Capacitors are made of two conducing plates separated by a dielectric or insulator. As voltage is applied, the capacitor charges up and stores energy across the dielectric. That energy is then released at a pre-determined discharge time, which is controlled by the circuit time constant. However, the dielectric between the two plates is not perfect. As a consequence, a small amount of current leaks out prematurely. This current is the leakage current.
Find the capacitance, "C," and the rated voltage, "V," associated with the capacitor. Refer to the capacitor specification sheet. As an example, assume C is 12 uf and V is 50 volts. The capacitance associated with capacitors is measured in units of farads, or "f." The unit "uf " stands for microfarads, which is 1 x 10^-6 farads. Microfarads capacitor values are typical for most circuits.
Choose the voltage applied time, or "T," to calculate the leakage current. For example, the leakage current value changes depending on the amount of time that voltage is applied to the circuit. With a longer applied voltage time, the leakage current grows larger because the dielectric breaks down over time. For this reason, choose T for your leakage current calculation. As an example, assume T is 5 minutes.
Calculate the leakage current, or IL, using the formula IL < (T C V) + 3 uA. The formula reads leakage current is less than the product of voltage applied time, "T," the capacitor, "C," and the voltage, "V," plus 3 uA. 3uA stands for microamps or 3 x 10^-6 amps, which is a constant associated with this equation. The less than sign means the leakage current will always be less than the value calculated. Using the example numbers:
IL < (5 12 x 10^-6 50) + 3uA = 3000uA + 3uA = 3003 uA or 3.003 mA.
The results in this example states that the leakage current will always be less than 3003 microamps or 3.003 milliamps. You divide microamps by 1000 to get milliamps.
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