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[FAQ] TPS2660: How much time does the eFuse takes to thermal shutdown during overcurrent fault ?

Guru* 94506 points
Part Number: TPS2660

How much time does the eFuse takes to thermal shutdown during overcurrent fault ?

  • Background:

    During overcurrent fault, the eFuse limits the fault current to a value set by the RILIM resistor at the ILIM pin. As the eFuse maintains constant output current, the output voltage VOUT is determined by the effective load impedance and hence VOUT will be lower than the supply voltage VIN. This causes power dissipation in the current limit mode, PD = ILIMIT x (VIN - VOUT).

    The power dissipation rises the device junction temperature TJ = TA + θJA x PD (θJA is the thermal resistance) which causes TPS2660 eFuse to enter into thermal shutdown once TJ crosses 157°C

    Example with TPS2660 eFuse:

    Let us say current limit is set to 2A and if the load tries to take more current than 2A, TPS2660 will limit current and VOUT will drop based on the power drawn by the load.

    For 24V VIN, let us assume the power drawn by the load is 24W i.e., 1A load (effective RLOAD of 24Ω). If the load changes to 60W i.e., 2.5A (effective RLOAD of 9.6 Ω), TPS2660 will limit current to 2A and VOUT will droop to

    VOUT = ILIMIT x RLOAD_EFF = 2 A x 9.6 Ω = 19.2 V

    PD = 2A x (24V-19.2V) = 9.6W

    Time to thermal shutdown (TSD):

    Time taken for the junction temperature to rise is based on the power dissipation and the operating temperature of the device as shown in the below “Thermal Shutdown Time vs Power Dissipation” figure. Also, note that 

    For example, at 25°C ambient operating temperature, with 10W of PD the device takes 472ms to turn off due to TSD.

    In case the output is short-circuited VOUT = 0V and then PD = 2A (24V-0V) = 48W. In this case, the device takes just 4ms to turn off due to TSD.