Three-phase bridge inverter, 120° conduction

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

     +-----+--------+--------+
     |     |        |        |
     |    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.

($V = V_{dc}$.)

Firing sequence and waveforms

 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.