# What will be the output ripple current per capacitor buck regulator?

I have a existing buck regulator design. Converter part number: TPS54340Q1.

Input voltage = 5 V. Output voltage = 3.3 V

RMS ripple current is calculated as = 108 mA. (Don't worry too much about value. Conider the value for mathematical analysis)

I have two decoupling capacitors at the output:

a) Cout1 = 1 uF, ESR1 - 110 mOhm

b) Cout2 = 4.7 uF, ESR2 - 120 mOhm

Switching frequency = 500 kHz.

Question 1

I just need to calculate how ripple current will be split between Cout1 and Cout 2. I mean how will be ratio.

Can we do the current splitting just as we do with two parallel resistors for ESR1 and ESR2?

Question 2

Assume we got the value for each capacitor ripple current. How can make sure that the existing decoupling capacitor is enough to withstand the ripple current passing through it. ( I checked for rated current value in capacitors datasheet, there is no such parameters with ceramic capacitors but yes, there is rated current for tantalum capacitors. My output capacitor is ceramic type)

Done some research on internet from my end,but there was not much I could not find.

EDIT: