# DC current at input and ouput capacitance of buck

I'm working on this website

http://www.onmyphd.com/?p=voltage.regulators.buck.step.down.converter

Could someone help me to understand those two statements:

1) "since the DC current of the inductor cannot flow through the output capacitor, then it can only flow to the load." But I can also read that "It is easy to see that during the off state, the capacitor is charged with current Iin".

My question is why the output DC current can't go inside the output capacitance but the input DC current can go inside input capacitance. I heard so many time that DC current can't go through a capacitance but at the input of buck it's what happen.

2) "That in turn means that only the ripple of the inductor current, centered around zero, flows through the capacitor. Then we have negative portions of the ripple that remove charge from the capacitor and positive portions of the ripple that add charge to the capacitor. Both must transfer the same amount of charge in equilibrium."

What way the decharging current of the output capacitance take. If I don't do any mistake : the only way I see is to decharge via load resistor. But that mean the load current is not so DC.

• Copy and paste the relevant section into your question. – Andy aka Apr 7 '20 at 11:38

1. Don't confuse voltage with current, and charge/discharge currents (alternating directions) with a continuous current flow in one direction (DC). Your main problem here is misunderstanding of two similar words that have different meaning, namely IN and THROUGH.
You are right that DC current can't flow THROUGH the capacitor, but you are confused with the mentioned text about the current flowing IN(to) the capacitor.

There is no contradiction here, as we all know that a current will flow INTO a capacitor when it is charging and will stop when it is charged up (meaning that the voltage across the capacitor is equal to the voltage of the source applied to the capacitor).
Likewise, the current will flow OUT of the capacitor when the capacitor's voltage is higher than the voltage of the circuit it is connected across, and will stop once they're equal.

1. "That in turn means that only the ripple of the inductor current, centered around zero, flows through the capacitor. Then we have negative portions of the ripple that remove charge from the capacitor and positive portions of the ripple that add charge to the capacitor. Both must transfer the same amount of charge in equilibrium."

The "equilibrium" mentioned here is when there is an apparent constant voltage across the load and the capacitor in parallel and an apparent (somewhat constant) DC current flow through the load. This apparent constant voltage is not perfectly flat, smooth, but it is a mostly DC with a small AC voltage on top of it (superimposed), so it keeps going up and down (its frequency is equal to the converter's switching frequency).
"Centered around zero" - the "zero" here is the middle point of this small AC voltage on top of the DC voltage, in other words, the middle between its maximum and minimum value (which would be its positive and negative peaks if we isolate the AC component alone).
The capacitor maintains the average, half-point or zero-point voltage between these maximum and minimum variations, and it does that by absorbing the charges/current (charging) during the maximum/peak point (voltage positive compared to the zero-point) and releasing the charges-current (discharging) during the minimum/valley point (voltage negative compared to the zero-point).

The image below should illustrate my points here ("Mean, Vdc" is the same as the "centered around zero" or the "zero-point" mentioned above):

• Thank you very much for taking your time to anwser . I think it's more clear now. For the first point i should mention it : When people say that DC current do not flow into the capacitor, we suppose of course that voltage between the capacitor is constante! else is not totally true. Often i just heard that " DC current can't flow into capacitance without giving any precision and it make me frustrated in my learning process :). For the second point : if i understood what you say: the decharging current of the capacitance discharge through the load right ? – kem Apr 13 '20 at 21:10
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