LM1084 5A/5V heating problem

[Dhanushka Jayaweera]

hey guys,

I'm using a LM1084 5V 5A in a circuit which has a 16V 5A power input.
Data sheet - url

LM1084 gets heated when my 4A load is connected to it.
I'm not sure how much temperature it generates in my LM1084, but it gets too hot to be touched.

When I read the data sheet, it says that LM1084 can be heated up to 125 Celsius.

I'm already using a 3cm * 1cm heat sink. (Used heat sink compound as well).

I was wondering if my heat sink is too small to handle the heat generated by the LM1084?


Circuit diagram is attached

 

[FvM]

Heat sinks can be characterized by a K/W thermal resistance. The shown heatsink might have 20 - 30 K/W. The power disspation with 4 A output current will be 11 V * 4 A = 44 W, you need less than 2 K/W to keep the voltage regulator temperature within an acceptable range, at least the tenfold heatsink size required.

- - - Updated - - -

The calculations didn't consider internal thermal resistance, so the required heatsink thermal resistance will be even lower. All-in-all it's not a good idea to operate the regulator with 11V voltage difference and 4A output current.

 

[Dhanushka Jayaweera]

Thanks a lot for your explanation about the LM1084's heating issue. I looked into LMZ12008.

It looks like a far better solution for my scenario. I have few concerns about the LMZ12008. (Vin - 16V, Vout - 5V, Current - 5A)

Is it necessary to use a fan and a heat sink?
is it necessary to use any BASIC type protection circuitry ?

Circuit attached.

 

[chuckey]

If you use a linear regulator, then its got to dissipate (Vin - Vout) X Iout. Your are dropping 11V at 4 amps, this is 44W any linear regulator will dissipate this power as heat. You could put a large wire wound resistor in series with the input to the regulator. The maths goes, leave 2V for the regulator to work with, so Vin = 7V, Vsupply = 16, so the resistor must drop 9V at 4 A = 36 W, but the regulator now has to dissipate (7V -5V) X 4A = 8W. your heat sink could be big enough for 8W.
The sophisticated thing would be to buy a buck regulator. This is a switched mode power supply in a chip, that needs an inductance and a few extra components to get from 16 V to 5 V at 4 A with a power loss of, say 5W or less.
Frank

 

[Dhanushka Jayaweera]

thanks a lot Frank

 

[FvM]

LMZ12008 is a buck regulator.