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Original question:

LM5576: Vin 48-57v,Vout 5V ,Iout 2A, LM5576 temperature is too high (160°F)

LM5576: Total Power Loss Estimation

Ella KIM
Expert 4212 points
Part Number: LM5576

Hello,

My customer is trying to estimate the approximate thermal performance of LM5576. The operating conditions is as below : 

input 48V 

ouput 28V/0.98A

27.36W(0.98A)

I have two questions regarding this.

1) Per fig. 9 in datasheet, the approximate efficiency would be 79%. Then,

Total loss : 27.36 * 21% =  5.74W

Inductor loss : DCR * 0.98A * 0.98A = 0.096 W

Diode loss : Vf * 0.98A = 0.48W

IC loss : 5.74W - 0.096W - 0.48W = 5.164W

Looks IC loss is too large. Can you check if the estimation above is valid?

2) I ran a Webench simulation and noticed that the efficiency is 96.8% and the total loss is only 0.89W. Can you let me know how this discrepancy between the simulation and datasheet comes and which value is correct?

WBDesign_LM5576_48Vto28V@0.98A..pdf

Thank you!

  • +1
    TI__Mastermind 27470 points

    Hi

    Webench result is right, normally in fixed Vin and Io, the loss will be similar, it is have little relation with Vo, so for 48Vin,  5V 1A, the power loss is 1.4W. when 28V output, the efficiency will be around 28W/(28W+1.4W)=95.2%

    Thanks 

  • 0 in reply to Daniel Li14
    TI__Expert 4212 points

    Hi Daniel,

    Thank you for comments but it's quite confusing.

    Daniel Li14 said:
    so for 48Vin,  5V 1A, the power loss is 1.4W

    Where can I get this information? I cannot find it in neither datasheet nor webench report.

    Daniel Li14 said:
    when 28V output, the efficiency will be around 28W/(28W+1.4W)=95.2%

    Webench report that I attached presents 0.89W of total loss and 96.85% of efficiency for 48Vin 28Vout. Could you please give me more information on your calculation compared to the webench report?

  • +1 in reply to Ella KIM
    TI__Mastermind 27470 points

    Kim:

    1 the efficiency curve you get from datasheet is EVM which is 5V output, for 5V 1A, the efficiency is 78%-79%, so loss will be (5*1/0.78)*0.22=1.4W

    2 my calculation is based on the loss is constant, but actually it will be slightly different by inductor and duty , so my result is a rough result, webench will be more accurate. 

    Thanks


    Daniel Li

  • 0 in reply to Daniel Li14
    TI__Expert 4212 points

    Hi Daniel,

    1) While discussing with customer regarding the discrepency in your calculation in datasheet (1.4W) and Webench report (0.9W), I find out Vout may contribute to power loss because the diode Pd is calculated with Vf*Io*(1-D).

    So customer operating condition consumes much less power than 1.4W which includes 0.75V*1A*(1-5V/48V) But the customer should calculate 0.75V*1A*(1-28V/48V).

    Daniel Li14 said:
    normally in fixed Vin and Io, the loss will be similar, it is have little relation with Vo, so for 48Vin,  5V 1A,

    Could you please advise if my thought is wrong?

    2) In the Webench report, the Pd items are as below : 

    Inductor Pd : 0.09W

    Diode Pd : 0.34W

    IC Pd : 0.49W

    Total : 0.89W

    Can you let me know IC Pd has margin or not? 0.49W is much lower than the customer's initial expectation, so they're worried about the case that the actual thermal is worse than the calculation when they fabricate the board.

    Please help to get this information so customer selects LM5576 for multiple designs. Thank you!

  • +1 in reply to Ella KIM
    TI__Mastermind 27470 points

    HI KIM:

    As I said, it will be slightly different by duty .

    You calculation for diode is right, and high side mosfet conduction loss = Io^2*D*rdson, diodes loss are Vf*Io*(1-D).. So when you decrease the diode loss, you will increase the conduction loss of high side fet, the total loss will be change a little bit.

    And switching loss is only related to VIn,Io ,inductor current ripple,  it is not related on Vo.

    the main loss is contribute by switching loss, conduction loss, diode loss , inductor loss. that's why the total loss will be close(but not same) in same Vin, Io but different duty condition.

    when customer firstly calculated the efficiency, they use 79% as efficiency which actually is at 5V 1A condition, that's why it is wrong. it means customer already think at 28V 1A it will have same efficiency . You can see our buck converter datasheet- higher Vo will have higher efficiency from half to full load. Because when you increase the output power and power loss only change a little , that will make the efficiency increasing. 

    Thanks