Three-phase single-way controlled rectifier, α = 30°

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. The load is highly inductive, so the load current is continuous and constant.

          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.

Waveform for α = 30°

 phase:      va           vb           vc
 fired at:   60           180          300   (deg)
 ends at:    180          300          420
 each phase is used from 60 to 180 deg of its
 own sine; vo just touches zero at the end
 of each pulse (boundary of continuity)

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

The thyristor currents are $120^\circ$ blocks of the constant load current $I_o$.

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 $V_{dc} = V_{dm}\cos\alpha$.

RMS output voltage

$$\begin{aligned} V_{rms}^2 &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m^2\sin^2\omega t\,d(\omega t) = \frac{3V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\pi/6+\alpha}^{5\pi/6+\alpha} \\ &= \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} - \frac{\sin\left(\frac{5\pi}{3}+2\alpha\right) - \sin\left(\frac{\pi}{3}+2\alpha\right)}{2}\right] = \frac{3V_m^2}{4\pi}\left[\frac{2\pi}{3} + \frac{\sqrt3}{2}\cos2\alpha\right] \end{aligned}$$

(using $\sin\left(\frac{5\pi}{3}+2\alpha\right) - \sin\left(\frac{\pi}{3}+2\alpha\right) = 2\cos(\pi+2\alpha)\sin\frac{2\pi}{3} = -\sqrt3\cos2\alpha$)

$$V_{rms} = V_m\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\cos2\alpha\right]^{1/2}$$

Calculation for α = 30°

Assumption: phase voltage 230 V (400 V line), $V_m = \sqrt2 \times 230 = 325.27$ V.

$$\begin{aligned} V_{dc} &= \frac{3\sqrt3 \times 325.27}{2\pi}\cos30^\circ = 268.99 \times 0.866 = 232.96\ \text{V} \\ V_{rms} &= 325.27\left[0.5 + \frac{3\sqrt3}{8\pi}\cos60^\circ\right]^{1/2} = 325.27 \times 0.7768 = 252.66\ \text{V} \end{aligned}$$

Answer: $V_{dc} = 0.716V_m$ = 232.96 V (for 230 V per phase).