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TL431ACDBZR and LM339APWR thermal question

kevin wu
Intellectual 2660 points
Other Parts Discussed in Thread: TL431, LM339

Hi

can you provide me Tj data of LM339APWR and TL431ACDBZR?

I try to calculate by Tj = Theta Ja * Pd + Ta

but for those part, i do not know how to get Pd.

So please let me know how to get Tj data.

thanks.

  • 0
    TI__Guru*** 140026 points

    Kevin,

    Pd the TL431 is VKA * IKA
    VKA and IKA are set by application.


    Pd for LM339 is sum of VCC * ICC and IOL * VOL for each low output.
    VCC and IOL is set by application.
    ICC and VOL values can be found in data sheet.

  • 0 in reply to Ron Michallick
    TI__Intellectual 2660 points

    Hi Ron,

    I calculate it, and can you help me to double confirm that?

    thanks.

    Pd the TL431 is VKA * IKA
    VKA and IKA are set by application.

    Check datasheet Recommended operation condition

    Vka=36v

    Ika=100mA

    TL431ACDBZR Theta JA=206 C/W

    1. 36x100=3.6w

    2. 206x3.6=741

    3.  70+741=Ta

    So Ta=811? is it correct?


    Pd for LM339 is sum of VCC * ICC and IOL * VOL for each low output.
    VCC and IOL is set by application.
    ICC and VOL values can be found in data sheet.

    VCC=5V

    ICC=2mA

    5X2=0.01W

    IOL=16mA

    VOL=400mV

    16X400=0.0064w

    0.01+0.0064=0.0164w

    LM339APWRTheta JA=113 C/W

    Tj= 25+113x0.0164=26.85

  • 0 in reply to kevin wu
    TI__Guru*** 140026 points

    Kevin,

    Pd the TL431 is VKA * IKA
    VKA and IKA are set by application.

    Check datasheet Recommended operation condition

    Vka=36v Ika=100mA

    While 36V and 100mA is allowed. It is not practical to do both at same time.
    For example, application has VKA = 5V and IKA = 10mA.

    TL431ACDBZR Theta JA=206 C/W

    1. 5V x 10mA= 50mW

    2. 206C/W x 50mW =10.3C

    3.  Ta + 10.3C = Tj

    Pd for LM339 is sum of VCC * ICC and IOL * VOL for each low output.
    VCC and IOL is set by application.
    ICC and VOL values can be found in data sheet.

    VCC=5V; ICC[max]=2mA

    P = 5V X 2mA =0.01W

    IOL=4 * 4mA; VOL=400mV

    P = 16mA X400mV =0.0064w

    0.01+0.0064=0.0164w

    LM339APWRTheta JA=113 C/W

    Tj= 25+113x0.0164=26.85

    Correct