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TPS54362 question

Kei Sato73562
Expert 2310 points

Hello

I have question of TPS54362 start up.

Data sheet say, p16-p17 " EXTERNAL LPM OPERATION "

" Part can ONLY power-up in LPM/DCM if, VReg < 5.5V AND VIN-VReg > 2.5V.

In active mode. the part powers-up when VIN > 3.6V (min)."

Does a power supply not start up with specifications below?

Vin : 8V - 16V

Vout : 6V

Iout : 1.6A

fSW : 400kHz

 

  • 0
    Expert 2310 points

    I have some additional questions.

    1. In the case of the following specifications, where should Vreg pin(16pin) connect to?

    Spec:

    LPM : Open (LPM mode)

    Vin : 8V - 32V

    Vout : 6V

    Iout : 0.3A(typ) 1.6A(max)

    fSW : 400kHz

    2. I am setting the Vreg pin voltage to 5V by adjusting the voltage divider. (Please see attached)

    Is there a problem with this method?

    3. When Vreg is equal to Vout and it's higher than 5.5V, Is it necessary to start up in continuous current mode?

    I would like to have some answer by the end of today.

    TPS54362 Vreg Voltage divider.pptx
  • 0 in reply to Kei Sato73562
    TI__Mastermind 24740 points

    Answer1: series resistor between Vreg to output voltage should make it. Vreg max is 5.5V so 0.5V is the drop across the series resistor. The ratio is 10:1 so 10 ohm series resistor should be OK. The LPM is disabled if VIN is below 8.5V. In this case the Vreg pull down resistor is activated. There is no problem for the Spec.

    Answer 2: Vreg has a pull down resistor to GND in some cases so it is not a high impedance pin. The series resistor is 10Ohm and I do not see the need of voltage divider.

    Answer 3: It is not necessary to start in continuous mode. At start up until 8.5V the Vreg 100 ohm resistor is active to provide charge for the boot capacitor. with aditinal 1 0ohm resistor the total pull down is 110 ohms 

    I do not see any problem for this application becasue the targeted output is only 0.5V above the recommended 5.5V and 10 ohm series resistor should make it.

  • 0 in reply to Mahmoud Harmouch
    Expert 2310 points

    Hello Harmouch-san,

    Thank you for your support.

    I have additional question.

    question 1: your advice is add 10Ω between Vout to Vreg. Can I use resistor several Ω?

    When it is 10 Ω, Vreg becomes 5.45V. I think that it's small margin.

    question 2: When input voltage is 8V, Does it is not startup?

    Vout is 6V, but Vreg is 5.45V. I think that it meets the following condition.

    "VIN-VReg > 2.5V"

    TPS54362 Vreg.pptx
  • 0 in reply to Kei Sato73562
    TI__Mastermind 24740 points

    question 1: your advice is add 10Ω between Vout to Vreg. Can I use resistor several Ω?

    When it is 10 Ω, Vreg becomes 5.45V. I think that it's small margin.

    Answer1: 12 ohm gives 5.35V. We should not use big resistor becasue the boot cap may not have enough charge in LPM mode

    question 2: When input voltage is 8V, Does it is not startup?

    Answer2: It does start

    Vout is 6V, but Vreg is 5.45V. I think that it meets the following condition.

    "VIN-VReg > 2.5V"

    Answer: the 100 ohm pull down resistor is active if VIN - Vreg < 2.5V. If VIN - Vreg > 2.5V the pull down is not necessary to be active becasue the NMOS has enough gate charge

  • 0 in reply to Mahmoud Harmouch
    TI__Mastermind 24740 points
    The questions were answered back in 2013. Is there any open question left?