Full converter, 15 A, fundamental input current

A single-phase full converter is a fully controlled bridge of four thyristors. With a highly inductive load the load current is continuous and ripple-free ($I_o = 15$ A), each thyristor pair conducts for $180^\circ$, 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 between a and b (is into a)
 Load (R-L, Io) between P and N

Operation

• 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 $\alpha$.

Output voltage (continuous conduction):

$$V_{dc} = \frac{1}{\pi}\int_{\alpha}^{\pi+\alpha}V_m\sin\omega t\,d(\omega t) = \frac{2V_m}{\pi}\cos\alpha, \qquad V_{rms} = \frac{V_m}{\sqrt2}$$

• $0 < \alpha < 90^\circ$: $V_{dc} > 0$, rectifier mode (power ac to dc).
• $90^\circ < \alpha < 180^\circ$: $V_{dc} < 0$, inverter mode (needs a dc source in the load).

Input current waveform

$i_s = +15$ A from $\alpha$ to $\pi+\alpha$ and $-15$ A from $\pi+\alpha$ to $2\pi+\alpha$: a square wave lagging $v_s$ by $\alpha$.

 vs     ___               ___
       /   \             /
 -----/-----\-----------/------
             \___/
 is
 +15 A    +---------+         +--
          |         |         |
  0 ------+         |         |
                    +---------+
 -15 A
          a      180+a     360+a

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

$$I_{s1} = \frac{4I_o}{\sqrt2\,\pi} = \frac{2\sqrt2}{\pi}I_o = 0.9003\,I_o, \quad \text{lagging } v_s \text{ by } \phi_1 = \alpha$$ $$\begin{aligned} I_{s1,peak} &= \frac{4 \times 15}{\pi} = 19.10\ \text{A} \\ I_{s1} &= \frac{2\sqrt2}{\pi} \times 15 = 0.9003 \times 15 = 13.50\ \text{A (rms)} \end{aligned}$$

The rms value of the total input current is $I_s = I_o = 15$ A, so the harmonic factor is $\sqrt{(I_s/I_{s1})^2 - 1} = 48.3\%$, and the input power factor is $0.9003\cos\alpha$ (lagging).

Answer: $I_{s1} = 13.50$ A (rms), i.e. $i_{s1} = 19.10\sin(\omega t - \alpha)$ A, lagging the supply voltage by $\alpha$.