# Extruded heatsink question

I have few questions about a general extruded heatsink. As example, we can use this from Farnell.

Let's say it is long 20cm with a given shape, sold to have a Thermal resistance of 10°C/W.

Q1: (theoretical question) if I apply a single TO-220 in the middle of such heatsink long 20cm, can I consider it to have a Th of 10°C/W?

Q2: (theoretical question) if I cut it in half along its extrusion dimension, I have (roughly) 2 heatsinks of 20°C/W each long 10cm?

Q3: What about the heatsink in the link? Can I use it with a single, "small" TO-220 and pretend to have around 3°C/W?

EDIT: Here a picture of 2 kind of extruded heatsinks.

In the center of the flat side of the grey one, I will put a circular component (it is a metalized PCB, nothing standard) which will be inside the two holes of the grey one. Now the update of the question:

Q4: conceptually, assuming that the two heatsinks in the picture are providing the same thermal resistance, is it better the black one, since it is squared? (More regularity ecc). And, conceptually again, is it better the grey one to dissipate the heat from two components of half power (wrt the single component placed at center of the black one) placed near the two holes?

Because from the comments I had understood the concept of the extruded dimensioning, but not so well on how to use the longer ones with single components, and if it is valid to use them in such way.

Q1: (theoretical question) if I apply a single TO-220 in the middle of such heatsink long 20cm, can I consider it to have a Th of 10°C/W?

Yes, but remember that that's the temperature rise of the heatsink. The component mounted on it will have thermal resistance to the heatsink and the chip inside the component will have thermal resistance to it's case. These all add to the chip temperature rise and will be higher than the heatsink temperature.

Q2: (theoretical question) if I cut it in half along its extrusion dimension, I have (roughly) 2 heatsinks of 20°C/W each long 10cm?

That would be the expected result.

Q3: What about the heatsink in the link? Can I use it with a single, "small" TO-220 and pretend to have around 3°C/W?

We don't really pretend in this business. The specification is 3.03°C/W so it looks OK.

For a full thermal analysis you need to sum all the thermal resistances and calculate the temperature rise for a given power flow. This is similar to Ohm's law. See the first link below.

Table 1. Comparison between electrical and thermal resistance calculations.

Electrical                                    Thermal
Voltage (potential difference) [V]            Temperature difference [K] or [°C]
Current [A]                                   Heat flow (power) [J/s] = [W]
Electrical resistance [Ω] = [V/A]             Thermal resistance [K/W] or [°C/W]


• Regarding Q3, considering all the additional resistances, I thought that a small contact surface of the component was influencing the efficiency of the heatsink: in other words, I thought that would increase the thermal resistance of the heatsink, or at least the th resistance of the case to heatsink. This because many heatsink are designed to work with a given surface area that "touches" the hotspot, in the sense that they have a provided surface to be in touch with a chip of given dimension. Commented Aug 21, 2016 at 13:45
• Yes, but if it's a TO220 heatsink, for example, you could expect them to have tested with a heatsource of size comparable to the TO220 tab. It's not going to affect the heatsink itself but you are correct that the $R_{TH}$ of the chip to heatsink will increase and that is likely to be the problem. Commented Aug 21, 2016 at 13:57
• I made an edit on the main question, showing what I did not understood from what you mean "that is likely to be the problem". Hope to be clear. Commented Aug 22, 2016 at 21:51
• Q4) I imagine that your intuition is correct and that the long narrow heatsink would perform better if the power was split between two components mounted 1/4 way in from each edge. Commented Aug 22, 2016 at 21:57
• ...But to be sure, ones shall simulate. Frustrating. Commented Aug 23, 2016 at 16:24

Surface area is a somewhat linear factor to thermal reistance if there is unrestricted convection flow.

AIr flow has a large improvement such vortex influence in heat pipes or orientation of fins and flow restrictions.

Forced air as little as 1 m/s over the surface can reduce thermal resistance in half.

The rated specs will be best case. Each minor contributing factor to restriction of air or conduction loss on the interface, adds a series thermal resistance.

Humidity will also improve thermal resistance.

Thus test verification is essential to confirm your assumptions.