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Here's what I know about inductors:

As per Lenz's law, inductors sort of maintain the direction of current passing through an inductor. So whenever there is an interruption on the source from which the inductor is getting current, the inductor tries to maintain the direction of current by flipping its own polarity or in other words, it goes from a component that sinks current to a component that sources current. In illustration:

enter image description here

So this is where I can't seem to get it:

I'm trying to wrap my head around the circuit analysis of buck converters at steady-state, particularly when the high-side transistor goes from closed to open. The equivalent circuit can be drawn like this:

enter image description here

The way I understand it with KVL, the equation for inductor voltage should be:

enter image description here

However, textbooks don't seem to consider the fact that the inductor changes its voltage polarity whenever there is a source voltage interruption which manifests as the transistor going from closed to open in buck converters.

Is there any explanation for this? Or is there a gap in my understanding of inductors and/or buck converter circuit analysis. Your help will be greatly appreciated.

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  • \$\begingroup\$ I am confused reading about your confusion. So I'm probably of no help. But it is more the case (at a high level of view) that the sign of the voltage across the inductor (however you are looking at it before and after) changes because the dI/dt changes. In short, when the voltage supply is no longer able to sustain an increase in current and the inductor is then declining its current, the sign changes. I look at this as when energy stored in the magnetic field is increasing then the sign is one way and when energy stored in the magnetic field is decreasing then the sign is opposite. \$\endgroup\$ Dec 6, 2023 at 23:02
  • \$\begingroup\$ Do we have prolems with image server? The images aren't shown properly. \$\endgroup\$
    – internet
    Dec 8, 2023 at 7:26

1 Answer 1

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When the high-side fet , the current flow from Vs to Vo so the voltage across the inductor is VL=Vs-Vo and so the current slope in the inductor is (Vs-Vo)/L …

When highside turnoff and low-Side is now conductive , the current keep the same direction toward Vo I mean because of the magnetic field of the inductor that try to prevent the current flowing into it from changing instantaneously... and so the voltage across the inductor is VL=0V-Vo and so the current slope is -Vo/L

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