# Inductor and capacitors in the Buck converter

#### F123

Joined Oct 26, 2018
12
Here is what I know about the inductors and capacitors in the buck converter starting from transient.

Inductor and capacitor : during the switch is on, the inductor current will increase from zero with a constant slope that is Vg-0/L, (Vo = 0 at first) during DTs. Then this will charge the capacitor to some point. During the switch is off, the inductor current will decrease with a constant slope of -Vo/L with Vo being close to zero, meaning a small decrease in current during Ts. during many period Ts, we will reach steady state, which the increase in current in the inductor is the same a the decrease during Ts. This also the same for the capacitor voltage.

Since the net change in inductor voltage is zero during Ts and how net change in the capacitor current is zero?

How this can explain the energy transfer?

Based on that, how do know the steady state given this circuit? do we have to get it from observing this second order LC filter?

I got things mixed and got confused a lot. Would any please refer me to a link that explain all this or can anyone explain this to me please?

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#### analogspiceman

Joined Oct 29, 2018
8
At steady state, over one complete switching cycle, there can be no average voltage across the inductor and no average current through the capacitor. Also, an implicit assumption for this type of switching converter is that the switching frequency is much higher than the LC filter resonant frequency. To simplify matters, assume that the inductor and capacitor are lossless. The LC filter is an ideal low pass filter. AC plus DC voltage is applied to the input side and only DC voltage appears on the output side. Just as importantly, but often overlooked, AC plus DC current is drawn from the input side (but this does not matter when you are only considering output DC voltage and the input source is ideal).

Now, ask yourself, what is the average voltage appearing at the switching node? That is what the steady state output voltage will be.