# decoupling capacitors for DC-DC converters

I am using a DC to DC converter (Murata MEJ2S0505SC (datasheet)) (5 V to 5 V at 500 mA) for isolation purposes. Now I want to calculate the input decoupling capacitor.

Can anyone please suggest any methodology to calculate it? (Note: It is not mentioned in the data sheet.)

• Have you tried simulating it? – winny May 25 '20 at 11:57
• Now i added data sheet,and i did not simulated and i need theritical explanation. – yugi May 25 '20 at 12:03
• The capacitor is usually associated with an inductive element to form a front-end $LC$ filter. You dimension these components based on a maximum allowable input ripple current at a certain frequency. Usually a small $L$ is adopted as it supports the dc current while a large capacitor can cope with the high-frequency pulses absorbed by the dc-dc module. Have a look at a seminar on filters I taught at an APEC conference in 2017. The example starts slide 80. Don't forget to damp the filter. – Verbal Kint May 25 '20 at 12:12
• Did you look through their application notes, DCAN-68 "Low power isolated DC-DC application notes"? – SteveSh May 25 '20 at 12:18
• murata.com/-/media/webrenewal/products/power/appnote/… – Andy aka May 25 '20 at 12:53

The input filter specifications depend on: (1) your source impedance and (2) dynamic frequency of the DC-DC converter (3) the spectrum of your step load. (4) radiated EMI of source current and conducted noise as a result of (1,2,3)

This is a 2-way filter spec. If you require load regulation noise on the source, the unbalanced load impedance being the input to the DC-DC can generate noise on the source.

Consider the input cap. as a LPF to the source impedance including path inductance, both DM and CM. Ideally, you want 0 Ohms and expect a well-designed DC-DC converter from Murata will have dampened chokes with suitable Q's or L/R ratios.

## Design thought experiment

• Consider the voltage divider relationship of a repetitive pulse with load impedance and the ESR of your input cap