Can I measure thermal resistance of a cheap heatsink with a thermocouple and resistor?

Thread Starter

seanspotatobusiness

Joined Sep 17, 2016
206
I have some cheap TO-220 heatsinks with no specifications and I was wondering whether it's possible to attach a TO-220 resistor and a thermocouple and use measurements at different power disipations to get an estimate of the thermal resistance (in °C/W)?
 

jpanhalt

Joined Jan 18, 2008
10,221
That seems like would give an approximation, but remember there is thermal resistance between the resistor and heat sink. You probably want to use a good thermal paste at that contact.

Other error sources:
1) Still air
2) Ambient temp
3) Uneven temp in heat sink
4) Heat dissipation by the resistor itself (connecting wires too, but probably much less)

EDIT: You can buy resistors in TO-220 packages. That might better simulate actual use conditions: https://www.digikey.com/product-detail/en/ohmite/TCH35P5R60JE/TCH35P5R60JE-ND/808233
 

AlbertHall

Joined Jun 4, 2014
10,390
I have some cheap TO-220 heatsinks with no specifications and I was wondering whether it's possible to attach a TO-220 resistor and a thermocouple and use measurements at different power disipations to get an estimate of the thermal resistance (in °C/W)?
For better accuracy make sure the heatsink is located in the same conditions as you intend to use it.
You really only need one measurement as the temperature above ambient is proportional to the power being dissipated.
°C/W = (heatsink temperature - external temperature) / power in resistor
 

Thread Starter

seanspotatobusiness

Joined Sep 17, 2016
206
When I model the situation intuitively in my head, I expect the relationship between power and temperature change to be non-linear. I think as the temperature gets higher, the rate of heat transfer will increase because of the higher temperature gradient.

I will be sure to use thermal compound between the heatsink and the resistor but one problem that is insurmountable is that the temperature probe itself will interfere with the conduction of heat from the resistor to the heatsink. I might use a rotary tool to carve a space into the metal on the resistor for the thermocouple to rest inside since I figure that will affect the result a bit less than a gap would. Maybe I can cut through one resistor first to find out where the resistor is so I can cut close to it and not into it.
 

jpanhalt

Joined Jan 18, 2008
10,221
Why not mount the probe on the back of the heatsink, maybe under a small spring? Thicknss is less than 1/16" (1.6 mm) and shouldn't make much difference. The TO-222 being attached is much larger.
 

AlbertHall

Joined Jun 4, 2014
10,390
I think as the temperature gets higher, the rate of heat transfer will increase because of the higher temperature gradient.
Once the temperature has stabilised, the heatsink gets rid of heat depending on the difference in temperature between the heatsink and the air around it. That loss of heat must be equal to the heat generated by the power in the resistor at equilibrium. So it is a linear relation.
 
I use a TO-220 junkbox transistor wired up as a two-wire CC source with variable PSU, to generate a few watts of heat. If you know the transistor's current and voltage, you know heat input.
A thermocouple (with cyanoacrylate glue) works but it loses a few degrees due to conduction heat loss in the wire. A thermal imaging camera works best to measure temperature.
Factors that give you poor results are the room's air currents/drafts from air conditioning, and the fact that heatsinks are non-linear. Convection-current turbulence and radiation vary with temperature and the fin design.
Some do better at say 100°C than at 50°C, while I have seen others do the opposite.
I have also seen people estimate the heatsinks rating with math, by calculating its surface area and fin pitch.
 

DickCappels

Joined Aug 21, 2008
6,584
As the temperature of the heatsink rises the heatsink will radiate more power into the surroundings. I've just used the formula posted by @AlbertHall for practical work for many decades "°C/W = (heatsink temperature - external temperature) / power in resistor" and always had satisfactory results.
 

ci139

Joined Jul 11, 2016
1,696
how the humidity , accumulated dust and the sink temperature will affect the convection around the last might be nonlinear and random , but for a short time span test - only the convection has the effective influence

? maybe there some dependency graph.-s
https://www.google.com/search?q=convection+range+heatsink+modelled+finite+element+linearity
https://www.google.com/search?q=convection+range+heatsink+modelled+finite+element+linearity&tbm=isch&sa=X
https://www.google.com/search?q=convection+range+heatsink+modelled+finite+element+linearity&tbm=isch&tbs=ic:gray

from http://www.iieta.org/journals/i2m/paper/10.18280/i2m.180216
 
Last edited:
I have some cheap TO-220 heatsinks with no specifications and I was wondering whether it's possible to attach a TO-220 resistor and a thermocouple and use measurements at different power disipations to get an estimate of the thermal resistance (in °C/W)?
Thermal insulance has the units square metre kelvins per watt (m2⋅K/W) in SI units or square foot degree Fahrenheit hours per British thermal unit (ft2⋅°F⋅h/Btu) in imperial units. It is the thermal resistance of unit area of a material. In terms of insulation, it is measured by the R-value.
 

MrAl

Joined Jun 17, 2014
7,803
When I model the situation intuitively in my head, I expect the relationship between power and temperature change to be non-linear. I think as the temperature gets higher, the rate of heat transfer will increase because of the higher temperature gradient.

