*IEC 60384-14* specifies that X1/X2-rated capacitors shall be tested to withstand an * impulse voltage *of

*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!**Σ

*TechOverflow calculators:*

You can enter values with SI suffixes like

**12.2m**(equivalent to 0.012) or

**14k**(14000) or

**32u**(0.000032).

The results are calculated

*while you type*and shown directly below the calculator, so there is no need to press

*return*or click on a

*Calculate*button. Just make sure that all inputs are

**green**by entering valid values.

F ⚠

#### 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

- For X1-class:

#### Why is the impulse withstand voltage lower for larger capacitors?

The rationale behind the *derating *of 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^2is kept constant (i.e. at the same value as for a equivalent capacitor of *1 μF*).