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In Buck Converter, when switch in ON, Vi(I/P voltage)=Vl(inductor Voltage)+Vo(O/P voltage)when we switch it On, current increases linearly(according to the wikipedia) then Vl=Ldi/dt=constant and Vo(=RI) increases linearly. So Vl+Vo increases but then it must be equal to Vi which in constant in ON cycle. How is it possible and why does the current increases linearly?

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  • \$\begingroup\$ This question has already been answered. Please read "Operation of a Buck Converter" under the "Related" column on the right side of the page. \$\endgroup\$ – Sparky256 Jun 9 '16 at 23:36
  • \$\begingroup\$ @Sparky256 i read it. But i don't understand why the current through Inductor is linear? Shouldn't it be exponential ? \$\endgroup\$ – tracy00 Jun 9 '16 at 23:40
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An inductor has inductive properties plus some DC resistance, but by itself it is a linear device, much like a resistor is a linear device by itself. The current through the inductor is linear because the voltage is either fully ON or OFF at the mosfet switch. The inductor core can only magnetize or demagnetize at a fixed maximum rate. Only a non-linear device such as a diode or transistor (bjt) would have non-linear behavior by themselves.

A non-linear rise or fall time is accomplished by combining resistors and capacitors and/or inductors together to form a time-constant. This does not include circuits with constant-current sources or sinks, as they would intentionally create a linear rise and fall time. Notice that in your equations you did not insert any non-linear properties.

There are many post on this site and websites involving "Switch mode power supplies" that offer books of data and pages of math to help in understanding the unique but predictable properties of inductors.

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When a buck regulator is operating correctly, the short amount of time that the transistor switch is closed does not significantly raise the output voltage. Yes, this is an approximation to reality but it holds good for very decent estimations. Consider that the output has a large value capacitor - it will not change its terminal voltage very quickly compared to the time base of the switching frequency.

If it did significantly change the output voltage in each switching cycle then it is a really crappy buck regulator.

The inductor and output capacitor form a 2nd order low pass filter with a cut-off frequency much lower than the switching frequency. Typically for a 100 kHz switching frequency using a 33 uH inductor and 100 uF capacitor the cut-off frequency is: -

Fc = \$\dfrac{1}{2\pi\sqrt{LC}}\$ = 2770 Hz

At 100 kHz the attenuation is 40 dB/decade from 10 kHz to 100 kHz plus about another 12 dB between 2770 Hz and 10 kHz. Total attenuation is 52 dB or about 400:1.

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