Bridge cycloconverter numerical (one-third frequency)

In a bridge-type single-phase cycloconverter, the positive (P) bridge supplies the positive half of the output and the negative (N) bridge the negative half. For $f_o = f_i/3$, each bridge works for 3 input half cycles in turn.

   P-bridge (io > 0)     N-bridge (io < 0)
   +--P1----P3--+        +--N1----N3--+
 ~ |            |        |            |
 vs+            +-A [R] B+            +
   |            |        |            |
   +--P4----P2--+        +--N4----N2--+
 P: current A -> B     N: current B -> A

Given: $V_s = 230$ V, 50 Hz, $R = 10\ \Omega$, $f_o = 50/3 = 16.67$ Hz, $\alpha = 30° = \pi/6$.

i) RMS output voltage

Each output half cycle consists of 3 identical half-sine pieces from $\alpha$ to $\pi$, so the rms value is the same as for one piece:

$$\begin{aligned} V_o &= V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} \\ &= 230\sqrt{\frac{1}{\pi}\left(\pi - \frac{\pi}{6} + \frac{\sin 60°}{2}\right)} \\ &= 230\sqrt{\frac{2.6180 + 0.4330}{\pi}} = 230\sqrt{0.9712} \\ &= 230 \times 0.9855 = 226.66\ \text{V} \end{aligned}$$

ii) RMS output current

$$I_o = \frac{V_o}{R} = \frac{226.66}{10} = 22.67\ \text{A}$$

iii) RMS current of each converter

Each bridge carries the load current for one half of the output period:

$$I_{conv} = \frac{I_o}{\sqrt{2}} = \frac{22.67}{\sqrt{2}} = 16.03\ \text{A}$$

iv) RMS current of each thyristor

Within its bridge, the two thyristor pairs (P1-P2 and P3-P4) conduct on alternate input half cycles. Taking each pair as conducting for half of its bridge's conduction time (the usual textbook assumption):

$$I_T = \frac{I_{conv}}{\sqrt{2}} = \frac{I_o}{2} = \frac{22.67}{2} = 11.33\ \text{A}$$

Conduction pattern ($f_o = f_i/3$)

Exact thyristor currents: for $f_o = f_i/3$ this pattern repeats exactly every output cycle (6 input half cycles), so the pairs do not share equally. P1, P2 (and N3, N4) conduct in 2 of every 6 input half cycles, while P3, P4 (and N1, N2) conduct in only 1:

$$I_{T(P1,P2,N3,N4)} = I_o\sqrt{\frac{2}{6}} = \frac{22.67}{\sqrt{3}} = 13.09\ \text{A}, \qquad I_{T(P3,P4,N1,N2)} = I_o\sqrt{\frac{1}{6}} = \frac{22.67}{\sqrt{6}} = 9.25\ \text{A}$$

The textbook value $I_o/2 = 11.33$ A is the rms average of these two, since $\frac{1}{2}\left(\frac{1}{3} + \frac{1}{6}\right) = \frac{1}{4}$.

Answer: $V_o = 226.66$ V, $I_o = 22.67$ A, $I_{conv} = 16.03$ A, $I_T = 11.33$ A (textbook value; exactly 13.09 A for P1, P2, N3, N4 and 9.25 A for P3, P4, N1, N2).