Full-wave thyristor rectifier, 220 V, 50 Ω, 45°

A single-phase full-wave thyristor rectifier (fully controlled bridge) with a resistive load conducts from $\alpha$ to $\pi$ in each half cycle. With an R load the current becomes zero at $\pi$, so the thyristors turn off naturally and $v_o$ never goes negative.

              P (+)
        +-------+-------+
        |               |
       T1              T3
        |               |
 a o----+               +----o b
        |               |
       T4              T2
        |               |
        +-------+-------+
              N (-)
 ac supply vs between a and b (is into a)
 Load R = 50 ohm between P and N

Operation and waveforms

• $0 < \omega t < \alpha$: all thyristors OFF, $v_o = 0$.
• $\alpha$ to $\pi$: T1, T2 fired, $v_o = v_s$, $i_o = v_s/R$.
• $\pi$ to $\pi+\alpha$: all OFF, $v_o = 0$.
• $\pi+\alpha$ to $2\pi$: T3, T4 fired, $v_o = -v_s$ (positive).
• The load current has the same shape as $v_o$.

 vs   ___
     /   \             /
 ---/-----\-----------/---
           \___/
 vo     __          __
       |  \        |  \
 ------+---\-------+---\---
 0    45   180   225   360   (deg)
 (chopped half-sine pulses, same for io)

Expressions

$$V_{dc} = \frac{1}{\pi}\int_{\alpha}^{\pi} V_m\sin\omega t\, d(\omega t) = \frac{V_m}{\pi}(1+\cos\alpha)$$ $$V_{rms} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin2\alpha}{2}\right)\right]^{1/2}$$

Calculation

$V_m = \sqrt2 \times 220 = 311.13$ V, $\alpha = 45^\circ = 0.7854$ rad.

$$\begin{aligned} V_{dc} &= \frac{311.13}{\pi}(1 + \cos45^\circ) = 99.04 \times 1.7071 = 169.06\ \text{V} \\ V_{rms} &= 220\left[\frac{1}{\pi}\left(\pi - 0.7854 + \frac{\sin 90^\circ}{2}\right)\right]^{1/2} \\ &= 220\left[\frac{2.8562}{\pi}\right]^{1/2} = 220 \times 0.9535 = 209.77\ \text{V} \end{aligned}$$

Load currents: $I_{dc} = 169.06/50 = 3.38$ A, $I_{rms} = 209.77/50 = 4.20$ A.

Answer: $V_{dc} = 169.06$ V, $V_{rms} = 209.77$ V.