Hi I went through some topics on choosing the output capacitor values.

They mentioned to look at two solutions

Cout=inductor_ripple/(8 * Frequency* Vripple)..............................(1)


and

necessary capacitance for desired manixum overshoot is

Cout=(max. current change)^2* inductor/(2*vout*Vovershoot).............(2)

which formula should be given more priority while choosing caps..some suggestion please

 

Inductor ripple current and turn on overshoot are two parameters, you decide which matters more in your design. As a rule of thumb, inductor ripple current should not exceed about 25% of average inductor current. Turn on overshoot should not be more than maybe 10%.

 

Often the output cap is selected for its ESR using:

Vripple = ILpp x ESR (just ohms law, E = IR)

- Vripple is the amount of ouput ripple voltage.

- ILpp is the peak to peak inductor ripple current.

- ESR is the capacitor ESR

Also, remember to consider using two or more caps in parallel to reduce the ESR.

Edit for clarification:
The procedure is to choose how much ripple you want to allow, then select a capacitors (or capacitors) to get an ESR that will get you there.

 

You can also calculate the voltage droop of the output of the power supply during the time when the output cap must hold up the load. For example in a boost when the switch is turned on.

I = C dV/dt

I = output load current

C = output cap

dV = amount of voltage droop you will allow

dt = time when the cap must hold up the load. For example, in a boost this is the on-time.

Usually when you calculate this, you'll find that the capacitance value required is driven by the ESR value calculated in my prior post. In other words, the ESR requirement for output ripple forces a larger cap than this calculation requires.