Three-phase bridge inverter 180° and 120° numerical

For a star-connected resistive load, the 180° mode gives a six-step phase voltage ($V/3$, $2V/3$) and the 120° mode gives a quasi-square phase voltage ($\pm V/2$ for 120°, zero for 60°).

Given: $V_{dc} = 450$ V, $R = 10\ \Omega$ per phase, star connection.

(a) 180° conduction mode

Phase voltage is $V_{dc}/3 = 150$ V and $2V_{dc}/3 = 300$ V in 60° steps.

(i) RMS load (phase) current

$$\begin{aligned} V_{ph,rms} &= \left[\frac{1}{\pi}\left(\Big(\tfrac{V}{3}\Big)^2\tfrac{\pi}{3}\cdot2 + \Big(\tfrac{2V}{3}\Big)^2\tfrac{\pi}{3}\right)\right]^{1/2} = \frac{\sqrt2}{3}V_{dc} \\ &= 0.4714\times450 = 212.13\ \text{V} \\ I_{L,rms} &= \frac{212.13}{10} = 21.21\ \text{A} \end{aligned}$$

(ii) RMS thyristor current — each thyristor carries the phase current for 180° of every 360°:

$$I_{T,rms} = \frac{I_{L,rms}}{\sqrt2} = \frac{21.21}{\sqrt2} = 15.0\ \text{A}$$

(iii) Load power

$$P = 3I_{L,rms}^2R = 3\times21.21^2\times10 = 13500\ \text{W} = 13.5\ \text{kW}$$

(b) 120° conduction mode

Only two thyristors conduct at a time; two phases are in series across $V_{dc}$, the third is open. So during conduction:

$$i = \frac{V_{dc}}{2R} = \frac{450}{20} = 22.5\ \text{A}, \qquad v_{ph} = \pm\frac{V_{dc}}{2} = \pm225\ \text{V}$$

Each phase conducts for 120° out of every 180°.

(i) RMS load current

$$\begin{aligned} V_{ph,rms} &= \frac{V_{dc}}{2}\sqrt{\frac{2}{3}} = \frac{V_{dc}}{\sqrt6} = 183.71\ \text{V} \\ I_{L,rms} &= 22.5\sqrt{\tfrac{2}{3}} = 18.37\ \text{A} \end{aligned}$$

(ii) RMS thyristor current — each thyristor carries 22.5 A for 120° of 360°:

$$I_{T,rms} = 22.5\sqrt{\tfrac{120}{360}} = \frac{22.5}{\sqrt3} = 12.99\ \text{A}$$

(iii) Load power

$$P = 3\times18.37^2\times10 = 10125\ \text{W} = 10.125\ \text{kW}$$

The 180° mode delivers more power (higher switch utilisation), while the 120° mode has an idle 60° in each leg that reduces the risk of shoot-through.