Three-phase single-way controlled rectifier, R load

A three-phase single-way controlled rectifier has three thyristors, one per phase of a star-connected supply, with the load between the common cathode and the neutral. With a purely resistive load the current has the same shape as the voltage, so a thyristor turns off as soon as its phase voltage reaches zero. Two cases arise: continuous conduction for $\alpha \le 30^\circ$ and discontinuous conduction for $\alpha > 30^\circ$.

          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

• $\alpha$ is measured from the natural commutation point ($30^\circ$ after the phase zero crossing): T1 is fired at $30^\circ+\alpha$, T2 at $150^\circ+\alpha$, T3 at $270^\circ+\alpha$.
• $\alpha \le 30^\circ$: when T2 is fired, $v_a$ is still positive but $v_b > v_a$, so T1 is commutated. Each thyristor conducts for $120^\circ$ and $v_o$ stays positive (continuous).
• $\alpha > 30^\circ$: $v_a$ reaches zero at $180^\circ$ before T2 is fired; the current falls to zero and T1 turns off. $v_o = 0$ until the next firing (discontinuous). Each thyristor conducts from $30^\circ + \alpha$ to $180^\circ$, i.e. for $(150^\circ - \alpha)$.

Waveforms

 vo (alpha = 30, R load)
        .-.          .-.          .-.
      /    \       /    \       /    \
      |     \      |     \      |     \
 -----+------+-----+------+-----+------+--
      60    180   180    300   300    420
 each pulse: 0.866 Vm -> Vm (at 90) -> 0

 vo (alpha = 60, R load)
      |`-.           |`-.           |`-.
      |   `-.        |   `-.        |   `-.
 -----+------`+------+------`+------+------`+
      90     180    210     300    330     420
 each pulse: Vm (at 90) -> 0 (at 180), then
 vo = 0 until the next thyristor is fired

Average and RMS output voltage

Case 1: $0 \le \alpha \le 30^\circ$ (T1 conducts from $\pi/6+\alpha$ to $5\pi/6+\alpha$):

$$\begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{5\pi/6+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[\cos\left(\frac{\pi}{6}+\alpha\right) - \cos\left(\frac{5\pi}{6}+\alpha\right)\right] \\ &= \frac{3V_m}{2\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{3} = \frac{3V_m}{2\pi}\cdot\sqrt3\cos\alpha \end{aligned}$$ $$V_{dc} = \frac{3\sqrt3\,V_m}{2\pi}\cos\alpha = 0.827\,V_m\cos\alpha$$ $$V_{rms} = V_m\left[\frac{1}{2} + \frac{3\sqrt3}{8\pi}\cos2\alpha\right]^{1/2}$$

Case 2: $30^\circ < \alpha \le 150^\circ$ (T1 conducts from $\pi/6+\alpha$ to $\pi$):

$$\begin{aligned} V_{dc} &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{\pi}V_m\sin\omega t\,d(\omega t) = \frac{3V_m}{2\pi}\left[1 + \cos\left(\alpha + \frac{\pi}{6}\right)\right] \end{aligned}$$ $$\begin{aligned} V_{rms}^2 &= \frac{3}{2\pi}\int_{\pi/6+\alpha}^{\pi}V_m^2\sin^2\omega t\,d(\omega t) = \frac{3V_m^2}{4\pi}\left[\omega t - \frac{\sin2\omega t}{2}\right]_{\pi/6+\alpha}^{\pi} \\ &= \frac{3V_m^2}{4\pi}\left[\frac{5\pi}{6} - \alpha + \frac{1}{2}\sin\left(\frac{\pi}{3} + 2\alpha\right)\right] \end{aligned}$$ $$V_{rms} = V_m\left[\frac{3}{4\pi}\left(\frac{5\pi}{6} - \alpha + \frac{1}{2}\sin\left(\frac{\pi}{3}+2\alpha\right)\right)\right]^{1/2}$$

Numerical values

Assumption: no voltage is given, so a 400 V line supply is taken, i.e. $V_{ph} = 230$ V, $V_m = 325.27$ V.

$$\begin{aligned} \alpha = 30^\circ: \quad V_{dc} &= 0.827 \times 325.27 \times \cos30^\circ = 232.96\ \text{V} \\ V_{rms} &= 325.27\left[0.5 + 0.2067 \times 0.5\right]^{1/2} = 252.66\ \text{V} \\ \alpha = 60^\circ: \quad V_{dc} &= \frac{3 \times 325.27}{2\pi}\left[1 + \cos90^\circ\right] = 155.30\ \text{V} \\ V_{rms} &= 325.27\left[\frac{3}{4\pi}\left(\frac{5\pi}{6} - \frac{\pi}{3} + \frac{1}{2}\sin\pi\right)\right]^{1/2} = 325.27\sqrt{0.375} = 199.19\ \text{V} \end{aligned}$$

Answer: $V_{dc} = 0.716V_m$ (232.96 V) at $\alpha = 30^\circ$ and $0.477V_m$ (155.30 V) at $\alpha = 60^\circ$.