Hunan Adio Electronic Technology Co., Ltd.

RF Current and Power of Capacitor

2020-11-13

This article mainly discusses the relationship between the loading current, power loss, operating voltage and maximum rated voltage of multilayer ceramic capacitors. The maximum current through the capacitor is mainly limited by the maximum rated voltage and the maximum power loss. The capacitance and the operating frequency of the capacitor determine which of the two is the dominant one. For a capacitor with a lower capacitance value at a given frequency or a given capacitor operating at a lower frequency, their highest voltage limit is generally reached faster than the limit of maximum power loss.

The maximum voltage rating is determined by the impedance (Z) of the capacitor, just as the power loss is determined by the impedance of the resistor, or the equivalent resistance (ESR) of the capacitor.

　　Z=1/[2pFC]

Here F is the frequency, the unit is Hz;C is the capacity, the unit is F.

The maximum peak current allowed to flow through the capacitor without exceeding the rated voltage of the capacitor is calculated as follows:

　　I=Ur/Z

Ur here is the rated voltage of the capacitor, the current is the peak current, and the unit is A.

The actual current flowing through the capacitor is calculated as follows:

　　I=R/I

Ua here is the applied voltage or the actual operating voltage.

The following examples explain how different capacitor variables affect the voltage and current limits at a fixed frequency.

Example 1: A 0.2 pF,500V capacitor is used at a frequency of 1000MHZ:

Capacitive impedance: Z = 1/[2(3.14)(1000 × 10)⁶)(0.2x10^-12)]=1591 ohms

Peak current: I = 500/795.5=0.629 A_peak

If this current is exceeded, the operating voltage will exceed the rated voltage.

Example 2: A 2.0 pF, 500 V capacitor is used at a frequency of 1000MHZ:

Capacitive impedance: Z = 1/[2(3.14)(1000 × 10)⁶)(2.0x10^-12)]=159 ohms

Peak current: I = 500/79.5=6.29 A_peak

If this current is exceeded, the operating voltage will exceed the rated voltage.

Example 3: A 20 pF, 500 V capacitor is used at a frequency of 1000MHZ:

Capacitive impedance: Z = 1/[2(3.14)(1000 × 10)⁶)(20x10^-12)]=15.9 ohms

Peak current: I = 500/7.95=62.9 A_peak

If this current is exceeded, the operating voltage will exceed the rated voltage.

Important Note: The maximum power loss value is calculated under the assumption that the end of the capacitor is an infinite heat sink. The amount of heat transferred to the air is negligible. A 10pF, 500V capacitor operates at a frequency of 1000MHZ, the peak current at the power limit is 7A, and the average current is about 5 Arms. At this operating current, the temperature of the capacitor will rise to 125°C. In order to work stably, its actual maximum working current is 2 Arms, if the heat dissipation effect of the end is very good, it can reach 3Arms.

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