Skip to main content

Cycloconverter doubling time period, rms output

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°.

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.

Doubling the time period means To=2TiT_o = 2T_i, i.e. fo=fi/2f_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)

          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 α\alpha)

Input half cycleSupply (vaOv_{aO})SCR ONLoad voltage
1: 00–π\pi+P1 at α\alpha+
2: π\pi–2π2\pi−P2 at π+α\pi+\alpha+
3: 2π2\pi–3π3\pi+N2 at 2π+α2\pi+\alpha−
4: 3π3\pi–4π4\pi−N1 at 3π+α3\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

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

With firing delay, each hump starts at α\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 vs=2Vssin⁡ωtv_s = \sqrt{2}V_s\sin\omega t (per half-winding):

Vo=[1π∫απ2Vs2sin⁡2ωt  d(ωt)]1/2=Vs1π(π−α+sin⁡2α2)\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 α=0\alpha = 0: Vo=VsV_o = V_s.
  • Example (assumed values): Vs=230V_s = 230 V, α=30°\alpha = 30°:
Vo=2301π(π−π6+sin⁡60°2)=230×0.9855=226.66 VV_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: Vo=Vs1π(π−α+sin⁡2α2)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 α=30°\alpha = 30°; output frequency 25 Hz from 50 Hz.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Power Electronics chapter 6 questions →