AC voltage controller on-off cycle control numerical

In integral cycle (on-off) control, the thyristors are fired at the zero crossings so that complete cycles are passed to the load for $n$ cycles and blocked for $m$ cycles. The duty ratio is

$$k = \frac{n}{n+m} = \frac{6}{6+4} = 0.6$$

Given: $V_s = 230$ V, $f = 50$ Hz, $R = 15\ \Omega$, $V_m = \sqrt{2}\times 230 = 325.27$ V.

(i) RMS output voltage

$$V_o = V_s\sqrt{k} = 230\sqrt{0.6} = 230 \times 0.7746 = 178.16\ \text{V}$$

Load current $I_o = V_o/R = 178.16/15 = 11.88$ A; load power $P = V_o^2/R = 2116$ W.

(ii) Input power factor

$$pf = \frac{P}{V_s I_o} = \frac{V_o^2/R}{V_s\,V_o/R} = \frac{V_o}{V_s} = \sqrt{k} = \sqrt{0.6} = 0.775$$

(iii) Thyristor currents

Peak current $I_m = V_m/R = 325.27/15 = 21.68$ A. Each thyristor conducts one half-cycle in each ON cycle.

$$\begin{aligned} I_{T,avg} &= \frac{k\,I_m}{\pi} = \frac{0.6 \times 21.68}{\pi} = 4.14\ \text{A} \\ I_{T,rms} &= \frac{I_m}{2}\sqrt{k} = \frac{21.68}{2}\times 0.7746 = 8.40\ \text{A} \end{aligned}$$

(iv) Waveforms

 vs  /\  /\  /\  /\  /\  /\  /\  /\  /\  /\
      \/  \/  \/  \/  \/  \/  \/  \/  \/  \/
     |<-------- 10 cycles = 200 ms -------->|

 vo  /\  /\  /\  /\  /\  /\
      \/  \/  \/  \/  \/  \/ ________________
     |<----- 6 ON (120 ms) ->|<- 4 OFF 80ms->|

• Input current has the same shape as $v_o$ (R load), flowing only in the 6 ON cycles.
• Thyristors switch at zero voltage, so little EMI; suited to heaters with large thermal time constants.

Answer: $V_o = 178.16$ V, pf = 0.775, $I_{T,avg} = 4.14$ A, $I_{T,rms} = 8.40$ A.