Bridge with diode, α-dependent output and input current

The circuit is a single-phase half-controlled bridge (semiconverter): two thyristors and two diodes. Assumption: the figure's labels are read as thyristors T1, T3 in the top (common-cathode) positions and diodes D4, D2 in the bottom positions; any valid half-controlled arrangement gives the same $v_o$ and $i_s$.

              P (+)
        +-------+-------+
        |               |
       T1              T3
        |               |
 a o----+               +----o b
        |               |
       D4              D2
        |               |
        +-------+-------+
              N (-)
 T1, T3: thyristors (common cathode)
 D4, D2: diodes (common anode)
 Highly inductive load (io = 10 A) between P and N

Operation and device waveforms

$V_m = \sqrt2 \times 230 = 325.27$ V, $I_o = 10$ A (constant).

• At $\pi$, $v_s$ reverses and D4 becomes forward biased; the load current freewheels through T1–D4, so $v_o$ never goes negative.
• Each thyristor conducts for $180^\circ$ ($\alpha$ to $\pi+\alpha$, i.e. 10 A pulses of $180^\circ$).
• Each diode also conducts for $180^\circ$, but from $0$ to $\pi$ (D2) and $\pi$ to $2\pi$ (D4), i.e. in step with the supply.

 vo    __          __
      |  \        |  \       (vs pieces from a to 180,
 -----+---\-------+---\---    zero from 180 to 180+a)
 iT1  +-----------+            10 A, a to 180+a
 iD2 +--------+                10 A, 0 to 180
 is
 +Io      +-----+          +--
          |     |          |
  0 ------+     +--+     +-+
                   |     |
 -Io               +-----+
          a    180 180+a  360

Average and RMS output voltage

$$V_{dc} = \frac{1}{\pi}\int_{\alpha}^{\pi} V_m\sin\omega t\, d(\omega t) = \frac{V_m}{\pi}(1+\cos\alpha) = 103.54\,(1+\cos\alpha)\ \text{V}$$ $$V_{rms} = \frac{V_m}{\sqrt2}\left[\frac{1}{\pi}\left(\pi-\alpha+\frac{\sin2\alpha}{2}\right)\right]^{1/2} = 230\left[\frac{1}{\pi}\left(\pi-\alpha+\frac{\sin2\alpha}{2}\right)\right]^{1/2}\ \text{V}$$

Fundamental component of input current

$i_s$ is $+I_o$ from $\alpha$ to $\pi$ and $-I_o$ from $\pi+\alpha$ to $2\pi$ (pulse width $\pi-\alpha$, centred at $(\pi+\alpha)/2$). Fourier analysis gives:

$$\begin{aligned} a_1 &= \frac{2}{\pi}\int_{\alpha}^{\pi} I_o\cos\omega t\, d(\omega t) = -\frac{2I_o}{\pi}\sin\alpha \\ b_1 &= \frac{2}{\pi}\int_{\alpha}^{\pi} I_o\sin\omega t\, d(\omega t) = \frac{2I_o}{\pi}(1+\cos\alpha) \\ I_{s1,peak} &= \sqrt{a_1^2+b_1^2} = \frac{4I_o}{\pi}\cos\frac{\alpha}{2}, \qquad \phi_1 = -\frac{\alpha}{2} \end{aligned}$$ $$\begin{aligned} I_{s1} &= \frac{2\sqrt2\,I_o}{\pi}\cos\frac{\alpha}{2} = \frac{2\sqrt2 \times 10}{\pi}\cos\frac{\alpha}{2} \\ &= 9.003\cos\frac{\alpha}{2}\ \text{A (rms)}, \quad \text{lagging } v_s \text{ by } \frac{\alpha}{2} \end{aligned}$$

Example: at $\alpha = 60^\circ$, $I_{s1} = 9.003 \times \cos30^\circ = 7.80$ A, lagging by $30^\circ$.

Answer: $I_{s1} = 9.003\cos(\alpha/2)$ A rms, phase $\alpha/2$ lagging. The displacement angle is only half of that of a full converter, so the semiconverter has a better input power factor.