I am designing a board for my project which is powered from 4.2V Li-Ion Battery. I am using a buck-boost converter ADP2504 for powering the board which outputs 4.0V constant voltage. I have buzzer on my board for alarm applications. Whenever I turn on my buzzer I get a huge drop of 1.4V in my regulator output (attached is my drop waveform) and comes back to 4.0V in 33 microseconds. 
When I turn on buzzer, current shoots up by 200mA (buzzer current = 200mA). Question is
How to calculate a capacitor to handle this transient with the above input? Currently I don't have any capacitor in my board. Above observations are without capacitor.
If the calculated value is very high in 100's of micro Farads, how to handle inrush current caused by that?
Best Answer
A correctly designed regulator based on ADP2504 should be able to supply 200mA indefinitely.
In general, every buck regulator has to have some output capacitance, but in many applications the capacitor does not have to be particularly large.
However, if your load increases abruptly, the regulator needs some time to ramp up the current in the inductor and, for that time window, the capacitor should be able to hold the line without significant voltage droop.
Based on your diagram, the regulator needs 33uS to catch up. Say, you want to limit the voltage drop during this time to 0.1V. Then the minimum capacitance could be calculated as C=dQ/dV=200mA*33us/0.1V=66uF.
You can reduce this time window and therefore the requirement for the capacitance by reducing the size of the inductor, but that would require more detailed analysis of the circuit.