--- title: "Switch states of three-phase VSI, 180° mode" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:vsi-180-switch-states" source: https://www.bctengineer.com/q/semester_7/power_electronics/vsi-180-switch-states site: BCT Engineering --- # Switch states of three-phase VSI, 180° mode Power Electronics · Chapter 5: Inverter - Asked in: 2079 Baishakh, 8 marks - Video: [Three-Phase Inverter Switch States](https://www.youtube.com/watch?v=CmwV2E1DRtA) — Engineering with Prof. Kim - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/vsi-180-switch-states - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question Obtain the switch states for three phase VSI of 180 degree conduction mode. Draw the waveform of output phase voltages of same inverter showing suitable modes of operation. ## Short answer In 180° conduction, three switches are on at any time and firing order is S1,S2,...,S6 every 60°. The phase voltage is a six-step wave with levels Vdc/3, 2Vdc/3, -Vdc/3, -2Vdc/3; each switch conducts 180°. ## Answer In a three-phase VSI with 180° conduction, each switch conducts for half a cycle, the two switches of a leg are complementary, and three switches (one per leg) are ON at any time. Gate signals are given in the order S1, S2, …, S6 every 60°, giving six switch states (modes) per cycle. #### 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) ``` #### Switch states Let the switching function of a leg be 1 when its upper switch is ON (phase on $+V_{dc}$) and 0 when its lower switch is ON. | Mode | $\omega t$ | ON switches | State (a b c) | $v_{an}$ | $v_{bn}$ | $v_{cn}$ | |---|---|---|---|---|---|---| | 1 | 0–60° | S5, S6, S1 | 1 0 1 | $V/3$ | $-2V/3$ | $V/3$ | | 2 | 60–120° | S6, S1, S2 | 1 0 0 | $2V/3$ | $-V/3$ | $-V/3$ | | 3 | 120–180° | S1, S2, S3 | 1 1 0 | $V/3$ | $V/3$ | $-2V/3$ | | 4 | 180–240° | S2, S3, S4 | 0 1 0 | $-V/3$ | $2V/3$ | $-V/3$ | | 5 | 240–300° | S3, S4, S5 | 0 1 1 | $-2V/3$ | $V/3$ | $V/3$ | | 6 | 300–360° | S4, S5, S6 | 0 0 1 | $-V/3$ | $-V/3$ | $2V/3$ | ($V = V_{dc}$.) The states 000 and 111 (zero states) are not used in six-step operation. #### Phase voltage in a mode (star resistive load) Mode 1: a and c are connected to $+V$, and b to the negative bus. Equivalent circuit: $R\parallel R = R/2$ in series with $R$: $$ i_b = \frac{V}{R/2 + R} = \frac{2V}{3R}, \quad v_{bn} = -i_bR = -\frac{2V}{3}, \quad v_{an} = v_{cn} = \frac{V}{3} $$ Mode 2: only a is on the positive bus; b and c are in parallel on the negative bus, giving $v_{an} = 2V/3$, $v_{bn} = v_{cn} = -V/3$. The other modes follow in the same way. In general, using the pole voltages $v_{aO}$, $v_{bO}$, $v_{cO}$ ($\pm V/2$): $$ v_{an} = \frac{2v_{aO} - v_{bO} - v_{cO}}{3} $$ #### Phase voltage waveforms ```text mode 1 2 3 4 5 6 wt 0 60 120 180 240 300 360 van 2V/3 ____ V/3 _____| |____ 0 | -V/3 |____ ____ -2V/3 |____| vbn 2V/3 ____ V/3 ____| |____ 0 | | -V/3 ____| |____ -2V/3 _____| vcn 2V/3 ____ V/3 _____ ____| 0 | | -V/3 |____ ____| -2V/3 |____| ``` Each phase voltage is a six-step wave ($V/3$, $2V/3$, $V/3$, $-V/3$, $-2V/3$, $-V/3$), and $v_{bn}$ and $v_{cn}$ lag $v_{an}$ by 120° and 240°. Results: $V_{ph,rms} = \frac{\sqrt2}{3}V_{dc} = 0.471V_{dc}$; fundamental peak $= \frac{2V_{dc}}{\pi}$; no triplen harmonics. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the state '1 0 1' (a b c) mean? A. Upper switches of legs a and c ON, lower of b ON B. All upper switches ON C. Lower switches of a and c ON D. Only leg b is ON Answer: A. 1 means the upper switch of that leg is ON (phase on +Vdc), so a and c are on the positive rail and b on the negative. (Key point: Switch state code) 2. Which states are not used in six-step 180° operation? A. 100 and 011 B. 101 and 010 C. 000 and 111 D. 110 and 001 Answer: C. The zero states 000 and 111 are unused; six active states appear each cycle. (Key point: Unused states) 3. In 60-120° (S6, S1, S2 ON), what are v_an and v_bn? A. V/3 and -2V/3 B. 2V/3 and -V/3 C. V/2 and -V/2 D. V and 0 Answer: B. Only a is on the positive bus and b, c in parallel on the negative, so v_an = 2V/3 and v_bn = v_cn = -V/3. (Key point: Mode 2 voltages) 4. Which expression gives phase voltage from pole voltages? A. v_an = v_aO - v_bO B. v_an = (v_aO + v_bO + v_cO)/3 C. v_an = (2v_aO - v_bO - v_cO)/3 D. v_an = v_aO/2 Answer: C. For a balanced star load, v_an = (2v_aO - v_bO - v_cO)/3. 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