A step up chopper has an input voltage of 150V. The voltage output needed is 450V. Given, that the thyristor has a conducting time of 150μseconds. Calculate the chopping frequency.

Solution −

The chopping frequency (f)

$f=frac{1}{T}$

Where T – Chopping time period = $T_{ON}+T_{OFF}$

Given − $V_{S}=150V$$V_{0}=450V$ $T_{ON}=150mu sec$

$V_{0}=V_{Sleft ( frac{T}{T-T_{ON}} right )}$

$450=150frac{T}{T-150^{-6}}$ $T=225mu sec$

Therfore, $f=frac{1}{225ast 10^{-6}}=4.44KHz$

The new voltage output, on condition that the operation is at constant frequency after the halving the pulse width.

Halving the pulse width gives −

$$T_{ON}=frac{150times 10^{-6}}{2}=75mu sec$$

The frequency is constant thus,

$$f=4.44KHz$$
$$T=frac{1}{f}=150mu sec$$

The voltage output is given by −

$$V_{0}=V_{S}left ( frac{T}{T-T_{ON}} right )=150times left ( frac{150times 10^{-6}}{left ( 150-75 right )times 10^{-6}} right )=300Volts$$