Temperature of component under thermal compound?

Thread Starter

smooth_jamie

Joined Jan 4, 2017
107
Hi All,

Hypothetically, I have a resistor that will get very hot. It is underneath a block of thermally conductive resin , 1 mm thick, 10 cm x 10 cm area which is intended to dissipate the heat. The resin / compound has a maximum exposure temperature of 200°C, and thermal conductivity of 1.3 w/m.k. I also know the ambient temperature could reach +40°C. If I calculated the power dissipated by the resistor is it possible to calculate the temperature of the component without knowing it's thermal resistance? The idea is that I want to make sure the heat sinking is sufficient enough so that the resin does not burn.

To answer this I re-arranged the heat transfer equation to give Temperature on the hot side, I have attached my excel calculations. Is this method OK to determine the temperature? My calcs suggest the resistor would need to dissipate 2kW to exceed 200°C, and at 100W it would only rise the temperature by about 7K
 

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Last edited:

wayneh

Joined Sep 9, 2010
17,498
I see one problem with your calcs - the area. 10cm^2 is not the same as a square 10x10cm, which of course is 100cm^2.
 

Thread Starter

smooth_jamie

Joined Jan 4, 2017
107
I see one problem with your calcs - the area. 10cm^2 is not the same as a square 10x10cm, which of course is 100cm^2.
Technically it was my post which was wrong (I will edit my post to reflect my question correctly). Just to clarify its a hypothetical area 10 cm x 10 cm which you have correctly indicated as 100 cm^2.

I am really asking for someone to tell me if my approach is correct. Do you agree with this approach?
 

wayneh

Joined Sep 9, 2010
17,498
Ah, OK.

I'm missing a piece: How will the 'cool' side of the resin transfer heat to the ambient air? It looks like you're using k properly for the layer of resin, but you need the total k between the heat source and the ambient air, including, typically, a heat sink. The sink supplier will have data to characterize how well the sink sheds heat into ambient air.

I haven't tried to reproduce and thoroughly validate your formula and worksheet, but it looks OK at a glance.
 

Thread Starter

smooth_jamie

Joined Jan 4, 2017
107
Ah, OK.

I'm missing a piece: How will the 'cool' side of the resin transfer heat to the ambient air? It looks like you're using k properly for the layer of resin, but you need the total k between the heat source and the ambient air, including, typically, a heat sink. The sink supplier will have data to characterize how well the sink sheds heat into ambient air.

I haven't tried to reproduce and thoroughly validate your formula and worksheet, but it looks OK at a glance.
It doubles up as ingress protection as well as a heat sink.

It wont have a metallic heat sink, the compound is designed to replace it since there isnt room for a finned metallic part. Therefore it will rely on the flat exposed area to dissipate into the ambient air.

I'm just trying to see if it's practical. If nobody can see any caveats then it must make sense.
 

wayneh

Joined Sep 9, 2010
17,498
I'm just trying to see if it's practical. If nobody can see any caveats then it must make sense.
You still need a k value for the transfer across the face to the ambient air. Without a finned sink, it won't be great and will be a lot smaller than the thermal conductivity.

Remember, you transfer heat via radiation, conduction and convection. You've accounted for conduction thru the resin, but not for the convection of heat away from it. Radiation will contribute a little but can probably be ignored. What little there is will provide a bit of a safety factor.
 

Thread Starter

smooth_jamie

Joined Jan 4, 2017
107
You still need a k value for the transfer across the face to the ambient air. Without a finned sink, it won't be great and will be a lot smaller than the thermal conductivity.

Remember, you transfer heat via radiation, conduction and convection. You've accounted for conduction thru the resin, but not for the convection of heat away from it. Radiation will contribute a little but can probably be ignored. What little there is will provide a bit of a safety factor.
Ok but will these factors significantly affect the temperature on the hot side?

If yes, then do I need to account for convection and find thermal conductivity surface-to-air? How do I calculate the effect of convection and where can I find the thermal conductivity surface-to-air?
 

wayneh

Joined Sep 9, 2010
17,498
Ok but will these factors significantly affect the temperature on the hot side?
Yes, it will be the rate-limiting step. When you rerun the calculations you'll find that the surface of the resin is just slightly cooler than the hot side. In other words, the resin will conduct the heat quite well, but then the heat "gets stuck" waiting to convect away. There will be very little ∆T gradient across the resin.
If yes, then do I need to account for convection and find thermal conductivity surface-to-air? How do I calculate the effect of convection and where can I find the thermal conductivity surface-to-air?
The manufacturer may have information on that but you could probably get close using a generic flat plate calculation, since the convection depends a lot more on the properties of air than on the composition of the flat surface. I can't find a great reference for you but here's a start.
http://people.csail.mit.edu/jaffer/SimRoof/Convection/
 

Thread Starter

smooth_jamie

Joined Jan 4, 2017
107
Ok so in my calcs I've calculated a 7K rise for 100W dissipated. This seems more generous than you hint it should be?

What you've said makes sense but, if i make the resin area larger in the calcs the resulting temperature rise is less not accounting for convection.

I'm a bit confused.
 

crutschow

Joined Mar 14, 2008
34,692
I've calculated a 7K rise for 100W dissipated
That is way too low for 100 cm² radiating area to air.
For example, the heat-sink below with 387 cm² radiating area has a thermal resistance to air of 5°C/W, so for 100W dissipation the temperature rise would be 500°C. :eek:

upload_2018-10-12_13-57-37.png
 

Thread Starter

smooth_jamie

Joined Jan 4, 2017
107
That is way too low for 100 cm² radiating area to air.
For example, the heat-sink below with 387 cm² radiating area has a thermal resistance to air of 5°C/W, so for 100W dissipation the temperature rise would be 500°C. :eek:

View attachment 161437
So would you mind having a look at my calcs and tell me what's wrong? So the method is not appropriate?

If i knew the thermal resistance I wouldn't of posted the question. All I have is the thermal conductivity of the compound W/m.k. which is dependant on geometry.

My goal is to find the surface temperature of the component under the resin. Btw 100W was just some arbitrary figure, I'm playing with numbers just to see if this makes sense.
 

Thread Starter

smooth_jamie

Joined Jan 4, 2017
107
I know this is a little old now but if anyone wants to know the solution to this, this is what I used and it's working for me.

Rth = L / (A x K)
Where:
Rth = Thermal resistance (K/W)
L = path parallel to Q, thickness of potting compound in my case (m)
A = Cross sectional area through where Q is passing, the exposed surface area of the compound (mm^2)
K = Thermal conductivity from the datasheet (W/mk)
 
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