Capacitor calculation at the input and output of the DC-DC converters.

 What is the need for the capacitor at the input and output of the DC-DC converters?

The first objective in selecting input capacitors is to reduce the ripple voltage amplitude seen at the input of the module. This reduces the RMS ripple current to a level that can be handled by bulk capacitors. Ceramic capacitors placed right at the input of the regulator reduce ripple voltage amplitude. Only ceramics have the extremely low ESR that is needed to reduce the ripple voltage amplitude. These capacitors must be placed close to the regulator input pins to be effective. Even a few nano henries of stray inductance in the capacitor current path raise the impedance at the switching frequency to levels that negate their effectiveness. Large bulk capacitors do not reduce ripple voltage. The ESR of aluminum electrolytic and most tantalums are too high to allow for effective ripple reduction. Large input ripple voltage can cause large amounts of ripple current to flow in the bulk capacitors, causing excessive power dissipation in the ESR parasitic. To reduce the RMS current in the bulk capacitors the ripple voltage amplitude must be reduced using ceramic capacitors. As a general rule of thumb, keeping the peak-to-peak ripple amplitude below 75 mV keeps the RMS currents in the bulk capacitors within acceptable limits.

How to calculate the capacitor for value Ceramic capacitor?

To determine the amount of ceramic capacitance required to reduce the ripple voltage amplitude to acceptable levels is given by equation.

Example for the ceramic calculation

Converting Vpp to Vrms 

75mVpp = 22mVrms

Ohm's law can be used to determine the ripple current throw the 35mohm to the ESR input buck Converter.

22mV/35mohm = 628mA

so the buck converter power dissipation can be given as = 13.8mW

The calculation for the Output buck capacitor

Given 

Current = 1A

dV = 0.3V (Ripple Voltage)

dt = 250us

C = I/(dV/dt) = 1/(0.3/250us) = 833.34mF

C = 833.34uF

As an alternative, we can also calculate the capacitance with the current and capacitor charge equation.

I = Q/t ---- (1) ; 

Q = CV ---- (2);

Using (1) we get Q

Q = 1*250us = 250uC

Now using (2)

C = 250us/0.3

C = 833.34uF