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heat sink Problems again!

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tia_design

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There are three voltage regulator in my design, they generated 3.3V (LT1962),
6V (MC78M06) and -6V(MC7906), so does any one give me ideas how to do choose and connect heat sink?
For LT1962 and MC78M06 I chose surface mounted and use copper area as heat sink. For MC7906 I chose through hole version and use 574602B03700 as heat sink. Thanks again!
 

You have to study the datasheets for the regulators regarding the heat losses with the load you have got. Considering the environment where your design is suppose to work you can calculate the size of the heatsink necessary for the design. There are several things to consider here like temperature of the surrounding air, the exchange rate of the air in the enclosure and so on.
 

Things you must know to get started:
. Tamb Maximum ambient air temperature of the environment
. Tjunc Maximum allowable junction temperature of the pass element.
. RthetaJC Rated thermal resistance from junction to case of the
. pass element (regulator).
. Pmax Maximum power that will be dissipated by the regulator.
.
Now, Let
. RthetaJA = Maximum allowable Thermal resistance from junction to ambient air
. (You are going to calculate this below).
. Rthetahs = Maximum allowable thermal resistance of the heat sink.
. (You are going to calculate this below).
. Tj = Tamb + (Pmax X RthetaJA)
. Solve for RthetaJA:
. RthetaJA = (Tj - Tamb) / Pmax
. Solve for Rthetahs:
. Rthetahs = RthetaJA - RthetaJC
. (This is the maximum heat sink thermal resistance that you can use).
.
If you are using a commercially available heat sink, the Rthetahs value will be a published value.
.
If you are using the copper on a PCB, things are a little harder. You can do one of the following:
. Use published tables of thermal resistance vs surface area
. Measure the copper thermal resistance by:
. Mounting a resistor on the copper, applying a known power P, and measuring
. the temperature rise Delta T. Then
. Rthetacopper = (Delta T)/P Deg/Watt.
.
I hope this helps,
Kral
 

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