SimScale CAE Forum

Thermal resistance


#1

Hello everyone,

i have a question regarding the use of SimScale:

I have an electronic component with a defined thermal resistance (junction to ambient - junction to case). I know that this component will dissipate x W. I don’t know how to specify the thermal resistance of the component (i don’t know the material of the component). Is it possible to do that?

Thank you very much,

Andrea


#2

Hi @agaudi
You can assign the wattage value to the component in this way. Just do some maths and find the W/m^3 value.

Regards


#3

Hi @HamzaBaig,

yes I can do like this. But the main problem is that I do not know the material’s parameters of the component but only the thermal resistance. How can I specify those parameters?

Thank you very much,

Andrea


#4

So you are saying you know the amount of Watt of heat generated and you know the thermal resistance of your component but not its material and you are not sure what to do?

Well the thermal resistance is the inverse of the thermal conductance. Assuming you only care about the steady state, you can choose any material and then edit its thermal conductivity accordingly. The other parameters (density and specific heat) are only interesting in a transient analysis as they effect how long it takes for the temperature to change but not what temperature is achieved.

Keep in mind that electronic components tend not to be a single material, they have different layers of material and so any thermal resistance value you have been given is approximate. This should be good enough as in electronics cooling the aim is usually to prevent temperature from climbing too high rather than trying to reach a specific temperature. I suppose in certain sensing applications there could be a requirement to achieve a specific temperature.


#5

Hi @roy_g,

thank you very much for your answer. You understood the problem. I only care about the final temperature of the components and not in the transient analysis. I have still one doubt:

  • The thermal resistance is expressed as K/W but the thermal conductance is expressed as W/mK, so they are not only the inverse of each other. Do you agree?

Have a nice day,

Andrea


#6

Hi @agaudi,
Thermal resistance has a unit of K/W and thermal conductance has a unit of W/K ,whereas thermal conductivity is expressed W/K.m. You just made a little mistake.
For understanding more about it, go through the following link.
https://ctherm.com/products/tci_thermal_conductivity/helpful_links_tools/thermal_resistance_thermal_conductance/

Regards


#7

Hi @HamzaBaig,

you are right. So i need the thermal conductivity of my device. Supposing that the thermal resistance of my device is 42 K/W so the thermal conductance is: 0.023W/K. If the tickness of my device is 0.5mm is it correct to say that the thermal conductivity of the material is about 47.61W/Km?

Thank you very much for the support,

Andrea


#8

(Thermal conductivity) K= (Q.L)/(A.Delta T)

Where

  • K is the thermal conductivity in W/m.K
  • Q is the amount of heat transferred through the material in Joules/second or Watts
  • L is the distance between the two isothermal planes
  • A is the area of the surface in square meters
  • ΔT is the difference in temperature in Kelvin

Make your calculation according to this formula.


#9

Hi @agaudi

Just thought I would chime in here, we do have resistance network modelling in SimScale, where you can define a power rating, a resistance for case, board and sides and further a thermal interface material. This is available in the advance concepts in the conjugate heat transfer analysis type. Note, you will need junction to case not junction to ambient I think.

Best,
Darren


#10

https://www.simscale.com/blog/2019/01/product-update-run-continuation/

See this post about the feature release :slight_smile:


#11

Hi @1318980,

thank you very much for your answer. I will take a look at this feature.

Andrea


#12

Hi @HamzaBaig,

for some reasons I’m not able to get the point. I need to find the final temperature of the component and, in order to do so, i need to know the thermal conductivity. But, in order to compute K, i need the temperature difference…

I really do not understand how to proceed.

From your link above i understood that:

R = L/K

Where L is the Thickness of the specimen [m].

Is my above calculation valid?

Thank you very much,

Andrea


#13

@agaudi
Share all the details of your project, the more the better also share project link if already created. You can tell here or PM me.
I will try to hit my best.
Regards


#14

@HamzaBaig

Thank you very much for your kindness.

Currently I have still not created the project but here are the complete details:

I would like to know the final component and PCB temperatures but my problem is to find the thermal conductivity used in SimScale simulation from my input data.

Regards,

Andrea


#15

Hi @agaudi

You have to wait a little, working on it.


#16

Hi @HamzaBaig,

thank you very much.

Andrea