--- title: "Three-phase inverter from three single-phase inverters" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:three-single-phase-inverters-120" source: https://www.bctengineer.com/q/semester_7/power_electronics/three-single-phase-inverters-120 site: BCT Engineering --- # Three-phase inverter from three single-phase inverters Power Electronics · Chapter 5: Inverter - Asked in: 2072 Kartik, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/three-single-phase-inverters-120 - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question Explain the operation of three phase inverter consisting of three sets of single phase inverter and each set is conducted with a phase difference of 120° to each other. Take the loads to be purely resistive. Draw the waveform of each phase voltage (VR, VY and VB) along with the neutral voltage (VN). Also draw the waveform of VRN. ## Short answer Three half-bridge inverters produce square waves of ±Vdc/2, 120° apart. The neutral voltage VN = (VR+VY+VB)/3 is a square wave of ±Vdc/6 at triple frequency. The phase voltage VRN is a six-step wave with steps ±Vdc/3 and ±2Vdc/3, containing no triplen harmonics. ## Answer A three-phase inverter can be built from three single-phase (half-bridge) inverters, one per phase, all fed from a common dc source with mid-point O. Each one produces a square wave of $\pm V_{dc}/2$, and their gating is shifted by 120°, giving a balanced three-phase output. This is the 180° conduction bridge seen leg by leg. (Assumed: half-bridge legs from a centre-tapped source, and a balanced star resistive load whose neutral N is not connected to O.) #### Circuit ```text +------+---------+---------+ | | | | Vdc/2 S1 S3 S5 | R Y B O S4 S6 S2 | | | | Vdc/2 +---------+---------+ | R, Y, B -> star +---(-) load, neutral N ``` #### Operation - Phase R: S1 is ON for 0–180° ($V_R = V_{RO} = +V_{dc}/2$) and S4 is ON for 180–360° ($-V_{dc}/2$). - Phase Y: the same, delayed by 120°; phase B: delayed by 240°. - Load neutral voltage (balanced star, $R$ in each phase): $$ V_N = V_{NO} = \frac{V_{RO} + V_{YO} + V_{BO}}{3}, \qquad V_{RN} = V_{RO} - V_{NO} $$ | $\omega t$ | $V_R$ | $V_Y$ | $V_B$ | $V_N$ | $V_{RN}$ | |---|---|---|---|---|---| | 0–60° | $+V/2$ | $-V/2$ | $+V/2$ | $+V/6$ | $V/3$ | | 60–120° | $+V/2$ | $-V/2$ | $-V/2$ | $-V/6$ | $2V/3$ | | 120–180° | $+V/2$ | $+V/2$ | $-V/2$ | $+V/6$ | $V/3$ | | 180–240° | $-V/2$ | $+V/2$ | $-V/2$ | $-V/6$ | $-V/3$ | | 240–300° | $-V/2$ | $+V/2$ | $+V/2$ | $+V/6$ | $-2V/3$ | | 300–360° | $-V/2$ | $-V/2$ | $+V/2$ | $-V/6$ | $-V/3$ | ($V = V_{dc}$.) #### Waveforms ```text wt 0 60 120 180 240 300 360 VR V/2 _______________ -V/2 |______________ VY V/2 ______________ -V/2 __________| |____ VB V/2 _____ _________ -V/2 |______________| VN V/6 _____ ____ ____ -V/6 |____| |____| |____ VRN 2V/3 ____ V/3 _____| |____ 0 | -V/3 |____ ____ -2V/3 |____| ``` - $V_R$, $V_Y$, $V_B$: square waves of $\pm V_{dc}/2$, 120° apart. - $V_N$: square wave of $\pm V_{dc}/6$ at three times the output frequency (it contains only triplen harmonics). - $V_{RN}$: six-step wave $V/3$, $2V/3$, $V/3$, $-V/3$, $-2V/3$, $-V/3$; it has no triplen harmonics, a fundamental peak of $2V_{dc}/\pi$ and an rms value of $0.4714V_{dc}$. The line voltage $V_{RY} = V_R - V_Y$ is $\pm V_{dc}$ for 120° and zero for 60°. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this three-phase inverter built from three single-phase half-bridge inverters, what does each phase produce and how are they related? A. Square waves of +/-Vdc/2 with gating shifted by 120 degrees B. Square waves of +/-Vdc with gating shifted by 60 degrees C. Sine waves shifted by 120 degrees D. Square waves of +/-Vdc/2 in phase with each other Answer: A. Each half-bridge gives +/-Vdc/2 and the gating is delayed 120 degrees per phase for a balanced three-phase output. (Key point: Construction) 2. How is the load neutral voltage VN found for a balanced star load? A. VN = VRO B. VN = 0 always C. VN = (VRO + VYO + VBO)/3 D. VN = VRO - VYO Answer: C. For a balanced star load with N not connected to O, VN = VNO = average of the three pole voltages. (Key point: Neutral voltage) 3. What does the neutral voltage VN look like? A. A sine wave of Vdc/6 at the output frequency B. A square wave of +/-Vdc/6 at three times the output frequency (triplen harmonics only) C. Zero D. A six-step wave of Vdc/3, 2Vdc/3 Answer: B. VN alternates between +Vdc/6 and -Vdc/6 every 60 degrees, so it is at three times the output frequency and contains only triplens. (Key point: VN waveform) 4. What is the shape of VRN = VRO - VNO? A. Square wave +/-Vdc/2 B. Quasi-square wave of 120 degrees C. Six-step wave: V/3, 2V/3, V/3, -V/3, -2V/3, -V/3, with no triplen harmonics D. Square wave +/-Vdc/6 Answer: C. Subtracting the triplen neutral shift from the square pole voltage leaves the six-step wave. 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