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TPS709: TPS70918DBVR can not see the dropout spec.

JC Huang
Prodigy 110 points
Part Number: TPS709


Hi 

I will use TPS70919DBVR in project  I see the loading & dropout; is 1.5V max dropout, typical ~1000 mV for the TPS70912 part,

 Bit tPS70918 dropout isn’t spec’ed in the datasheet I found, please help it.

         Thank

        JC Huang

  • 0
    TI__Mastermind 23361 points
    Hi JC,

    On the datasheet, dropout is measured with VIN = 0.98 × VOUT(nom). Dropout is only valid when VOUT ≥ 2.8 V because of the minimum input voltage limits.

    What's the input voltage you plan to have in your application? TPS709 is a high voltage device; if you need to have your Vin very close to your Vout, this may not be the part for your application. If you could let me know your vin and vout and current load, I may be able to suggest you with a another device.

    Regards,
    Jason Song
  • 0 in reply to Jason Song
    Prodigy 110 points

    Hi Jason

          I only use Vin=3.3V , Vout is 1.8V  (about 10mA~20mA). for  ITE 8230  IC use.

                                Thank

              JC  Huang

       

  • 0 in reply to JC Huang
    Prodigy 110 points

    Hi All

      Any update my question？？

    　ＪＣ　ＨＵＡＮＧ

  • 0 in reply to JC Huang
    TI__Guru 50190 points

    HI JC,

    Maybe I can help clarify for you. When Vin is 3.3V they dropout voltage is shown on Figure 9. At a TJ of 125C and 150mA load the drop-out is 1400mV.

    So with 3.3VIN-1.4V=1.9V when means even at the full 150mA of load current you have 100mV of margin. You can think of VDO as an equivalent resistance meaning if VDO is 1.4V at 150mA, then the effective resistance of the drop is 9.33Ohms. If you have a 20mA load, this means  the LDO would be only dropping 186mV due to its internal resistance and therefore have to modulate the gate of the pass FET to drop an additional 1.313V to properly regulate 1.8VOUT from 3.3VIN (3.3-1.313-0.186=1.8).

    Does this make sense?