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LM25118-Q1: LM25118 power dissipation

Rana LM-S
Prodigy 210 points
Part Number: LM25118-Q1

Hello,

I would like to know how to calculate the power dissipation of LM25118

this is my estimation:

Pdiss = Pvcc + PIn + Penable

Pvcc=(VIN-Vout)*(Qg_HO+Qg_LO)*Fsw

PIn=VIN*IQ                                         where  Quiescent Current  IQ=10 uA 

Penable = I_EN*V_EN(on)                 I_EN=1 uA

I still don't know exactly how to estimate the power dissipation in the SS pin and FB pin ??

Best Regards

Rana 

  • 0
    TI__Guru 68495 points
    The main losses are in the driver which you got right in Pvcc. The rest losses can be calculated by non-switching Ivin, which is about 5mA per datasheet. These are internal logic losses, and all current comes from VIN pin: Vin x Ivcc = Vin x 5mA. The SS current is included in this Ivcc. There are barely dissipation inside the FB pin because it is a very high impedance at the input of error amplifier.
  • 0 in reply to Youhao Xi
    Prodigy 210 points
    thank you for your support.

    Best Regards,
    Rana
  • 0 in reply to Rana LM-S
    TI__Guru 68495 points
    One correction: For PVCC, it should be VIN x (Qg_HO+Qg_LO) x Fsw. The driver current is supplied by the internal VCC regulator and which also comes from VIN, so the total losses due to switching is VIN times the averaged driver current.

    In short, the total power dissipation inside the IC is:

    VIN x [ 5mA + (Qg_HO+Qg_LO) x Fsw ]

    Sorry for the confusion.
  • 0 in reply to Youhao Xi
    Prodigy 210 points
    Hello,

    After my research I've come across another formula to calculate the amount of current needed to fully charge the gate capacitance of the power mosfet.

    IG = QG/t(transition) where t(transition) = desired transition time (ton of MOSFET )

    with the first formula mentioned above IG=QG *Fsw = 28nC*275Khz= 7.7 mA

    with the second formula IG = QG/Tr = 28nC/15ns=1.86 A Where Tr= Rise Time of MOSFET

    Could please confirm witch one I need to use to calculate the gate current?

    also in my schematic I use resistance = 4.7Ohm between the gate of MOSFET and the pin HO, and I want to calculate the power dissipation of this resistance. Do I need calculate the IG(avg ) or use the peak gate current ??

    Best Regards
    Rana
  • 0 in reply to Rana LM-S
    Prodigy 210 points
    kind remember
  • 0 in reply to Rana LM-S
    TI__Guru 68495 points

    The 7.7mA is the average current that you use to calculate the power losses.  The 1.86A is the peak charging current that you need to evaluate the driving capability.

    Your 4.7Ohm gate resistor shares the losses with the internal driver, as they are in series. It is greater than the internal driver impedance (about 2 Ohm), so it will share a larger part of what you calculated above.