Buck Converter - 2

Electricity and Magnetism Level 3

The picture above is an example of a buck converter. It converts a high voltage DC signal to a low voltage high current DC signal with great efficiency.

During the ON state, the inductor is charged until the supply is cut off. When that happens the OFF state is achieved, and the inductor supplies the load in turn.

The magnitude of the output voltage is controlled by varying the time of the ON state and OFF state of the circuit. That is, the longer the time of the ON state, the larger the magnitude of $V_o$ .

For a $25\: kHz$ triggering signal, how long (in $\mu S$ ) should current flow in the diode in every cycle if I want the magnitude of the output voltage to be $10$ % of the supply voltage? Ignore capacitance effects and the internal resistances of the components.

The answer is 36.

1 solution

Efren Medallo
Mar 8, 2016

Let $T_{off}$ be the time allotted for current to flow in the diode. Let $T_{on}$ be the time the current won't flow in the diode. Essentially, $T_{on}$ will also represent the charging time of the inductor, and $T_{off}$ the discharging time. Let us now analyze.

$\large V_{L} \: = \: L \frac{di}{dt}$

$\large \Delta i \: = \: \frac {V_{L}}{L} \Delta t$

when $\Delta t \: = T_{on}$ , $V_{L} \: = \: V_i - V_o$

when $\Delta t \: = T_{off}$ , $V_{L} \: = \: V_o$

Substituting the values of $V_{L}$ for corresponding values of $\Delta t$ will yield two equations for $\Delta i$ . Equating them gives us

$\large \frac{V_i - V_o}{L} \cdot T_{on} = \frac {V_o}{L} \cdot T_{off}$

Rearranging the terms, we get

$\large \frac {V_o}{V_i} = \frac {T_{on}}{T_{on} + T_{off}}$

since we want $V_o$ to be 10% of $V_i$ , this will simplify to

$\frac {T_{on}}{T_{on} + T_{off}} = 0.1$

from here we will see that $T_{off} = 9T_{on}$ . But $T_{on} + T_{off} = T = \frac{1}{f} = 40 \mu S$ .

So we now know that $T_{off} = 36 \mu S$ .

Buck Converter - 2

The answer is 36.

1 solution

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