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TPS54531 Thermal impedance and design rating

Pao Wu
Intellectual 920 points

Hello,

We'd like to Package Dissipation Ratings of TPS54531.
On the data sheet, following values show.


Condition : Vin = 20V. Output = 15V/3A, Schottky diode Vf = 0.45 (SS54), Inductor loss = 0.8W.

Vin Iin Pin Vout Iout Pout Eff
19.76 2.3837 47.10191 14.99 3.005 45.04495 95.63295


<Question1>

How to Calculate IC junction temperature ?? (When no measure Tc or Ta).

We need to know design assessment.

<Question2>

If measure Tc = 66 degC.  Ambiant temp = 25 degC

a. Diode power loss = 0.45 * Low side Avg current = 0.32W

b. Inductor loss = 0.8W

47.1-45 = 2.1W

2.1W - 0.32 - 0.8 = 0.98W


Tj = 66 + (63.2 * 0.98) = 66 + 61.936 = 127.936W <= Is this right??

  • 0
    TI__Guru**** 191445 points

    Webench indicates a power dissipation of 1.24 W for your conditions.  Assuming 50 C/W theta ja, the temp rise is 62 degrees for a junction temp of 92 degrees at 30 degrees ambient.  The wetherm simulation for the EVM board has the same power dissipation at 1.24 W.  The EVM is a 4 layer board with really a lot of heat dissipating copper.  The junction temperature is 60.6 deg. for an effective theta j-a of (60.6 - 30) / 1.24 = 25.68 C/W.

    Thermal performance is greatly dependent on PCB design.  The datasheet thermal parameters are based on JEDEC standard coupon and are for comparison purposes. you can use the Theta JC(bottom) parameter to model your PCB design if you have the appropriate software.

    Your actual design should fall somewhere in between those two extreme cases.

  • 0 in reply to JohnTucker
    Intellectual 920 points

    Dear John,

    Thank you for your support.

    If measure Tc = 66 degC.  Ambiant temp = 25 degC

    Vin Iin Pin Vout Iout Pout Eff
    19.76 2.3837 47.10191 14.99 3.005 45.04495 95.63295

    a. Diode power loss = 0.45 * Low side Avg current = 0.32W

    b. Inductor loss = 0.8W

    47.1-45 = 2.1W

    2.1W - 0.32 - 0.8 = 0.98W

    Tj = 66 + (63.2 * 0.98) = 66 + 61.936 = 127.936W <= Is this right??

  • 0 in reply to Pao Wu
    TI__Intellectual 1100 points

    Pao,

    Typically, if you get the Tc, please use psi(JT) to calculate the junction temperature.

    Tj = 66 + (14.9 * 0.98) = 66 + 14.602 = 80.602C

    Abull