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In section (a) in the picture given below, while the switch is conducting the capacitor is charging so the ic= (il) - (v/R). While the switch is not conducting, section b, the capacitor should be discharging isn't it? Based on that I have 2 question as below:

  • In both circuits in the picture (a, b) the capacitor current is shown as it is charging because the the way of the current is shown same. Why?
  • Considering the capacitor is discharging in (b), ic = (v/R)- (il), is this node equation correct?

enter image description here
Source: https://elprivod.nmu.org.ua/files/converters/Robert_Erikson_fundamentals-of-power-electronics-3n_2020.pdf

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  • \$\begingroup\$ The "sense" of the "defining" current ic(t) is the same. It is not said that the current is the "same" (charging) ... \$\endgroup\$
    – Antonio51
    Commented Nov 18 at 16:57

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The current equation is exactly the same for both scenarios i.e. \$I_C = I_L - \frac{V}{R}\$. Or, putting it another way, \$I_L = I_C + \frac{V}{R}\$ (more clearer).

Any part of the equation does not assume a polarity hence, when you ask this: -

In both circuits in the picture (a, b) the capacitor current is shown as it is charging because the the way of the current is shown same. Why?

You might be missing the detail of what is happening...

For a standard buck converter using a flyback diode, the inductor current (if non-zero) is always feeding the capacitor/load with energy. This happens in scenario (a) and scenario (b), except that in scenario (b), once the current from the inductor falls to zero, the capacitor is exclusively supplying current to the load. Inductor current won't go negative.

Considering the capacitor is discharging in (b)...

As soon as scenario (b) is entered the inductor continues charging the capacitor/load until it's energy store is depleted. For synchronous buck converters, the inductor current will reverse of course and the capacitor will discharge.

But, none of this changes the equation.

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  • \$\begingroup\$ Isn’t it so, that when the inductor current falls below Io (average inductor current), it is the capacitor which is discharging and supplying current to the load? \$\endgroup\$
    – Tyassin
    Commented Nov 18 at 19:45
  • \$\begingroup\$ If capacitor is charged in both cases and node equations are same , a question about charge balance arise which is here : electronics.stackexchange.com/questions/731192/… @Andy aka \$\endgroup\$
    – Mhan
    Commented Nov 18 at 20:16
  • \$\begingroup\$ @Tyassin are you trying to hint that I've said something incorrect in my answer? \$\endgroup\$
    – Andy aka
    Commented Nov 18 at 21:03
  • \$\begingroup\$ @Mhan I haven't exactly said that so, what is your comment about? Why are you linking me to a new question and implying that the new question is somehow connected. I'm answering your question above and, if I find the time I will try and answer your newer question. \$\endgroup\$
    – Andy aka
    Commented Nov 18 at 21:03
  • \$\begingroup\$ @Andy aka Nope…..I should have stated it more clearly as a question. \$\endgroup\$
    – Tyassin
    Commented Nov 18 at 21:28

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