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TPS92518: TPS92518

Gustavo Garcia
Prodigy 40 points
Part Number: TPS92518

Hello, I would like to know if I can get  2.75Amp per channel and Vout: 35V wiht TPS92518, I need 95W per channel?

Thanks

Gustavo Garcia

  • 0
    TI__Mastermind 25500 points

    Hello Gustavo,

    Yes, it can do that much current per channel. You just need to keep in mind the VCC gate drive capability and thermals.

    Ivccx will be equal to Qg*Fsw and the current limit if Ivccx can be as low as 25mA. So make sure the Qg and/or switching frequency is low enough not to exceed 25mA.

    The power dissipation in the IC will be (Ivcc1 + Ivcc2)*VIN. At 35V input it could be significant so it may require an even lower Fsw and/or Qg to keep the device from going into thermal shutdown.

    But if designed properly it can do that just fine.

    Regards,

    Clint

  • 0 in reply to Clinton Jensen
    Prodigy 40 points
    Thanks!! Clint foro YouTube support
  • 0 in reply to Gustavo Garcia
    Prodigy 40 points
    Sorry for the typing, my keyboard in Spanish.

    The input voltage will be 65V and 35V output / 2.75Amp, so I think thermal is going to be a big issue, is there any risk if I use a Totem Pole for the gate and high frequency 400kHz? or could you recommend me other solution?


    Thanks in advance

    Gustavo
  • 0 in reply to Gustavo Garcia
    TI__Mastermind 25500 points

    Hello Gustavo,

    If you choose the FETs carefully it can be done as is. For example, if you choose FETs with a low gate charge you might be fine. If they are 10nC gate charge FETs then the IC dissipation would be (400kHz*10nC*65V)*2 which would be about 1/2W which is fine for that package.

    But 10nC is pretty low and may be difficult to find. In that case an external supply could offload the power dissipation from the IC, but it will still get dissipated in the external supply. In that case the best option is not a totem pole (that would require and external source at least 5V higher than VINmax). It would be better to supply an external 8V and connect the anode of the BOOTx diode to that source rather than VCC. Then the power is dissipated by the external 8V regulator rather than the IC internally.

    Regards,

    Clint