# DC to DC Converters Solved Example

$f=\frac{1}{T}$
Where T – Chopping time period = $T_{ON}+T_{OFF}$
Given − $V_{S}=150V$$V_{0}=450V T_{ON}=150\mu sec V_{0}=V_{S\left ( \frac{T}{T-T_{ON}} \right )} 450=150\frac{T}{T-150^{-6}} T=225\mu sec Therfore, f=\frac{1}{225\ast 10^{-6}}=4.44KHz The new voltage output, on condition that the operation is at constant frequency after the halving the pulse width. Halving the pulse width gives −$$T_{ON}=\frac{150\times 10^{-6}}{2}=75\mu sec$$The frequency is constant thus,$$f=4.44KHzT=\frac{1}{f}=150\mu sec$$The voltage output is given by −$$V_{0}=V_{S}\left ( \frac{T}{T-T_{ON}} \right )=150\times \left ( \frac{150\times 10^{-6}}{\left ( 150-75 \right )\times 10^{-6}} \right )=300Volts$\$