A single-phase AC voltage converter has the following details −
ON time = 6 min, OFF time = 4 min, frequency = 50Hz, and
Voltage source Vo = 110V
Calculate the following.
Solution −
$T=2\times \left ( T_{ON}+T_{OFF} \right )$ but $f=50Hz,$ $T=2\times \left ( 6+4 \right )=20mins$
$360^{\circ}=20min,$ $1min=18^{\circ}$
Therefore, $T_{OFF}=4min$
Then,
$$\alpha =\frac{4}{0.1}\times 1.8=72^{\circ}$$Solution −
$$V_{0}=\left ( V_{S}\times D \right ),\quad where \quad D=\frac{T_{ON}}{T_{ON}+T_{OFF}}=\frac{6}{10}=0.6$$ $$T_{ON}=6min,\quad T_{OFF}=4 min,\quad V_{S}=110V$$ $$V_{0}\left ( Voltage Output \right )=V_{S}\times D=110\times 0.6=66Volts$$