- #1
CoolDude420
- 201
- 9
I am reading a book 'Fundamentals of Power Electronics' - Erickson. The book begins with a buck converter and the assumptions that we use so that we do not have to deal with 2nd order equations. I will first explain what the book says
The book says that the LC filter of a buck conveter should be designed so that the cut-off is lower than the switching frequency of the converter. This is to remove any ripples at the switching frequency or its harmonics. Of course, you cannot fully eliminate them, you can only attenuate with the LC filter, so there will be a small ripple. The small ripple approximation assumes that the output of the LC filter (= output of buck converter) is just the DC output voltage and that we can neglect these ripples. Ok, makes sense. Let's move on.
The book now draws the inductor current waveform. It provides an equation for the inductor ripple current. The DC portion of this current goes to the resistive load, thereby generating a fixed output voltage. The AC portion goes to the output capacitor.
I am very confused. We started with an assumption that the output voltage is fixed, we used that assumption to get the equation for the inductor current. The AC part of this current flows to the cap, which creates an AC ripple on the output. This is a contradiction.
Neglect ESR's, ESLs, on-resistances, parasitics, etc.
What the book says?
The book says that the LC filter of a buck conveter should be designed so that the cut-off is lower than the switching frequency of the converter. This is to remove any ripples at the switching frequency or its harmonics. Of course, you cannot fully eliminate them, you can only attenuate with the LC filter, so there will be a small ripple. The small ripple approximation assumes that the output of the LC filter (= output of buck converter) is just the DC output voltage and that we can neglect these ripples. Ok, makes sense. Let's move on.
The book now draws the inductor current waveform. It provides an equation for the inductor ripple current. The DC portion of this current goes to the resistive load, thereby generating a fixed output voltage. The AC portion goes to the output capacitor.
Why am I confused?
The AC portion that goes to output capacitor will cause the output capacitor voltage to increase. The output capacitor voltage IS the output voltage of the converter, hence there will be a ripple on the output voltage due to the AC current of the inductor flowing into the cap. This contradicts the small-ripple approximation we made at the start, that the output voltage is fixed.I am very confused. We started with an assumption that the output voltage is fixed, we used that assumption to get the equation for the inductor current. The AC part of this current flows to the cap, which creates an AC ripple on the output. This is a contradiction.
Neglect ESR's, ESLs, on-resistances, parasitics, etc.