Skip to main content

Three-phase inverter from three single-phase inverters

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.

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.

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 ±Vdc/2\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

  +------+---------+---------+
  |      |         |         |
 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° (VR=VRO=+Vdc/2V_R = V_{RO} = +V_{dc}/2) and S4 is ON for 180–360° (−Vdc/2-V_{dc}/2).
  • Phase Y: the same, delayed by 120°; phase B: delayed by 240°.
  • Load neutral voltage (balanced star, RR in each phase):
VN=VNO=VRO+VYO+VBO3,VRN=VRO−VNOV_N = V_{NO} = \frac{V_{RO} + V_{YO} + V_{BO}}{3}, \qquad V_{RN} = V_{RO} - V_{NO}
ωt\omega tVRV_RVYV_YVBV_BVNV_NVRNV_{RN}
0–60°+V/2+V/2−V/2-V/2+V/2+V/2+V/6+V/6V/3V/3
60–120°+V/2+V/2−V/2-V/2−V/2-V/2−V/6-V/62V/32V/3
120–180°+V/2+V/2+V/2+V/2−V/2-V/2+V/6+V/6V/3V/3
180–240°−V/2-V/2+V/2+V/2−V/2-V/2−V/6-V/6−V/3-V/3
240–300°−V/2-V/2+V/2+V/2+V/2+V/2+V/6+V/6−2V/3-2V/3
300–360°−V/2-V/2−V/2-V/2+V/2+V/2−V/6-V/6−V/3-V/3

(V=VdcV = V_{dc}.)

Waveforms

 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                     |____|
  • VRV_R, VYV_Y, VBV_B: square waves of ±Vdc/2\pm V_{dc}/2, 120° apart.
  • VNV_N: square wave of ±Vdc/6\pm V_{dc}/6 at three times the output frequency (it contains only triplen harmonics).
  • VRNV_{RN}: six-step wave V/3V/3, 2V/32V/3, V/3V/3, −V/3-V/3, −2V/3-2V/3, −V/3-V/3; it has no triplen harmonics, a fundamental peak of 2Vdc/π2V_{dc}/\pi and an rms value of 0.4714Vdc0.4714V_{dc}.

The line voltage VRY=VR−VYV_{RY} = V_R - V_Y is ±Vdc\pm V_{dc} for 120° and zero for 60°.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Power Electronics chapter 5 questions →