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WEBENCH® Tools/LM5118: LM5118 Power Dissipation

Part Number: LM5118

Tool/software: WEBENCH® Design Tools

Dear Team,

My customer is designing:

Vin:40~60V

Vo:54V

Io:1.5A

Base on Webench, they find out 1.78W power dissipation, I am not sure how to calculate power dissipation on LM5118.

Can you kindly tell me where is 1.78W come from, how to calculate it?

WebenchReportsServlet_thermal.pdf

BR

Kevin

  • Hi Kuan Wei,

    I think the webench considered the following losses inside the LM5118:

    (1) gate drive current: Igate=2 x Qg x Fsw, where Qg is each MOSFET's total gate charge, and Fsw is the switching frequency
    (2) the VCC regulator: It is an LDO, and its input current would be Igate, so the total losses = Vinmax x Igate
    (3) Other IC power dissipation = Vinmax x Iq, where Iq is the IC quiescent current.

    So the total IC power dissipation is Vinmax x (Igate +Iq). Since Vinmax=60V, the IC power dissipation can be high (1.8w in this case).

    To lower the losses, you can employ an external linear circuit (an NPN base follower, with a 10V Zener at the base, and the Emitter to connect VCC pin, and the Collector to tie to the VIN pin, and use a large resistor to feed VIN to the 10V Zener). This will turn off the internal LDO and the IC losses would be only limited to Vinmax x Iq. If your system has a 10V supply you can tie that to VCC to save the said linear circuit.

    Hope this clarifies.

    Youhao Xi, Applications Engineering
  • Dear Youhao,

    Thanks for quick reply.
    Do you have circuit can directly implment to lower the VCC pin loss?
    Customer is lookinf for it.

    BR
    Kevin
  •  Hi Kevin,

    Something like the attached.   Choose a 80V 200mA NPN.  The zener can also be 11V.  The shadowed diode across BE may be needed to protect the BE junction.

    Thanks

    Youhao