LM1085: Getting Hot

Hello Pavan,

Linear regulators will get hot as the output see's an increase in load.
The first step you will want to do is identify the input and output voltages, and what the output current is.

Then use the thermal information in the datasheet to estimate the temperature rise in your component.

Example:

Vin: 6V
Vout: 3V
Iout: 500mA
Tja: 40.6 C/W
Ambient Temp: Room temperature = 25C

So the temperature rise is:

(Vin-Vout)*Iout*Tja = (6V-3V)*500mA*40.6C/W = 60.9C

Total junction temperature is 25C + 60.9C = 85.9C.

If I have resolved your question, please click the "Resolved" link.

Thanks,

- Stephen

Hi Pavan,

Your math is correct but it is unlikely that the component is getting to that temperature.
What is probably different is the thermal resistance for your specific board.
I've copied the table from the datasheet with the various thermal resistances.
Ultimately the temperature your linear regulator will rise to depends on the amount of copper in the PCB and the thermal path from the linear regulator to a thermal heat sink.  A great resource on measuring the thermal impedance for your specific application is TI Application Note SLVA422, "Measuring the Thermal Impedance of LDOs in Situ".

Ultimately it will be hard to drop (24V-3.3V) = 20.7V across a linear regulator and provide any significant output current.
If all you need is to provide bias power, which only requires milliamps of current, then your design is a possible solution.
If you need 300mA, then you will need to choose a different package (larger package) with lower thermal impedance, or update your topology to include another component to dissipate the heat.  This other component can be another LDO in series with the LM1085, a DC dropping resistor in series with the LM1085, etc.

If I have resolved your question, please click the "Resolved" link.

Thanks,

- Stephen