos\tfrac{n\pi}{3} - \cos\tfrac{2n\pi}{3}\Big] \\ b_1 &= \frac{2V}{3\pi}\big[2 + 0.5 + 0.5\big] = \frac{2V_{dc}}{\pi} \end{aligned} $$ Method 2 — from the line voltage: the 120° quasi-square $v_{ab}$ has fundamental peak $\frac{4V_{dc}}{\pi}\cos 30^\circ = \frac{2\sqrt3V_{dc}}{\pi}$; dividing by $\sqrt3$ gives the same phase value. $$ \begin{aligned} \hat V_{ph1} &= \frac{2V_{dc}}{\pi} = 0.6366\,V_{dc} \\ V_{ph1,rms} &= \frac{\sqrt2\,V_{dc}}{\pi} = 0.4502\,V_{dc} \end{aligned} $$ Full series: $v_{an} = \sum_{n=6k\pm1}\frac{2V_{dc}}{n\pi}\sin n\omega t$ (triplen harmonics absent). Total rms phase voltage $= \frac{\sqrt2}{3}V_{dc} = 0.4714V_{dc}$. Example: $V_{dc} = 600$ V gives a fundamental of 381.97 V peak (270.09 V rms). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In 180 degree conduction of a three-phase bridge, how many switches conduct at any instant and how often does the sequence advance? A. Two switches, every 60 degrees B. Three switches, every 60 degrees C. Three switches, every 120 degrees D. Six switches, every 180 degrees Answer: B. Each switch conducts 180 degrees, one per leg, so three are ON at a time with firing at 60 degree intervals. (Key point: 180 degree conduction) 2. In mode 1 (S5, S6, S1 ON) with a star resistive load, what are van, vbn, vcn? A. V/3, -2V/3, V/3 B. 2V/3, -V/3, -V/3 C. V/2, -V/2, V/2 D. V/3, -V/3, 0 Answer: A. a and c are in parallel (R/2) in series with b (R): total 1.5R, giving vbn = -2V/3 and van = vcn = V/3. (Key point: Phase voltage levels) 3. What is the peak fundamental phase voltage? A. Vdc/pi B. 4Vdc/pi C. 2Vdc/pi = 0.6366 Vdc D. sqrt(2)Vdc/pi = 0.4502 Vdc Answer: C. b1 = (2Vdc/3pi)(2+0.5+0.5) = 2Vdc/pi; 0.4502 Vdc is the rms. (Key point: Fundamental phase voltage) 4. How can the phase fundamental be obtained from the line voltage? A. Multiply the line fundamental peak by sqrt(3) B. Take the line fundamental, 2 sqrt(3)Vdc/pi, and divide by sqrt(3) C. Take the line fundamental and divide by 2 D. Line and phase fundamentals are equal Answer: B. The 120 degree quasi-square line voltage has fundamental peak 2 sqrt(3)Vdc/pi; dividing by sqrt(3) gives 2Vdc/pi. 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