Three-phase diode bridge rectifier

A three-phase full-wave bridge rectifier uses six diodes: D1, D3, D5 form the positive (common-cathode) group and D4, D6, D2 the negative (common-anode) group. At any instant the diode of the most positive phase in the top group and the diode of the most negative phase in the bottom group conduct, so the load gets the highest line voltage.

        P (+) o-----+------+------+-----+
                    |      |      |     |
                   D1     D3     D5     |
                    |      |      |   Load
 a o----------------+      |      |     |
 b o-----------------------+      |     |
 c o------------------------------+     |
                    |      |      |     |
                   D4     D6     D2     |
                    |      |      |     |
        N (-) o-----+------+------+-----+
 (phase a joins D1-D4, b joins D3-D6, c joins D5-D2)

Operation

• Each diode conducts for $120^\circ$; a new pair takes over every $60^\circ$. Conduction sequence: D1D2, D2D3, D3D4, D4D5, D5D6, D6D1 (numbering = order of conduction).
• Output voltage is the upper envelope of the six line voltages $v_{ab}, v_{ac}, v_{bc}, v_{ba}, v_{ca}, v_{cb}$: a six-pulse waveform with ripple frequency $6f = 300$ Hz.
• Each segment lasts $60^\circ$, varying from $\frac{\sqrt3}{2}V_{mL}$ to $V_{mL}$ and back.

 vo  vab  vac  vbc  vba  vca  vcb
    _/\__/\__/\__/\__/\__/\_   <- peak VmL
                                  min 0.866 VmL
    30   90   150  210  270  330  390 (deg)
    six 60-degree caps per cycle

Average output voltage

Let $V_{mL} = \sqrt3 V_m$ be the peak line voltage. Taking one $60^\circ$ segment of a line voltage $V_{mL}\sin\theta$ from $\theta = \pi/3$ to $2\pi/3$:

$$\begin{aligned} V_{dc} &= \frac{6}{2\pi}\int_{\pi/3}^{2\pi/3}V_{mL}\sin\theta\,d\theta = \frac{3V_{mL}}{\pi}\left[\cos\frac{\pi}{3} - \cos\frac{2\pi}{3}\right] \\ &= \frac{3V_{mL}}{\pi} = \frac{3\sqrt3\,V_m}{\pi} = 1.654\,V_m = 1.35\,V_L \end{aligned}$$

RMS output voltage

$$\begin{aligned} V_{rms}^2 &= \frac{3}{\pi}\int_{\pi/3}^{2\pi/3}V_{mL}^2\sin^2\theta\,d\theta = \frac{3V_{mL}^2}{2\pi}\left[\theta - \frac{\sin2\theta}{2}\right]_{\pi/3}^{2\pi/3} \\ &= \frac{3V_{mL}^2}{2\pi}\left[\frac{\pi}{3} + \frac{\sqrt3}{2}\right] = 3V_m^2\left[\frac{1}{2} + \frac{3\sqrt3}{4\pi}\right] \end{aligned}$$ $$V_{rms} = V_m\left[\frac{3}{2} + \frac{9\sqrt3}{4\pi}\right]^{1/2} = 1.6554\,V_m$$

The rms value is almost equal to the average (form factor $= 1.6554/1.6540 = 1.0009$, ripple factor about 4%), so the output is very smooth.

Example: for a 400 V line supply, $V_{dc} = 1.35 \times 400 = 540$ V.