Description
To test capacitors using lock-in thermography, a parallel resistor (bypass resistor) is required to shunt current through the bypass resistor when voltage is in the low state. Without the bypass resistor, due to the existing capacitor charge during the voltage high state, current would continue to pass through the leakage site, reducing lock-in test sensitivity.
Capacitor Charge Time
The time it takes for a source meter to fully charge a capacitor depends on the RC time constant of the circuit. A capacitor is considered fully charged when it reaches approximately 99% of its target voltage, which typically requires approximately five time constants. The time constant (τ) is calculated as the product of the resistance (R) and capacitance (C): τ = R [ohm]* C [farad].
When sourcing voltage, source meters typically have low output impedance (~1 ohm) and therefore the time constant, and therefore time to charge a capacitor, is relatively short.
For example, for a source meter with 1 ohm output impedance and 1 µF capacitor, the time constant is equal to:
τ = 1 * 10E-6 = 0.00001 sec (0.01 ms)
5 * τ = 0.05 ms
Therefore, the 1 µF capacitor will be fully charged in a fraction of the shortest lock-in cycle time.
Capacitor Discharge Time
The discharge time, however, may be longer than the charging time, depending on the fault resistance and bypass resistor values and can be calculated using the time constant equation above. This must be taken into account when selecting the lock-in cycle time to allow adequate time for the capacitor to discharge during the voltage low state.
If adequate discharge time is too short and the capacitor storage charge does not decrease during the voltage low state, then a similar level of current will flow through the fault during low state as during the high state, significantly reducing lock-in test sensitivity.