--- title: "180° inverter waveforms VRY, VYB, VBR, VNO, VRN" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:inverter-180-line-neutral-waveforms" source: https://www.bctengineer.com/q/semester_7/power_electronics/inverter-180-line-neutral-waveforms site: BCT Engineering --- # 180° inverter waveforms VRY, VYB, VBR, VNO, VRN Power Electronics · Chapter 5: Inverter - Asked in: 2078 Bhadra, 8 marks - Video: [[Lec 12] 3 Phase Inverter (Power Electronics)](https://www.youtube.com/watch?v=fjPh7oAgkks) — PE Movies - Nagoya University - - 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-line-neutral-waveforms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question For three phase 180 degree conduction inverter, draw the output waveforms for VRY, VYB, VBR, VNO, and VRN, where the notation has their usual meaning. ## Short answer In 180 degree conduction, line voltages are quasi-square waves of plus or minus Vdc for 120 degrees with 60 degree zero intervals, shifted 120 degrees. VNO is a square wave of plus or minus Vdc/6 at triple frequency. Phase voltage VRN is a six-step wave with levels Vdc/3, 2Vdc/3, Vdc/3, minus Vdc/3, minus 2Vdc/3, minus Vdc/3. ## Answer In the three-phase 180° conduction inverter, each pole (leg) output, measured from the dc mid-point O, is a square wave of $\pm V_{dc}/2$, and the three pole voltages are shifted by 120°. The line voltages, the neutral-point voltage $V_{NO}$ and the phase voltages are obtained from these pole voltages. #### Relations used $$ \begin{aligned} V_{RY} &= V_{RO} - V_{YO}, \quad V_{YB} = V_{YO} - V_{BO}, \quad V_{BR} = V_{BO} - V_{RO} \\ V_{NO} &= \frac{V_{RO} + V_{YO} + V_{BO}}{3} \quad (\text{balanced star load}) \\ V_{RN} &= V_{RO} - V_{NO} \end{aligned} $$ #### Values in each 60° interval ($V = V_{dc}$) | $\omega t$ | $V_{RO}$ | $V_{YO}$ | $V_{BO}$ | $V_{RY}$ | $V_{YB}$ | $V_{BR}$ | $V_{NO}$ | $V_{RN}$ | |---|---|---|---|---|---|---|---|---| | 0–60° | $+V/2$ | $-V/2$ | $+V/2$ | $V$ | $-V$ | 0 | $V/6$ | $V/3$ | | 60–120° | $+V/2$ | $-V/2$ | $-V/2$ | $V$ | 0 | $-V$ | $-V/6$ | $2V/3$ | | 120–180° | $+V/2$ | $+V/2$ | $-V/2$ | 0 | $V$ | $-V$ | $V/6$ | $V/3$ | | 180–240° | $-V/2$ | $+V/2$ | $-V/2$ | $-V$ | $V$ | 0 | $-V/6$ | $-V/3$ | | 240–300° | $-V/2$ | $+V/2$ | $+V/2$ | $-V$ | 0 | $V$ | $V/6$ | $-2V/3$ | | 300–360° | $-V/2$ | $-V/2$ | $+V/2$ | 0 | $-V$ | $V$ | $-V/6$ | $-V/3$ | #### Waveforms ```text wt 0 60 120 180 240 300 360 VRY V __________ 0 |____ ____ -V |_________| VYB V _________ 0 ____| |____ -V _____| |____ VBR V _________ 0 _____ ____| -V |_________| VNO V/6 _____ ____ ____ -V/6 |____| |____| |____ VRN 2V/3 ____ V/3 _____| |____ 0 | -V/3 |____ ____ -2V/3 |____| ``` - **Line voltages** $V_{RY}$, $V_{YB}$, $V_{BR}$: quasi-square waves, $\pm V_{dc}$ for 120° and zero for 60°, mutually displaced by 120°. - **Neutral voltage** $V_{NO}$: square wave of $\pm V_{dc}/6$ at three times the output frequency (it carries all the triplen harmonics). - **Phase voltage** $V_{RN}$: six-step wave with levels $V/3$, $2V/3$, $V/3$, $-V/3$, $-2V/3$, $-V/3$. Rms values: $V_{RY} = 0.8165V_{dc}$, $V_{RN} = 0.4714V_{dc}$, $V_{NO} = V_{dc}/6$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the pole voltages v_RO measured from the dc mid-point? A. Square waves of ±Vdc/2, shifted 120° from each other B. Six-step waves C. Square waves of ±Vdc D. Quasi-square waves with zero intervals Answer: A. Each leg outputs ±Vdc/2 and the three legs are 120° apart. (Key point: Pole voltages) 2. What is the shape and frequency of V_NO? A. Sine at output frequency B. Zero at all times C. Square of ±Vdc/6 at three times the output frequency D. Square of ±Vdc/2 at the output frequency Answer: C. V_NO = (V_RO + V_YO + V_BO)/3 alternates ±Vdc/6 three times per cycle and carries the triplen harmonics. (Key point: Neutral voltage) 3. What does V_RY look like? A. Six-step with V/3, 2V/3 B. Quasi-square: ±Vdc for 120°, zero for 60° C. Square of ±Vdc/2 D. Triangular Answer: B. V_RY = V_RO - V_YO is ±Vdc for 120° and zero for 60°. (Key point: Line voltage) 4. In 0-60° with V_RO = +V/2, V_YO = -V/2, V_BO = +V/2, what is V_RN? A. V/6 B. 2V/3 C. V/3 D. V Answer: C. V_NO = V/6, so V_RN = V/2 - V/6 = V/3. 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