Three-phase fully controlled bridge with R load

A three-phase fully controlled bridge rectifier (six-pulse full converter) uses six thyristors. With a resistive load the output current follows the output voltage, so conduction is continuous for $\alpha \le 60^\circ$ and discontinuous for $\alpha > 60^\circ$ (the line voltage reaches zero before the next thyristor is fired).

        P (+) o-----+------+------+-----+
                    |      |      |     |
                   T1     T3     T5     |
                    |      |      |     R
 a o----------------+      |      |     |
 b o-----------------------+      |     |
 c o------------------------------+     |
                    |      |      |     |
                   T4     T6     T2     |
                    |      |      |     |
        N (-) o-----+------+------+-----+
 (a joins T1-T4, b joins T3-T6, c joins T5-T2)

Operation (3 marks)

• T1, T3, T5 form the positive group and T4, T6, T2 the negative group. Thyristors are fired in the sequence T1, T2, T3, T4, T5, T6 at $60^\circ$ intervals; two thyristors (one from each group) conduct at a time, so the load sees a line voltage.
• $\alpha$ is measured from the natural commutation point, $30^\circ$ after the zero crossing of the phase voltage: T1 is fired at $\omega t = 30^\circ + \alpha$ (with respect to $v_a$).
• Each thyristor needs a double pulse (or a $60^\circ$+ wide pulse) so that the incoming thyristor and its partner are both on at start-up.

• With R load, for $\alpha \le 60^\circ$ each segment of $v_o$ stays positive and each thyristor conducts $120^\circ$.
• For $\alpha > 60^\circ$, the line voltage becomes zero inside the segment, the current drops to zero and the thyristors turn off; $v_o = 0$ until the next pair is fired.

Average and RMS output voltage (3 marks)

For $0 \le \alpha \le 60^\circ$ (continuous):

Using a line voltage $v_{ab} = V_{mL}\sin\theta$ ($V_{mL} = \sqrt3V_m$); the $60^\circ$ segment runs from $\theta = \pi/3 + \alpha$ to $2\pi/3 + \alpha$:

$$\begin{aligned} V_{dc} &= \frac{3}{\pi}\int_{\pi/3+\alpha}^{2\pi/3+\alpha}V_{mL}\sin\theta\,d\theta = \frac{3V_{mL}}{\pi}\left[\cos\left(\frac{\pi}{3}+\alpha\right) - \cos\left(\frac{2\pi}{3}+\alpha\right)\right] \\ &= \frac{3V_{mL}}{\pi}\cdot 2\sin\left(\frac{\pi}{2}+\alpha\right)\sin\frac{\pi}{6} = \frac{3V_{mL}}{\pi}\cos\alpha = \frac{3\sqrt3V_m}{\pi}\cos\alpha \end{aligned}$$ $$\begin{aligned} V_{rms}^2 &= \frac{3}{\pi}\int_{\pi/3+\alpha}^{2\pi/3+\alpha}V_{mL}^2\sin^2\theta\,d\theta = \frac{3V_{mL}^2}{2\pi}\left[\frac{\pi}{3} - \frac{\sin\left(\frac{4\pi}{3}+2\alpha\right) - \sin\left(\frac{2\pi}{3}+2\alpha\right)}{2}\right] \\ &= \frac{3V_{mL}^2}{2\pi}\left[\frac{\pi}{3} + \frac{\sqrt3}{2}\cos2\alpha\right] \end{aligned}$$ $$V_{rms} = V_{mL}\left[\frac{1}{2} + \frac{3\sqrt3}{4\pi}\cos2\alpha\right]^{1/2} = \sqrt3V_m\left[\frac{1}{2} + \frac{3\sqrt3}{4\pi}\cos2\alpha\right]^{1/2}$$

For $60^\circ < \alpha \le 120^\circ$ (discontinuous): each segment runs from $\pi/3 + \alpha$ to $\pi$:

$$V_{dc} = \frac{3V_{mL}}{\pi}\left[1 + \cos\left(\alpha + \frac{\pi}{3}\right)\right], \qquad V_{rms} = V_{mL}\left[\frac{3}{2\pi}\left(\frac{2\pi}{3} - \alpha + \frac{1}{2}\sin\left(2\alpha + \frac{2\pi}{3}\right)\right)\right]^{1/2}$$

Output becomes zero at $\alpha = 120^\circ$.

Waveforms for α = 0° and α = 30° (2 marks)

 alpha = 0 (same as diode bridge)
 vo  vab  vac  vbc  vba  vca  vcb
    _/\__/\__/\__/\__/\__/\_    peak VmL,
   30   90   150  210  270  330  min 0.866 VmL

 alpha = 30
 vo  each 60-deg segment starts at VmL (peak of
     the line voltage) and falls to 0.5 VmL:
    |`-.|`-.|`-.|`-.|`-.|`-.
   60   120  180  240  300  360  (deg)

• $\alpha = 0^\circ$: $V_{dc} = \frac{3V_{mL}}{\pi} = 1.35V_L$ (same as diode bridge); six pulses per cycle, ripple at 300 Hz.
• $\alpha = 30^\circ$: each pulse is the falling part of a line voltage from its peak $V_{mL}$ to $0.5V_{mL}$; $V_{dc} = 1.35V_L\cos30^\circ = 1.169V_L$.