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Centre-tapped cycloconverter, one-third frequency

Short answer

A centre-tapped cycloconverter uses four thyristors P1, P2, N1, N2. For fo = fi/3, each output half-cycle uses three input half-cycles: P1, P2, P1 for positive output, then N1, N2, N1 for negative output, repeating every six input half-cycles.

Draw the circuit of a single phase cycloconverter using a center tapped transformer; Prepare a table showing the conduction pattern of thyristors and current paths for an output frequency of one-third the input frequency.

A single-phase cycloconverter with a centre-tapped transformer uses four thyristors: P1, P2 (positive group) and N1, N2 (negative group). For one-third frequency (fo=fi/3f_o = f_i/3, e.g. 50 Hz to 16.67 Hz), each output half cycle is built from three input half cycles.

Circuit

          a o----+---P1->|---+
            |    +---|<-N1---+
  ~ vs    Tr|                |
            O o----[ R ]-----+ K
            |                |
          b o----+---P2->|---+
                 +---|<-N2---+
 vo = v_KO; positive when current flows K -> O
  • vaOv_{aO} positive = positive supply half cycle (then bb is negative w.r.t. O).
  • P1 conducts when aa is positive; P2 when bb is positive.
  • N1 conducts when aa is negative; N2 when bb is negative.

Conduction pattern and current paths (fo=fi/3f_o = f_i/3, R load)

Input half cycleSupplySCR ONCurrent pathvov_o
1+ (aa +)P1a → P1 → K → R → O+
2− (bb +)P2b → P2 → K → R → O+
3+ (aa +)P1a → P1 → K → R → O+
4− (aa −)N1O → R → K → N1 → a−
5+ (bb −)N2O → R → K → N2 → b−
6− (aa −)N1O → R → K → N1 → a−

The pattern repeats every 6 input half cycles (3 input cycles = 1 output cycle).

Waveforms

 vs  /\    /\    /\
    /  \  /  \  /  \
 --/----\/----\/----\/----\/----\/--
     1   2   3   4   5   6  (halves)
 vo  /\  /\  /\
    /  \/  \/  \
 --/------------\------------/--
                 \  /\  /\  /
                  \/  \/  \/
    P1  P2  P1   N1  N2  N1
    |<----- To = 3 Ti ----->|

Remarks

  • Each SCR turns off naturally at the supply zero because the load is resistive.
  • The output frequency is fi/3f_i/3 and, with firing angle α\alpha, the rms output is Vo=Vs1π(π−α+sin⁡2α2)V_o = V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)}, where VsV_s is the rms voltage of each half winding.
  • For an inductive load, longer gate pulses and a dead time at changeover between P and N groups are required.

✓ Answer checked by Claude Opus 5.5 (high effort)

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