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.

PTH12020L: Power loss

Part Number: PTH12020L
Other Parts Discussed in Thread: TPS7A94EVM-046

Tool/software:

Hi team, 

1) we are using P/N: PTH12020 in our design,

input voltage:12V

output voltage: 0.85V

Load Current: 13.745A

efficiency:80% (as per datasheet)

for Power loss we got 2.921W using formula (o/p power*((1-efficiency)/efficiency)), is ok to get 2.9W power loss???, we feel it is more than usual, does it required any heatsink??

2) Similarly, for P/N: TPS7A9401DSCR

input voltage:6.5V

output voltage: 2.5V

Load Current: 0.530A

for Power loss P = (Vin-Vout)*Iout we got 2.12W is this acceptable?? will it generate more heat??

thanks in advance,

Regards

Haritha

  • Hi Haritha,

    1) we are using P/N: PTH12020 in our design,

    input voltage:12V

    output voltage: 0.85V

    Load Current: 13.745A

    efficiency:80% (as per datasheet)

    for Power loss we got 2.921W using formula (o/p power*((1-efficiency)/efficiency)), is ok to get 2.9W power loss???, we feel it is more than usual, does it required any heatsink??

    Your calculation is correct. No, you do not need a heatsink or forced airflow as long as your ambient temperature is below 80ºC. As you can see in the temperature derating vs output current plot, natural convection can support up to 15A at 80ºC of ambient temperature. 

    2) Similarly, for P/N: TPS7A9401DSCR

    input voltage:6.5V

    output voltage: 2.5V

    Load Current: 0.530A

    for Power loss P = (Vin-Vout)*Iout we got 2.12W is this acceptable?? will it generate more heat??

    I apologize for the inconvenience, but could you please create a new thread to ensure that it reaches the appropriate support contact? My opinion will be as follows:

    Upon reviewing the datasheet, the thermal resistance is 25.6ºC/W (assuming your PCB layout is similar to the TPS7A94EVM-046). This means that 2.12W x 25.6ºC/W = 54.3ºC of temperature rise. Considering that the recommended maximum junction temperature is 125ºC, as long as your ambient temperature remains below 70.7ºC (based on 125ºC - 54.3ºC) everything should be fine.

    Thank you,
    Tomoya