i am looking for a heatsink. I am new to this henc i am not sure on how to calculate.

Input supply ( from solar cells) : 15.6v, 1.33A ( max)

ouput of the regulator is 12.85V.

Power dissipation: ( 15.6-12.85)* load current ( approx 1.3) = 3.575w

Everytime i run the lm350 it gets very hot. I had tested on the dc supply , the regulator switch to its thermal protection by limiting the current.

jc- junction to case, cs- case to heatsink, sa- heatsink to ambient

the junction temperature must not exceed 125 degrees.

can i neglect Rcs?

T(junction)- T(ambient) = Power loss ( Rjc + Rcs + Rsa)

125- ( how do i determine?) = 3.575( 3+ Rsa)

Rsa is to determine he minimum thermal resistance of the heatsink?.

 

Too much math. Take a 5W or 10W power resistor, the ceramic kind that has an aluminum flange for mounting on a flat surface. Or better yet, one like this. Temporarily screw it to a 4" square of aluminum, about 1/8" thick or thicker. Hook the resistor to a power supply, adjust the voltage until the resistor is dissipating ~4W. E=sqrt(4/R).

In the final application, can you mount the heatsink vertically or horizontally? Set the aluminum plate in the same orientation. Let it reach thermal equilibrium (might take a few minutes). Measure the temperature of resistor body and the plate near where the resistor is mounted.

If the body of the resistor body is hotter than ~75 degC, you need a bigger heatsink, or it needs to be placed vertically instead of horizontally. If the temperature is < 50 degC, then you can make the plate smaller.

If you have a heatsink in the junk box, you can use this test on that, too.

 

Hello,

Here is a link to the heatsink page from ESP:
http://sound.westhost.com/heatsinks.htm

Greetings,
Bertus

 

thanks for the link..will try to understand the concept.

 

I had shortlist a few heatsinks but not sure whether i had done it correctly.

http://singapore.rs-online.com/web/search/searchBrowseAction.html?method=getProduct&R=2342299

Tj = Ta + P( Rjc + thermal resistance of heatsink) < 125 degrees.

Ta- operating temperature of air eg sunny day take as 34 degrees

Tj = 34+ 3.575(3+9.9) = 50 plus degrees < 125 therefore it will not activate the thermal shutdown of the ic chip.

correct me if i am wrong..thx...

 

Hello,

Keep in mind that the thermal connection between the IC and heatsink is to be calculated too.
This can be upto 1 °C/Watt.
You can use a sheet of mica with some thermal pasta to isolate the IC from the heatsink.

Take a heatsink with a lower resistance like this one:
http://singapore.rs-online.com/web/search/searchBrowseAction.html?method=getProduct&R=0403190

Greetings,
Bertus

 

noted..had purchased some mica.. any negatives to screw the ic with a nut to the heatsink?.with the mica in between to provide insulation?..

will be switching the heatsink once i completed my testing on my breadboard..thx..

 

Hello,

It can be that the adjust pin is connected to the housing.
The regulator will give the wrong outpout when this is grounded.
There are isolation rings to mount it isolated from the heatsink.

Greetings,
Bertus

 

noted thanks