Three-phase single-way uncontrolled rectifier

A three-phase single-way (half-wave, three-pulse) uncontrolled rectifier uses three diodes, one in each phase of a star-connected supply (or transformer secondary), with the load connected between the common cathode point and the neutral. It is the simplest three-phase rectifier and is also called the three-pulse midpoint (M3) rectifier.

          D1
 a o------>|-----+
          D2     |
 b o------>|-----+------o P (+)
          D3     |        |
 c o------>|-----+       Load
                          |
 n o----------------------o N (-)
 a, b, c, n: star-connected secondary
 va = Vm sin wt, vb lags 120, vc lags 240

Operation

• Each diode's anode is connected to one phase; the cathodes are joined to form the positive output. The diode connected to the most positive phase conducts and the other two are reverse biased.
• Measured from the zero crossing of $v_a$, $v_a$ is the highest from $30^\circ$ to $150^\circ$, $v_b$ from $150^\circ$ to $270^\circ$ and $v_c$ from $270^\circ$ to $390^\circ$. These crossover points ($30^\circ$, $150^\circ$, $270^\circ$) are the points of natural commutation.
• Each diode conducts for $120^\circ$; the output is the upper envelope of the three phase voltages, with three pulses per cycle (ripple frequency $3f = 150$ Hz), never falling below $V_m/2$.
• Load current returns through the neutral; each diode must block a peak inverse voltage of $\sqrt3 V_m$ (peak line voltage).

 vo     va        vb        vc
      _----_    _----_    _----_
     /      \__/      \__/      \
    /  D1     ^   D2    ^   D3    \
   30        150       270       390  (deg)
 vo = upper envelope; minimum Vm/2 at
 crossover points, peak Vm

Expression for average output voltage

Since the three pulses are identical, the average over one pulse ($2\pi/3$) equals the average over a cycle:

$$\begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6}^{5\pi/6}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\frac{\pi}{6} - \cos\frac{5\pi}{6}\right] \\ &= \frac{3V_m}{2\pi}\left[\frac{\sqrt3}{2} + \frac{\sqrt3}{2}\right] = \frac{3\sqrt3\,V_m}{2\pi} = 0.827\,V_m \end{aligned}$$

In terms of rms values: $V_{dc} = \frac{3\sqrt3}{2\pi}\sqrt2\,V_{ph} = 1.17\,V_{ph} = 0.675\,V_L$.

Remarks

• Each diode carries the full load current for $120^\circ$: $I_{D,avg} = I_o/3$, $I_{D,rms} = I_o/\sqrt3$ (inductive load).
• Peak inverse voltage per diode $= \sqrt3V_m$.
• The transformer secondary carries dc current (unidirectional), which can saturate the core; a zig-zag secondary or a bridge circuit avoids this.
• Example: $V_{ph} = 230$ V gives $V_{dc} = 0.827 \times 325.27 = 268.99$ V.