--- title: "Three-phase bridge inverter, 120° conduction" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:three-phase-inverter-120-conduction" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-inverter-120-conduction site: BCT Engineering --- # Three-phase bridge inverter, 120° conduction Power Electronics · Chapter 5: Inverter - Asked in: 2081 Bhadra, 8 marks - Video: [Three Phase Inverter - 120 degree operation with Voltage Graphs](https://www.youtube.com/watch?v=l60p4G7Bdi0) — Xplanator - 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-120-conduction - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question Discuss the principle of working of a three-phase bridge inverter with an appropriate circuit diagram. Draw phase and line voltage waveforms on the assumption that each thyristor conducts for 120° and the resistive load is star-connected. The sequence of firing of various SCRs should also be indicated in the diagram. ## Short answer In 120° conduction mode, each thyristor conducts for 120° and only two thyristors (one upper, one lower in different legs) conduct at any time. Firing sequence is T1 to T6 at 60° intervals, giving phase voltages +Vdc/2, 0, -Vdc/2 and six-step line voltages with levels Vdc, Vdc/2, -Vdc/2, -Vdc. ## Answer A three-phase bridge inverter converts dc into three-phase ac using six thyristors (with commutation circuits) or self-commutated switches, and six feedback diodes. In the **120° conduction mode** each thyristor conducts for 120°, only two thyristors (one upper and one lower, in different legs) conduct at any time, and a new thyristor is fired every 60° in the order T1, T2, T3, T4, T5, T6. #### Circuit ```text +-----+--------+--------+ | | | | | T1 D1 T3 D3 T5 D5 Vdc | | | | +--a +--b +--c | | | | | T4 D4 T6 D6 T2 D2 | | | | +-----+--------+--------+ a, b, c -> star load (R per phase) ``` T1, T3, T5 connect phases a, b, c to the positive bus; T4, T6, T2 connect them to the negative bus. The numbering gives the firing sequence T1 → T2 → … → T6 at 60° intervals. #### Principle of working - In each 60° interval, one phase is connected to the positive bus, one to the negative bus, and the third is left open (floating). - The two connected phases are in series across $V_{dc}$, so each carries $V_{dc}/(2R)$, and their phase voltages are $+V_{dc}/2$ and $-V_{dc}/2$; the open phase has 0 V (resistive load). - There is a 60° gap between the turn-off of the upper switch and the turn-on of the lower switch of the same leg, so the risk of shoot-through is low. | Mode | $\omega t$ | Conducting | $v_{an}$ | $v_{bn}$ | $v_{cn}$ | $v_{ab}$ | |---|---|---|---|---|---|---| | 1 | 0–60° | T6, T1 | $V/2$ | $-V/2$ | 0 | $V$ | | 2 | 60–120° | T1, T2 | $V/2$ | 0 | $-V/2$ | $V/2$ | | 3 | 120–180° | T2, T3 | 0 | $V/2$ | $-V/2$ | $-V/2$ | | 4 | 180–240° | T3, T4 | $-V/2$ | $V/2$ | 0 | $-V$ | | 5 | 240–300° | T4, T5 | $-V/2$ | 0 | $V/2$ | $-V/2$ | | 6 | 300–360° | T5, T6 | 0 | $-V/2$ | $V/2$ | $V/2$ | ($V = V_{dc}$.) #### Firing sequence and waveforms ```text wt 0 60 120 180 240 300 360 T1 ########## T2 ########## T3 ########## T4 ########## T5 ########## T6 ##### ##### van V/2 __________ 0 |____ ____ -V/2 |_________| vbn V/2 _________ 0 ____| |____ -V/2 _____| |____ vcn V/2 _________ 0 _____ ____| -V/2 |_________| vab V _____ V/2 |____ ____ 0 | | -V/2 |____ ____| -V |____| ``` - **Phase voltage** $v_{an}$: $+V/2$ for 120°, 0 for 60°, $-V/2$ for 120°, 0 for 60° (quasi-square). $v_{bn}$ and $v_{cn}$ lag it by 120° and 240°. - **Line voltage** $v_{ab} = v_{an} - v_{bn}$: six-step wave with levels $V$, $V/2$, $-V/2$, $-V$, $-V/2$, $V/2$. #### Main results $$ \begin{aligned} V_{ph,rms} &= \frac{V_{dc}}{\sqrt6} = 0.408\,V_{dc}, \qquad V_{L,rms} = \frac{V_{dc}}{\sqrt2} = 0.707\,V_{dc} \\ \hat V_{an1} &= \frac{4}{\pi}\cdot\frac{V_{dc}}{2}\cos30^\circ = \frac{\sqrt3\,V_{dc}}{\pi} = 0.551\,V_{dc} \end{aligned} $$ Only harmonics of order $6k\pm1$ are present. Compared with the 180° mode, switch utilisation and output voltage are lower, and with an inductive load the voltage of the open phase is not zero (it depends on the load), which is a drawback. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the 120° conduction mode, how many thyristors conduct at one time and what happens to the third phase? A. Three conduct; no phase is open B. Two conduct; the third phase is open (floating) C. Two conduct; the third phase is shorted D. Four conduct Answer: B. One upper and one lower thyristor conduct, so one phase is on +, one on -, and the third is open. (Key point: 120° operation) 2. With a star resistive load, what is the voltage of a conducting phase? A. ±Vdc/3 B. ±2Vdc/3 C. ±Vdc D. ±Vdc/2 Answer: D. Two phases are in series across Vdc, each carrying Vdc/(2R) so each has ±Vdc/2. (Key point: Phase voltage level) 3. What is the line voltage v_ab in 0-60° (T6, T1 conducting)? A. V/2 B. V C. 0 D. -V Answer: B. v_an = V/2 and v_bn = -V/2 so v_ab = V. (Key point: Line voltage) 4. What is a benefit of the 60° gap between turning off an upper switch and turning on the lower switch of the same leg? A. Higher output voltage B. No harmonics C. Low risk of shoot-through D. Better switch utilisation Answer: C. The idle gap keeps both switches of a leg from conducting together. Utilisation and output voltage are lower than in 180° mode. 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