Common heatsink Calculation

[Bjtpower]

I want to use the common heatsink for the Bridge rectifier,IGBT and Boost Diode.

I have calculated the Each of the Dissipation

P_DIS_Bridge=18w

P_Dis_igbt=24w

P_boost_diode=6w

RJC_BRIDGE=0.6DEGREE/W
RJA_BRIDGE=1.63 DEGREE/W

RJC_IGBT=0.58 DEGREE/W
RJA_IGBT=1.66 DEGREE/W

RJC_Diode=0.85 DEGREE/W
RJA_Diode=6.66 DEGREE/W
TJ=90 DEGREE
TA=50 DEGREE

How to calculate the common heatsink for all 3 power components...??

[KlausST]

Hi,

Any heatsink calculaton document will give this information. Read through them.

RJA is not valid for your situation, because you use a heatsink.
(RJA is for use without heatsink)

RJC is for use with heatsink.

RJC x power_dissipation gives the difference temperature between junction and case.
Calculate this for each device.
Take the max value of those three.

Take TJ - max_vlaue to get max_heatsink_temp at full load. (add some margin for difference temperature between device_temp and heatsink_temp)

Then calculate the difference between TA and max_heatsink_temp = allowed heatsink_temperature_rise.

Heatsink value = allowed heatsink_temperature_rise / total power dissipation.

Klaus

[Bjtpower]

Hello KlauuST

I have calculated a per your suggestions.

Normally all datasheet have provided data of R(J-C) IN degree/w but the attached ipm provided it in kelvin/w.

Now it hard to handle.
Can you pls advice how to convert the same in degree/W

I google 3k/w in kelvin to degree Calculator and it is coming -270degree/w and whole result is getting wrong.

Pls advice on the same

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Here is the attached image

[KlausST]

Hi,

For temperature rise, which is a relative change in temperature K and °C is the same.

Lets say the temperature rises from 25°C to 60°C.
The temperature rise in °C is: (60°C - 25°C) = 35°C
25°C = (25 + 273)K = 298K
60°C = (60 + 273)K = 333K
The temperature rise in K is: (333K - 298K) = 35K

For all relative temperature calculations you may use the same value in K as in °C.

5K/W for relative calculations is the same as 5°C/W.

Klaus

[FvM]

No need to convert between degree C and K for a linear heat dissipation problem.

It's the simple point that thermal resistance has the commonly agreed unit K/W, although degree C/W is the same number because you are measuring temperature differences.

[Bjtpower]

Hello Guys..

Understand..

Now what problem i am facing here is.
Power dissipation per igbt =15.68 (There are 6 igbt inside the ipm module)

TJ=90 Degree
TA=60 Degree
Junction-to-Ambient Thermal Resistance=(TJ-TA)/Power dissipation across single igbt)=(90-60)/15.68=1.97degree/Watt

Now,
R(heatsink to ambient)=ØJ-A-(ØJ-C+ØC-H)=1.97-(3K/W+0.2)=-0.8

This is the problem i am facing..?

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For The final calculation
RSA_IPM=(TJ1-TA)-P_IPM(RJC4+RCS4)/(Total heat dissipated)

TJ1=90
TA=60
P_IPM=Heat dissipated in ipm=15w
RJC4=3K/W
RCS4=0.2D/W (For silpad)

RSA_IPM=((90-60)-15(3+0.2))/140=-363degree/w

[KlausST]

Hi,

This is the problem i am facing..?

--> reread post#2

RJC4=3K/W

I doubt this value. It should be much smaller. Although you showed the datasheet picture... could you give the link to the complete datasheet?
* maybe there is additional information or there is an errata...

And there is a mistake in your calcualtion.
The result of your formula should be -0,129K/W
And were do the 140W come from?

Klaus

[Bjtpower]

https://www.mitsubishielectric-mesh....S92F6-AG_N.pdf

The Addition of all heat dissipated.

1) Bridge Rectifier:18w
2) IGBT_PFC=24W
3)Boost=6W
4) IPM=15*6=90W

TJ1-TA-P_IPM*6*(RJC4+RCS4)/(Total heat dissipated)

YES IT IS 0.129K/W

[KlausST]

Hi,

according datasheet the 3K/W is true.
So with a disspated power of 15W per IGBT you already get a temperature difference of 15W * 3K/W = 45K between junction and IBGT case.
This means if you want to limit junction temperature to 90°C (why that low? datasheet says 150°C) then the IGBT_case temparature needs to be less than 45°C.
This is impossible with 60C ambient temperature.

Klaus

[Bjtpower]

Yes.. now by the calculation

Thermal resistance of the Heatsink is 0.133°C/W

How heatsink area calculation works..??

How to design customised heatsink for Aluminium type.

Regards
Marx

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Yes.. now by the calculation

Thermal resistance of the Heatsink is 0.133°C/W

How heatsink area calculation works..??

How to design customised heatsink for Aluminium type.

Regards
Marx