Single-phase full converter, 10 A, α = 30°

A single-phase full converter is a fully controlled bridge of four thyristors. With a highly inductive load, the load current is continuous and constant ($I_o$), and the output voltage can be positive or negative (two-quadrant converter).

              P (+)
        +-------+-------+
        |               |
       T1              T3
        |               |
 a o----+               +----o b
        |               |
       T4              T2
        |               |
        +-------+-------+
              N (-)
 ac supply vs = 230 V, 50 Hz between a and b (is into a)
 Highly inductive load (Io = 10 A) between P and N

Operation

1. $\alpha$ to $\pi + \alpha$: in the positive half cycle, $T_1$ and $T_2$ are fired at $\omega t = \alpha$. The load current flows $T_1 \to$ load $\to T_2$, so $v_o = v_s$ and $i_s = +I_o$. After $\omega t = \pi$, $v_s$ is negative, but $T_1, T_2$ keep conducting because the inductance maintains the current, so $v_o$ becomes negative.
2. At $\pi + \alpha$: $T_3$ and $T_4$ are fired. The supply voltage reverse biases $T_1, T_2$, which turn OFF (line commutation). Now $v_o = -v_s$ and $i_s = -I_o$.
3. The cycle repeats. The average output voltage is

$$V_o = \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha} V_m \sin\omega t \, d(\omega t) = \frac{2V_m}{\pi}\cos\alpha$$

For $\alpha < 90^\circ$ the converter rectifies; for $\alpha > 90^\circ$ (with a suitable d.c. source in the load) it inverts.

Waveforms for $\alpha = 30^\circ$

 vs    .-.         .-.
      /   \       /   \
 ----/-----\-----/-----\---> wt
             '-'         '-'
 vo    _          _
      | '.       | '.
 -----+---'._.---+---'._.--> wt
      a    pi    pi+a  (negative
     30         210     part)
 is   +Io_________
      |           |
 -----+           |      +---> wt
                  |______|
                  -Io
      T1,T2 ON    T3,T4 ON

The input current $i_s$ is a square wave of amplitude $I_o = 10$ A, lagging the supply voltage by $\alpha$.

Average output voltage at $\alpha = 30^\circ$

$$\begin{aligned} V_m &= \sqrt{2} \times 230 = 325.27\ \text{V} \\ V_o &= \frac{2V_m}{\pi}\cos\alpha = \frac{2 \times 325.27}{\pi}\cos 30^\circ = 207.07 \times 0.866 = 179.33\ \text{V} \end{aligned}$$

Other values: with $I_o = 10$ A, the load power is $179.33 \times 10 = 1793.3$ W, and the rms input current is $I_s = I_o = 10$ A.

Answer: $V_o = 179.33$ V.