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TPS54531: Characterization of Thermal Performance in situ

rojo-mojo
Prodigy 250 points
Part Number: TPS54531

I was reviewing the TI white paper "Measuring the Thermal Performance of theTPS54620" (http://www.ti.com/lit/an/slva397/slva397.pdf). This white paper suggests using the FET body diode characterized over temperature (unpowered) to directly measure the Junction Temperature.

It would appear I can apply this method to the TPS54531 to determine junction temperature as the part is used in my circuit/PCB by measuring the voltage (under constant current) from "PH" (pin 8) to "Vin" (pin 2). I can also use the data sheet to estimate the power in the SMPS IC at various load currents.

What is NOT clear to me is how (assuming a constant ambient temperature ... e.g. room temperature) you load the circuit in operation AND continue to measure the forward drop across the internal FET body diode in order to relate that to temperature.

  • 0
    TI__Expert 8245 points
    Hi

    Notice that the FET used on this paper is between Power Good pin and GND, it offers an open-drain output to indicate whether the IC is operating nomarally. No current on this FET when IC's working if not used, so it's different from that one between "PH" (pin 8) to "Vin" (pin 2) from TPS54531, which is the high-side power MOSFET of buck topology. Actually, the voltage of PH is fixed (almost Vin when high-side MOS's on or 0 when off) when load's heavy. You may still find oppotunities to apply an enxternal voltage higher than Vin on PH when load is light, but it's less meaningful and easy to break the IC.
  • 0 in reply to Iven Su
    Prodigy 250 points
    Could I apply power to the circuit with the body FET constant current source disconnected and run at a load until I reach thermal stability, then turn OFF the load and switch in my constant current source quickly to measure the temperature? I realize as soon as I remove the load, the temperature will decrease rapidly, but maybe I can be quick enough to get close.
  • 0 in reply to rojo-mojo
    TI__Expert 8245 points
    Hi,

    You also need to turn off the input power and the temperature will no doubt decrease rapidly. You can compare the measured temp with surface temp observed by a radiation thermometer.