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LM1086: Heat issue for 24v supply.

MURUGESH HM
Prodigy 20 points
Part Number: LM1086
Other Parts Discussed in Thread: LM2576

Hi,

there is a heat issue for using LM1086 3.3V linear regulator 

i am giving 24v and i need output of 3.3v

primarily i had given 24v and got 3.3v but now there is problem in hardware that if i give 24v its getting heat and regulator goes bad,

circuit is as below

PIN1 GND

PIN2 Output connected with 0.1uF capacitor

PIN3 Input Connected with0.1uF Capacitor.

SCHEMATIC1 _ POWER.pdf

  • 0
    TI__Guru 55940 points
    Hi Murugesh,

    How much current do you have going through the LDO? Is your entire load the LED?

    We can calculate the amount of power dissipated in the LDO as Pd = (Vin - Vout) x Iout = (24 V - 3.3 V) x Iout = 20.7 V x Iout. The dissipated power is converted to heat. Due to your large amount headroom (Vin - Vout), even a small amount of current will generate a lot of heat.

    You can reduce the amount of power dissipated in the LDO by reducing the headroom. This can be accomplished by placing a DCDC in front of the LDO. This will provide a more energy efficient step-down conversion. Alternatively you could place a second LDO or even a power resistor in front of U1 in order to spread the power dissipation.

    Very Respectfully,
    Ryan
  • 0 in reply to Ryan Eslinger
    Prodigy 20 points
    Hi Ryan,

    Thanks for your quick response,

    Worst case 400mA current i needed through the LDO.

    Can you guide me how to reduce the headroom and how to place DCDC in LDO or how much value of Power Resistor should be placed in front of U1.

    i am new in this field please guide me to solve this issue,

    Respectfully,
    Murugesh HM
  • 0 in reply to MURUGESH HM
    TI__Guru 55940 points

    Hi Murugesh,

    At 400 mA your application as designed is dissipating 8.28 W per the calculation in my previous post.  This is why you are having thermal issues.  Per the Thermal Information table we can see that on a JEDEC High-k board ThetaJA is 23.0 C/W.  This means we can estimate your Junction temperature to be ~190 C above your ambient temperature.

    Per the Electrical Characteristics table we find that the maximum dropout specification for LM1086 is 1.5 V.  This means that the minimum input voltage for your application is 4.8 V (3.3 V + 1.5 V).  A higher input voltage will help improve some ac performance of the LDO such as PSRR; however, it will come at the cost of higher power dissipation and the benefit becomes marginal after ~3 V headroom (Vin = 6.3 V for your application).  As such, you could use a DCDC similar to how you are using LM2576 in your application to buck the 24 V input to between 4.8 V and 6.3 V at the LM1086 input.  

    If you choose to use another LDO, you will probably want to split the power dissipation between the two LDOs evenly.  As such you would want half of the voltage drop over each LDO; therefore, you would want the first LDO to drop 24 V to ~10.3 V and then use your current LDO to drop the remaining voltage.

    If you choose to use a power resistor, you will need to be careful to select a resistor with an appropriate power rating.  You would use Ohm's law to calculate the voltage drop across the resistor for your 400 mA maximum load.  The purpose of this resistor is to reduce the input voltage to the LDO by dissipating power outside of the LDO.

    Very Respectfully,

    Ryan