Three-phase single-way controlled rectifier, α = 90°

A three-phase single-way (half-wave, three-pulse) controlled rectifier uses three thyristors, one per phase, connected to a star-connected supply (or transformer secondary) with the load returned to the neutral. The output voltage is controlled by delaying the firing of each thyristor by $\alpha$ from its natural commutation point.

          T1
 a o------>|-----+
          T2     |
 b o------>|-----+------o P (+)
          T3     |        |
 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 thyristor's anode goes to one phase and the cathodes are joined to the positive output. A thyristor can conduct only when its phase is the most positive and it has been fired.
• The firing angle $\alpha$ is measured from the natural commutation point (the instant a diode would start conducting), i.e. from $30^\circ$ after the zero crossing of its phase voltage. So T1 is fired at $\omega t = 30^\circ + \alpha$, T2 at $150^\circ + \alpha$, T3 at $270^\circ + \alpha$.
• When T2 is fired, $v_b > v_a$, so T1 becomes reverse biased and turns off (natural/line commutation).
• With a highly inductive load, the load current is continuous and constant, so each thyristor conducts for exactly $120^\circ$, even if its phase voltage becomes negative. The output follows the phase of the conducting thyristor.

Average output voltage

Taking the conduction of T1 (the waveform repeats every $2\pi/3$):

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

where $V_m$ is the peak phase voltage. Maximum value (at $\alpha = 0$) is $V_{dm} = 0.827V_m$, so $V_{dc} = V_{dm}\cos\alpha$.

• $0 \le \alpha < 90^\circ$: $V_{dc} > 0$, rectifier mode.
• $\alpha = 90^\circ$: $V_{dc} = 0$.
• $90^\circ < \alpha < 180^\circ$: $V_{dc} < 0$, inverter mode (needs a dc source in the load).

Waveform for α = 90°

 phase:      va           vb           vc
 fired at:   120          240          360   (deg)
 ends at:    240          360          480
 each phase is used from 120 to 240 deg of its
 own sine; positive and negative areas are
 equal, so Vdc = 0

 vo
      |`-.            |`-.            |`-.
 -----+---`-.-----+---+---`-.-----+---+---`-.
              `-. |           `-. |
      120  180  240           360
 each pulse: +0.866 Vm -> 0 (at 180)
 -> -0.866 Vm (at 240); equal areas

At $\alpha = 90^\circ$: $V_{dc} = 0.827V_m\cos90^\circ = 0$. The output is a train of $120^\circ$ pulses, each going from $+0.866V_m$ to $0$ to $-0.866V_m$; the positive and negative areas cancel, so the average output voltage and the average power are zero, although the load current still flows (it is sustained by the inductance and, in practice, requires a dc source in the load).