Leakage Current Calculator:
Enter the values of voltage, V(V) and capacitance, C(F) to determine the value of Leakage current, LC(A).
Leakage Current Formula:
Leakage current is the small amount of electrical current that flows through an insulating material or dielectric, even when there should ideally be no current.
This current can affect the performance and efficiency of electronic circuits and systems, making it crucial to monitor and minimize leakage.
Leakage current is particularly significant in high-frequency circuits and is often associated with capacitors and insulators. Leakage current is caused by insulation imperfection, voltage stress, environmental factors and aging which causes power loss, heat generation and safety risks.
- DC Leakage Current: This occurs in circuits where a direct current (DC) voltage is applied. It is usually stable over time unless affected by changes in temperature or aging of materials. • AC Leakage Current: Occurs in alternating current (AC) circuits, often associated with capacitive and inductive coupling. It can vary with the frequency of the AC signal.
Implication in semiconductor devices:
- MOSFETs: Leakage current in MOSFETs, especially in deep sub-micron technologies, can lead to significant power dissipation. Techniques like using high-k dielectrics and metal gates are employed to reduce leakage.
- CMOS Circuits: Standby leakage current in CMOS technology is a major concern for battery-operated devices. Techniques such as power gating and body biasing are used to manage leakage.
Leakage current, LC(A) in amperes is calculated by the product of 377, voltage, V(V) in volts and capacitance, C(F) in farads.
Leakage current, LC(A) = 377 * V(V) * C(F)
LC(A) = leakage current in amperes, A.
V(V) = voltage in volts, V.
C(F) = capacitance in farads, F.
Leakage Current Calculation:
- Calculate the leakage current for a circuit with a voltage of 10 volts and a capacitance of 2 microfarads.
Given: V(V) = 10V, C(F) = 2 * 10-6.
Leakage current, LC(A) = 377 * V(V) * C(F)
LC(A) = 377 * 10 * 2 * 10-6
LC(A) = 0.00754A.
- Suppose a voltage source has a capacitance of 5 microfarad and a leakage current of 0.00377 amperes. Calculate the voltage:
Given: LC(A) = 0.00377A, C(F) = 5 * 10-6.
Leakage current, LC( A) = 377 * V(V) * C(F)
V(V) = LC( A) / 377 * C(F)
V(V) = 0.00377 / 377 * 5 * 10-6
V(V) = 0.00377 / 1.885 * 10-3
V(V) = 2V.