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LM76003-Q1: LM76003-Q1 IC power dissipation

Part Number: LM76003-Q1
Other Parts Discussed in Thread: LM76003

Hello all, 

We are going to use LM76003 in one of our designs and we want to be sure that there will not have any thermal problem.

I can easily calculate conduction losses based on the Rds_on values provided in datasheet but don't know how to proceed with switching losses without having traise and tfall.

In the datasheet page 46, it's mentioned that losses are 3W for a case of Vin=12V, Vout=5V Iout=3,5A and fsw=2100KHz.

It means efficiency in this case would be (5*3,5) / [(5*3,5)+3] = 85%.., If we check Efficiency graph for fsw  2200KHz in page 32 we see it should be 90%. So I cannot conclude from where these 3W are coming and how they were calculated since it's not matching with the efficiency graph...

In our design, we have Vin_max = 16V Vout = 3,8V Iout= 2,1A and fsw = 620KHz and I would like to know the total IC Pd in order to be able to calculate de thermal resistance I need to be safe. 

Could you please tell me how to arrive to this expected Pd in my application?

Thanks and best regards,

Manel

 

  • Hi Manel,

    I am not too sure where the 3W comes from, according to the description it's probably related to the power dissipation at 125 deg C.

    One way to estimate your power dissipation is to use webench with your operating ambient temperature and see how hot the IC gets. 

    This value is an estimation based on EVM layout with specific Rtheta JA.

    You can follow the layout guidelines in the datasheet for getting the best thermal performance. 

    https://www.ti.com/product/LM76003

    Thanks

    -Arief