Full controlled rectifier, 25 A, α = 30°, derivation

A single-phase full controlled rectifier (full converter) uses four thyristors in a bridge. With a highly inductive load the load current is continuous and constant ($I_o = 25$ A), so each thyristor pair conducts for $180^\circ$.

              P (+)
        +-------+-------+
        |               |
       T1              T3
        |               |
 a o----+               +----o b
        |               |
       T4              T2
        |               |
        +-------+-------+
              N (-)
 ac supply vs between a and b (is into a)
 Load (R-L, Io) between P and N

i) Operation for α = 30°

Data: $V_s = 230$ V, $V_m = \sqrt2 \times 230 = 325.27$ V, $\alpha = 30^\circ$.

• At $\omega t = 30^\circ$, T1 and T2 are fired; the load is connected to $v_s$.
• After $180^\circ$, $v_s$ becomes negative but T1, T2 keep conducting because the inductance keeps the current flowing; $v_o$ goes negative for $30^\circ$.
• At $210^\circ$, T3 and T4 are fired. They apply reverse voltage to T1, T2, which turn off (natural/line commutation). The cycle repeats.
• Input current $i_s$ is a square wave of $\pm 25$ A, lagging $v_s$ by $\alpha = 30^\circ$.

 vs       ___               ___
         /   \             /
 -------/-----\-----------/----
               \___/
 vo: follows vs from 30 to 210 deg, then
     -vs from 210 to 390 deg (goes below
     zero for 30 deg in each half cycle)
 io: constant 25 A
 is
 +Io      +---------+         +---
          |         |         |
  0 ------+         |         |
                    +---------+
 -Io
         30        210       390   (deg)

ii) Average and RMS output voltage

Output repeats every $\pi$; $v_o = V_m\sin\omega t$ from $\alpha$ to $\pi + \alpha$:

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

So the rms output equals the supply rms (230 V) for any $\alpha$ in continuous conduction.

iii) Average output voltage

$$\begin{aligned} V_{dc} &= \frac{2 \times 325.27}{\pi}\cos 30^\circ = 207.07 \times 0.866 \\ &= 179.33\ \text{V} \end{aligned}$$

Answer: $V_{dc} = 179.33$ V, $V_{rms} = 230$ V.