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TPS61040: TPS61040 SW pin and SCH consulting

Fandy
Expert 2285 points
Part Number: TPS61040

Hello

I have some questions, Please help to solve,thanks

1.How much negative voltage/min volatge does SW pin could understand ?


2. I use TINA simulate the Figure17 SCH in datasheet, below is the TINA-SCH and waveform, I find the SW pin will be drop to -6.55V , I don’t know how that happened, do you have some comments here ? and could -6.55V possibly damage the device ?




 

3.Output capacitor C2 could palce at the behind of PNP ? I place it behind of PNP, and all waveform have some abnormal. especial for SW,max 133V,min -14V.  Could you explain it ?



4.Actually my customer SCH is show below, Vin 5V, Vout 6.5V. load current maybe need dozens of mA, I place 300ohm, simiulate 20mA load. And I find SW Pin also have -6.6V, is there any risk?



  • 0
    Expert 2285 points

    Hello

    By the way,My customer draw wrong SCH, place output voltage at the behind of PNP, the TINA simulate the waveform seems very dangerous, this simulate reaule reliable?

  • 0 in reply to Fandy
    TI__Mastermind 34201 points

    Hi Fandy,

    1. The min spec for SW pin -0.3V means the minimum DC voltage applied at SW pin to GND. Once DC voltage is less than -0.3V, the internal MOSFET may break down.

    2. This is normal behavior. During switching operation, the negative rating is exceeded due to the operating principles of a boost converter. This does not violate the absolute maximum rating.

    3. Place output capacitor C2 before the PNP transistor. When lower switch in on, the voltage across inductor L1 is Vin, inductor current rises thus energy is stored in L1. When lower switch turns off, inductor current will go through high side diode SD1 and charge output capacitor C2. C2 voltage then goes high until the PNP transistor turns on. The output voltage feedback resistors R1, R3 monitor Vo so that the internal feedback loop starts working. 

    If output capacitor C2 is placed after PNP transistor, when lower switch is turned off, the inductor current has no where to go, there SW pin voltage may go very high to damage the internal MOSFET.

    4. It's ok.

  • 0 in reply to Zack Liu
    Expert 2285 points

    Hi Zack,

    thanks,

    For the second point, I don't understand "the negative rating is exceeded due to the operating principles of a boost converter. This does not violate the absolute maximum rating", could you please explain it more clear? thank you so much

  • 0 in reply to Fandy
    TI__Mastermind 34201 points

    Hi Fandy,

    When lower switch turns on, inductor current will go through lower switch, not diode any more. During this period, current of diode drops down to zero and the diode has reverse current. Parasitic inductors of power loop and output capacitor of MOSFET, diode starts ringing.