Three-phase single-way thyristor rectifier: average

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. A highly inductive load is assumed so that the load current is continuous.

          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.

Conduction table

Waveforms

For $\alpha = 30^\circ$ the output just reaches zero at the end of each pulse:

 vo
        .-.          .-.          .-.
      /    \       /    \       /    \
      |     \      |     \      |     \
 -----+------+-----+------+-----+------+--
      60    180   180    300   300    420
 each pulse: 0.866 Vm -> Vm (at 90) -> 0

For $\alpha = 60^\circ$ part of each pulse is negative:

 vo
      |`-.           |`-.           |`-.
      |   `-.        |   `-.        |   `-.
 -----+------`-.-----+------`-.-----+------`-.
                `-.  |         `-.  |         `-.
      90       180 210         330           450
 each pulse: Vm (at 90) -> 0 (at 180)
 -> -0.5 Vm (at 210), then jumps to next phase

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).

In terms of rms phase voltage ($V_m = \sqrt2V_{ph}$), $V_{dc} = 1.17V_{ph}\cos\alpha$ (more exactly $1.1695V_{ph}\cos\alpha$). For example, with $V_{ph} = 230$ V and $\alpha = 30^\circ$, $V_{dc} = 1.1695 \times 230 \times 0.866 = 232.96$ V.