Full converter, 25 A, α = 30°, fundamental input current

A single-phase full converter uses four thyristors in a bridge. With a highly inductive load the current is continuous and constant ($I_o = 25$ A), so it can work as a rectifier ($\alpha < 90^\circ$) or inverter ($\alpha > 90^\circ$, with a dc source). The figure's bottom pair is taken as T4 (left) and T2 (right).

              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 = 30°

Data: $V_s = 230$ V, $V_m = 325.27$ V, $I_o = 25$ A.

• 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

 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)

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

$$I_{s1} = \frac{4I_o}{\sqrt2\,\pi} = \frac{2\sqrt2}{\pi}I_o = 0.9003\,I_o, \quad \text{lagging } v_s \text{ by } \phi_1 = \alpha$$ $$\begin{aligned} I_{s1} &= 0.9003 \times 25 = 22.51\ \text{A (rms)} \\ I_{s1,peak} &= \frac{4 \times 25}{\pi} = 31.83\ \text{A} \end{aligned}$$

So $i_{s1} = 31.83\sin(\omega t - 30^\circ)$ A.

Also, $V_{dc} = \frac{2V_m}{\pi}\cos\alpha = 207.07 \times 0.866 = 179.33$ V.

Answer: Fundamental input current $I_{s1} = 22.51$ A (rms), lagging $v_s$ by $30^\circ$.