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TPS7B86-Q1: Vin = 13.5V ; Vout = 3.3V ; Iout = 0.5A, PD = 5.1W

Roy Chou1
Genius 15035 points
Part Number: TPS7B86-Q1

Hi Team,

I would like to select the LDO for below condition

Vin = 13.5V

Vo=3.3V

Io=0.5A

PD= (13.5-3.3)*0.5=5.1W

thedaJA=29.7C/W

rise temp will be 5.1*29.7=150C

does it mean we couldn't use this part to achieve my condition?

Roy

  • 0
    TI__Mastermind 36565 points

    Hi Roy,

    The ambient temperature would need to be zero degrees for the device to be used at this power dissipation and prevent it from entering thermal shutdown.  You have 2 options: spread the thermal dissipation to another component(s), or use a thermal heat sink.

    Option 1a: Spread the dissipation to another component (resistor)
    A common option is to add a dropping resistor in front of the LDO to drop power dissipation at max load.  This moves power dissipation from the LDO to the dropping resistor.  The dropping resistor can be placed further away at the optimum heat sink location (such as near chassis mounts).  Lets assume we want 2/3 of the dissipation on the dropping resistor and 1/3 of the dissipation on the LDO.  Then 0.66 * (13.5 - 3.3) * 0.5 = 3.366W will be dissipated across the dropping resistor(s), which will have a value of 13.5 ohms.  The input to the LDO will be (13.5V - (13.5 ohms)*(0.5A)) = 6.75V.  So the drop across the LDO is now just (3.45V)(0.5A) = 1.725W which is much more manageable from a thermal perspective.

    Option 1b: Spread the dissipation to another component (LDO)
    Another option is to use the TPS7B86-Q1 twice to spread the heat across 2 LDO's.  Spreading the dissipation evenly will achieve 75C rise across both devices.  This has many added benefits, such as significant improvements in PSRR at the output of the second LDO.  PSRR has a cumulative effect, so if you have 70dB of PSRR on one LDO, you will have 140 dB of rejection for two LDO's in series.

    Option 2: Thermal heat sink
    You may want to investigate using a heat sink, if you have access to the additional space.  You may also need forced air cooling.  Using a heat sink the thermal resistance will not be 29.7 C/W, it will be the summation of the: (junction to case bottom) + (heat sink compound) + (heat sink).  The package you are looking at has a very low junction to case thermal resistance.  I will assume 55C ambient temperature and 125C maximum as an example.  Then the thermal resistance would want to be (125C - 55C) / 5.1W = 13.725 C/W.  Subtracting the KVU Junction to Case thermal resistance gives 12.225 C/W for the thermal compound and heat sink.  Again, at these levels you may need forced air to keep everything cool.  Check the heat sink datasheets.

    Thanks,

    - Stephen