# RMS Voltage change of MLCC Capacitor with Frequency

#### SiCEngineer

Joined May 22, 2019
270
Hi all.

So I know that the impedance of the capacitor changes with frequency. However, stumbling across a document by KEMET and their new stacked MLCC capacitors which allow higher ripple current, https://www.mouser.co.uk/datasheet/2/212/1/KEM_C1105_KONNEKT_KC_Link_C0G-1855982.pdf

There are some figures which show that the AC RMS Voltage of the capacitor is reducing with frequency, on page 4 for example.

Now, I see that these types of capacitors are always used in LLC resonant converters - at frequencies close to a 1MHz and even higher sometimes. How is this possible if the permitted AC voltage in the capacitor is very low at higher frequencies? Or am I reading the data sheet wrong!?

#### Ian0

Joined Aug 7, 2020
759
Seems like it might have a maximum current rating.

• SiCEngineer

#### SiCEngineer

Joined May 22, 2019
270
Seems like it might have a maximum current rating.
I believe so, that’s why they’re stacked to increase the ripple current. However it still doesn’t explain how the achievable voltage reduces with frequency. You can get 100’s volts out of the capacitors just because the frequency goes above a few hundred kilohertz... never seen that before. So I think maybe I’m not understanding what the data sheet is attempting to show. The rates DC voltage is very high so I don’t see why this should be the case.

#### Ian0

Joined Aug 7, 2020
759
Reactance = 1/(2 x PI x frequency x Capacitance)
Voltage = current x reactance
so Voltage = current/(2 x PI x frequency x Capacitance)

So if there is a maximum current, which is probably the maximum amount that the metallisation can withstand) then it follows that there is also a maximum voltage that is inversely proportional to frequency.

Ever wondered why a low-ohm Power resistor has a maximum voltage rating equal to SQRT(P x R) ?

• SiCEngineer

#### SiCEngineer

Joined May 22, 2019
270
Reactance = 1/(2 x PI x frequency x Capacitance)
Voltage = current x reactance
so Voltage = current/(2 x PI x frequency x Capacitance)

So if there is a maximum current, which is probably the maximum amount that the metallisation can withstand) then it follows that there is also a maximum voltage that is inversely proportional to frequency.

Ever wondered why a low-ohm Power resistor has a maximum voltage rating equal to SQRT(P x R) ?
Okay, that makes sense. So if I want to operate a resonant tank with a voltage or say 300V at 500kHz it won’t work? I should find different capacitors?

#### Ian0

Joined Aug 7, 2020
759
You didn’t specify the capacitance.
If I take 100nF because it is in the middle of the range.
X= 3.18Ω,
So current must be 94 Amps with 300V @ 500kHz across the capacitor.
I think it will melt.