I will be sure to use thermal compound between the heatsink and the resistor but one problem that is insurmountable is that the temperature probe itself will interfere with the conduction of heat from the resistor to the heatsink. I might use a rotary tool to carve a space into the metal on the resistor for the thermocouple to rest inside since I figure that will affect the result a bit less than a gap would. Maybe I can cut through one resistor first to find out where the resistor is so I can cut close to it and not into it.
Hi,

You can check out Newtons Law of Cooling
dT/dt=k*[T(t)-T(amb)]

which says that the rate of change of cooling is the difference between the temperature T at a time t and the temperature of the ambient surroundings.
There are implications to this which you can look up on the web.
You can also check out the Heat Equation
pu/pt=alpha*Laplacian
where
pu/pt is the rate of change of a function u,
alpha is a diffusivity constant,
the Laplacian involves the squared physical dimensions x,y,z but often just x and y (or x and z).
This is useful if you want to study the temperature at every point on the surface of the heat sink but may not be needed for a simpler study of how the temperature of just the die changes.

The right way to measure the case temperature is to drill a very small hole in the bottom of the heat sink (if it is large enough of course) and push the probe up into it.
Today though we have IR temperature guns that could be used although you have to take into account the radiation effects of the surface of the metal. I had gotten some interesting results measuring the temperature of free air power resistors with different power levels doing it that way. It may not be as accurate as a probe though.
Once you get the case temperature (or rather the temperature of the case plus the thermal paste) you can calculate the die temperature if needed.
Note that if the thermal paste is very very thin as it is usually applied it does not matter much. This is interesting by itself because manufacturers often tout their product as being so much better than others when it does not make nearly as big a difference as the heatsink itself and the manner of mounting. Very thin layers of even lower thermal conductivity materials conduct very well just because they are so thin.
 

DickCappels

Joined Aug 21, 2008
6,584
how the humidity , accumulated dust and the sink temperature will affect the convection around the last might be nonlinear and random , but for a short time span test - only the convection has the effective influence
(some content removed for clarity)
Depending upon temperature the percentage of power radiated is often near or above 30% of the total, with the rest going via convection.
 

Chris65536

Joined Nov 11, 2019
224
Depending upon temperature the percentage of power radiated is often near or above 30% of the total, with the rest going via convection.
I have always understood that the conducted and convected heat would always be proportional to the delta-T. But what about the heat radiated as LWIR? Is that proportional too, or is it a higher order function? The higher the temperature, the more of a factor that would be.
 

OBW0549

Joined Mar 2, 2015
3,546
I have some cheap TO-220 heatsinks with no specifications and I was wondering whether it's possible to attach a TO-220 resistor and a thermocouple and use measurements at different power disipations to get an estimate of the thermal resistance (in °C/W)?
Setting aside all the thermodynamic fine points, the answer is "yes, that method will get you an estimate."

It won't be precise, but it should be good enough for most practical purposes.
 
Surface temperature is always tough. I've used really low mass thermocouples, but there is always limits of error in the wire itself. T-type is good for room temperature. Repeatability is easy. The absolute value is very difficult. +-1 deg. C is probably achievable, but barely.

I've had lots of issues with measuring equipment especially cold junction compensation.
 

DickCappels

Joined Aug 21, 2008
6,584
quo
I have always understood that the conducted and convected heat would always be proportional to the delta-T. But what about the heat radiated as LWIR? Is that proportional too, or is it a higher order function? The higher the temperature, the more of a factor that would be.
I don't know, but there is an offset. If the heatsinks were to be in an environment that happened to be exactly the same temperature as the heatsink there would be no net transfer via radiation.

As @OBW0549 said, "Setting aside all the thermodynamic fine points, the answer is 'yes, that method will get you an estimate."
 

Thread Starter

seanspotatobusiness

Joined Sep 17, 2016
206
I made some temperature measurements on a 10R resistor. I used a K-type thermocouple pressed against the top of the mounting tab so the real temperature will be higher than measured. As you can see, the heatsink is pretty bad. I guess it's okay up to about 3 W for things that don't mind being toasty. The value for °C/W goes down as the power increases. I'm not sure how heatsink manufacturers arrive at their values.

The ambient temperature which was subtracted from the measured temperature before dividing by the power was 20.6 °C.

V (V)​
C (A)​
P (W)​
T (°C)​
°C/W​
3.0​
0.3​
0.9​
62​
46​
4.3​
0.425​
1.8​
94​
40​
5.3​
0.52​
2.8​
118​
35​
6.0​
0.595​
3.6​
142​
34​
 
Can you post a pic of the heatsink? Your numbers look fine. I did measure some of the generic extrusions that china offers for TO-220 parts. There's maybe 3 styles with no specs given. Example is 20x10x15mm.

20x10x15mm_heatsink.jpg
 

Thread Starter

seanspotatobusiness

Joined Sep 17, 2016
206
The one in question looks just like this (see below). I am going to use one which looks more like the one you posted because I may need to dissipate 5 W from my transistor. I will have to test it to be sure though because I might be pushing it in that case.

I've seen one that looks similar to mine but has specification 24 °C/W: https://www.digikey.com/product-detail/en/aavid-thermal-division-of-boyd-corporation/507302B00000G/HS115-ND/5849 - it looks somewhat wider and maybe has a better anodisation.

You know, I'm also now wondering whether I even mounted it correctly. I mounted the resistor inside the cubby hole made by the fins but that means that some heat was radiating off the insides of the fins and back onto the resistor. It would probably perform a little bit better if I put the resistor on the other side, although it would take up a bit more space on a circuit board.

 
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