Thank you, MrAl.
To make sure I know what you mean, I will write my understanding below of yours.
The large capacitor handles the lower frequencies simply because it has a lot of capacitance and can hold the voltage over the time of the cycle or half cycle.
I think this is related to time constant τ= 1/RC. With large capacitance, τ=1/RC will be small.
Discharging equation of capacitor:
V = V(0)*e^(-1/RC)*t
where V(0) is the initial voltage of the capacitor.
τ=1/RC is small therefore the electrolytic capacitor will hold its voltage over a long time.
The smaller cap cant hold the voltage long enough to filter properly.
The same goes for ceramic capacitor, because its capacitance is small; it cannot hold the voltage long enough to filter properly.
However, I am confused about ESR of each capacitors.
Electrolytic capactitor has a large ESR while ESR of ceramic capacitor is very small.
And this seems not appropriate for charging of capacitor.
Charging equation of capacitor.
V = Vs*[1- e^(-t/RC)]
where Vs is the source voltage.
Electrolytic capacitors have a big resistance => constant time τ=1/RC is small and the capacitor will take a long time to be charged to source voltage, Vs.
This means that the capacitor voltage will cannot follow exactly source voltage in the charging process.
Is this a problem?
The small cap can handle the higher frequency components because it has less inductance.
At high frequency, Zc = 1/ωC and Zl= ωL will be very small and this capacitor can be considered as a short and the output voltage (the voltage across the capacitor) will be zero.
Is this the mechanism that ceramic capacitors cancel out high frequencies?
The higher the frequency the lower the amplitude of that component, so the diodes dont blow out.
I don't get that. Can you explain it more?