Hey,
It was said in Wikipedia that
"As we consider that the converter operates in steady-state conditions, the amount of energy stored in each of its components has to be the same at the beginning and at the end of a commutation cycle".

Could some please explain to me why its correct?

For example, if the cycle duration is 10sec, and the duty cycle is 0.01,
then it means that the output capacitor will provide the load with energy for 9.9 seconds before it will reach a 0.1sec recharing period.
How can the capacitor get in 0.1sec, all voltage it lost after holding up the load for 9.9sec without being recharged during that time?

Thank you.

 

For example, if the cycle duration is 10sec, and the duty cycle is 0.01,

10 sec cycle duration? Usually, the DC-DC converters have higher switching frequency to minimize the inductors size. Maybe you wanted to say 10 usec?

How can the capacitor get in 0.1sec, all voltage it lost after holding up the load for 9.9sec without being recharged during that time?

A capacitor can be charged quickly and discharged slowly depending on the charge/discharge current. The equation is Q = I*t where Q is the charge (in Coulombs), I is the charge/discharge current and t is time. Since the charge is the same, the lower the current the longer the time.

However, your capacitor question does not apply here, since the waveform you saw is the current through the converter inductor. The inductor is the storage device. The output capacitor is just a filter, so that the voltage on load (the output resistor) is the average of the sawtooth like current times load.

 

Thanks for the help.
Theoretically, why in steady state considitions, the amount of energy stored in each component remaines the same at the begninnig and at the end of each cycle?

 

Yes, that is correct. Each cycle has a charging and discharging time. The amount of energy that goes in and out is the same, only the rates of charge/discharge differ.

 

Thank you
What does it mean that the boost converter is in steady state conditions?
When the switch is open, then steady state means that the capacitor's voltage equals Vin (assuming Vdiode=0V), and the inductor's current equals Vin/R, but that is certainly not the case, so what does it mean?