Ideal Buck Converter

Thread Starter

tim510

Joined Jan 11, 2012
16
Hello, I'm new to the forum but I've been lurking around for a few months.
It's a great forum and hope to contribute in the future.


My hw question is discussing a non ideal buck converter.


I'm looking at an ideal buck converter and trying to come up with the current for the capacitor during one period (steady state).


courtesy of mathworks.com

Current-second balance is needed.
<ic> = 0

During Switch on time: D
ic = iL - Vo/R

During Switch off time: D'
ic = - Vo/R

This is what I came up with. Does it look correct? The one that I was a little confused with is during off time. Wasn't sure what the inductor is doing at this time. Does it create a back EMF with that sudden change of the switch but is it not accounted for supplying current during off time because it is so little?

Does this sound correct?

Thanks
 

crutschow

Joined Mar 14, 2008
24,746
During the on time, the inductor current is increasing by di = ΔV/L dt where di is the amount of current increase, ΔV is the voltage across the inductor (Vin - Vo), and dt is the time the switch is on. Energy is stored in the inductance with a value of 1/2 LI\(^{2}\).

During the off (grounded) time the stored inductor energy keeps the current flowing but the current is decreasing as it losses energy, again by a value of di = ΔV/L dt where di is the amount of current decrease, ΔV now equals Vo, and dt is the time the switch is off.

There is thus a ripple in the inductor current that is smoothed by the output capacitor. The larger the capacitor, the smaller the output ripple voltage.

Since the switch rapidly changes position, the inductive spike is not seen. If you opened the switch without connecting it to ground, then there would be a large voltage spike.
 
Last edited:

Thread Starter

tim510

Joined Jan 11, 2012
16
Thanks for the input Crutschow I think this answers my question about the inductor. Not sure if this verifies my equation for the capacitor current though.
 
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