---
title: "Cycloconverter doubling time period, rms output"
subject: "Power Electronics"
chapter: 6
key: "semester_7/power_electronics:cyclo-double-time-period-rms"
source: https://www.bctengineer.com/q/semester_7/power_electronics/cyclo-double-time-period-rms
site: BCT Engineering
---

# Cycloconverter doubling time period, rms output

Power Electronics · Chapter 6: AC voltage controller

- Asked in: 2080 Bhadra, 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/cyclo-double-time-period-rms
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-6

## Question

Explain the working of single phase cyclo-converter which will double the time period of input AC voltage with suitable circuit diagram and waveforms. Also calculate the rms value of output voltage.

## Short answer

To double the time period, each output half cycle uses two input half cycles, giving output frequency half of input. RMS output voltage is V_s sqrt((1/pi)(pi - alpha + sin(2 alpha)/2)), e.g., 226.66 V for 230 V and alpha=30°.

## Answer

Doubling the time period means $T_o = 2T_i$, i.e. $f_o = f_i/2$ (50 Hz to 25 Hz). This is done by a single-phase step-down cycloconverter in which each output half cycle is made from two input half cycles.

#### Circuit (centre-tapped transformer type)

```text
          a o----+---P1->|---+
            |    +---|<-N1---+
  ~ vs    Tr|                |
  (input) O o----[ Load ]----+  (common point K)
            |                |
          b o----+---P2->|---+
                 +---|<-N2---+
```

P1, P2 form the positive group (load current K to O) and N1, N2 the negative group (load current O to K).

#### Working (R load, firing angle $\alpha$)

| Input half cycle | Supply ($v_{aO}$) | SCR ON | Load voltage |
|---|---|---|---|
| 1: $0$–$\pi$ | + | P1 at $\alpha$ | + |
| 2: $\pi$–$2\pi$ | − | P2 at $\pi+\alpha$ | + |
| 3: $2\pi$–$3\pi$ | + | N2 at $2\pi+\alpha$ | − |
| 4: $3\pi$–$4\pi$ | − | N1 at $3\pi+\alpha$ | − |

- In half cycles 1 and 2 the positive group acts as a full-wave rectifier, so two positive humps appear.
- In half cycles 3 and 4 the negative group gives two negative humps.
- Each SCR turns off naturally at the supply zero (R load).

#### Waveforms

```text
 vs  /\    /\
    /  \  /  \
 --/----\/----\/----\/---
             (alternate humps -ve)
 vo  /\  /\
    /  \/  \
 --/--------\--------/---
             \  /\  /
              \/  \/
    | P1 | P2 | N2 | N1 |
    |<--- To = 2 Ti --->|
```

With firing delay, each hump starts at $\alpha$ after the zero crossing.

#### RMS output voltage

Every output half cycle is made of two identical chopped half-sine humps, so the rms over one hump equals the rms over the whole output. With $v_s = \sqrt{2}V_s\sin\omega t$ (per half-winding):

$$
\begin{aligned}
V_o &= \left[\frac{1}{\pi}\int_{\alpha}^{\pi} 2V_s^2\sin^2\omega t\; d(\omega t)\right]^{1/2} \\
&= V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}
\end{aligned}
$$

- For $\alpha = 0$: $V_o = V_s$.
- Example (assumed values): $V_s = 230$ V, $\alpha = 30°$:

$$
V_o = 230\sqrt{\frac{1}{\pi}\left(\pi - \frac{\pi}{6} + \frac{\sin 60°}{2}\right)} = 230 \times 0.9855 = 226.66\ \text{V}
$$

**Answer: $V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}$, e.g. 226.66 V for 230 V and $\alpha = 30°$; output frequency 25 Hz from 50 Hz.**

## Check yourself

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

1. A cycloconverter that doubles the time period of the input gives what output from 50 Hz?
   A. 100 Hz
   B. 25 Hz
   C. 12.5 Hz
   D. 50 Hz
   Answer: B. To = 2Ti means fo = fi/2 = 25 Hz. (Key point: Output frequency)

2. Which SCR sequence over the four input half cycles gives this operation?
   A. P1, N1, P2, N2
   B. P1, P2, N1, N2
   C. P1, P2, N2, N1
   D. N2, N1, P1, P2
   Answer: C. P1 and P2 give two positive humps, then N2 and N1 give two negative humps. (Key point: Conduction sequence)

3. Why can the rms over the whole output be found from a single hump?
   A. Because the humps are all positive
   B. Because alpha is zero
   C. Because the load is inductive
   D. Because every hump, positive or negative, is identical in shape and squares to the same value
   Answer: D. Every output half cycle consists of two identical chopped half-sine humps, so the rms of one hump equals the rms of the whole. (Key point: Why rms equals one hump)

4. For alpha = 0, what is Vo in terms of Vs?
   A. Vs
   B. Vs/2
   C. Vs/sqrt(2)
   D. 0
   Answer: A. At alpha = 0, the formula gives Vs sqrt((1/pi)(pi)) = Vs. (Key point: Example numerical)

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