X1/X2/Y1/Y2/Y4 impulse withstand rating voltage calculator (IEC 60384-14)

IEC 60384-14 specifies that X1/X2-rated capacitors shall be tested to withstand an impulse voltageof 4 kV (X1), 2.5 kV (X2, Y4), 8 kV (Y1) or 5 kV (Y2).

However these values only apply for a capacitance $$\leq 1 μF$$ (except for Y1/Y4 capacitors). Use this calculator for X1/X2/Y2 capacitances $$> 1 μF$$!

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[calculator-input name=“c” label=“capacitance” unit=“F”][/calculator-input][calculator-expression name=“ratingX1” formula=“c <= 1e-6 ? 4e3 : 4000.0/sqrt(c/1e-6)” unit=“V”][calculator-expression name=“ratingX2” formula=“c <= 1e-6 ? 2.5e3 : 2500.0/sqrt(c/1e-6)” unit=“V”][calculator-expression name=“ratingY2” formula=“c <= 1e-6 ? 5e3 : 5000.0/sqrt(c/1e-6)” unit=“V”][calculator-output name=“impulse withstand rating” unit=“kV”] A X1-rated capacitor of <%= format(c, “F”) %> has an impulse withstand rating of <%= format(ratingX1, “V”) %>

A X2-rated capacitor of <%= format(c, “F”) %> has an impulse withstand rating of <%= format(ratingX2, “V”) %>

A Y1-rated capacitor of <%= format(c, “F”) %> has an impulse withstand rating of <%= format(8e3, “V”) %>

A Y2-rated capacitor of <%= format(c, “F”) %> has an impulse withstand rating of <%= format(ratingY2, “V”) %>

A Y4-rated capacitor of <%= format(c, “F”) %> has an impulse withstand rating of <%= format(2.5e3, “V”) %>

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Formula:

$$Up = \frac{Up_{\leq 1 μF}}{\Large\sqrt{\frac{C}{1\,000\,000\frac{μF}{F}}}}$$

where:

• $Up$ is the impulse withstand voltage rating
• $C$ is the capacitance in Farads
• $Up_{\leq 1 μF}$ is the voltage rating for that capacitor class with a capacitance of $\leq 1 μF$:
  • For X1-class: 4 kV
  • For X2-class: 2.5 kV
  • Y1-class impulse withstand voltage is always 8 kV no matter what capacitance
  • For Y2-class: 5 kV
  • Y4-class impulse withstand voltage is always 2.5 kV no matter what capacitance

Why is the impulse withstand voltage lower for larger capacitors?

The rationale behind the deratingof the impulse withstand voltage is that larger capacitances will have sufficient capacitance so that a given overvoltage doesn’t cause a large voltage spike in the capacitor.

The formula (see above) is chosen so that the energy in the capacitor:

$$E = \frac{1}{2}\cdot{}C\cdot{}U_p^2$$

is kept constant (i.e. at the same value as for a equivalent capacitor of 1 μF).