INA215: INA215 temperature problem question.

Part Number: INA215

Hi.

INA215 DCK package is in use. I would like to know the criteria that judge a junction to be destroyed when the case temperature of the INA215 reaches a certain level.
I find it very difficult to calculate using the formula provided by TI.
Can I get a rough estimate temperature?

Ask for help.

Thanks.


  • Hi Chun,

    Thank you for posting on the E2E foruns.

    This is a complex subject, which I will not be able to fully answer without knowing more about your system, where the part will be placed, the temperatures, currents, operating voltages, altitude, etc. I will however try to help you get a better understanding of what to expect and what is required to calculate this.

    Ultimately, all energy is lost as heat, therefore the worst case scenario would happen if the device would not be able to dissipate any power through the PCB, not heatsinked, operated at maximum allowable voltage (26V) output short circuited, very high environmental temperature and very high altitude.

    Figure 10, on the typical characteristics section of the datasheet shows that, at a junction temperature of 125C, the device can sink/source a max of 8mA. Under the circumstances above 8mA*26V = 208mW. If we then add the quiescent current (100uA) and input bias current (28uA*2) we get a total input power of 212mW.

    Using the table on section 6.4, Thermal information, we know that for the required package and, again, assuming the device was not heatsinked and not able to dissipate any thermal power through the PCB the junction to ambient resistance is given as 223C per Watt.

    The formula above gives the worst case scenario, which would occur in situations where no power dissipation would be sinked or sourced to the system board and only take the surrounding environmental temperatures in account. In practice, we can assume that EITHER the ambient temperature OR the PCB temperature (whichever highest) as our limiting factor, for example, if the ambient temperature is 25C, but there is a linear regulator heatsinked to the PCB, and therefore heating it, we should take the board temperature into account instead. In practice it gets more complicated since, for example, the junction to board temperature is lower, at only 72W/C, but being the worse case scenario ensures, as engineers, we can guarantee our product to be protected.

    As we have calculated earlier, the worst case disspitation is 212mW, therefore, the application of the formula above would result in TJ = TA + (227*0.212) which we can relate to a junction temperature 47C+Ambient. As such, if the ambient temperature is 40C, the junction temperature is 87C, similarly, if the ambient temperature is 90C, the junction temperature is now 137C. What you can conclude from this is that if you plan to operate the part at maximum case temperature, specified on the datasheet (125C), the output current must be limited, otherwise the junction will exceed the 150C limit and it will be destroyed, on the other hand, under normal operating condition, where the output is not short circuited, the part quiescent current and input bias current only represent about 4mW (at 26V) and therefore the device temperature is only limited by the package, to 125C.

    One last thing to have in consideration is altitude. As air starts to become less dense, power dissipation to the environment goes down. For example, commercial airliners operate with a cabin pressure equivalent to that at 7000ft altitude. At this pressure the 47C value would be multiplied by 1.2, and hence TJ becomes 57C + Ambient (C).

    As you can see, there are several factors to have into consideration and a complete analysis is only possible once all system details are known, but I hope this information will help you better understand what is involved in calculating the thermal limitations of a device and how to make sure your designs are protected against failure.

    Kind Regards,

    Carlos S., TI Sensing Products, Applications Support

    TI makes no warranties and assumes no liability for applications assistance or customer product design. You are fully responsible for all design decisions and engineering with regard to your products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning your designs.