Full converter, 10 A, α = 45°

A single-phase full converter is a bridge of four thyristors (T1–T4). With a highly inductive load the current is continuous and constant ($I_o = 10$ A), and each thyristor pair conducts for $180^\circ$. Data: $V_s = 230$ V, 50 Hz, $V_m = 325.27$ V, $\alpha = 45^\circ$.

              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

Operation for α = 45°

• Positive half cycle: T1 and T2 are forward biased and are fired at $\omega t = \alpha$. Load is connected to the supply, $v_o = v_s$, $i_s = +I_o$.
• At $\omega t = \pi$, $v_s$ reverses, but the large inductance keeps $I_o$ flowing, so T1, T2 continue to conduct and $v_o$ becomes negative from $\pi$ to $\pi+\alpha$ (energy returns from the inductance to the supply).
• At $\omega t = \pi+\alpha$, T3 and T4 are fired. Since $v_s < 0$, they apply reverse voltage across T1, T2, which turn off by natural (line) commutation. Now $v_o = -v_s$, $i_s = -I_o$.
• At $2\pi + \alpha$ T1, T2 are fired again and the cycle repeats.
• Each pair conducts for $180^\circ$; the input current is a square wave of amplitude $I_o$ lagging the voltage by $\alpha$.

Waveforms of $v_s$, $v_o$ and $i_s$

 vs       ___               ___
         /   \             /
 -------/-----\-----------/----
               \___/
 vo: follows vs from 45 to 225 deg, then
     -vs from 225 to 405 deg (goes below
     zero for 45 deg in each half cycle)
 io: constant 10 A
 is
 +Io      +---------+         +---
          |         |         |
  0 ------+         |         |
                    +---------+
 -Io
         45        225       405   (deg)

Average output voltage:

$$V_{dc} = \frac{2V_m}{\pi}\cos\alpha = \frac{2 \times 325.27}{\pi}\cos45^\circ = 207.07 \times 0.7071 = 146.42\ \text{V}$$

Fundamental component of input current

The input current is a square wave of amplitude $I_o$, displaced by $\alpha$ from $v_s$. Its Fourier series is

$$i_s(t) = \sum_{n=1,3,5,\ldots}\frac{4I_o}{n\pi}\sin(n\omega t - n\alpha)$$

so the fundamental has peak $\frac{4I_o}{\pi}$ and rms value

$$I_{s1} = \frac{4I_o}{\sqrt2\,\pi} = \frac{2\sqrt2}{\pi}I_o = 0.9003\,I_o, \quad \text{lagging } v_s \text{ by } \phi_1 = \alpha$$ $$\begin{aligned} I_{s1,peak} &= \frac{4 \times 10}{\pi} = 12.73\ \text{A} \\ I_{s1} &= 0.9003 \times 10 = 9.00\ \text{A (rms)} \end{aligned}$$

So $i_{s1} = 12.73\sin(\omega t - 45^\circ)$ A. Displacement factor $= \cos45^\circ = 0.707$; input PF $= 0.9003 \times 0.707 = 0.637$ lagging.

Answer: $I_{s1} = 9.00$ A (rms), lagging $v_s$ by $45^\circ$.