So the concrete answer ended up being it depends
SITUATION 1: If you have a small load connected to the buck converter, it will draw a very small current. If this is the case the capacitor will be the main current provider while the transistor is OFF.
This is because the buck converter will work in DCM or Discontinious conduction mode.
This means that because the load is small
- It draws very little current from the buck converter
- Then the current through the inductor is small when the transistor is ON
- That small current cant completely charge up the magnetic field of the inductor when the transistor is ON.
- This means that when the transistor turns off, the inductor does not have much (or almost any) energy to provide, so it cant provide a current to the load.
- At this point the capacitor provides the current needed, since the
amount of current needed is small and it needs to provide it for a small amount of time (beccause the frecuency is high), there is no problem and the output is continious.
At this mode, the inductor is just limiting the speed at which the capacitor recharges, while the capacitor is the main current provider.
The following simulation illustrates this situation
SITUATION 2: If you have a big load connected to that very same buck converter, that load will draw a larger amount of current. In this case the inductor will be the main current provider while the transistor is OFF and the capacitor will just be there to smooth the output
This is because the buck converter will work in CCM or Continious Conduction Mode, this means that because the load is large
- It draws more current from the buck converter
- then the current through the inductor in big when the transistor is ON
- That bigger current charges up the magnetic field in the inductor when the transistor is ON
- This means that when the transistor turns off, the inductor has a lot of energy to provide, so it provides current to the load while also mantaining the charge of the capacitor
- The capacitor then just smooths the output.
At this mode, the inductor is the main power provider, while the capacitor just smooths it's output.
The following simulation illustrates this situation, using the same buck converter with a larger load.
SITUATION 3: The load is somewhere in the middle, the inductor charges up some, but decays quickly during the transistor OFF time and the capacitor takes over, this is still DCM, however here the role of main current provider is distributed in both elements up to some degree.