# Selecting best Heat Sink for TEG (Thermoelectric generator)

I am trying to solve a puzzle about selecting the right heat sink for my TEG, but I am really confused and lost with lots of equations. Here is the Specification of my TEG (TG12-2.5-01L)

• Module Width = 29.972 mm
• Module Length = 34.036 mm
• Module Height = 3.937 mm
• Max Temperature (Tmax) = 250 °C
• Th = 230 °C
• Tc = 50 °C
• Efficiency = 5.02 %
• Power = 2.71 W
• Open circuit voltage (Voc) = 9.56 V
• Thermal resistance = 3.33 °C/W

What I want to do is maintain a temperature difference of max 50 °C. So in order to do this I need to have a good heat Sink which will radiate the heat in the air faster. So how can I calculate which heat sink I actually need? What do I need in order to calculate this? Which heat sink is the best and how do I know if it is good or bad?

Any information related to this will be really helpful.

The thermal resistance of the module is 3.33°C/W, which means that at a ΔT of 50°C, there will be about 50/3.33 = 15 W of heat flowing through the device.

What you need to decide is how much of your ΔT you're willing to "waste" across your heatsinks. In other words, you have a thermal circuit that consists of three resistances in series:

heat source
|
| thermal resistance of "hot side" heatsink
|
-----------
|
| thermal resistance of TEG
|
-----------
|
| thermal resistance of "cold side" heatsink
|
ambient temperature


Your overall ΔT will be distributed across these thermal resistances in proportion to their values, just like voltage is distributed across a series of electrical resistors in proportion to their values.

If you want, say, 90% of your total ΔT to appear across the TEG, you need to select heatsinks that each have a thermal resistance that's less than 5% of the value of the TEG itself, or 0.1666°C/W, and they need to be large enough to handle 15W of heat flow.

Note that fans on one or both heatsinks can considerably reduce their overall thermal resistance, but the power consumed by the fans will cut into your overall electrical efficiency, so you'll have to take that into account. You could use a Stirling cycle motor to power your fans directly from the heat, rather than using the electricity.

• So if the thermal resistance was 10 then the heat flow would be 5W right, this is better than 15 W? I thought less the resistance better the Model is, am I wrong in this? – Electronic Curious Mar 17 '14 at 13:30
• You entered a response to my answer as another answer, but that's not how this site works. I've converted it to a series of comments here, where I'll respond to them individually. If the thermal resistance is higher, then yes, the heat flow will be less, but this also means that you'll get less current (electrical power) out of the device for a given delta-T. Remember, this device is only converting about 5% of the heat power to electricity. "Better" would presumably be more current for a given delta-T, which would mean less thermal resistance. – Dave Tweed Mar 17 '14 at 14:31
• "You said that if I want 90% of total ΔT across TEG then the heat sink would be of 0.166°C/W, how did you calculate this?" The two heatsinks together must add up to less than 10% of the total thermal resistance. I just split that up into 5% each, but you could choose differently. – Dave Tweed Mar 17 '14 at 14:33
• "I want to maintain ΔT=50, is this the temperature between the hot side of the TEG and the cold side which is the heat sink. For example if Thot = 200 then Tcold(Theatsink) = 150 is this correct?" Yes. If the cold side of your TEG is going to be considerably above ambient, then you can afford to have a relatively high-resistance cold-side heatsink. – Dave Tweed Mar 17 '14 at 14:37
• Thank you for explaining. I think I am getting somewhere now :). now I am trying to solve this let me know if I am doing it correct. A module has Re=3.85, Rcs=1, N= 127, Q=44, seebeck=0.049, Tair=50, ΔT=50, Thot=150. So now I calculated Rjc=ΔT/Q=1.13°C/W Rsa=(150-50)/44-Rjc-Rsa = 0.136°C/W (is this the required heat sink to maintain ΔT?) I also tried to calculate the output power as, P= (127*0.049*0.049*50*50)/(4*3.85)= 48.5W is this correct, or I am doing something wrong? – Electronic Curious Mar 18 '14 at 15:05