---
title: "Bridge cycloconverter numerical (one-third frequency)"
subject: "Power Electronics"
chapter: 6
key: "semester_7/power_electronics:bridge-cyclo-numerical"
source: https://www.bctengineer.com/q/semester_7/power_electronics/bridge-cyclo-numerical
site: BCT Engineering
---

# Bridge cycloconverter numerical (one-third frequency)

Power Electronics · Chapter 6: AC voltage controller

- Asked in: 2071 Shrawan, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/bridge-cyclo-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-6

## Question

A single phase bridge type cyclo-converter has input voltage of 230V, 50Hz and load of R = 10Ω. The output frequency is one third of input frequency. For firing angle of 30°, calculate: i) rms value of output voltage ii) rms value of output current iii) rms current of each converter iv) rms current of each thyristor

## Short answer

Use Vo = Vs sqrt[(1/pi)(pi - alpha + sin2alpha/2)] = 226.66 V; Io = Vo/R = 22.67 A; each converter current = Io/sqrt2 = 16.03 A; thyristor current = Io/2 = 11.33 A (textbook average; exact values 13.09 A and 9.25 A depending on position).

## Answer

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.

```text
   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 ($f_o = f_i/3$)

| Input half | Supply | Bridge | SCRs | $v_o$ |
|---|---|---|---|---|
| 1 | + | P | P1, P2 | + |
| 2 | − | P | P3, P4 | + |
| 3 | + | P | P1, P2 | + |
| 4 | − | N | N3, N4 | − |
| 5 | + | N | N1, N2 | − |
| 6 | − | N | N3, N4 | − |

**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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why can the cycloconverter's rms output voltage be found with the same formula as the AC voltage controller?
   A. Because the output is DC
   B. Because each output half cycle is made of identical half-sine pieces from alpha to pi
   C. Because the frequency is unchanged
   D. Because the load is inductive
   Answer: B. All the pieces have the same shape, so the rms over the whole output equals the rms of one piece: Vs*sqrt((1/pi)(pi - alpha + sin2alpha/2)) = 226.66 V. (Key point: RMS output voltage)

2. For fo = fi/3 with a 50 Hz input, how is each bridge operated?
   A. For 3 input half cycles in turn, then the other bridge for 3
   B. For 1 input half cycle
   C. For 6 input half cycles continuously
   D. For 3 input cycles then off
   Answer: A. Each output half cycle uses 3 input half cycles, so P conducts for 3 halves and N for the next 3, giving fo = 16.67 Hz. (Key point: Output frequency)

3. Why is the rms current of each converter bridge Io/sqrt(2) = 16.03 A?
   A. It carries current for only half of the output period
   B. It carries half the voltage
   C. It has twice the resistance
   D. The input is halved
   Answer: A. P carries the positive half of the output and N the negative half, so each carries load current for half the time, giving Io/sqrt(2). (Key point: Converter rms current)

4. What is the rms current of each thyristor in the textbook solution?
   A. 22.67 A
   B. 16.03 A
   C. 11.33 A
   D. 5.67 A
   Answer: C. Each thyristor conducts about half of its bridge's time, so I_T = I_conv/sqrt(2) = Io/2 = 11.33 A. (Key point: Thyristor rms current)

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