# 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:

[latex display=“true”]Up = \frac{Up_{\leq 1 μF}}{\Large\sqrt{\frac{C}{1,000,000\frac{μF}{F}}}}[/latex]

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:

[latex display=“true”]E = \frac{1}{2}\cdot{}C\cdot{}U_p^2[/latex]

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