Capacitor Impedance Calculator | Capacitive Reactance Xc = 1/(2πfC)

How to use capacitor impedance calculator

To use the calculator enter two values. The blank field will be automatically calculated. If you fill in all three fields, the last calculated value will be recalculated. You can modify the units using the selectors.

Impedance of a capacitor

A capacitor is a passive circuit component that stores charge for short periods of time. The quantity that describes a capacitor is the capacitance, and is given in Farads [F]. For 'small' electronics, one Farad is a very large value, so it is usual to give it in 'millis', 'micros', 'nanos'... Farads indicate the amount of charge that can be stored in a capacitor for each volt between the terminals. The main characteristic of a capacitor is that its impedance varies with frequency. In a way, you can understand it as a resistor that changes its value according to the frequency of the wave in the circuit. This last sentence hides some little lies, but, in general, you can see it like this. For very low frequencies the impedance is huge. Actually, it is so big that it behaves like an open circuit in DC. For high frequencies, the opposite happens, its impedance is very low. At high enough frequency, you can almost consider it as a short circuit. By combining capacitors and other elements, filters can be made that select or reject certain frequencies. The equation describing the impedance of a capacitor as a function of frequency is:

$$Z[Ω] = \frac{1}{2\cdot \pi \cdot f \cdot C}$$

A note on terminology: the theoretical element of a circuit is called capacitance. The physical object is called capacitor. By abuse of language, sometimes they are interchanged, but depending on the context, it may be necessary to speak properly.

If you are choosing this capacitor for a filter, the low-pass filter calculator helps you see the cutoff, and the capacitor impedance graph shows where ESR and ESL start to matter.

Example

What is the capacitance a capacitor must have if I want its impedance to be 0.5[Ω] at 50 k[Hz]?

• To calculate the capacitance, we use the above equation but we solve for this variable: $$C = \frac{1}{2\cdot \pi \cdot f \cdot Z} = $$ $$= \frac{1}{2\cdot \pi \cdot 50000 \cdot 0.5} = $$ $$= 0.0000063 [F] = 6.3 µ[F]$$

Frequently Asked Questions

• How does frequency affect capacitor impedance?
  As frequency increases, impedance decreases. This is why capacitors block DC (zero frequency) but pass AC signals. Ideally, at infinite frequency, impedance is zero. Quick trick: at DC, capacitor is an open circuit; at very high frequencies, it behaves like a short circuit.
• Why does capacitor impedance decrease with frequency?
  When a voltage is applied across a capacitor, it takes time to charge due to the electric field building between its plates (charges go to the plates). When voltage is reversed, a wave propagates from one plate to the other through the circuit. At higher frequencies, the capacitor has less time to charge fully before the voltage changes direction, resulting in lower impedance. When voltage varies rapidly, charges are always coming and going, so the capacitor appears to offer less opposition to current flow.
• Why is the calculator result ideal?
  This calculator uses the ideal capacitor formula. Real capacitors have parasitic inductance (ESL) and resistance (ESR) which affect impedance at high frequencies. Use our Capacitor Impedance Graph for real-world modeling.