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180° three-phase inverter and fundamental phase voltage

Short answer

In 180° conduction, each switch conducts for half a cycle, producing a six-step phase voltage with levels ±Vdc/3 and ±2Vdc/3. The fundamental component per phase is obtained by Fourier analysis: peak value = 2Vdc/π, rms = √2 Vdc/π.

Explain the operation of a three phase inverter for 180 degree conduction with neat circuit diagram and waveforms. How the fundamental component of output (per phase) voltage can be calculated?

In 180° conduction each switch of the six-switch bridge conducts for half a cycle, three switches are ON at a time (one per leg), and firing is in the order S1 to S6 at 60° intervals. With a star resistive load this gives a six-step phase voltage.

Circuit

     +-----+--------+--------+
     |     |        |        |
     |    S1 D1    S3 D3    S5 D5
    Vdc    |        |        |
     |     +--a     +--b     +--c
     |     |        |        |
     |    S4 D4    S6 D6    S2 D2
     |     |        |        |
     +-----+--------+--------+
      a, b, c -> star load (R per phase)

Operation

Modeωt\omega tON switchesvanv_{an}vbnv_{bn}vcnv_{cn}
10–60°S5, S6, S1V/3V/3−2V/3-2V/3V/3V/3
260–120°S6, S1, S22V/32V/3−V/3-V/3−V/3-V/3
3120–180°S1, S2, S3V/3V/3V/3V/3−2V/3-2V/3
4180–240°S2, S3, S4−V/3-V/32V/32V/3−V/3-V/3
5240–300°S3, S4, S5−2V/3-2V/3V/3V/3V/3V/3
6300–360°S4, S5, S6−V/3-V/3−V/3-V/32V/32V/3

In mode 1, a and c are on the positive bus and b on the negative bus: total resistance R/2+R=1.5RR/2 + R = 1.5R, so vbn=−2V/3v_{bn} = -2V/3 and van=vcn=V/3v_{an} = v_{cn} = V/3. Other modes follow similarly.

 van 2V/3    ___
      V/3 __|   |__
        0 ----------+----------- wt
     -V/3           |__     __
    -2V/3              |___|
 vab   V  _______
        0        |___       ___
       -V            |_____|
          0  60 120 180 240 300 360

Line voltage vabv_{ab}: +V+V for 0–120°, 0 for 120–180°, −V-V for 180–300°, 0 for 300–360°.

Fundamental component of phase voltage

Method 1 — Fourier analysis of the six-step wave (odd, half-wave symmetric):

bn=2π∫0πvansin⁡nθ dθ=2V3nπ[2+cos⁡nπ3−cos⁡2nπ3]b1=2V3π[2+0.5+0.5]=2Vdcπ\begin{aligned} b_n &= \frac{2}{\pi}\int_0^{\pi}v_{an}\sin n\theta\,d\theta = \frac{2V}{3n\pi}\Big[2 + \cos\tfrac{n\pi}{3} - \cos\tfrac{2n\pi}{3}\Big] \\ b_1 &= \frac{2V}{3\pi}\big[2 + 0.5 + 0.5\big] = \frac{2V_{dc}}{\pi} \end{aligned}

Method 2 — from the line voltage: the 120° quasi-square vabv_{ab} has fundamental peak 4Vdcπcos⁡30∘=23Vdcπ\frac{4V_{dc}}{\pi}\cos 30^\circ = \frac{2\sqrt3V_{dc}}{\pi}; dividing by 3\sqrt3 gives the same phase value.

V^ph1=2Vdcπ=0.6366 VdcVph1,rms=2 Vdcπ=0.4502 Vdc\begin{aligned} \hat V_{ph1} &= \frac{2V_{dc}}{\pi} = 0.6366\,V_{dc} \\ V_{ph1,rms} &= \frac{\sqrt2\,V_{dc}}{\pi} = 0.4502\,V_{dc} \end{aligned}

Full series: van=∑n=6k±12Vdcnπsin⁡nωtv_{an} = \sum_{n=6k\pm1}\frac{2V_{dc}}{n\pi}\sin n\omega t (triplen harmonics absent). Total rms phase voltage =23Vdc=0.4714Vdc= \frac{\sqrt2}{3}V_{dc} = 0.4714V_{dc}. Example: Vdc=600V_{dc} = 600 V gives a fundamental of 381.97 V peak (270.09 V rms).

✓ Answer checked by Claude Opus 5.5 (high effort)

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