• State Resolved
  • Locked Locked
  • Replies 5 replies
  • Subscribers 23 subscribers
  • Views 1052 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

TPS65320-Q1 the function of VIN_LOD pin

Atsushi Yamauchi
Genius 4810 points
Other Parts Discussed in Thread: TPS65320-Q1, TPS54362-Q1

I would like to confirm the function of VIN_LDO of TPS65320-Q1.

In the datasheet, when LDO and Buck converter is enabled, LDO input is switched the buck converter
output.

In the case of LDO and Buck converter is enabled, cannot we the input of "Vin" connected the battery
as LDO input? Do we need Buck converter output > LDO input?

The customer needs 1.5V/3A output (buck) and 3.3V/100mA.

Best regards,
Atsushi Yamauchi

  • 0
    TI__Genius 9655 points

    Atsushi-san,

    Indeed, Output voltage of the Buck regulator has to be higher than LDO-output voltage. One can NOT tie VIN to VIN_LDO, as VIN_LDO has insufficient protection against potential surges on the battery-supply (abs.maximum 20V). In the intended configuration, the LDO-input is protected via the VIN-protection in case the buck is disabled or out of regulation. Note: even if you would ensure to not violate abs-max-ratings on VIN, the LDO would stay OFF repectively track Buck-output-voltage if Buck-output-voltage is insufficient.

    One option is to use another supply rail, in your case if you'd have e.g. a 5V rail availble, you can use this to supply VIN_LDO. 

    Best regards,

    Frank

  • 0 in reply to Frank Dehmelt
    Genius 4810 points

    Frank-san,

    Thank you for your reply!

    We need to the input voltage for LDO of TPS65320-Q1 on "VIN_LDO" for all conditions.
    And VIN is the input for Buck converter only.
    Are they correct?

    Best regards,
    Atsushi Yamauchi

  • 0 in reply to Atsushi Yamauchi
    TI__Genius 9655 points

    Atsushi-san,

    if the Buck is disabled, the LDO will take it's source from VIN, even if there is a sufficient supply voltage on VIN_LDO from another source.

    Therefore, yes, you always need a sufficient (external) supply on VIN_LDO as long as Buck is enabled and in regulation.

    Your 2nd statement that VIN is input for buck converter only is only partially correct: If the buck is enabled and inregulation, it is correct (LDO is sourced from VIN_LDO). If Buck is disabled or out of regulation, VIN will be the input for LDO.

    BR,

    Frank

  • 0 in reply to Frank Dehmelt
    Genius 4810 points

    Frank-san,

    I got your response.

    I have questions for Vin_LDO.

    If buck and LDO are enable and Vin_LDO is less than 20V, can we use Vin_LDO?
    And if the condition is under this sentense, can we realize the output by TPS65320-Q1

    - Vin = 12V by the battery, Vout(buck) = 1.5V, Iout(buck) = 3A
    - Vin_LDO = 12V with the countermeasurement against the surge (ex: TVS diode)
    - Vout(LOD) = 3.3V, Iout(LOD) = 100mA


    I think the output realizes by the operation of TPS65320-Q1, but I'm afraid the heat.
    Can TPS65320-Q1 accord the heat by this condition?

    Best regards,
    Atsushi Yamauchi

  • 0 in reply to Atsushi Yamauchi
    TI__Genius 9655 points

    Atsushi-san,

    indeed, thermal consideration is necessary:

    using a 12V (and this may have a tolerance as well?) to generate 3.3V, 100mA with a linear regulator will dissipate 870mW ((12V-3.3V)*0.1A). With the thermal dissipation of the part of 49.9oC/W (see datasheet page 3), this would cause a delta-temperature of 43.5oC, with a maximum junction temperature of 150oC, this reduces the allowable ambient temperature to ~105oC, provided you have a good thermal design (plenty of vias on the power-pad, large PCB-area to dissipate the heat, no other components that generate heat).

    Furthermore, you are going to use the Buck of the same component driving a 3A load. This will generate switching-losses and some losses due to quiescent current. This will increase the power-dissipation and further reduce the allowable ambient-temperature.

    I'm afraid, but I can not recommend to use the part in this use-case, unless you can provide another source of lower voltage to drive VIN_LDO. Note, the dissipation is the same for any LDO, but there are some stand-alone ones with different thermal performance (~30oC/W), allowing for higher ambient temperatures, e.g. TPS7A6033QKVU. Also this part would not share the same silicon with the buck, so you need to consider the losses of the linear-regulator only on IC-level, but note that the PCB still needs to dissipate the heat of both, TPS65320-buck and TPS7A6033-LDO.

    If not using the internal LDO, other buck-converters, e.g. TPS54362-Q1 may be an option as well.

    Best regards,

    Frank