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I am using MCP73213 IC for charging my 2S Li Ion battery. I want to use this IC for 1 ampere charging during constant current charge cycle. My specifications are as follows. Input source voltage : 9 volts(10 % accurate) charge current : 1 ampere during constant charging cycle Max power dissipation for such application is as follows : (9.9 volts -6 volts)* 1 ampere = 3.9 W Thermal resistance of IC : 62 C/W So rise of temperature during max power dissipation : 62*3.9 = 241 centigrade. But according to datasheet, IC will die at 150 centigrade. Datasheet of MCP73213 : http://ww1.microchip.com/downloads/en/DeviceDoc/20002190C.pdf I would like to place PCB heat sink below MCP73213 IC to allow power dissipation. What should be the dimension of PCB heat sink. I am really confused. Please help me in this regard.

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Texas Instruments has this handy guide for PCB heatsinks. AN-2020 Thermal Design By Insight, Not Hindsight

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  • \$\begingroup\$ I love how the app note number is "twenty twenty" in the context of hindsight. :) \$\endgroup\$ – Wossname Oct 8 '18 at 7:21
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Yup, You are looking at a decent sized grill.

Because the MCP73213 is internally limited, keeping its temperature down that wattage is essential to achieve a steady current output. I might be wrong, because heat sink design is not my speciality but here's my two cents:

To calculate the thermal resistance of the "profil" needed to keep your IC cool we use:

Thermal resistance

T_J max = 125 C T_A (the ambient temp.) = 25 (You can do a plot of this function using T_A as variable, and see how the demands change). Theta_JC = 20.4 Theta_CS = Depends on how you mount the IC and how well it couples with your PCB trace. Using a Thermal compound will keep this value low. ((125 - 25 )C/3.9W)-(20.5 C/W +1 C/W) = 4.14 C/W

4.14 C/W is a large grill!

Formular from the app note:
http://ww1.microchip.com/downloads/en/AppNotes/Design%20Considerations%20for%205V%20to%203.3V%20Pass%20Regulators.pdf

The datasheet is relatively sparse regarding info on general efficiency and seems to have a flat temp. respons when charging. However, we do not see graphs for more than 500 mA. For 1000 mA continous, you should consider the following: PCB trace with lots of copper to dissipate heat - even use vias if possible and use back plate to dissipate heat. Mount a large grill and consider adding a fan to generate flow. This vastly improves the thermal capabilities of any grill.

General info on thermal: http://www.designworldonline.com/how-to-select-a-suitable-heat-sink/ Great resource for understanding different parameters.

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Standard copper foil (default 1 ounce/foot^2 foil, 1.4mils thick, 35 micron thick), has thermal resistance of 70 degree centigrade per watt of lateral heat flow per square of foil. For any size square of foil.

Can you use a small fan to cool the foil?

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If you want to dissipate 3.9W and need to keep die temperature at 150°C, then you have to have thermal resistance 38.4°C/W or less.
62°C/W is a property of that device/case, you won't change that.
You have to choose different case, not 3x3 DFN. Or put heatsink on top of that chip (PCB heatsinking is not enough in this case). Or choose different chip or redo your design (you can add LDO to your circuit, before this IC, to drop voltage down to say 5V).

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Given the following...

Ta = 25C

Tj_max = 150C

Theta_Jc = 20.5C/W

P = 3.9W

The maximum allowed junction to ambient thermal resistance is...

Theta_Ja_max = (Tj_Max - Ta) / P = 32°C/W.

The the total thermal resistance (Theta_Ja) is the sum of the junction to case thermal resistance (Theta_Jc) and the case to ambient thermal resistance (Theta_ca).

Theta_Ja = Theta_Jc + Theta_ca

Therefore...

Theta_ca = Theta_Ja - Theta_Jc = 32°C/W - 20.5°C/W = 11.5C/W.

In still air with natural convection a 1 square inch pad of copper typically gives Theta_ca of approximately 40C/W. To get 11.5C/W you would be looking at a very large pad. Also you would need multiple layers of copper just to be able to conduct the heat out into a wide enough surface area.

If you insist on using this part you would most likely need one or more of the following...

  • Using a fan to increase airflow will drastically reduce the junction to ambient thermal resistance. The fan would be placed directly over the part blowing down onto your 1 square inch copper pad.
  • A PCB mount finned heat sink could be used. But that method is going to be expensive and even then you may not be able to get it to work.
    • To get the heat to the heat-sink, the heat sink would need to be placed directly on the opposite side of the PCB. You would need to move the heat through the board with a grid of thermal vias placed directly under the pad.
    • Vias themselves don't have very good thermal conductivity so you would need to fill them with silver or some other high conductivity material.
    • The pad under the part is not very large so you may not even be able to fit enough vias to make this method work.

My advice would be to find a different part.

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