DC to DC Converters Solved Example

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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$ $T_{ON}=150\mu sec$

$V_{0}=V_{S\left ( \frac{T}{T-T_{ON}} \right )}$

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

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,

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

The voltage output is given by −

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