This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

TPS7A87: TPS7A87 Thermal Temp

Part Number: TPS7A87

Dear TI Team,

When calculating the thermal temp for the TPS7A87, do I need to calculate the sum of power for out1 / 2?

Or should I calculate the higher of the two outputs?

Thanks.

  • Hello Chun,

    You should calculate the power for each LDO channel, sum them, and multiply the power loss by the thermal resistance to obtain the thermal increase in temperature.

    Note 1 on page 3 states:

    (1)  Lowercase x indicates that the specification under consideration applies to both channel 1 and channel 2, one channel at a time.

    Page 29 describes the Power Dissipation (Pd) of the linear regulator.
    Equation (8) states:

    Please let me know if you have any further questions.
    If I answered your question, please click the "Resolved" link at the bottom of this post.

    Thanks,

    - Stephen

  • Dear Stephen,

    Thank you for Answer.

    We want to use 3.3V 110mA, 3.3V 120mA as the output specifications of LDO1, LDO 2.

    In this case, I would like to know if I can calculate the power of the two outputs together.

    ex) 25+ (33x3.3 * 0.23)

    Thanks.

  • Hello Chun,

    Let's say you have 5V input for both internal LDO's.
    And let's say both outputs are 3.3V, but one is at 110mA and the other is at 120mA.
    We can neglect the internal LDO ground current and quiescent current draw because it is so small compared to your output load (110mA and 120mA respectively).

    So the power dissipation of the first LDO would be:

    (Vin - Vout) * Iout = (5V - 3.3V) * 110mA = 187mW

    The power dissipation of the second LDO would be:

    (Vin - Vout) * Iout = (5V - 3.3V) * 120mA = 204mW

    The total power dissipation of the component is 187mW + 204mW = 391mW.

    You can then multiply this by your thermal impedance to obtain the thermal rise of the component.
    Table 6.4 in the datasheet provides the thermal resistances of the component.

    Let's say you are operating at 25C. 
    Let's further say that the junction to ambient thermal impedance is an accurate reflection of your PCB thermal impedance. 
    Then according to table 6.4 your thermal resistance is 33 C/W. 

    The temperature increase becomes:
    0.391mW * 33 C/W = 12.903C rise

    The total temperature becomes:
    25C + 12.903C = 37.903 C.

    Let me know if this does not answer your question.

    Thanks again,

    - Stephen