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How much power must be dissipated in a low resistance short circuit to detect using lock-in thermography?
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Lock-in thermography tests involve repeatedly applying voltage to the circuit containing the resistive short or leakage. When testing a device or board with a very low resistance, usually the power supply voltage is reduced to limit current to prevent damage to the device/board or altering the fault. Dead shorts (shorts with very low resistance) can be difficult to detect because there is very little power dissipated in the short at low current flow.
Calculate the power dissipated in the fault using the equations V = I * R and P = I * V to determine if there is adequate power dissipation to be detected. Make sure to use the actual voltage applied (after any current limiting) to calculate actual power dissipation.
In order to detect a low resistance short, following are useful guidelines. For bare or decapsulated devices, a minimum of approximately 10µW dissipation in the fault is required to detect within 10-15 minutes. 100µW can typically be detected within seconds. Less than 10µW may require several hours or even overnight testing. Heating of 1mK (0.001°C) can be detected within 5 minutes. If a test is run for approximately 20 hours, heating of 0.1mk (0.0001°C) can be detected.