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In my project I am using 12V DC from external power supply and EUT will load max 2A; in which I am using EMI EMC filter. I have considered Input MOV’s for common mode and differential modes, differential noise filter, common mode choke. Can I need to use X capacitor too my circuit to comply the EMI EMC testing?? If yes how I can calculate capacitor value?

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X-type capacitors are designed to be across an AC line and they are tested and manufactured to more stringent requirements in order that they do not short out the AC, take plenty of amps and cause a fire.

This is unlike (say) ceramic caps - once they break down they can go short circuit and you get a fire if the power source is fairly unlimited (like an AC line)!

If your external supply is 12V DC and can only supply a current of a few amps then I don't see the need to use X-type capacitors - basic ceramic types of 25 V rating with a decent dielectric will do the job and probably better in terms of reducing EMI.

Regards the calculation of a value, there is no information in the question to solve this.

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  • \$\begingroup\$ Hi Andy aka, Thanks for your answere and clarifying my questions. \$\endgroup\$
    – Arumugam
    May 28 '15 at 5:00
  • \$\begingroup\$ @ Andy aka we have selected 4.7uF/50V capacitor; now need to go for pre complaints testing. I have concern for Damped oscillatory testing both slow and fast damped oscillatory testing as per EN61000-4-18. Will selected ceramic 4.7uF/50V capacitor pre complaints pass ?? or need to go back for X rated capacitor with higher voltage ?? I assume the damped oscillatory wave should be sine oscillation ?? \$\endgroup\$
    – Arumugam
    Sep 8 '15 at 7:41
  • \$\begingroup\$ This is all beyond the scope of your original question. \$\endgroup\$
    – Andy aka
    Sep 8 '15 at 9:35
  • \$\begingroup\$ Yes you’re correct; Now we are trying to qualify the product for pre complaints testing as per requirement; Since i am seeking your advice to go further. \$\endgroup\$
    – Arumugam
    Sep 8 '15 at 11:12
  • \$\begingroup\$ Please raise this as a new question if you are happy that the original question has been adequately answered (requires acceptance of the answer). \$\endgroup\$
    – Andy aka
    Sep 8 '15 at 11:36

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