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!
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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.
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.
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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 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).