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DS90UB948-Q1: Thermal calculation (IC ambient)

Roy Chou1
Genius 16210 points
Part Number: DS90UB948-Q1

Hi Team,

I would like to know how to calculate IC ambient temp.

If my VDD12 total current is 2.2A, VDD33 total current is 0.15A. how could I  calculate the IC ambient temp?

Regards,

Roy

  • 0
    TI__Mastermind 19720 points

    Hi Roy,

    Ambient temp is the temperature of the environment that the board is in. For example of the room the board is in is at 25C then the ambient temp is 25C.

    If you are asking about the temperature of the chip itself then you are talking about junction temp. Junction temp is calculated using the following table.

    for example of the ambient temp is 25C and the chip is using 3.135 Watts(1.2V*2.2A+3.3V*.15A) then the junction temp will be 25C + ( 24.8C/W * 3.135 W) = 102.75 C.

    Regards,

    Michael W.

  • 0 in reply to Michael.Walker
    TI__Genius 16210 points

    Hi  Michael,

    It seem like your calculation is ambient to junction. But I agree with your calculation.

    Could you let me know how to calculate the IC surface temp?(top case)

    Roy

  • 0 in reply to Roy Chou1
    TI__Mastermind 19720 points

    Hi Roy,

    the formula to calculate the Top of case temp is below for the following document: http://www.ti.com/lit/an/spra953c/spra953c.pdf?ts=1590681912765

    So the top of case temp will be 102.75 C - (0.1 C/W *3.135 W) = 102.44 C

    Regards,

    Michael W.

  • 0 in reply to Michael.Walker
    TI__Genius 16210 points

    Hi Michael,

    Below is my power consumption. 

    Deserializer

    DS90UB9xx

    　

    　

    　

    　

    　

    power

    V

    A

    P(W)

    θJA(℃/W)

    Rise(˚C)

    Ta=25℃

    VDD18

    1.8

    0.2

    0.36

    　

    　

    　

    VDD

    1.1

    2.2

    2.42

    　

    　

    　

    Total

    　

    　

    2.78

    18.1

    50.318

    75.318

    For my application : If my ambient temp is 25C and then junction temp will be 25+2.78*18.7=75.318C (IC internal junction temp)

    And IC top case temp, The result is 75.318C-0.1*2.78 = 75.04C. But we are afraid that when ambient temp is 75C, the IC top case will be 75+50=125C The IC will be thermal shutdown. The SPEC is -40~105℃.

    Does it make sense to you?

    Roy

  • 0 in reply to Roy Chou1
    TI__Mastermind 19720 points

    Hi Roy,

    That 105C temperature on the datasheet is the max ambient air temp the device can operate in not the max the junction temp can be.

     Regards,

    Michael W.

  • 0 in reply to Michael.Walker
    TI__Genius 16210 points

    Hi Michael,

    Do we have spec for max junction temp of 948 and 984?

    Regards,

    Roy

  • 0 in reply to Roy Chou1
    TI__Guru** 103215 points

    Roy,

    The abs max junction temp for FPD-Link devices is 150C which is stated in the datasheet. Please ensure that you are calculating this correctly though - the equations above that you posted are assuming Tja can be used to calculate the junction temp which is not accurate. The app note that Mike posted discusses this in detail in section 1.1

    Please check that app note thoroughly to understand how PCB design affects thermal dissipation 

    Best Regards,

    Casey